Titlepage List of symbols Thermodynamic concepts...

Lecture notes

in

Thermodynamic methods TKP4175

Tore Haug-Warberg

Department of Chemical EngineeringNTNU (Norway)

02 September 2013

Exercise 1 Table of contents Exercise 156

Tore Haug-Warberg (Chem Eng, NTNU) Termodynamic methods TKP4175 02 September 2013 1 / 1776

−200 −160 −120 −800

100

200

300

400

500

600

700

800

900

Temperature [F]

Pre

ssur

e[p

sia]

c.p.

N = 2913

minS ,N

(H)p ⇔ minS ,V ,N

(U)


Part-ContentsTitlepage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1 List of symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Thermodynamic concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3 Prelude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

4 The Legendre transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

5 Euler’s Theorem on Homogeneous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

6 Postulates and Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250

7 Equations of state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267

8 State changes at constant composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334

9 Closed control volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401

10 Dynamic systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453

11 Open control volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481


Part-Contents (2)12 Gas dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 642

13 Departure Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 708

14 Simple vapour–liquid equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 740

15 Multicomponent phase equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 790

16 Chemical equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 850

17 Simultaneous reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 879

18 Heat engines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 937

19 Entropy production and available work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 983

20 Plug flow reactor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1194

21 Material Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1311

22 Thermofluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1367

25 T , s and p, v-Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1524


Part-Contents (3)26 SI units and Universal Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1531

27 Newton–Raphson iteration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1546

28 Direct Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1559

29 Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1567

30 Nine Concepts of Mathematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1585

31 Code Snippets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1602


News Olds

Part 2

Thermodynamic concepts

see also Part-Contents


News Olds

2 Thermodynamic . . .New concepts

1 System

2 State

3 Process

4 Property

5 Extensive vs. intensive

6 Equilibrium

7 Heat and work


News Olds

2 Thermodynamic . . .ThermodynamicsENGLISH text is missing.The language1 Language shapes our ability to think independently and to commu-

nicate and collaborate with other people.

2 It has enabled us to go far beyond simply meeting our basic needsassociated with survival and social interaction.

3 Our ability to observe, describe and record our thoughts aboutphysical phenomena is vital to a subject like thermodynamics.

4 For us it is particularly important to reach a common physical un-derstanding of abstract concepts such as energy and entropy[a], but aslanguages are always evolving, the premises for a shared conceptionare constantly shifting.

5 All authors have to contend with that dilemma, and few of us havethe good fortune to be writing for future generations.

6 The benefit of writing about a phenomenological subject such asthermodynamics is that as long as the observed phenomena remainunchanged, the subject will endure.

7 The challenge you face with a classical subject is combining thepreservation of an intellectual heritage with the injection of new ideas,and it is rather utopian to believe that the latter is possible in our case.

8 Since the rise of thermodynamics over the period 1840–1880, theoriginal beliefs have been replaced by a more (ope)rational interpreta-tion.

9 Meanwhile, the content has become so well-established and thor-oughly tested that there is little room for innovation, although there isstill scope to inject fresh life into old ideas.

22 / 1776

[a] As illustrated by Professor Gustav Lorentzen’s article: Bør Stortinget oppheve2. hovedsetning? (Should the Norwegian parliament abolish the second law of ther-modynamics?), Ingeniør-nytt, 23(72), 1987 — written as a contribution to the debateon thermal power stations in Norway.

ENGLISH text is missing.Notation1 Our understanding of thermodynamics depends on some vital, and

precisely defined, concepts that form part of a timeless vocabulary.

2 Accurate use of language improves our insight into the subject,thereby reducing the risk of misunderstandings.

3 However, an exaggerated emphasis on precision may overwhelmreaders and thus be counterproductive.

4 In this chapter we will revise and explain the key concepts in ther-modynamic analysis[a] [a] in accurate, but nevertheless informal, terms.

5 Some of the concepts are concrete, whilst others are abstract, andit is by no means easy to understand them on a first read-through.

6 However, the aim is to convince readers that a good grasp of thebasic concepts is within reach, and that such a grasp is a prerequisitefor mastering the remaining chapters of the book.

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[a] James A. Beattie and Irwin Oppenheim. Principles of Thermodynamics. Elsevier,1979.

[a] Rubin Battino, Laurence E. Strong, and Scott E. Wood. J. Chem. Eng. Educ.,74(3):304–305, mar 1997.

1 Thermodynamics describes natural phenomena in idealised terms.

2 The scope of the description can vary, and must be adapted to ourneeds at any given time.

4 3 The more details we want to understand, the more information wemust include in the description. The abstract concept of a “system” is at the heart of our under-

standing.

5 A thermodynamic system simplifies physical reality into a mathe-matical model.

7 6 We can then use the model to perform thought experiments thatreveal certain characteristics of the system’s behaviour. Note that the physical size and shape of the system is quite irrele-

vant— all simple thermodynamic systems have uniform properties andcan always be described by a single set of state variables regardless ofthe actual geometry. 8 That is an absolute principle.


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2 Thermodynamic . . .Properties

1 The state of a thermodynamic system is determined by the sys-tem’s properties— and vice versa.

3 2 The logic is circular, but experimentally the state is defined when allof the thermodynamic properties of the system have been measured. For example, the energy of the system is given by the formula

U = (∂U/∂X1)X1 + (∂U/∂X2)X2 + · · · + (∂U/∂Xn)Xn once the statevariables Xi and the derived properties (∂U/∂Xi) are known.

6 4 This determines the value of U in one specific state.5 If we need to know the value of U in a number of different states, it

is more efficient to base our calculations on a mathematical model. (∂U/∂Xi) can then be calculated from the functionU(X1,X2, . . . ,XN) through partial differentiation with respect to

Xi . 7 The measurements are no longer directly visible to us, but are in-stead encoded in the shape of model parameters (that describe the ob-servations to some degree at least).8 Normally it is only the first and secord derivatives of the model thatare relevant for our calculations, but the study of so-called critical pointsrequires derivatives up to the third or maybe even the fifth order.9 Such calculations put great demands on the physical foundation ofthe model.


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2 Thermodynamic . . .§ 001 § 156 § 2

syste

mboundary

system

environment

Explain in your own words what is meant by the concepts:System, boundary and environment or surroundings.


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2 Thermodynamic . . .§ 001 System

1 A system is a limited part of the universe with a boundary thatmay be a mathematical surface of zero thickness, or a physical barrierbetween it and the surroundings Expanded footnote [a]:

[a] We must be able to either actively control the system’s mass and composition,impulse (or volume) and energy, or passively observe and calculate the conservedquantities at all times. All transport properties ρ must satisfy limx→0+ ρ = limx→0− ρ

so that ρ is the same for the system and the surroundings across the control surface.In physics, this is called a continuum description. The exception to this is the abruptchange in state variables observed across a shock front. Well-known examples ofthis include high-explosive detonations and the instantaneous boiling of superheatedliquids. It is important to note that the concept of phase boundary in Theme 4 does notsatisfy the continuity requirement and is therefore unsuitable as a control surface. .

2 A reservoir is a system that can interact with other systems withoutundergoing any change in its state variables.

6 3 Our physical surroundings such as the air, sea, lakes and bedrockrepresent unlimited reservoirs for individual human beings, but not forthe entire world population[a].4 Similarly, a hydropower reservoir represents a thermodynamicreservoir for the turbine, but not for the power company.5 This may appear trivial, but it is still worth trying to justify the abovestatements to yourself or a fellow student. These reservoirs, and othersimple systems, have properties that are spatially and directionally uni-form (the system is homogeneous with isotropic properties).

[a] The first proof that the oceans were polluted by man came from Thor Heyerdahl’sRa II expedition in 1970. A real-world system will never be perfectly uniform, nor completely

unaffected by its surroundings, but it is nevertheless possible to makesome useful simplifying assumptions.

7 Thus, a closed system, in contrast to an open one, is unable toexchange matter with its surroundings.

8 An isolated system can neither exchange matter nor energy. 9 Note that the term control volume is used synonymously with opensystem.10 The boundary is then called a control surface.11 In this context an adiabatic boundary is equivalent to a perfect in-sulator, and a diabatic boundary is equivalent to a perfect conductor.


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2 Thermodynamic . . .State variables

1 State variables only relate to the current state of the system.

4 2 In other words, they are independent of the path taken by the sys-tem to reach that state.3 The number of state variables may vary, but for each (independent)interaction that exists between the system and its surroundings theremust be one associated state variable. In the simplest case, there are C + 2 such variables, where C is

the number of independent chemical components and the number 2represents temperature and pressure. 5 The situation determines the number of chemical composition vari-

ables required.6 For example, natural fresh water can be described using C = 1component (gross chemical formula H2O) in a normal steam boiler, orusing C = 5 components involving 9 chemical compounds in an isotopeenrichment plant[a].

[a] Naturally occurring water consists of a chemical equilibrium mixture of 1 H 1 H 16 O+2 H 2 H 16 O = 2(1 H 2 H 16 O) and others, formed by the isotopes 1 H, 2 H, 16 O, 17 O and18 O.


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2 Thermodynamic . . .Processes

1 The state of the system can be changed through what we call aprocess.

2 Terms such as isothermal, isobaric, isochoric, isentropic, isen-thalpic, isopiestic and isotonic are often used to describe simple physi-cal processes. 3 Without considering how to bring about these changes in practice,

the terms refer to various state variables that are kept constant, suchthat the state change takes place at constant temperature, pressure,volume, entropy, enthalpy, (vapour) pressure or osmotic pressure re-spectively.4 Generically the lines that describe all of these processes are some-times called isopleths.5 Another generic expression, used in situations where the proper-ties of ideal gases are of prime importance (aerodynamics, combus-tion, detonation, and compression) is the polytropic equation of statep2/p1 = (V1/V2)γ.6 Here γ ∈ R can take values such as 0, 1 and cp/cv , which respec-tively describe the isobaric, isothermal and isentropic transformation ofthe gas.


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2 Thermodynamic . . .§ 002 § 1 § 3

Explain in your own words what is meant by the terms:State, property, process and path.


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2 Thermodynamic . . .§ 002 The state concept

1 A thermodynamic state is only fully defined once all of the relevantthermodynamic properties are known.

3 2 In this context, a thermodynamic property is synonymous with astate variable that is independent of the path taken by the system[a].

[a] In simple systems, viscosity, conductivity and diffusivity also depend on the currentthermodynamic state, in contrast to rheology, which describes the flow properties of asystem with memory. Two examples of the latter are the plastic deformation of metalsand the viscosity of paint and other thixotropic liquids. In simple systems, the properties are (by definition) independent of

location and direction, but correctly identifying the system’s state vari-ables is nevertheless one of the main challenges in thermodynamics.

9 4 One aid in revealing the properties of the system is the process,which is used to describe the change that takes place along a givenpath from one state to another.5 In this context, a path is a complete description of the history of the

process, or of the sequence of state changes, if you like.6 Thermodynamic changes always take time and the paths are there-

fore time-dependent, but for a steady (flow) process the time is unim-portant and the path is reduced to a static state description of the inputand output states.7 The same simplification applies to a process that has unlimited time

at its disposal.8 The final state is then the equilibrium state, which is what is of

prime concern in thermodynamic analysis.

y

x

A cycle is the same as a closed path. The cycle can either betemporal (a periodic process) or spatial (a cyclic process).

11 10 This choice greatly affects the system description. In a steadystate, the variables do not change with time, whereas in dynamic sys-tems they change over time. Between these two extremes you have aquasi-static state: The state changes as a function of time, but in sucha way that the system is at all times in thermodynamic equilibrium, asdescribed in greater detail in Theme 4. A state that is in thermodynamic equilibrium appears static at the

macroscopic level, because we only observe the average properties ofa large number of particles, but it is nontheless dynamic at the molecu-lar level.

12 This means that we must reformulate the equilibrium principlewhen the system is small, i.e. when the number of particles n→ 0.13 At this extreme, intensive properties such as temperature, pressureand chemical potential will become statistical variables with some de-gree of uncertainty.

1±ε kg

1 kg+

−

14 In a theoretical reversible process, itis possible to reverse any change of stateby making a small change to the system’sinteraction with its surroundings. Thus,two equal weights connected by a stringrunning over a frictionless pulley will un-dergo a reversible change of state if thesurroundings add an infinitely small massto one of the weights. This is an idealised

model that does not apply in practice to an irreversible system, where afinite force will be required to reverse the process.

15 Let us assume, for example, that themovement creates friction in the pulley.You will then need a small, but measur-able, change in mass in order to set theweights in motion one way or the other.16 The total energy of the system is con-

served, but the mechanical energy is converted into internal (thermal)energy in the process— so it is irreversible.17 If the friction is suddenly eliminated the weights will accelerate.This change is neither reversible nor irreversible. Instead, it is referredto as lossless, which means that the mechanical energy is conservedwithout it necessarily being possible to reverse the process. For that tohappen, the direction of the gravitational field would also need to be re-versed.


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2 Thermodynamic . . .Extensive–intensive

1 2 In thermodynamics there are usually many (N > 2) state variables,as well as an infinite number of derived properties (partial derivatives). Experimentally it has been shown that the size of a system is pro-

portional to some of its properties. These are called extensive proper-ties, and include volume, mass, energy, entropy, etc.

3 Intensive properties, meanwhile, are independent of the size of thesystem, and include temperature, pressure and chemical potential. 4 Mathematically these properties are defined by Euler functions of

the 1st and 0th degree respectively, as described in the separate chap-ter on the topic.5 There are also properties that behave differently.6 For example, if you increase the radius of a sphere, its surface areaand volume increase by r2 and r3 respectively.7 This contrasts with the circumference, which increases linearly withthe radius (it is extensive), and the ratio between the circumference andthe radius, which always remains 2π (it is intensive).


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2 Thermodynamic . . .Gibbs phase ruleENGLISH text is missing.Mass and mole1 The system’s mass is a fundamental quantity, which is closely re-

lated to inertia, acceleration and energy.

2 An alternative way of measuring mass is by looking at the numberof moles of the various chemical compounds that make up the system.

3 A mol is defined as the number of atoms[a] which constitute exactly0.012 kg of the carbon isotope 12 C, generally known as the Avogadroconstant NA = 6.022136(7)1023.

4 In this context, molality and molarity are two measures of concen-tration stated in moles per kilogramme of solvent and moles per litre ofsolution respectively.

5 These measures are not of fundamental importance to thermody-namics, but they are important concepts in physical chemistry, and aresometimes used in thermodynamic modelling.

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[a] A mole is defined in terms of the prototype kilogramme in Paris and not vice-versa.Atomic mass is measured in atomic mass units, where the mass of one atom of the12 C isotope is defined as 12 amu. The 1 amu was originally defined as being equalin mass to one hydrogen atom, which deviates a little from the current definition. Thedeviation is due to a difference in nuclear energy between the two nuclei and is relatedto Einstein’s relation E = ∆mc2.

1 Systems of variable mass contain a minimum number of indepen-dent components that together make up the chemical composition ofthe system.

3 2 A component should here be considered a degree of freedom inGibbs’ phase rule F = C+2−P, where C is the number of componentsand P is the number of phases at equilibrium (see Theme 4). Only the composition at equilibrium can be described in these sim-

ple terms, which is because we are forced to specify as few as possiblevariables that depend on mass or the number of moles. 4 This restriction influences our choice of components, and in prac-

tice the question is whether chemical reactions are taking place in thesystem, although the conventions of the applied discipline are equallyimportant.5 Generally components are selected from the system’s chemicalconstituents or species, or from its reactants, whereas for electrolytesand salt mixtures it is natural to use the ions that make up the system’schemical composition as components[a].

[a] Components do not necessarily represent physical constituents. One example isternary recipocral systems (salt mixtures) of the type NaCl+KBr = NaBr+KCl, whichhave 4 possible substances (salts), but only 3 independent components. For example,the salt NaCl can be described by the vector ( 1 0 0 0 ), or ( 0 −1 1 1 ),or any linear combination of these two vectors. This illustrates the use of a seeminglynon-physical (in this case negative) number of moles for one of the components. Thespecification does make sense, however, because the true amounts of each of the ionsin the mixture are non-negative, and hence real physical quantitites.


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2 Thermodynamic . . .§ 004 § 3 § 5 § 31 Explain in your own words what is meant by a property being

intensive or extensive. If you divide an extensive property by the numberof moles in the system or its mass, you get a molar or specific propertyrespectively. Show that the property obtained is intensive.

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§ 3 Extensiveness1 In thermodynamics, size is not only a measure of the volume of a

system, but also of any properties related to its mass.

2 This implies that two systems with identical state descriptions be-come a system of double the size when combined.

3 Properties that can be doubled in this way, such as entropy S,volume V and the number of moles N, are proportional to the size ofthe system, and are referred to as extensive variables.

4 This means that all of the extensive variables must be increased bythe same factor.

5 You cannot simply double the volume while keeping the number ofmoles constant— entropy, volume and the number of moles must all bedoubled together.

6 At the same time, other derived extensive properties like energy,total heat capacity, etc. are scaled the same amount.

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§ 3 Intensiveness1 Another group of properties is not affected by any change in the

size of the system.

2 These properties are referred to as intensive [a] properties.

3 Well-known examples include temperature T , pressure p andchemical potential µ.

4 Certain pairs of extensive and intensive variables combine to forma product with a common unit (most commonly energy), and feature inimportant relationships such as U = TS − pV + µ1N1 + µ2N2 + · · · +µnNn.

5 These pairs of T and S, p and V , and µi and Ni are called conjugatevariables.

6 A mechanical analogy is the flow rate ρ~V (extensive) and the grav-itational potential g∆z (intensive) at a hydropower station.

7 The power generated by the turbine can be written g∆zρ~V , whichis the product of an intensive and extensive variable.

8 Equivalent analogies can be found in electrical and mechanicalengineering.

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[a] The vague definitions of extensive and intensive used here will later be replaced byones involving Euler’s homogeneous functions of the 1st and 0th degree respectively.

§ 3 Densities1 Dividing one extensive property by another gives you a new, inten-

sive variable that represents the ratio between the two properties.

2 Let e.g. f = ax and g = bx be two extensive properties expressedas functions of x. Then ρ = f/g = a/b is independent of x, in otherwords ρ is intensive.

3 These kinds of variables express generic densities and are widelyused to describe systems in a way that does not refer to their size, butmany fields have different practices, and it is often unclear whether thedefinition is being expressed on a mass or mole basis (which is themost common source of misunderstanding).

4 The difference between a specific and a molar quantity is that oneis expressed per kilogramme and the other per mole of the substance(or mixture).

5 Common examples include specific and molar heat capacity, spe-cific and molar volume, etc.

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ENGLISH text is missing.Degrees of freedom1 Let us consider a thermodynamic system that does not change

with time, which means that it must be in equilibrium.

2 This minimises the degrees of freedom we have to specify.

3 At high temperature chemical equilibrium, for example, it is suffi-cient to state the quantity of each of the atoms present.

4 The distribution of the atoms amongst the many substances in themixture occurs by chemical reaction which is determined by the princi-ple of equilibrium (see below) and the system’s equation of state.

5 In the case of phase equilibria, the chemical substances are sim-ilarly distributed across the system’s phase boundaries and it is suffi-cient to specify the total composition of the entire system.

6 As a rule of thumb these problems are easy to specifiy, but theydo nevertheless require numerical solution by iteration which in manycases is a challenging task.

7 If the system has no internal degrees of freedom in the form ofchemical reactions or phase transformations, the equilibrium state willbe determined by an ordinary (but multivariate) function, generally with-out iteration.

8 The general principles of equilibrium mean that the energy of thesystem will be minimised with respect to all the degrees of freedom thatform the basis for the system description.

9 Alternatively the entropy is maximised with respect to its systemvariables.

10 The degrees of freedom are at all times controlled by the physicalnature of the system, and this determines which extremal principle toapply.

11 The theorectical foundations for this topic are discussed in a laterchapter on Legendre transformations.

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ENGLISH text is missing.Kinetics1 True equilibrium does not assert itself instantaneously, and from

our perspective it is not possible to judge whether equilibrium can beattained, or whether the system has kinetic limitations that prevent this.

2 All that thermodynamics does is describe the state of the systemat equilibrium, and not how long the process takes.

3 To describe in detail how the system changes over time we needan understanding of kinetics and transport theory.

4 Kinetics is the study of forces and the motion of bodies, whilst reac-tion kinetics looks specifically at rates of chemical reactions and phasetransitions.

5 Transport theory is a concept taken from nonequilibrium thermody-namics that can be used to describe the changes that take place in asystem until it reaches equilibrium.

6 As a general rule, all transport problems must be formulated aspartial differential equations, whereas thermodynamic equilibrium prob-lems can always be expressed using algebraic equations[a].

7 This difference in mathematical treatment reflects the gradients inthe system.

8 If they are not significant, it is sufficient to determine one represen-tative value for each of the scalar fields of temperature, pressure andchemical potential.

9 This distinguishes a thermodynamic problem from a transportproblem, where the scalar fields must be determined simultaneouslythroughout the space.

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[a] This is the difference between a distributed and lumped description of the system.

syste

mboundary

environment

phase b’ndary

α

βγ

Explain in your own words the following terms associatedwith equilibrium and equilibrium states: Phase, phaseboundary, aggregate state, equilibrium and stability.


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2 Thermodynamic . . .§ 004 The phase concept

12 A phase is defined as being a homogeneous, macroscopic subsys-tem separated from the rest of the system by a phase boundary.3 It must be possible to separate a phase from other phases throughmechanical means alone.4 This is an important prerequisite.5 The system’s equilibrium phases are often designated by the Greekletters α, β, γ, etc.6 The entropy density, energy density and mass density are con-stant within the phases, but they vary discontinuously across the phaseboundaries.7 Temperature, pressure and chemical potential, meanwhile, areconstant throughout the entire system (assuming equilibrium).8 Spatially a phase can be discretely distributed across the availablevolume, cf. fog particles in air and drops of fat in milk.9 If the whole system only consists of one phase, it is said to behomogeneous.10 Otherwise, it is heterogenous.11 A phase is referred to as incompressible if the volume is dependenton the pressure, but in reality all phases are compressible to a greateror lesser extent (particularly gases).12 In principle a phase has two possible states of matter: Crystallineand non-crystalline (glass and fluids).13 For practical purposes a fluid may sometimes be referred to asa gas without a specific volume, a liquid with surface tension, or anelectrically conductive plasma, but in thermodynamics there is a gradualtransition between these terms, and there are no strict criteria as to whatfalls within which category.

stable

metastable

unstable

Equilibrium is the state attained by the system when t →∞.

15 14 If the system returns to the same equilibrium state after exposureto a large random perturbation (disturbance), the equilibrium is said tobe stable. A metastable equilibrium is stable when exposed to minor pertur-

bations, but it becomes unstable in the event of major displacements.

19 16 An unstable equilibrium is a mathematical limit case.17 It describes a material state that breaks down if exposed to infinitelysmall perturbations.18 It is impossible to physically create this kind of equilibrium, but it isnevertheless very important from a theoretical point of view, as it sets alimit on what can physically exist based on simple laws of physics andmathematics. If the stability limit is exceeded, the system will split into two or

more equilibrium phases (from the mechanical analogy in the figure itis equally likely that the ball will fall to the right as to the left).


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2 Thermodynamic . . .(In)exact differential ENGLISH text is missing.Surroundings1 The assumptions and limitations set out in this chapter, and in thebook as a whole, in relation to thermodynamic system analysis, aresufficient for the purposes of analysing the effects of a large number ofchanges of state, but they do not tell us anything about how the systemrelates to its surroundings.

2 In order to analyse that, we must introduce two further terms,namely work W and heat Q.

3 These two properties control the system’s path as a function of itsphysical relationship to its surroundings.

4 On the one hand you have the abstract analysis of the system,which deals with state functions and mathematical formulae, and on theother you have an equally idealised interpretation of the surroundings.

5 Within this context, heat and work are defined as two differentmodes of exchanging energy.

6 It is important to note that heat and work are not state variablesof the system: They simply describe two different mechanisms for ex-changing energy between the system and its surroundings.

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1 State variables are variables that form part of a state function.

2 A state function always produces an exact differential, but not alldifferentials in physics are exact Expanded footnote [a]:

[a] Let e.g. f(x, y) = xy + c be a state function with x and y as its state variables. Heredf = y dx + x dy is the total differential of the function. The right side of the equationis then called exact. If as a pure thought experiment we change the plus operator onthe right side to minus, the differential becomes non-exact. It is possible to transformthe left-hand side into y2 dg = y δx − x δy, where y2 is an integrating factor for thedifferential and dg is the total differential of g(x, y) = xy -1 + c, but the new differentialy δx − x δy remains non-exact, since it cannot be expressed as the total differential ofany known function. .

3 One example from thermodynamics is (dU)n = δQ − δW , whichdescribes the energy balance for a closed system. Here the energy Uis a state function with the total differential (dU)n.

4 For any change δQ − δW there is a unique value of (dU)n.

6 5 However, the reverse is not true. For any change (dU)n there are in principle an infinite number ofcombinations of δQ and δW , since only the difference between heatand work is observable in the system.


News Olds

2 Thermodynamic . . .Das Ding an sich § 5

1

δW

δQ

dU=

δQ− δW

Explain in your own words the meaning of the terms: Heat, work andenergy. Are all three state variables?

45 / 1776

§ 5 Heat and work1 Heat and work are two closely related mechanisms for transporting

energy between the system and its surroundings.

2 Transporting energy affects the state of the system, but the heatand work are not themselves accumulated in the system.

3 Work involves moving a macroscopic mass, or elementary parti-cles, against an external force.

4 A moving piston, a rotating axle, electrons in an electric circuit andwater flowing through a turbine are all examples of this.

5 Heat results from large numbers of random, microscopic move-ments that do not result in any net movement of mass.

6 For heat to be converted into work, the microscopic movementsmust first be coordinated.

7 The 2nd law of thermodynamics then dictates that some of theheat will be lost to a thermal reservoir of the same temperature as thesurroundings.

8 The term energy dissipation is used to emphasise the fact thatspontaneous processes always result in a reduction in available energy.

9 The convention is for work done and heat supplied to be expressedas positive values.

10 Individually, neither heat nor work are state variables, but the differ-ence between them gives us the change in the system’s energy, whichis a state function.

11 In other words, any given change in energy can be produced in aninfinite number of ways by varying the contributions made by heat andwork.

46 / 1776

ENGLISH text is missing.The purpose of the anal-

ysis1 Having defined the fundamental concepts, what we now must do is

turn theory into practice.

2 That will require a good understanding of the physics involved in theproblems that we will be looking at, some mathematics and in particulardifferential equations and linear algebra, and a selection of relevantdescriptions of substances, known as equations of state.

3 Last but not least, we need to know the purpose of our analysis.

4 It is often based in a wish to find a simple model to explain thechanges in the system’s state.

5 Thermodynamics is, in short, a subject that combines most of thethings taught in undergraduate chemistry, physics and mathematics atthe university.

47 / 1776

1 2 However, it is somewhat optimistic to believe that mechanical en-gineering, thermodynamics, electromagnetism and other calculation-heavy subjects can fully describe the world we live in, and that we arein a position to decide how detailed an answer we want.3 Mathematics provides us with a useful tool, but that does not mean

that models and reality are two sides of the same coin, and hence thatwe, by carefully eliminating “all” assumptions, can reach an absolutelytrue answer. Mathematical descriptions allow us to understand some character-

istic events that surround us, but they do not give us the whole pic-ture.

Duck or rabbit ?

4 Immanuel Kant Expanded footnote [a]:

[a] Immanuel Kant, 1724–1804. German philosopher and logician. unified the most im-portant strands of rationalism and empiri-cism with his interpretation of das Dingan sich. 5 He stressed that we can never be certain about the intrinsic, true

nature of anything, and that our understanding of the world is limited byour subjective experiences in time and space. Science is based on observable

events taking place in a world where timeand space are assumed to be indepen-dent of us, but from a philosophical pointof view Kant argues that we may be un-able to see all the aspects of whatever weare observing. 6 For example, does the picture on the left represent a duck or a

rabbit?7 With a bit of imagination we can spot both animals, but we can onlysee one of them at any given time.8 The full picture is not available to us, and at first glance we don’teven realise that there are two possibilities[a].

[a] At first glance, around half of the population will see the duck, whilst the other halfwill see the rabbit.


News Olds

2 Thermodynamic . . .Das Ding an sich II

1 2 Thermodynamics is really a subject that describes das Ding füruns [a] as opposed to das Ding an sich. The figure on the right illustratesthis distinction in a rather subtle manner.

[a] Referred to as Erscheinung in Kant’s thesis. The solid circles show calometric readings for the mixing enthalpy∆mıxh of the system H2O–D2O Expanded footnote [a]:

[a] D. V. Fenby and A. Chand. Aust. J. Chem., 31(2):241–245, 1978. at varying compositions (mole frac-tions x) of the two compounds.

b

b

bb b

bb

b

b

32 J mol-1

h

x

∆rxh or ∆mıxh ?

⊗

⊗

⊗⊗

⊗ ⊗

⊗

⊗

⊗

bC bC bCbC bC bCbC bC bC

4 3 The continuous line has been fitted using the semi-empiric modelax(1 − x), which gives a first-order approximation of ∆mıxh. The line fits the data points very well,

and we can at least temporarily concludethat this simple model adequately repre-sents the readings taken.

5 This posture is, in simple terms, howthe experiment look to us and what ismeant by das Ding für uns. The ques-tion is if this is the only rational explana-tion that fits the observations.


News Olds

2 Thermodynamic . . .Looking backENGLISH text is missing.Isotopes1 Based on our understanding of nature, isotopes are chemically

very similar, but in this case the mixing enthalpy measured is equivalentto a fall in temperature of 0.43 K for an equimolar mixture of the twoisotopes, which is much greater than expected.

2 The reason for this discrepancy is that the two components reactendothermically to form HDO.

3 The net reaction involves the protonation of H2O ⇔ H++OH−

and the deuteronation of D2O⇔ D++OD− which are fast reactions.

4 The result is that H2O+D2O ⇔ 2 HDO reaches equilibrium[a] ina virtually ideal mixture.

5 The last statement appears to be self-contradictory, as the compo-nents react chemically at the same time as they take part in an idealequilibrium.

6 This apparent contradiction arises because of the unclear bound-ary between a chemical reaction and a physical interaction.

7 For a thin gas there is an explanation for this, but for a liquid thereis no entirely satisfactory definition.

8 The reaction product HDO is not stable either since it decomposesinstantaneously to H2O and D2O, which prevents us from observingthe substance in its pure state.

9 Our understanding of nature is, in other words, based on evidencederived from indirect observations and mathematical models of thephysical properties in the water phase.

10 In reality all scientific knowledge is based on theoretical models ofone kind or the other, and therefore does not imply that we have anyexact understanding of das Ding an sich.

50 / 1776

[a] In the system CD3OD–CH3OD, covalent bonds form between C–D and C–H, pre-venting any isotopic reaction, meaning that the change in enthalpy is only 0.79 J mol-1

or 0.01 K; cf. T. Kimura et al. J. Therm. Anal. Cal., 64:231–241, 2001, shown as opencircles in the figure. Subsituting OD to OH changes the chemistry to CD3OD–CH3OH

which again allows for proton–deuteron exchange and the equimolar change in en-thalpy increases to an intermediate value of 8.4 J mol-1 or 0.1 K.

ENGLISH text is missing.Thermodynamic proofs1 The foundations of phenomenological thermodynamics are too

weak to state hard facts about the true nature of the systems it de-scribes, but strangely enough thermodynamic theory can still be usedto falsify claims that break the laws upon which it is founded[a].

2 The theoretical basis is also sufficient for deriving important rela-tionships between the state variables, but it does not constitute inde-pendent evidence in the mathematical sense.

3 Thermodynamic analysis is capable of confirming prior assump-tions, or of demonstrating new relationships between existing results,but the calculations are not necessarily correct even if the model ap-pears to correspond with reality.

4 Firstly, the result is limited to the sample space for the analysisand to the underlying assumptions and secondly, there may be severalequally good explanations for a single phenomenon (as shown in theenthalpy example above).

51 / 1776

[a] The perpetuum mobile is the most famous example of this. In Norway it is in factimpossible to apply for patent protection for a perpetual motion machine, cf. the Nor-wegian Industrial Property Office’s Guidelines for processing patent applications. Theonline (2010) version of Part C: Preliminary examination; Chapter II Contents of thepatent application, except requirements; 3.3.6 Insufficient clarity excludes inventionsthat “. . . are inherently impossible to produce, as doing so would require acceptedphysical laws to be broken— this applies e.g. to perpetual motion machines.”

ENGLISH text is missing.The simpler the better1 It is also worth remembering that even a small one-component

system has maybe 1015–1020 microscopic degrees of freedom that aremodelled with only three thermodynamic state variables[a].

2 This means that a significant amount of information about the sys-tem is lost along the way.

3 It is therefore necessary to develop our ability to recognise what isimportant for the modelling, so that we can perform the right calcula-tions, rather than trying to look (in vain) for the very accurate descrip-tion.

4 In keeping with that principle, we will mainly be looking at simplemodels such as ideal gas and the Van der Waals equation, but thatdoes mean that we will be very precise in our analysis of the ones thatwe do look at.

52 / 1776

[a] Systems with many components are obviously even more complex.

1 System: A part of the universe with uniform T , p, µ

2 State: U = (∂U/∂X1)X1 + (∂U/∂X2)X2 + · · · as a function of Xi

3 Process: The change from one state A to another state B

4 Property: (∂U/∂Xi), (∂2U/∂Xi∂Xj), . . . in N + 2 variables

5 Extensive: Mass proportional property U, V , S, H, . . .

6 Intensive: Mass independent property T , p, µ, ρ, . . .

7 Equilibrium: Time invariant state Ueq = min U

8 Heat is not work: Q 9W

9 Work means heat: W → Q


News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds

Part 4

The Legendre transform




4 The . . .New concepts

1 Legendre transformation

2 Manifolds

3 Canonical forms

4 Conjugated variables

5 Maxwell relations

6 Gibbs–Helmholtz equation



4 The . . .Energy

1 The formal definition of the canonical thermodynamic potentials Expanded footnote [a]:

[a] I.e. the functions U(S ,V ,N), H(S , p,N), A(T ,V ,N), G(T , p,N), etc. isone of the three fundamental pillars of thermodynamic theory, togetherwith the principle of equilibrium, and the somewhat opaque distinctionbetween heat and work Expanded footnote [a]:

[a] Herbert Callen. Thermodynamics and an Introduction to Thermostatistics. Wiley,2nd edition, 1985.

Expanded footnote [a]:

[a] Michael Modell and Robert C. Reid. Thermodynamics and Its Applications. PrenticeHall, 2nd edition, 1983. .

4 2 The theoretical background of the energy potentials and their mu-tual transformation properties form the background to what we will nowdiscuss, but for the moment we will refrain from examining to what ex-tent mathematical formulae are of practical relevance to the applicationsof the theory.3 Complications arise from the fact that there are several energyfunctions to choose from, and it can be difficult to know exactly whichfunction is best suited for a particular problem. From a pragmatic point of view it is sufficient to remember that U is

useful for dynamic simulation, H for steady state calculations, A for thesimulation of storage tanks, etc.



4 The . . .Energy (2)

6 5 If a feasible solution to the problem already exists then this rule-based understanding is perfectly adequate, but when seeking a newsolution we need a deeper theoretical understanding to see all the al-ternatives that exist. Here, the Legendre Expanded footnote [a]:

[a] Adrien-Marie Legendre, 1752–1833. French mathematician. transformation is the key to new insight, asit provides a clever yet simple formula that allows us replace some orall of the free variables of a function with their corresponding partialderivatives. For example, the variable V in internal energy U(S ,V ,N)

can be replaced by (∂U/∂V)S ,N, see Figure 4.1.




H1

H2

H3

H4

Hi = Ui − (∂Ui∂Vi

)Vi

= Ui + piVi

U1 U2 U3 U4

V1V2 V3 V4

Figure 4.1: Legendre transformation of internal energy to enthalpy.




vmode (bug): hmode expected 7 The new variable can be interpreted as the negative pressure π

and the resulting transformed function, called enthalpy H(S, π,N), is inmany cases more versatile than U itself. In fact, as we will see later, Hhas the same information content as U.8 This is one of the key reasons why the Legendre transformation iscentral to both the thermodynamic and mechanical theories.



4 The . . .Transformation rules

1 Mathematically, the Legendre transformation φi of the function f isdefined by:

φi(ξi , xj , xk , . . . , xn) = f(xi , xj , xk , . . . , xn) − ξixi ,

ξi =( ∂f∂xi

)xj ,xk ,...,xn

. (4.1)

2 As mentioned above, one example of this is the transformation ofinternal energy to enthalpy:

H = UV (S , π,N) = U(S ,V ,N) − πV

π =(∂U∂V

)S ,N = − p

Note that the volume derivative of U is the negative pressure π, becauseU diminishes when the system performs work on the surroundings —not vice versa.



4 The . . .Transformation rules (2)

3 To begin to get an understanding of the Legendre transform weshall first write out the differential

dφi = df − xi dξi − ξi dxi ,

and then substitute in the total differential of the initial function f ex-pressed as:df = ξi dxi +

∑nj,i (∂f/∂xj)xi ,xk ,...,xn

dxj .

4 The simplification is obvious:

dφi = −xi dξi +

n∑

j,i

( ∂f∂xj

)xi ,xk ,...,xn

dxj .



4 The . . .Transformation rules (3)

5 If we now consider φi as a function of the derivative ξi rather thanof the original variable xi , the total differential of φi can be written

dφi =(∂φi∂ξi

)xj ,xk ,...,xn

dξi +

n∑

j,i

(∂φi∂xj

)ξi ,xk ,...,xn

dxj .

6 Comparing the last two equations term-by-term gives us theimportant transformation properties (∂φi/∂ξi)xj ,xk ,...,xn

= −xi and(∂φi/∂xj)ξi ,xk ,...,xn

= (∂f/∂xj)xi ,xk ,...,xn= ξj .



4 The . . .Repeated transformation

1 The latter property shows that further transformation is straightfor-ward:

φij(ξi , ξj , xk , . . . , xn) =φi(ξi , xj , xk , . . . , xn) − ξjxj , (4.2)

ξj =(∂φi∂xj

)ξi ,xk ,...,xn

=( ∂f∂xj

)xi ,xk ,...,xn

.

2 Combining Eqs. 4.1 and 4.2 gives the alternative, and conceptuallysimpler, expression

φij(ξi , ξj , xk , . . . , xn) = f(xi , xj , xk , . . . , xn) − ξixi − ξjxj , (4.3)

where the sequential structure of the Legendre transforms in 4.1and 4.2 has been replaced by a simultaneous transformation of two



4 The . . .Repeated transformation (2)

(or more) variables. 3 The two alternative approaches are equivalent because ξj is thesame regardless of whether it is calculated directly from the transformas (∂φi/∂xj)ξi ,xk ,...,xn or from the original function as (∂f/∂xj)xi ,xk ,...,xn Thisis clear from the differential equation above, and from Theme 20 onpage 132. Knowing the initial function f and its derivatives is thereforesufficient to define any Legendre transform. Moreover, as the Legendre transform is indepen-

dent of the order of differentiation, we know that φij = φji . 4 Mathematically we say that the Legendre operator [a] commutes.5 The three sets of variables (xi , xj , . . . , xn), (ξi , xj , . . . , xn) and (ξi ,

ξj , . . . , xn) are particularly important, and are often referred to as thecanonical variables of the functions f , φi and φij = φji .6 Canonical means simple and in this context it puts emphasis on

state descriptions that contain all the information about the system.

[a] Mathematical operators are often allocated their own symbols, but in thermodynam-ics it is more usual to give the transformed property a new function symbol.



4 The . . .Canonical potentials

1 2 The legal system of the Roman Catholic Church, also called canonlaw, has medieval roots, and is based on a collection of texts that theChurch considers authoritative (the canon).3 Today the word canonical is used to refer to something that is or-

thodox or stated in a standard form.4 The latter definition is particularly used in mathematics, where e.g.

a polynomial written with the terms in order of descending powers issaid to be written in canonical form. In thermodynamics, we refer to canonical potentials, meaning

those that contain all of the thermodynamic information about the sys-tem.

6 5 Here we will show that the Legendre transforms of internal energygive us a canonical description of the thermodynamic state of a system. Essentially, what we need to show is that U(S ,V ,n), A(T ,V ,n),

H(S ,−p,n), etc. have the unique property that we can recreate all ofthe available information from any single one of them.

7 This is not trivial, as we will see that U(T ,V ,n) and H(T ,−p,n), forinstance, do not have this property.



4 The . . .§ 019 § 18 § 20

Derive all of the possible Legendre transforms ofinternal energy. State carefully the canonical variablesin each case. Use the definitions Expanded footnote [a]:

[a] Let τ and π denote temperature and negative pressure respectively. This is toemphasize that they are transformed quantities, like the chemical potential µ. In thisnotation, all intensive derivatives of internal energy are denoted by lower case Greekletters. τ = (∂U/∂S)V ,N,

π = (∂U/∂V)S ,N and µ = (∂U/∂N)S ,V to help you.



4 The . . .§ 019 Energy functions

1 2 For any given thermodynamic function with m = dim(n) + 2 vari-ables there are 2m − 1 Legendre transforms. For a single-component system this means that there are 23 −1 =

7 possible transformations.

3 By using Eq. 4.1 on each of the variables in turn we get three ofthe transforms:

A(τ,V ,N) = U(S ,V ,N) − (∂U∂S

)V ,N S = U − τS , (4.4)

H(S , π,N) = U(S ,V ,N) − (∂U∂V

)S ,N V = U − πV , (4.5)

X(S ,V , µ) = U(S ,V ,N) − (∂U∂N

)S ,V N = U − µN . (4.6)



4 The . . .§ 019 Energy functions (2)

4 By using Eq. 4.3 on pairs of variables we can obtain three moretransforms:

G(τ, π,N) = U(S ,V ,N) − (∂U∂V

)S ,N V − (∂U

∂S)V ,N S

= U − πV − τS , (4.7)

Y(S , π, µ) = U(S ,V ,N) − (∂U∂V

)S ,N V − (∂U

∂N)S ,V N

= U − πV − µN , (4.8)

Ω(τ,V , µ) = U(S ,V ,N) − (∂U∂S

)V ,N S − (∂U

∂N)S ,V N

= U − τS − µN . (4.9)



4 The . . .§ 019 Energy functions (3)

5 Finally, by using Eq. 4.2 on all three variables successivly we canobtain the null potential, which is also discussed on page 192 in Chap-ter 5:

O(τ, π, µ) = U(S ,V ,N) − (∂U∂V

)S ,N V − (∂U

∂S)V ,N S − (∂U

∂N)S ,V N

= U − πV − τS − µN

≡ 0 (4.10)



4 The . . .H A U G

1 2 Several of the Legendre transforms of energy have their ownnames: Internal energy U(S ,V ,N) is used when looking at changes to

closed systems, and is in many respects the fundamental relationshipof thermodynamics.

3 Helmholtz energy A(τ,V ,N) is central to describing the state prop-erties of multicomponent fluids.

4 Gibbs energy G(τ, π,N) has traditionally been the transform that isof most interest in chemical thermodynamics and physical metallurgy.

5 Enthalpy H(S , π,N) is important for describing thermodynamic pro-cesses in chemical engineering and fluid mechanics. 6 The grand canonical potential Ω(τ,V , µ) is used when describing

open systems in statistical mechanics.7 Meanwhile, the null potential O(τ, π, µ) has received less attentionthan it deserves in the literature, and has no internationally acceptedname, even though the function has several interesting properties, aswe shall later see.8 This just leaves the two functions X(S,V , µ) and Y(S, π, µ), whichare of no practical importance.9 However, it is worth mentioning that X is the only energy functionthat has two, and always two, extensive variables no matter how manychemical components there are in the mixture.10 This function is therefore extremely well suited to doing stabilityanalyses of multicomponent phase equilibria.11 In this case the basic geometry is simple, because the analysis isdone along a 2-dimensional manifold (a folded x , y plane) where thechemical potentials are kept constant.12 This topic is discussed further in a separate chapter on phase sta-bility.



4 The . . .§ 020 § 19 § 21

The variable ξj in Eq. 4.2 can be defined in two ways.Either as (∂φi/∂xj)ξi ,xk ,...,xn

or as (∂f/∂xj)xi ,xk ,...,xn. Use

implicit differentiation to prove that the two definitionsare equivalent.



4 The . . .§ 020 Differentiation I

1 2 Note that the variables xk , . . . , xn are common to both φi and f andmay hence be omitted for the sake of clarity. We will therefore limit ourcurrent analysis to the functions f(xi , xj) and φi(ξi , xj). It is natural to start with Eq. 4.1, which we differentiate with respect

to xj :(∂φi∂xj

)ξi=

(∂(f−ξixi)∂xj

)ξi, (4.11)

where (∂ξi/∂xj)ξiis by definition zero.

3 Hence, at constant ξi :

d(f − ξixi)ξi=

( ∂f∂xi

)xj

dxi +( ∂f∂xj

)xi

dxj − ξi dxi

= ξi dxi +( ∂f∂xj

)xi

dxj − ξi dxi

=( ∂f∂xj

)xi

dxj



4 The . . .§ 020 Differentiation I (2)

5 4 or, alternatively:(d(f−ξixi)dxj

)ξi =

( ∂f∂xj

)xi . (4.12) The full derivative takes the same value as the corresponding par-

tial derivative (only one degree of freedom). Substitution into Eq. 4.11yields

(∂φi∂xj

)ξi=

( ∂f∂xj

)xi= ξj (4.13)

leading to the conclusion that differentiation of φi with respect to theuntransformed variable xj gives the same derivative as for the originalfunction f .



4 The . . .§ 021 § 20 § 22 ENGLISH text is missing.Before or after?1 It is a matter of personal preference, therefore, whether we want tocalculate ξj as (∂f/∂xj)xi before the transformation is performed, or tocalculate (∂φi/∂xj)ξi after the transformation has been defined.

2 Normally it is easiest to obtain all of the derivatives of the originalfunction first — especially when dealing with an explicitly defined ana-lytic function— but in thermodynamic analyses it is nevertheless usefulto bear both approaches in mind.

3 The next paragraph shows the alternative definitions we have fortemperature, pressure and chemical potential.

135 / 1776

Use the result from Theme 20 on page 132 to show thatthe chemical potential has four equivalent definitions:µ = (∂U/∂N)S ,V = (∂H/∂N)S ,π = (∂A/∂N)τ,V =

(∂G/∂N)τ,π. Specify the equivalent alternative definitionsfor temperature τ and negative pressure π.



4 The . . .§ 021 Identities I

1 2 Let f = U(S ,V ,N) be the function to be transformed. The questionasks for the derivatives with respect to the mole number N and it istacitly implied that only S and V are to be transformed. From the Eqs. 4.4 and 4.5 we have φ1 = A(τ,V ,N) and φ2 =

H(S , π,N), which on substitution into Eq. 4.13 give:(∂A∂N

)τ,V =

(∂U∂N

)S ,V , (4.14)

(∂H∂N

)S ,π =

(∂U∂N

)S ,V . (4.15)

3 The transform φ12 = φ21 = G(τ, π,N) in Eq. 4.7 can be reachedeither via A or H.

4 Inserted into Eq. 4.13 the two alternatives become:(∂G∂N

)τ,π

=(∂A∂N

)τ,V , (4.16)

(∂G∂N

)τ,π

=(∂H∂N

)S ,π . (4.17)



4 The . . .§ 021 Identities I (2)

6 5 Note that all the Eqs. 4.14–4.17 have one variable in common onthe left and right hand sides (V , S, τ and π respectively). For multicom-ponent systems this variable will be a vector; cf. xk , . . . , xn in Theme 20. So, in conclusion, the following is true for any single-component

system:

µ =(∂A∂N

)τ,V =

(∂H∂N

)S ,π =

(∂G∂N

)τ,π

=(∂U∂N

)S ,V . (4.18)

7 By performing the same operations on temperature and negativepressure we obtain:

τ =(∂H∂S

)π,N =

(∂X∂S

)V ,µ =

(∂Y∂S

)π,µ

=(∂U∂S

)V ,N , (4.19)

π =(∂A∂V

)τ,N =

(∂X∂V

)S ,µ =

(∂Ω∂V

)τ,µ

=(∂U∂V

)S ,N (4.20)



4 The . . .§ 022 § 21 § 23 ENGLISH text is missing.Freedom to choose1 This shows very clearly that we are free to choose whatever co-ordinates we find most convenient for the description of the physicalproblem we want to solve.

2 For example, in chemical equilibrium theory it is important to knowthe chemical potentials of each component in the mixture.

3 If the external conditions are such that temperature and pressureare fixed, it is natural to use µi = (∂G/∂Ni)T ,p,Nj,i , but if the entropyand pressure are fixed, as is the case for reversible and adiabatic statechanges, then µi = (∂H/∂Ni)S ,p,Nj,i is a more appropriate choice.

139 / 1776

The Legendre transform was differentiated with respectto one of the orginal variables xj in Theme 20. However,the derivative with respect to the transformed variable ξi

remains to be determined. Show that (∂φi/∂ξi)xj ,xk ,...,xn

= −xi .



4 The . . .§ 022 Differentiation II

1 2 As in Theme 20, the variables xk , . . . , xn are common to both f andφi , and have therefore been omitted for the sake of clarity. Let us start with φi(ξi , xj) = f(xi , xj)−ξixi from Eq. 4.1 and differen-

tiate it with respect to ξi . Note that the chain rule of differentiation( ∂f∂ξi

)xj=

( ∂f∂xi

)xj

(∂xi∂ξi

)xj

has been used to obtain the last line below:(∂φi∂ξi

)xj=

( ∂f∂ξi

)xj− (∂(ξixi)

∂ξi

)xj

=( ∂f∂ξi

)xj− xi − ξi

(∂xi∂ξi

)xj

=( ∂f∂xi

)xj

(∂xi∂ξi

)xj− xi − ξi

(∂xi∂ξi

)xj. (4.21)



4 The . . .§ 022 Differentiation II (2)

3 From Eq. 4.3 we know that (∂f/∂xi)xj= ξi , which can easily be

substituted into Eq. 4.21 to produce(∂φi∂ξi

)xj= −xi , (4.22)

which leads to the following conclusion: The derivative of φi with respectto a transformed variable ξi is the original variable xi , but with the signreversed. 4 In other words, there is a special relationship between the variables

xi in f(xi , xj) and ξi in φi(ξi , xj) [a].5 Due to the simple relationship set out in Eq. 4.22, the variables (ξi ,

xj) are said to be the canonical variables of φi(ξi , xj).

[a] The variables x and ξ are said to be conjugate variables.



4 The . . .Non-canonical differentiation

1 2 The Eq. 4.22 is strikingly simple, and leads to a number of simplifi-cations in thermodynamics.3 In particular the properties explained in the previous paragraph are

of prime importance. Before we move on, however, we should investigate what happensif we do not describe the Legendre transform in terms of canonicalvariables.

4 Differentiating φi = f − ξixi with respect to the original variable xi

gives:(∂φi∂xi

)xj,i

=( ∂f∂xi

)xj,i− (∂ξi

∂xi

)xj,i

xi − ξi = −( ∂2f∂xi∂xi

)xj,i

xi ,



4 The . . .§ 023 § 22 § 24ENGLISH text is missing.Non-canonical differenti-

ation1 The expression on the right-hand side is not particularly complex,

but unfortunately it is not canonically related to f . It is nevertheless animportant result, because φi must by definition be a function with thesame variables as f itself.

2 The change in variable from xi to ξi is an abstract concept that willonly rarely produce an explicit expression stated in terms of ξi .

3 Hence, if we want to calculate the derivative of φi by means of anexplicit function, we have to go via f and its derivatives— there is simplyno way around it. A simple example shows how and why this is thecase:

(∂U∂T

)V ,N = − ( ∂2A

∂T∂T)V ,N T =

(∂S∂T

)V ,N T =

CV

T T = CV (T ,V ,N) .

4 Here the internal energy is, at least formally speaking, a function ofthe canonical variables S, V and N, but there are virtually no equationsof state that are based on these variables.

5 T , V and N are much more common in that context, so to lure Uinto revealing its secrets we must use T rather S.

6 Note, however, that in this case S = − (∂A/∂T)V ,N is a function tobe differentiated and not a free variable.

7 This illustrates in a nutshell the challenges we face in thermody-namics: Variables and functions are fragile entities with changing in-terpretations depending on whether we want to perform mathematicalanalyses or numerical calculations.

144 / 1776

Use the results obtained in Theme 22 to prove that thederivatives (∂H/∂π)S ,N, (∂G/∂π)τ,N and (∂Y/∂π)S ,µ

are three equivalent ways of expressing the volume Vof the system. Specify the corresponding expressions

for the entropy S and the mole number N.



4 The . . .§ 023 Identities II

1 2 Let us start out once more from f = U(S ,V ,N) and define[a] thetransform φ2 = H(S , π,N) = U − πV . Inserted into Eq. 4.22, this givesus (∂H/∂π)S ,N = −V .

[a] The symbol π = −p is used here as the pressure variable. This is quite deliberate,in order to avoid the eternal debate about the sign convention for p. As used here,τ, π and µ are subject to the same transformation rules. This means that the samerules apply to e.g. (∂H/∂π)S ,N = −V and (∂A/∂τ)V ,N = −S, whereas the traditionalapproach using (∂H/∂p)S ,N = V and (∂A/∂T)V ,N = −S involves different rules for pand T derivatives. Note, however, that it makes no difference whether −p or p is keptconstant during the differentiation. Systematically applying Eq. 4.22 to all of the energy functions in

Theme 19 on page 127 yields Expanded footnote [a]:

[a] Sharp students will note the absence of −S = (∂O/∂τ)π,µ, V = (∂O/∂π)τ,µ and−N = (∂O/∂µ)τ,p . These relations have no clear thermodynamic interpretation, how-ever, because experimentally τ, π, µ are dependent variables, see also Theme 26 onpage 153. :

−V =(∂H∂π

)S ,N =

(∂G∂π

)τ,N =

(∂Y∂π

)S ,µ , (4.23)

−S =(∂A∂τ

)V ,N =

(∂G∂τ

)π,N =

(∂Ω∂τ

)V ,µ , (4.24)

−N =(∂X∂µ

)S ,V

=(∂Y∂µ

)S ,π

=(∂Ω∂µ

)τ,V

. (4.25)



4 The . . .§ 024 § 23 § 25 ENGLISH text is missing.Systematics1 Knowing the properties of the Legendre transform, as expressed

by Eqs. 4.13 and 4.22, allows us to express all of the (energy) functionsU, H, A , . . . O that we have considered so far in terms of their canonicalvariables.

2 The pattern that lies hidden in these equations can easily be con-densed into a table of total differentials for each of the functions (seebelow).

147 / 1776

Use the results from Themes 21 and 23 on pages135 and 142 to find the total differentials of all the

energy functions mentioned in Theme 19.



4 The . . .§ 024 Canonical differentials

1 The total differentials of the energy functions can be stated by tak-ing the results from Eqs. 4.18–4.20 and 4.23–4.25 as a starting point:

dU (S ,V ,N) = τ dS + π dV + µ dN , (4.26)

dA (τ,V ,N) = −S dτ + π dV + µ dN , (4.27)

dH (S , π,N) = τ dS − V dπ + µ dN , (4.28)

dX (S ,V , µ) = τ dS + π dV −N dµ , (4.29)

dG (τ, π,N) = −S dτ − V dπ + µ dN , (4.30)

dY (S , π, µ) = τ dS − V dπ −N dµ , (4.31)

dΩ (τ,V , µ) = −S dτ + π dV −N dµ , (4.32)

dO (τ, π, µ) = −S dτ − V dπ −N dµ . (4.33)



4 The . . .Manifolds

1 In what has been written so far we note that all of the state variablesτ, S, π, V , µ and N appear in conjugate pairs such as τdS, −S dτ,π dV , −V dπ, µ dN or −N dµ.

3 2 This is an important property of the energy functions. The obvious symmetry reflects Eq. 4.22, which also implies thatthe Legendre transform is “its own inverse”.

4 However, this is only true if the function is either strictly convexor strictly concave, as the relationship breaks down when the secondderivative f(xi , xj , xk , . . . , xn) is zero somewhere within the domain ofdefinition of the free variables (see also Section 4.4).



4 The . . .Manifolds (2)

5 This is illustrated in Figure 4.2 based on the transformation of thethird order polynomial

f(x) = x(1 − x2) ⇒ ξ(x) =(∂f∂x

)= 1 − 3x2 .

The Legendre transformation of f to φ = f −ξx can be expressed in twodifferent ways:

φ(x) = 2x3 ⇒ φ(ξ) = ±2(

1−ξ3

)3/2. (4.34)

6 Moreover, both x and f can be expressed as functions of the trans-formation variable ξ:

x = ±(

1−ξ3

)1/2⇒ f(ξ) = ±

(1−ξ

3

)1/2 (2+ξ

3

),

7 This means that in total we have to consider three functions involv-ing x, and three involving ξ: f(x), φ(x), ξ(x), f(ξ), φ(ξ) and x(ξ).8 In order to retain the information contained in f(x), we need to know

either φ(x) and ξ(x), or f(ξ) and x(ξ), or simply φ(ξ). In principle, thelatter is undoubtedly the best option, and this is the immediate reasonwhy φ(ξ) is said to be in canonical form.9 Nevertheless, there is an inversion problem when φ is interpretedas a function of ξ rather than x. This is clearly demonstrated by theabove example, where x is an ambiguous function of ξ, which meansthat φ(ξ), and similarly f(ξ), are not functions in the normal sense.10 Instead, they are examples of what we call manifolds (looselyspeaking folded surfaces defined by a function) as illustrated in Fig-ures 4.2c–4.2d[a].

[a] A function is a point-to-point rule that connects a point in the domain of definitionwith a corresponding point in the function range.



Figure a–b). The function f = x(1 −x2) and its Legendre transform φ =

2x3 defined as the intersection of thetangent bundle of f and the ordinateaxis. Note that f shows a maximumand a minimum while φ has no ex-trema.

x

f(x)

a)

x

φ(x)

b)

ξ

f(ξ)

c)

ξ

φ(ξ)

d)

Figure c–d). The same two functionsshown as parametric curves with ξ =

(∂f/∂x) = 1−3x2 along the abscissa.The fold in the plane(s) is located atthe inflection point (∂2f/∂x∂x) = 0.

Figure 4.2: The Legendre transformation of f = x(1− x2) to φ = f − ξx whereξ = (∂f/∂x) = 1 − 3x2. The two domains of definition are x ∈ [−

√2/3,

√2/3]

and ξ ∈ [−1,1]. Together, the four graphs demonstrate how an explicit functionof x, i.e. f(x) or φ(x), is turned into an implicit manifold when expressed interms of ξ, i.e. f(ξ) or φ(ξ).



4 The . . .§ 025 § 24 § 26ENGLISH text is missing.Consistency require-

ments1 Finally, let us look at what differentiating φ with respect to ξ implies.From Theme 22, we know that the answer is −x, but

(∂φ

∂ξ

)= ∓

(1−ξ3

)1/2

initially gives us a manifold in ξ with two solutions, as shown in Fig-ure 4.2.

2 It is only when we introduce x(ξ) from Eq. 4.34 that the full pictureemerges: (∂φ/∂ξ) = −x.

3 This does not mean that we need x(ξ) to calculate the values of(∂φ/∂ξ).

4 However, the relationship between x, ξ and (∂φ/∂ξ) is needed forsystem identification.

5 If the three values have been independently measured, then therelation (∂φ/∂ξ) = −x can be used to test the experimental values forconsistency.

6 The existence of these kinds of tests, which can involve a variety ofphysical measurements, is one of the great strengths of thermodynam-ics.

152 / 1776

ENGLISH text is missing.Vapour–liquid1 The above example also has practical relevance. Many equations

of state, and particularly those for the fluid state, are explicit equationsof the form p = p(T ,V ,N). These equations of state can be used tocalculate the Helmholtz energy of the fluid.

2 The problem in relation to manifolds becomes clear when this en-ergy is transformed into Gibbs energy, which requires us to know theterm V = V(p) in the definition G(p) = A(V(p)) + pV(p). Along thesub-critical isotherms the relation V = V(p) is equivalent to a manifoldwith 3 roots as illustrated in Figure 14.3 on page 759.

3 This results in a characteristic folding of the energy surface as ex-amplified by the p, µ diagram in Figure 14.4 on page 767; where themolar volume v is used as a parameter for p(v) along the abscissa andfor µ(v) along the ordinate axis.

4 In view of these practical considerations, we cannot claim that Leg-endre transforms are globally invertible, but it is true to say that theyare locally invertible on curve segments where the sign of (∂2f/∂x∂x)remains unchanged. We shall therefore take the precaution of treatingeach curve segment as a separate function.

5 For instance, in vapor–liquid equilibrium calculations we shall des-ignate one of the curve segments as being vapour and the other as be-ing liquid, although there is no thermodynamic justification for this dis-tinction, other than the fact that (∂2A/∂V∂V)T ,N = 0 somewhere alongthe isotherm.

154 / 1776

ENGLISH text is missing.From φ to f ?1 In this section we will explain how, and under what conditions, itis possible to convert the Legendre transform φ back to the originalfunction f .

2 This operation will be based on the transformation rule set out inEq. 4.1

φ(y) = f − (∂f∂x

)x , (4.35)

where the properties of the variable y are, as yet, unknown.

3 What we need to know is whether the same transformation ruleapplies to the inverse transform of φ(y), such that

f ?= φ − (∂φ∂y

)y , (4.36)

and, if so, what y is (we have an inkling that y = (∂f/∂x), but we needto prove it).

4 Substituting Eq. 4.36 into Eq. 4.35, rearranging the expressionslightly, and applying the chain rule, gives us:

− (∂φ∂y

)y = − (∂φ

∂x) (∂x∂y

)y ?=

(∂f∂x

)x . (4.37)

5 Next we substitute Eq. 4.35 into Eq. 4.37:

−((∂f∂x

) − ( ∂2f∂x∂x

)x − (∂f

∂x)) (∂x

∂y)y ?=

(∂f∂x

)x

( ∂2f∂x∂x

)x(∂x∂y

)y ?=

(∂f∂x

)x

(∂x∂y

)y ?=

(∂f∂x

) ( ∂2f∂x∂x

)-1. (4.38)

155 / 1776

ENGLISH text is missing.Integration1 For the inverse transform to exist, it must be true that (∂2f/∂x∂x) ,

0. This is a necessary condition.

2 It is also worth adding that in general, partial differential equationsdo not have an analytical solution. However, thermodynamic problemsare often unusually simple, and that is also the case here.

3 Let us define ξ = (∂f/∂x), which gives us(∂x∂y

)y ?= ξ

(∂ξ∂x

)-1.

4 Introducing the logarithmic variables dln y = dy /y and dln ξ =

dξ /ξ produces an elegant solution to the problem (note how the chainrule is used in reverse in the second-last line):

( ∂x∂ ln y

)=

(∂ ln ξ∂x

)-1(∂ ln ξ∂x

) ( ∂x∂ ln y

)= 1

(∂ ln ξ∂ ln y

)= 1 (4.39)

156 / 1776

ENGLISH text is missing.Solution criteria1 The most general solution can be expressed as y = cξ =

c (∂f/∂x), where c is an arbitrary factor.

2 It is natural to choose c = 1, which means that y ≡ ξ, as we knowfrom our earlier discussion of the Legendre transform, but any value ofc ∈ R+ will give the same result, as the inverse transform in Eq. 4.36 isinsensitive to scaling.

3 We say that x and ξ are the natural (or canonical) variables of thefunctions f(x) and φ(ξ), as they allow transformation in both directionswithout the loss of any information.

4 As an aside, it should be mentioned that the function φ(x) does nothave this property, because y = x is not a solution to Eq. 4.36.

5 In the context of thermodynamics, this means that e.g.A(T ,V ,n) can be inverse transformed to U(S,V ,n), provided that(∂2U/∂S∂S)V ,n , 0.

6 However, the transformation rule is symmetrical, and the same re-quirement must also apply in the other direction: U(S ,V ,n) can only betransformed into A(T ,V ,n) if (∂2A/∂T∂T)V ,n , 0.

7 The two requirements may appear contradictory, but in actual factthey are overlapping, because:

( ∂2U∂S∂S

)V ,n =

(∂T∂S

)V ,n =

(∂S∂T

)-1V ,n = − ( ∂2A

∂T∂T)-1V ,n

8 In states in which one of the derivatives tends to zero, and the otherone tends to infinity, the system is on the verge of being unstable.

9 Depending on the variables involved, this relates to either thermal(T , S), mechanical (p, V) or chemical (µi, Ni) stability.

10 These kinds of states are always buried inside a region with two ormore phases, being found in what is referred to as the spinodal regionof the system (from the gr. spinode, meaning “cusp”).

157 / 1776

Use the result from Theme 22 on page 138 to showthat performing two Legendre transformations on f ,first with respect to xi and then with respect to ξi ,

returns the original function.



4 The . . .§ 025 Inverse transform

1 2 Given Definition 4.1 and Eq. 4.22, the inverse Legendre transfor-mation can be written

φi − (∂φi∂ξi

)xj ξi = φi − (−xi) ξi ≡ f , (4.40)

but it is not entirely clear what variable set should be used to define f . To understand the nature of the problem, let us consider the follow-ing example. From Eq. 4.4 it follows that:

U(S ,V ,N) − (∂U(S ,V ,N)∂S

)V ,N S = U(S ,V ,N) − τS = A(τ,V ,N) . (4.41)

3 To approach it from the opposite direction, we have to calculate theLegendre transform of A(τ,V ,N) with respect to the variable τ:

A(τ,V ,N)−(∂A(τ,V ,N)∂τ

)V ,N τ = A(τ,V ,N)−(−S)τ ≡ U(−S ,V ,N) . (4.42)



4 The . . .§ 025 Inverse transform (2)

5 4 This clearly returns the original function U, but the set of canonicalvariables has changed from S ,V ,N to −S,V ,N. In order to get back to where we started, we must perform two

further Legendre transformations:

U(−S ,V ,N) − (∂U(−S ,V ,N)∂(−S)

)V ,N

(−S) = U(−S ,V ,N) − (−τ)(−S)

= A(−τ,V ,N) , (4.43)

A(−τ,V ,N) − (∂A(−τ,V ,N)∂(−τ)

)V ,N

(−τ) = A(−τ,V ,N) − S(−τ)

≡ U(S ,V ,N) . (4.44)

6 In other words, performing repeated Legendre transformations re-veals a closed cycle, where the original information contained in U isretained, as shown in the figure below: Expanded footnote [a]:

[a] In practice it may be easier to use the canonical variables (x, y) = (− (∂φ/∂z)y , y)for the inverse transformation, rather than (x, y) = ((∂φ/∂z)y , y) as has been donehere.



4 The . . .§ 025 Inverse transform (3)

U(S ,V ,N)S−→ A(τ,V ,N)

↑ −τ ↓ τA(−τ,V ,N)

−S←− U(−S ,V ,N)



4 The . . .§ 026 § 25 § 27ENGLISH text is missing.Invertibility1 This example shows that the Legendre transform is (locally) invert-ible, which means that we can choose whatever energy function is mostsuitable for our purposes.

2 Moreover, Eq. 4.10 tells us that the Legendre transform of U withrespect to all of the state variables S, V and N is a function with veryspecial properties. The function, which is known as the null potential, isidentical to zero over the entire definition domain, and as such it has noinverse.

3 Our line of reasoning breaks down for this function, and we mustexpect O(T ,p, µ) to have properties that differ from those of U and theother energy functions.

4 Note that this is not a general mathematical property of the Leg-endre transform; rather it follows inevitably from the fact that U and allof the other extensive properties are described by first-order homoge-neous functions to which Euler’s theorem can be applied. These func-tions display a kind of linearity that results in their total Legendre trans-forms being identical to zero. Hence, the differentials are also zeroover the entire definition domain. In thermodynamics, this is called theGibbs–Duhem equation.

162 / 1776

Show that performing a Legendre transformationof internal energy U(S ,V ,N) with respect to all the

canonical variables S, V and N gives the null potentialO(τ, π, µ) = 0. Next, show that the differential of Ois identical to the Gibbs–Duhem equation; see also

Theme 38 in Chapter 5.



4 The . . .§ 026 Null potential

1 The differential of the null potential in Eq. 4.10 is of course 0.

4 2 However, since this result is valid for the entire domain of definitionit must have a bearing on the degrees of freedom of the system.3 Mathematically, the O-function forms a hypersurface in dim(n) + 2

dimensions. This becomes clear if we differentiate O(τ, π, µ) in Eq. 4.10: dU −τdS − S dτ − V dπ − π dV − µ dN − N dµ = 0.

5 If we spot that τdS +π dV +µ dN is the total differential of internalenergy, we can simplify the expression to:

S dτ+ V dπ+ N dµ = 0 . (4.45)

6 This result is identical to the Gibbs–Duhem Eq. 5.28, which playsan important role in checking the thermodynamic consistency of exper-imental data. 7 This is because Eq. 4.45 tells us about the relationship between τ,

π, µ.8 If a series of measurements is taken in order to obtain experimental

values for all of the variables τ, π, µ, then the Gibbs–Duhem equationallows us to check the quality of the measurements, given that Eq. 4.45,or Eq. 4.10 for that matter, must be fulfilled.9 In practice, this is possible because the chemical potential [a], can

be expressed as a function of temperature and negative pressure, oralternatively as f(τ, π, µ) = 0.

[a] For single component or multicomponent systems at fixed concentration.



4 The . . .§ 027 § 26 § 28 ENGLISH text is missing.Seven samurai1 We have in this chapter introduced seven energy functions: U, H,

A , G, X , Y and O, and six state variables: S, V , N, t , p and µ.

2 On reflection it is clear that only four of these variables are inde-pendent, because the discussion has relied entirely on the fact that in-ternal energy U is a function of S, V and N.

3 All of the other variables have been defined at a later stage eitheras partial derivatives or as transformed functions of U.

4 There is therefore a substantial degree of redundancy in the vari-ables that we are dealing with.

5 This redundancy becomes even clearer when the energy functionsare differentiated twice with respect to their canonical variables.

165 / 1776

Set out all of the Maxwell relations that can be derived byapplying the Schwarz Expanded footnote [a]:

[a] Hermann Amandus Schwarz, 1843–1921. German mathematician. theorem ∂2f/∂xi∂xj = ∂2f/∂xj∂xi

to the functions: U, A , H, Y , G, Ω, X and O, restrictingyourself to single-component systems.



4 The . . .§ 027 Maxwell relations

1 Let us begin by illustrating what is meant by a Maxwell Expanded footnote [a]:

[a] James Clerk Maxwell, 1831–1879. Scottish physicist. relationusing U(S ,V ,N) as our starting point. From the definitions of τ and π itfollows that:

( ∂2U∂S∂V

)N =

( ∂2U∂V∂S

)N ⇒

[∂∂S

(∂U∂V

)S ,N

]V ,N

=[∂∂V

(∂U∂S

)V ,N

]S ,N

⇒ (∂π∂S

)V ,N =

(∂τ∂V

)S ,N .

2 Similarly, cyclic permutation of the variables S ,V and N yields:( ∂2U∂S∂N

)V =

( ∂2U∂N∂S

)V ⇒ (∂µ

∂S)V ,N =

(∂τ∂N

)S ,V ,

( ∂2U∂V∂N

)S =

( ∂2U∂N∂V

)S ⇒ (∂µ

∂V)S ,N =

(∂π∂N

)S ,V .



4 The . . .§ 027 Maxwell relations (2)

3 By systematically comparing the second derivatives of all of theLegendre transforms mentioned in Theme 19 on page 127, we get theresults shown below.

7 4 Also see Theme 24 on page 144 for a complementary table of totaldifferentials and first derivatives.5 Note that the Maxwell relations which are derived from the null po-tential O(τ, π, µ) have no physical interpretation, as experimentally τ, πand µ are dependent variables.6 This means that e.g. (∂τ/∂π)µ is unpredictable because Eq. 4.45on integrated form is equivalent to a relation g such that g(τ, π, µ) = 0.These relations have therefore been omitted from the table.

(∂π∂S

)V ,N=

(∂τ∂V

)S ,N

(∂µ∂S

)V ,N=

(∂τ∂N

)S ,V

(∂µ∂V

)S ,N=

(∂π∂N

)S ,V

− (∂S∂V

)τ,N =

(∂π∂τ

)V ,N − (∂S

∂N)τ,V =

(∂µ∂τ

)V ,N

(∂π∂N

)τ,V =

(∂µ∂V

)τ,N

− (∂τ∂π

)S ,N=

(∂V∂S

)π,N

(∂τ∂N

)S ,π=

(∂µ∂S

)π,N − (∂V

∂N)S ,π=

(∂µ∂π

)S ,N(∂τ

∂V)S ,µ=

(∂π∂S

)V ,µ − (∂τ

∂µ

)S ,V

=(∂N∂S

)V ,µ − (∂π

∂µ

)S ,V

=(∂N∂V

)S ,µ

(∂S∂π

)τ,N =

(∂V∂τ

)π,N − (∂S

∂N)τ,π

=(∂µ∂τ

)π,N − (∂V

∂N)τ,π

=(∂µ∂π

)τ,N

− (∂τ∂π

)S ,µ =

(∂V∂S

)π,µ

− (∂τ∂µ

)S ,π

=(∂N∂S

)π,µ

(∂V∂µ

)S ,π

=(∂N∂π

)S ,µ



4 The . . .§ 027 Maxwell relations (3)

− (∂S∂V

)τ,µ

=(∂π∂τ

)V ,µ

(∂S∂µ

)τ,V

=(∂N∂τ

)V ,µ − (∂π

∂µ

)τ,V

=(∂N∂V

)τ,µ



4 The . . .Alternative procedure

1 We are familiar with the key characteristics of Legendre trans-forms, having looked at topics like transformation geometry, differen-tiation rules, inverse transforms and Maxwell relations.

3 2 In the derivations we have focused primarly on the simple rule usedin Figure 4.1, which shows that the Legendre transform is equal to theintersection of the tangent bundle of the original function and the ordi-nate axis. The subject becomes even more interesting now, as we are going

to prove another rule, which is even simpler, although not so easy toderive.

4 The key to this is a simple change in variables from y and x in(∂y/∂x) to y/x and 1/x in the expression (∂(y/x)/∂(1/x)). 5 This relationship is generally known as the Gibbs–Helmholtz equa-

tion.



4 The . . .§ 028 § 27 § 29

Verify the Gibbs–Helmholtz equation (∂(G/τ)/∂(1/τ))π,N = H.See if you can generalise this result.



4 The . . .§ 028 Gibbs–Helmholtz

1 2 By differentiating and then substituting (∂G/∂τ)π,N = −S and G +

TS = H, we can confirm that the equation is correct, but there is moreto it than that. From the mathematical identity

∂(y/x)∂(1/x) ≡ y + 1

x∂y

∂(1/x) ≡ y − x ∂y∂x

we can conclude that(∂(f/xi)∂(1/xi)

)xj ,xk ,...,xn

= f − ξixi = φi , (4.46)

cf. the Legendre transform in Eq. 4.1. The identity applies generally,including to φij and to any other derived transforms.

4 3 This means that Eq. 4.46 provides an alternative way of calculatingthe value of a Legendre transform, without this changing the originaldefinition. One of the classic applications of the Gibbs–Helmholtz equation

is in calculating the temperature derivative of the equilibrium constant,as shown in Chapter 16, where ln K = −∆rxG/RT is shown to be analmost linear function of 1/T with a slope of ∆rxh/R. 5 An equally useful equation is (∂(A/τ)/∂(1/τ))V ,N = U, and in fact

there are a whole series of similar equations that are based on the sameprinciple.



4 The . . .Looking back ENGLISH text is missing.Continued differentiation1 It remains to be shown that it is straightforward to differentiate the

Gibbs–Helmholtz equation. To illustrate this, let us use the derivative ofthe equation given in the worked example above:

(∂H∂π

)τ,N =

( ∂∂π

(∂(G/τ)∂(1/τ)

)π,N

)τ,N =

( ∂∂(1/τ)

(∂(G/τ)∂π

)τ,N

)π,N =

(∂(V/τ)∂(1/τ)

)π,N .

Note that the enthalpy has been differentiated with respect to nega-tive pressure, which is a canonical variable; the result follows an eas-ily recognisable pattern. We will come back to this form of the Gibbs–Helmholtz equation in Chapter 8, but there it will be used to integrateequations of state at a fixed composition.

2 Finally, it is worth adding that every time we derive an expression ofthe form y−x (∂y/∂x) we can choose to replace it with (∂(y/x)/∂(1/x)).These two alternatives look different, but they are nevertheless mathe-matically identical.

173 / 1776

1 Transformation rule: φi(ξi , xj , xk , . . . , xn) = f(xi , xj , xk , . . . , xn) − ξixi

2 E.g. H = U − (−p)V = U + pV

3 Differentiation I: (∂φi/∂xj)ξi ,xk ,...,xn= (∂f/∂xj)xi ,xk ,...,xn

4 E.g. (∂H/∂S)−p,N = (∂U/∂S)V ,N = T

5 Differentiation II: (∂φi/∂ξi)xj ,xk ,...,xn= −xi

6 E.g. (∂H/∂(−p))S ,N = −V

7 Canonical forms: U(S ,V ,N), H(S , π,N), A(T ,V ,N), G(τ, π,N), . . .

8 Conjugated variables: (τ,S), (π,V), (µ,N), . . .

9 Maxwell relations: (∂2U/∂S∂V)N = (∂2U/∂V∂S)N, . . .

10 Gibbs–Helmholtz: (∂(G/τ)/∂(1/τ))π,N = H


News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds

Part 5

Euler’s Theorem on Homogeneous Functions




5 Euler’s . . .New concepts

1 Homogeneity

2 Scaling laws

3 Extensive (order k = 1)

4 Intensive (order k = 0)

5 Hessian matrix

6 Euler’s 1st theorem (Euler integration)

7 Euler’s 2nd theorem



5 Euler’s . . .State description

1 The thermodynamic state of a so-called simple system is describedmathematically by the mapping f(x) : x ∈ Rn+2 → R where n ∈ Nstands for the number of indentifiable chemical components Expanded footnote [a]:

[a] An empty chamber is without internal mass and has therefore zero chemical com-ponents, but it nevertheless constitutes a thermodynamic system of electromagneticradiation with two degrees of freedom. in the

system. 2 For the system to be physically realisable, it must be macroscop-ically uniform and the properties of the system must not explicitly de-pend on time — which is to say the system is without memory of anykind.3 The domain of definition is mathematically very flexible despite thislimitation and the only formal requirement we ask is that n has a finitevalue.4 This general description fits all geometric surfaces in n + 3 dimen-

sions but the understanding of the thermodynamic systems is consid-erably simplified by the fact that its energy function f is linear along allstate vectors x starting at the origin.



5 Euler’s . . .Homogeneity

1 2 The practical consequences of this linearity[a] will be clarified whenwe start investigating the mathematical properties of f , but let us firstformalise the said linearity by stating that the function f(x1, . . . , xn) ishomogeneous of order k ∈ Z provided that the parametrised functionf(λx) is proportional to λk in the direction of x = (x1, . . . , xn).3 Here, the parameter λ ∈ R+ is a dimensionless measure of thedistance from the origin to the coordinate tuple λx ∈ Rn. We shall laterlearn that λ is also a measure for the physical size of the system.

[a] Herbert Callen. Thermodynamics and an Introduction to Thermostatistics. Wiley,2nd edition, 1985. More precisely, the function f(x1, . . . , xn, ξn+1, . . . , ξm) is homoge-

neous of order k in the variables x1, . . . , xn if the following criteria aresatisfied:

F(X1, . . . ,Xn, ξn+1, . . . , ξm) = λk f(x1, . . . , xn, ξn+1, . . . , ξm) , (5.1)

Xi = λxi . (5.2)



5 Euler’s . . .F(X) vs. f(x)

1

X

F k=−1

k=0

k=1

k=2

Note that F has exactly the same functiondefinition as f . Hence, it is only by conven-tion that we distinguish the two forms. Wemay therefore write:

FxX

f



5 Euler’s . . .F(X) vs. f(x) (2)

2 The distinction between F and f is a littlecontrived and can use a more thorough ex-planation:

3 In thermodynamics it is customary to use the same symbol forclosely related functions like e.g. S(T ,V ,N), S(T ,p,N) and S(H,p,N).4 The state functions are supposedly all the same but their free vari-ables differ so the function mappings must necessarily be different.5 The current situation represents the opposite: Two different sym-bols F and f are used to map two different states while referring to thesame function definition.6 The difference is only a matter of scaling of the system size[a].7 We can easily explain this with an example.

[a] Let X = λx indicate a scaling of the system by the dimensionless factor λ. Here,k = 1 is an important special case, under which the function F is considered ex-tensive. In thermodynamics it is common to write F = X〈f〉 in this case, whereX is an extensive physical quantity with an associated unit of measurement andwhere 〈f〉 is the properly normalized version of f . How this can be understood interms of Euler’s theorem becomes clearer if we consider a system with one singlegroup of homogeneous variables. We could look, for example, at the total numberof moles for all components in the system while it is kept at a constant composition(constant mass or mole fractions). In this case F(X) = λf(x) where X = λx andX = X [X]. Here, denotes the magnitude of X and [] denotes its unit of mea-surement. Let us now choose x = X/X i.e. λ = X. It is then possible to writeF(X) = X · f(X/X) = X [X] · f(X/X)/ [X] = X · f(X/X)/ [X]. The most commonexamples in thermodynamics are X = N [mol], X = M [kg] and X = V [m3]. Specif-ically, when X = N [mol] we also get F(N) = Nf(N = 1 mol)/ [mol] = N〈f〉. The ex-pression is the same as F = λf except that N is a physical quantity whilst λ is a di-mensionless number. This means that f has the unit of [F] while 〈f〉 is the associatedmolar quantity [F mol-1]. Subtle, yes, but nevertheless essential because in our day-to-day work we use 〈f〉 rather than f . It is for instance quite common to write the to-tal Gibbs energy for the system as G(τ, π,N) = Ng(τ, π, x) where x is a vector of molefractions, or, alternatively, as G = Ng(τ, π) if x = ( 1 ). The function g = 〈f〉 de-notes molar Gibbs energy in both cases. Correspondingly, we can write U = Nu(s, v),H = Nh(s, π) etc. for the other molar energies. Our Euler notation does not have one-to-one equivalents in other writings on the subject, but it is hard to avoid conflicts whenthe established standard does not fully reflect the physico-mathematical reality. Let f(x) = 3x2. Hence

F(X) = f(λx) = 3(λx)2 = λ2 · 3x2 = λ2f(x)

which proves that f(x) is homogeneous of 2nd order.

8 However, what is more important right now is that F(X) is just an-other way of writing f(λx). 9 The two function mappings are exactly the same although the func-

tion values (for any given x) are differerent.



5 Euler’s . . .Homogeneity ENGLISH text is missing.Differentiation of F(X)1 Just by curiosity we can go on and see what happens when f(x) is

differentiated to (∂f/∂x) = 6x. Simultaneously, for F(X):(∂F∂X

)=

(∂3(λx)2

∂(λx))= 6(λx) = λ · 6x = λ

(∂f∂x

)

2 There are two things to be noted here:

3 Firstly (∂F/∂X) has exactly the same function definition as (∂f/∂x)and secondly the derivative is no longer homogeneous of 2nd order butof 1st order.

4 The first observation is what should be expected because the dif-ferentiation operator is depending on the function definition only and notthe value of x.

5 The second observation is the prelude to a prolonged discussionthat will occupy the rest of this chapter.

181 / 1776

1 It has tacitly been assumed that ξn+1, . . . , ξm do not take part in thehomogeneity of f . 2 They are of course used in the function descriptions of F , and in all

the partial derivatives of F , but they do not take part in the scaling of xi . That means the scaling laws in Eq. 5.1–5.2 is validfor all choices of ξn+1, . . . , ξm.

5 3 Strange at first sight, maybe, but the grouping of variables into dis-joint subsets is quite natural in physics.4 For example the kinetic energy of an n-particle ensemble EK (m,

υ) = 12

∑ni miυ2i is homogeneous of order 1 with respect to themasses mi and homogeneous of order 2 with respect to the veloci-ties υi . The same function is homogenous of order 3 if mass and veloc-ity are considered simultaneously as independent variables. Taking this a step further, one can for instance say that the volume

v of a cube with dimensions x, y and z is a homogeneous function oforder 1 in the height z given the base area xy, and a homogeneousfunction of order 3 if all the dimensions x, y og z are scaled the sameamount.



5 Euler’s . . .From our daily lives

1 Of particular interest to us are the energy functions U, A , . . . , Owith state variables belonging to S, V , and N or τ, π, and µ.

2 The energy functions, and entropy, volume and mole number, arehomogeneous functions of order 1, while temperature, pressure Expanded footnote [a]:

[a] Here, τ and π are used for temperature and negative pressure respectively, asalready introduced in Chapter 4. In this context we want to stress that the properties(in common with the chemical potential µ) are intensive quantities. and

chemical potential are homogeneous functions of order 0.

3 In the context of thermodynamics these quantities are referred toas extensive (k = 1) and intensive (k = 0) state variables respectively Expanded footnote [a]:

[a] A physical quantity is extensive if it is proportional to the system size and intensiveif it is insensitive to the system size.

.



5 Euler’s . . .Differential in X and ξ

1 We are not initially aiming to embark on a general discussion ofthe multicomponent functions in Eq. 5.1. Instead, we will start with adetailed analysis of the simpler two-variable function F(X , ξ).

3 2 The results will later be generalised in Chapter General Theory,and we lose nothing by taking an easy approach here. In order to exploit the function properties, it makes sense to use

the total differential of F expressed in X and ξ-coordinates:

dF =(∂F∂X

)ξ

dX +(∂F∂ξ

)X

dξ .

4 The variable X is defined as a function of x and λ in Eq. 5.2, andby substituting the total differential of X , or more precisely dX = λdx +

x dλ, we obtain

dF =(∂F∂X

)ξ

x dλ+(∂F∂X

)ξλdx +

(∂F∂ξ

)X

dξ . (5.3)



5 Euler’s . . .Differential in λ and f

1 An alternative would be to make use of F = λk f from Eq. 5.1 asthe starting point for the derivation:

dF = kλk−1f dλ+ λk df .

2 Substitution of the total differential of f expressed in x and ξ-coor-dinates gives:

dF = kλk−1f dλ+ λk (∂f∂x

)ξ

dx + λk (∂f∂ξ

)x

dξ . (5.4)

3 Note that Eq. 5.3 and Eq. 5.4 are two alternative expressions forthe same differential dF (λ, x , ξ).

4 Comparing the equations term-by-term reveals three relations ofgreat importance to thermodynamic methodology:



5 Euler’s . . .Case dλ

1 Comparing the dλ terms reveals that (∂F/∂X)ξ x = kλk−1f . Multi-plying both sides by λ gives

(∂F∂X

)ξλx = kλk f ,

2 which if we substitute in X = λx from Definition 5.2 and F = λk ffrom Eq. 5.1 can be transformed into:

(∂F∂X

)ξ

X = kF . (5.5)

4 3 The closed form of Eq. 5.5 indicates that there is a general solutionto the indefinite integral F(X , ξ) = ∫ (dF)ξ =

∫ (∂F/∂X)ξ dX . The result is known as Euler Expanded footnote [a]:

[a] Leonhard Euler, 1707–1783. Swiss mathematician. s first theorem for homogeneousfunctions, or simply as the Euler integration of F .



5 Euler’s . . .Case dx

1 Comparing the dx terms reveals that (∂F/∂X)ξ λ = λk (∂f/∂x)ξ.Dividing each side by λ leads to:

(∂F∂X

)ξ= λk−1 (∂f

∂x)ξ. (5.6)

3 2 Because F has the same function definition as f (it is only thenames of the free variables that differ) it follows that ∂F/∂X and ∂f/∂xare equivalent functions. Hence, (∂F/∂X)ξ is a homogeneous function of order k − 1.

4 Differentiation with respect to X therefore reduces the order of ho-mogeneity of F by one.



5 Euler’s . . .Case dξ

1 Comparing the dξ-terms reveals that the homogeneity of thederivative with respect to ξ is unchanged:

(∂F∂ξ

)X= λk (∂f

∂ξ

)x, (5.7)

3 2 Using the same arguments as in the case above this implies thatthe derivative of F with respect to ξ is a new homogeneous function ofdegree k . Differentiation with respect to ξ therefore conserves the homogene-

ity in X .



5 Euler’s . . .Extended notation IENGLISH text is missing.Note1 It should be stressed that the Euler integration in Eq. 5.5 is not

limited to one particular interpretation of X .

2 In fact, any scaled variable x = λ-1X satisfies the equation:(∂f∂x

)ξ x = kf . (5.8)

3 The validity of this statement is confirmed by first combiningEqs. 5.5 and 5.6, and then substituting in Eq. 5.1 and Definition 5.2.

4 This result emphasises the practical importance of Euler’s theoremas outlined in Eq. 5.5.

5 Euler homogeneous functions define a small yet important functionclass in mechanics, and in thermodynamics in particular.

6 One of the reasons of their success is that the Legendre transformφi = f(x, ξ) − (∂f/∂xi) xi of f conserves the homogeneity of f . From thedifferentiation rules layed out above we already know that (∂f/∂xi) ishomogeneous of order k − 1. Hence, the multiplication with xi into theproduct (∂f/∂xi) xi regains the homogeneity of order k . Subtracting twohomogeneous functions with the same homogeneity order k does notchange the value of k and we can conclude that the Legendre transformof f with respect to xi conserves the Euler homogeneous properties off(x, ξ).

189 / 1776

1 The general properties of homogeneous functions will be ex-plained further in Chapter General Theory, but to get a sense of theoverall picture we shall briefly mention what changes are required inEqs. 5.5–5.7 to make them valid for multivariate functions:

n∑i=1

Xi(∂F∂Xi

)Xj,i ,ξl

= kF , (5.9)

(∂F∂Xi

)Xj,i ,ξl

= λk−1 ( ∂f∂xi

)xj,i ,ξl

, (5.10)( ∂F∂ξk

)Xj ,ξl,k

= λk ( ∂f∂ξk

)xj ,ξl,k

. (5.11)

2 In other words, increasing the dimensionality of x does not add tothe complexity of the homogeneous properties of f very much.



5 Euler’s . . .§ 029 § 28 § 30

Internal energy U = U(S ,V ,N) is an extensive functionin the variables S, V , and N. The total differential of Uis dU = τdS +π dV +µ dN. Explore the homogeneity

associated with the functions for τ, π and µ.



5 Euler’s . . .§ 029 Intensive functions

1 2 The functions τ, π and µ mentioned above must first be defined. Mathematically, the total differential of U(S ,V ,N) is:

dU =(∂U∂S

)V ,N dS +

(∂U∂V

)S ,N dV +

(∂U∂N

)S ,V dN .

4 3 Comparing this with the differential given in the text implies that τ =

(∂U/∂S)V ,N, π = (∂U/∂V)S,N and µ = (∂U/∂N)S,V , see also Theme 19on page 127 in Chapter 4. Substitution of k = 1, Xi = S, Xj ∈ V ,N and ξl = ∅ Expanded footnote [a]:

[a] Actually an empty set. Here it is used to denote a missing variable or an emptyvector. in Eq. 5.10

yields:

τ(S ,V ,N) =(∂U(S ,V ,N)

∂S)V ,N = λ0 (∂u(s,v ,n)

∂s)v ,n = τ(s, v ,n) . (5.12)

5 The function τ(S ,V ,N) is obviously homogeneous of order 0 in thevariables S ,V ,N because it is independent of the scaling factor λ.



5 Euler’s . . .§ 029 Intensive functions (2)

7 6 The temperature is commonly referred to as an intensive variableand is taken to be independent of the system size[a].

[a] Valid only when the system is sufficiently big i.e. when the number of particles islarge enough to fix the temperature. Similarly, differentiation of U with respect to V and N yields

π(S ,V ,N) =(∂U(S ,V ,N)

∂V)S ,N = λ0 (∂u(s,v ,n)

∂v)s,n = π(s, v ,n) , (5.13)

µ(S ,V ,N) =(∂U(S ,V ,N)

∂N)S ,V = λ0 (∂u(s,v ,n)

∂n)s,v = µ(s, v ,n) , (5.14)

where the (negative) pressure and the chemical potential are intensivevariables as well.



5 Euler’s . . .§ 030 § 29 § 31

ENGLISH text is missing.Partial pressure1 A prolonged experience with thermodynamic issues will slowly

strengthen our intuition with regard to the role of the intensive and ex-tensive state variables but even at this early stage we can give a strik-ing example on the importance of the Euler homogeneity in the mea-surement of physico-chemical quantities.

2 As all chemists know there is something called partial pressure, butthere are few of us, I believe, who have ever been given the opportunityto reflect on the exact nature of this quantity.

3 Let us by chance look at Eq. 5.13 which tells us that pressure is anintensive variable.

4 We can exploit this property further by substituting k = 0 intoEq. 5.9, see below:

V(∂p∂V

)T ,n +

n∑i=1

Ni( ∂p∂Ni

)T ,V ,Nj,i

= 0 · p = 0

5 Substituting ideal gas behaviour:

V −NRTV2 +

n∑i=1

Ni RTV = 0

6 Back-substituting pıg = NRT/V :

−pıg +n∑

i=1

Ni

Npıg = 0

7 After reorganisation and introduction of the mole fraction yi we get:

pıg =n∑

i=1yi · pıg =

n∑i=1

pi

8 To our surprise the partial pressure is not due to an independentdefinition of any kind but is actually an inevitable implication of the ho-mogeneity properties of pıg representing an ideal gas mixture.

9 In practise, pi also finds use in the description of non-ideal gases,which we to a certain extent may find useful, but only if we at the sametime make clear that the property then is not thermodynamically mea-sureable.

10 Vi cannot, therefore, define the partial pressure as e.g. “the fractionof the pressure that is due to a single component i in the mixture”.

11 That would in fact be quite meaningless because the pressure ina gas mixture is a communal property that is caused by all the compo-nents acting together.

12 Due to the mutual physical interactions between the molecules it isnot possible to sum up the contributions from each of the componentsin an independent manner.

13 The question is then: What is the correct interpretation of the partialpressure?

14 Well, by winding backwards through the list of derivations pi canalternatively be expressed on this form:

pi =( ∂pıg

∂ log Ni

)T ,V ,Nj,i

= Ni(∂pıg

∂Ni

)T ,V ,Nj,i

= Ni RTV = Ni

N · NRTV = yipıg

15 The conclusion is that the partial pressure pi can be understood asthe partial derivative of the pressure of an ideal gas mixture with respectto the logarithm of the mole number of component i .

16 On this ground the quantity is secured as a thermodynamic statevariable of the same kind as pressure itself. [a]

194 / 1776

[a] The discussion of the physical foundation of the partial pressure has a lot in com-mon with the discussion of pH which is defined as “the negative logarithm of the con-centration of H+ ions”. They do both refer to quantities that are not measureable bythermodynamic means alone. I mean: What would be the pH in a sample of methanol?Maybe 20 or even 40 since there are scarcely any H+ ions present in that medium.We instinctively feel there is a lack of stringency which leads us into making wrong as-sumptions about the medium’s true nature. The true nature of the problem becomes alittle clearer if we incorporate the ionization constant of the solvent into the discussionbut even then the definition is not proof because are only able to measure the concen-tration of the H+ ion indirectly (using conductivity for instance).

Internal energy U = U(S ,V ,N) is an extensive functionin the variables S, V and N. The total differential ofU is dU = τdS + π dV + µ dN. What is the correct

integral form of U? Explain the physical significanceof this integral.



5 Euler’s . . .§ 030 Euler integration

1 Substituting k = 1, Xi ∈ S ,V ,N and ξl = ∅ into Eq. 5.9, togetherwith the definitions from Theme 29 on page 177 i.e. τ = (∂U/∂S)V ,N,π = (∂U/∂V)S ,N and µ = (∂U/∂N)S ,V yields Euler’s equation applied tointernal energy:

U = τS + πV + µN . (5.15)

2 We are not being asked about multicomponent mixtures, butEq. 5.15 is quite general and can (by analogy with Eq. 5.9) be extendedto:

U = τS + πV +n∑

i=1µiNi = τS + πV + µTn (5.16)



5 Euler’s . . .§ 031 § 30 § 32ENGLISH text is missing.Remark1 The total differential of internal energy for a single component

system is dU = τdS + πdV + µ dN.

2 The differential can be integrated without actually solving anypartial differential equations because τ, π, µ are intensive (size-independent) variables.

3 Physically, this means that the system can be built up from zerosize[a] in a manner that keeps τ, π, µ constant during the process.

4 This is achieved by combining a large number of small systems atfixed temperature, (negative) pressure and chemical potential.

5 When these subsystems are merged into one big system there willbe no changes in the intensive properties, because the requirements forthermodynamic equilibrium are automatically fulfilled, see also Chap-ter 21 on page 1312.

197 / 1776

[a] By definition, a system of zero size has zero internal energy, i.e. U = 0. In thiscontext, “zero size” refers to zero volume, zero mass and zero entropy. Note, however,that it is not sufficient to assume zero mass, because even an evacuated volume willhave radiation energy proportional to T4!

Gibbs energy G(τ, π,N) is an extensive function ofthe mole number N at a given temperature τ and(negative) pressure π. The total differential of G is

dG = −S dτ−V dπ+ µ dN. Explore the homogeneityassociated with the functions S and V .



5 Euler’s . . .§ 031 Extensive functions

1 2 We will apply the same procedure as was used in Theme 29 onpage 177. First of all the functions S, V , and µ must be defined. Mathemati-

cally, the total differential of G(τ, π,N) is:

dG =(∂G∂τ

)π,N dτ+

(∂G∂π

)τ,N dπ+

(∂G∂N

)τ,π

dN .

3 Comparing this with the differential in the problem formulationyields −S(τ, π,N) = (∂G/∂τ)π,N, −V(τ, π,N) = (∂G/∂π)τ,N and µ =

(∂G/∂N)τ,π.

5 4 From Eq. 5.11 we can deduce that both volume and entropy arehomogeneous functions of order 1 in the mole number at specified tem-perature and (negative) pressure, i.e. they are extensive variables. From Eq. 5.10 it can be seen that the chemical potential is (still) a

homogeneous function of order 0, see also Eq. 5.14.



5 Euler’s . . .§ 033 § 32 § 34ENGLISH text is missing.Non-canonical variables1 Any further discussion of homogeneity should ideally be based on

the properly identified canonical state variables of the function.

2 However, this is not an absolute condition, as the thermodynamicstate of a system is uniquely determined when any of its independentvariable sets is fully specified.

3 For example U is extensive in S,V ,N while G is extensive in N at agiven τ and π, but we can alternatively state the opposite and say thatU is extensive in N at a given τ and π and that G is extensive in S,V ,N.

4 There are plenty of alternative ways to describe the system, but acombination of three arbitrary state variables is not always sufficient.

5 To demonstrate this, let us look at the specification H, τ,N.

6 For example, the ideal gas enthalpy can be written Hıg = H(τ,N).At a given τ,N the system is underspecified, because there is no waywe can determine the pressure.

7 If we try to specify H in addition to τ and N, this will give us aredundant (or even contradictory) statement.

8 To avoid problems of this kind we shall therefore make use ofcanonical variables unless otherwise stated.

200 / 1776

§ 321 Show that G(S ,V ,N) is really an extensive function of the vari-

ables S, V and N, thus verifying the conjecture set out in the introduc-tion to this chapter (on page 184).

201 / 1776

§ 32 Gibbs energy II1 From the definition G(τ, π,N) = U − τS − πV , where U =

U(S,V ,N), τ = (∂U/∂S)V ,N and π = (∂U/∂V)S ,N, it follows that wecan express Gibbs energy as a function of S, V and N, because theright-hand side of the equation only includes U and functions derivedfrom U:

G(S ,V ,N) = U(S ,V ,N) − τ(S,V ,N)S − π(S,V ,N)V .

2 We already know that τ and π are intensive variables, whereas U,S and V are extensive variables.

3 From this it follows that

G(S,V ,N) = λu(s, v ,n) − τ(s, v , n)λs − π(s, v ,n)λv

= λg(s, v ,n) ,

which clearly demonstrates that G is homogeneous, in accordance withthe conjecture.

4 Note that the last equation tacitly assumes the homogeneity rela-tions S = λs,V = λv and N = λn.

202 / 1776

The Gibbs energy of a binary mixture is given as ahomogeneous function of order 1 in the mole numbersN1 and N2 (at fixed T and p). Make a contour diagramillustrating the function G = aN1x2 +N1 ln x1 +N2 ln x2

where N2 is plotted along the ordinate axis and N1 along theabscissa. Let x1 = N1/(N1 +N2) and x2 = N2/(N1 +N2).

Show that the isopleths corresponding to constantG (the contour lines) are equidistant for a series of

evenly distributed Gibbs energy values. Use a = 2.4in your calculations.



5 Euler’s . . .§ 033 Gibbs energy I

1 The function is nonlinear in N1 and N2, and the isopleths must becalculated iteratively using e.g. Newton–Raphson’s method: N2,k+1 =

N2,k − (Gk −G)/µ2, where µ2 = ax21 + ln x2 is the partial derivative of G

with respect to N2.

3 2 A fixed value is selected for N1, and N2 is iterated until Gk→∞ hasconverged to G, see Program 1.4 in Appendix 31. The result obtainedis shown in Figure 5.1 on page 187. Note that each isopleth defines a non-convex region, which can be

interpreted as a fundamental thermodynamic instability.

4 The corresponding two-phase region (the symmetry of the modelreduces the phase equilibrium criterion to µ1 = µ2) can be calculatedfrom the total differential of G, rewritten here into the tangent of theisopleth:

(dN2dN1

)T ,p,G

= − µ1µ2.



5 Euler’s . . .§ 033 Gibbs energy I (2)

6 5 It can be proved (do it!) that the tangent intersects with the y and xaxes at G/µ2 and G/µ1 respectively. This indicates that the phase equilibrium condition is fulfilled when-

ever two points on the same isopleth have common tangents (remem-ber that G takes constant values along each isopleth such that the cri-terion is reduced to µ1 = µ2).



0 2 4 60

1

2

3

4

5

6

N1

N2

Gµ1

Gµ2

Figure 5.1: Contour diagram of Gibbs energy (solid lines). Equidistant values(open circles) are made visible along 3 rays (dotted lines) from the origin.The two-phase region is spanned by the two outermost rays. One of theisopleths indicates the phase equilibrium condition in the shape of a convexhull construction.



5 Euler’s . . .§ 034 § 33 § 35

Homogeneity causes a whole range of remarkableresults. One formula obtained by differentiating Eq. 5.5

is:

X d(∂F∂X

)ξ− k

(∂F∂ξ

)X

dξ = (k − 1)(∂F∂X

)ξ

dX ,

For k = 1 this implies that (∂2F/∂X∂X)ξ = 0 and(∂F/∂ξ)X = (∂2F/∂X∂ξ)X . Verify these results and

give a physical explanation for them.



5 Euler’s . . .§ 034 Euler’s 2nd theorem

1 The left-hand side of Eq. 5.5 is differentiated and the right-handside is replaced by the total differential of F :

X d(∂F∂X

)ξ+

(∂F∂X

)ξ

dX = k(∂F∂X

)ξ

dX + k(∂F∂ξ

)X

dξ .

2 For k = 1 the expression reduces to X d(∂F/∂X)ξ = (∂F/∂ξ)X dξ. 3 Note that X and dξ are arbitrary.

To proceed we need the differential of ∂F/∂X , and because F is a func-tion of X and ξ it can be written as the total differential

d(∂F∂X

)ξ=

( ∂2F∂X∂X

)ξ

dX +( ∂2F∂X∂ξ

)dξ ,

5 4 where dX is also arbitrary. Substituted into the equation above (letting k = 1), this yields theintermediate result:

X( ∂2F∂X∂X

)ξ

dX + X( ∂2F∂X∂ξ

)dξ =

(∂F∂ξ

)X

dξ . (5.17)



5 Euler’s . . .§ 034 Euler’s 2nd theorem (2)

7 6 The trick is to recognise that X ,dX and dξ are independent vari-ables. This implies that two non-trivial relations follow from Eq. 5.17 (one

equation in three variables gives two non-trivial relations), irrespectiveof the actual values of X , dX , ξ and dξ:

( ∂2F∂X∂X

)ξ= 0 , (5.18)

X( ∂2F∂X∂ξ

)=

(∂F∂ξ

)X

(5.19)



5 Euler’s . . .Example ENGLISH text is missing.Parametric form1 From a physical point of view any extensive function F(X , ξ) can

be expressed in the parametric form F = β(ξ)X .

2 This stems from the fact that F(0, ξ) = 0 in addition to ∂2F/∂X∂X =

0, see Eq. 5.18[a] The derivative of F with respect to ξ is therefore anextensive function β′(ξ)X where the second derivative of F with respectto both X and ξ is equal to the intensive parameter β′(ξ).

3 The numerical value of ∂F/∂ξ can easily be obtained by integrating∂2F/∂X∂ξ using the Euler method as shown in Eq. 5.19.

210 / 1776

[a] It should be noted that it is the second derivative function which is zero. It is notsufficient to say that the second derivative is zero at a given point.

ENGLISH text is missing.Extended notation II1 In the general case, x and ξ would be vector quantities.

2 If F(x, ξ) is extensive in x it can be shown that the differential inTheme 34 on page 188 takes the form:

xT d(∂F∂x

)ξ −

(∂F∂ξ

)T

x dξ = 0 , (5.20)

where the differential of ∂F/∂x is written:

d(∂F∂x

)ξ =

( ∂2F∂x∂x

)dx +

( ∂2F∂x∂ξ

)dξ . (5.21)

3 The two quantities dx and dξ are independent, and by substitutingthe differential 5.21 into Eq. 5.20 it follows that:

( ∂2F∂x∂x

)ξ x = 0 , (5.22)

( ∂2F∂ξ∂x

)x =

(∂F∂ξ

)x . (5.23)

4 From a logical point of view Eq. 5.23 is a true generalisation of 5.19,whereas Eq. 5.18 represents a specialisation of 5.22. This means thatthe second derivative of F(X , ξ) with respect to X is zero for all single-variable functions, while the corresponding Hessian F(x, ξ) for multi-variate systems is singular in the direction of x, where x is proportionalto one of the eigenvectors of the Hessian.

5 This is Euler’s second theorem for homogeneous functions.211 / 1776

§ 351 Use the result from Theme 34 on page 188 to determine the

second derivatives (∂2G/∂N∂N)τ,π, (∂2Y/∂S∂S)π,µ and (∂2Ω/∂V∂V)τ,µ,where G = G(τ, π,N) and Y = Y(S , π, µ) and Ω = Ω(τ,V , µ).

212 / 1776

§ 35 Linear potentials1 The three functions G, Y and Ω are extensive in N, S, and V

respectively, i.e. in one variable each.

2 This makes Eq. 5.18 valid and the substitution of the definitions forτ, π and µ yields:

( ∂2G∂N∂N

)τ,π =

(∂µ∂N

)τ,π = 0 , (5.24)

( ∂2Y∂S∂S

)π,µ =

(∂τ∂S

)π,µ = 0 , (5.25)

( ∂2Ω∂V∂V

)τ,µ =

(∂π∂V

)τ,µ = 0 . (5.26)

213 / 1776

§ 361 Show that the Hessian of A = A(τ,V ,N) has only one indepen-

dent element when the temperature τ is taken to be a constant param-eter. Use Eq. 5.22 as your starting point.

214 / 1776

1 Helmholtz energy is extensive in V and N at any given temperatureτ.

2 Bearing in mind Eq. 5.22 the Hessian of A can be expressed as afunction of only one independent variable because the matrix, in addi-tion to being symmetric, must satisfy the relation:

( ∂2A∂x∂x

)τ

x =

(∂π∂V

)τ,N

(∂π∂N

)τ,V

(∂µ∂V

)τ,N

(∂µ∂N

)τ,V

V

N

=

0

0

.

3 The fact that the matrix, like all Hessians, must be symmetric, inthis case leads to (∂π/∂N)τ,V = (∂µ/∂V)τ,N.

4 Altogether there are 4 matrix elements, with 3 associated relations.



5 Euler’s . . .§ 038 § 37 § 39§ 36 Hessians1 The solution of the homogeneous (!) system of equations can be

formulated in an infinite number of ways, one of which is:

NV

(∂µ∂N

)τ,V

NV −1−1 V

N

VN

=

00

. (5.27)

2 Note that the second derivative of Helmholtz energy with respectto the mole number i.e. the matrix element H2,2 = V/N is non-zero inEq. 5.27, while the corresponding second derivative of Gibbs energy iszero in Eq. 5.24. In the form of the chemical potential derivative we canwrite:

from 5.24:(∂µ∂N

)τ,π = 0 ,

and from 5.27:(∂µ∂N

)τ,V , 0 .

3 This emphasises the fact that it is important to have a proper un-derstanding of the many peculiarities of the energy functions, and it alsohighlights that it is essential to know which variables are held constantduring differentiation.

216 / 1776

§ 371 Find analytical expressions for (∂G/∂τ)π,N and (∂G/∂π)τ,N by

using the Euler method to integrate the partial molar entropy and partialmolar volume.

217 / 1776

§ 37 Partial molarity

1 Replace F by G(ξ,n) in Eq. 5.23, where ξT = ( τ π ) and nT =

( N1 N2 · · · Nn ). This gives, without much difficulty:

(∂G∂τ

)π,N

(∂G∂π

)τ,N

=

( ∂2G∂τ∂nT

)π

( ∂2G∂π∂nT

)τ

n = −

( ∂S∂nT

)τ,π

( ∂V∂nT

)τ,π

n = −

sTn

vTn

.

2 The partial derivatives of S and V on the right hand side are calledthe partial molar entropy and partial molar[a] volume respectively.

3 These and other partial molar quantities occur so frequently in thethermodynamic theory of mixtures that they have been given their owndifferential operator, see Theme 10 on page 81.

4 A more compact notation is therefore S = nTs and V = nTv.218 / 1776

[a] A partial molar quantity is defined as f = (∂F/∂n)τ,π, irrespective of whether or notF has τ, π,N1, . . . ,Nn as canonical variables (it is only for Gibbs energy that there is acorrespondence between the two sets of variables).

Substitute U = U(S ,V ,N) into Theme 34 on page 188and show that S dτ+V dπ+N dµ = 0. Do you knowthe name of this equation in thermodynamics? Doesit make any difference if you plug in G = G(τ, π,N)

rather than U(S ,V ,N)?



5 Euler’s . . .§ 038 Gibbs–Duhem

1 Substituting F = U(x, ξ) into Eq. 5.20 where xT = ( S V N )

and ξ = ∅, reduces the expression to xT d(∂U/∂x) = 0. From thedefinitions of τ, π and µ in Theme 29 on page 177 we can write

S dτ+ V dπ+ N dµ = 0 , (5.28)

3 2 which is better known as the Gibbs–Duhem equation. Alternatively, if F = G(x, ξ) where x = N and ξT = ( τ π ), thenEq. 5.20 takes the form:

N d(∂G∂N

)τ,π− (∂G

∂τ

)π,N dτ − (∂G

∂π

)τ,N dπ = 0 .

4 The partial derivatives of G with respect to N, τ and π have beenidentified as µ,−S and −V in Theme 31 on page 182.

5 The expression can therefore be reformulated as N dµ + S dτ +

V dπ = 0 which is identical to the Gibbs–Duhem equation.Tore Haug-Warberg (Chem Eng, NTNU) Termodynamic methods TKP4175 02 September 2013 220 / 1776


5 Euler’s . . .§ 038 Gibbs–Duhem (2)

6 In fact, all Legendre transforms of U end up giving the same Gibbs–Duhem equation.



5 Euler’s . . .§ 039 § 38 § 40ENGLISH text is missing.Comment1 We have not been asked for any extensions, but by analogy to

Eqs. 5.9 and 5.16 the Gibbs–Duhem equation may be extended to amulticomponent form:

S dτ+ V dπ+n∑

i=1Ni dµi = S dτ+ V dπ+ nT dµ = 0 . (5.29)

2 Note that the Gibbs–Duhem equation follows inevitably from theproperties of extensive functions when differentiated, which in turn re-flect their homogeneity.

3 This is true for all functions of this kind, and not only thermody-namic ones.

4 The homogeneity, which effectively removes one degree of free-dom in the function expression, shows up as a mutual dependency inthe partial derivatives (of U).

5 As a result, only n − 1 of the derived intensive state variables areindependent.

6 In single component systems this means that any arbitrary inten-sive variable can be expressed as a function of (at most) two other in-tensive variables, see also Theme 39.

222 / 1776

ENGLISH text is missing.Canonical forms1 It is important to realise that the information contained in U is

conserved during the Legendre transformation to H, A , . . . , O .

2 For example, knowing the Gibbs energy G(τ, π,N) really impliesfull knowledge of U(S ,V ,N), and vice versa.

3 The Gibbs–Duhem equation can therefore be derived from any ofthe energy functions.

4 In particular this also applies to the differential of the null-potentialO(τ, π, µ) which is identical to Eq. 5.28, see Theme 26 on page 153.

223 / 1776

Itensive state1 Understanding the intensive properties of a system is of fundamen-

tal importance for thermodynamic methodology.

2 This emerges clearly from Euler’s 1st theorem in Eqs. 5.8 and 5.5,and also from the Gibbs–Duhem equation 5.28.

3 The first two equations show that the extensive state of a systemis given as an analytic integral over the corresponding intensive proper-ties.

4 The last equation shows that τ, π and µ in a real system are notentirely independent but that they collectively must satisfy a differentialequation called the Gibbs–Duhem equation.

224 / 1776

Starting from Theme 35 on page 191, can you tellhow many thermodynamic variables are needed to

determine the intensive state of a system?



5 Euler’s . . .§ 039 The intensive state

1 A thermodynamic system is normally described by n + 2 indepen-dent state variables.

2 However, the intensive state can be determined once n + 1 (inten-sive) variables are known.

4 3 This fact is illustrated by Eqs. 5.24–5.26, where the derivative ofan intensive property with respect to an extensive variable is zero if theother variables are intensive, and are held constant during differentia-tion. In fact, the (single) extensive variable determines the system size

but has no influence on τ, π and µ.

5 For a single-component system we can describe the intensive statein three different ways:

τ = τ(π, µ) ,

π = π(τ, µ) ,

µ = µ(τ, π) .



5 Euler’s . . .§ 039 The intensive state (2)



5 Euler’s . . .§ 040 § 39 § 41

ENGLISH text is missing.Generalized Euler inte-

gration1 We shall now examine to which extent these results are of generalvalidity, or whether they only apply in the case of an extensive (energy)function F(X) = f(λx).

2 As a basis for the discussion we choose Euler’s theorem 5.8.

3 The functional dependencies are made clearer by explicitly spellingout the argument (x) where needed:

(∂f(x)∂x

)ξ x = kf(x) . (5.8)

4 The main challenge with our analysis so far is that the homogeneityof ∂f/∂x depends on the homogeneity of f .

5 It is only for k = 1 that the derivative is in intensive form.

6 Then what if this is not the case?

7 What is the significance of the intensive state for a function that ishomogeneous of order k , 1?

8 To answer this question we shall tentatively write the Euler equationon another form so that the derivative is always intensive.

9 A step in the right direction might be to multiply the left side ofEq. 5.8 with 1 ≡ x−(k−1) · xk−1.

10 The first factor can then compensate for the fact that the derivativeis homogeneous of order k − 1, tacitly understood that the function f ishomogeneous of order k , but it leads us nowhere unless we are able totransform the entire expression to a form that agrees with Eq. 5.8.

11 It is therefore of considerable interest to show that the variabletransformation

y(x) = xk

is doing exactly this.

12 The chain rule of differentiation gives(∂f(x)∂x

)ξ =

(∂f(x)∂y(x)

)ξ

(∂y(x)∂x

)ξ =

(∂f(x)∂y(x)

)ξ kxk−1 ,

which combined with Eq. 5.8 leads to(∂f(x)∂y(x)

)ξ kxk−1x = kf(x) ,

and thereafter to(∂f(x)∂y(x)

)ξ y(x) = f(x) ,

given the transformation rule xk−1x = y(x).

13 Because y(x) is homogeneous of order k just like f itself so mustthe partial derivative ∂f/∂y be homogeneous of order 0.

14 It is thus an intensive function!

15 That means we have managed to write Euler’s theorem in a generalform applicable to all homogeneous functions, mind except those whichbasically are intensive.

16 This result has far reaching consequences.

17 It means for example that all non-intensive thermodynamic proper-ties can be calculated with knowledge of an intensive property— in thegeneral case several such properties — multiplied by a correspondingcoordinate that says something about the size of the system.

18 It also means that each of these properties has its own, we couldalmost say private, Gibbs–Duhem equation that occupy one degree offreedom among the intensive properties, namely the partial derivatives∂f/∂y.

19 The scaling law does also apply still so that we by defining

X = λx and Y = λk xk ,

can write(∂f(λx)∂y(λx)

)ξ y(λx) = f(λx) ⇒ (∂F(X)

∂Y(X)

)ξ Y(X) = F(X) .

20 We note that the expression to the right of the implication sign is acontinuation of Eq. 5.5 from earlier in this chapter.

21 A generalization to several Y variables gives similarlyn∑

i=1Yi

(∂F(X1,X2,...,Xn)∂Yi(Xi)

)Yj,i ,ξl

= F(X1,X2, . . . ,Xn) , (5.30)

which is a continuation of Eq. 5.9.

22 Both of these equations are on the same form as those we derivedearlier and then must the derived properties be similar too.

23 Theme 36 for example, applies to any non-intensive thermody-namic property f(x) as defined in the text above.

24 We may even dare to say that the variable transformation from x toxk forms the basis for a generalization of Euler’s 1st theorem.

228 / 1776

We shall apply the generalization of the Euler’s 1st theoremonto the nth derivative of the Helmholtz energy for an

ideal gas— on the occasion written as:

A (n)V =

( ∂n(A ıg)∂V∂V ···∂V

)τ,N = (−1)n(n − 1)! · NRT

V · V1−n .

The function A (n)V is homogeneous of order k = 1− n

and by observing the Hessian matrix of this quantitywith respect to the variables Vk and Nk , we can verifythat this matrix is singular in exactly the same manneras the Hessian matrix of A with respect to V and N in

Eq. 5.27.



5 Euler’s . . .§ 040 The intensive state

1 To simplify the discussion, we write first A (n)V on a form that shows

the homogeneity k as an explicit parameter:

A (n)V = (−1)k+1k ! · NRT

V · Vk .

3 2 Here k ! must be understood as |k |! if k < 0. By introducing new state variables Y1 = Vk and Y2 = Nk we get:

A (n)V = (−1)k+1k ! · RT · Y1/k

2 · Y1−1/k1 .

4 The first order partial derivatives of A (n)V are

(∂A (n)V

∂Y1

)τ,Y2

= (−1)k+1k ! · RT · Y1/k2 ·

(1 − 1

k

)Y−1/k

1 ,

(∂A (n)V

∂Y2

)τ,Y1

= (−1)k+1k ! · RT ·(1k

)Y1/k−1

2 · Y1−1/k1 ,




6 5 and we can easily check that these two equations satisfy the gen-eralized form of the Euler’s 1st theorem in Eq. 5.30. We also realize that these derivatives are intensive because Y1

and Y2 appearantly have exponents of the same magnitude but withopposite signs.

7 The second order derivatives are elements in the Hessian matrixof A (n)

V which satisfies the homogeneity condition

(∂2(A (n)V )

∂y∂y

)τ

y = Y2Y1

h22

Y2Y1−1

−1 Y2Y1

Y1

Y2

=

0

0

, (5.31)

where the lower right matrix element h22 can be written:

h22 = (−1)k+1k ! · RT ·(1k

)Y1/k−2

2 ·(1k− 1

)Y1−1/k

1 . (5.32)




8 The matrices 5.31 and 5.27 have the exact same structure despitethe fact that Eq. 5.31 has free variables that are different from Eq. 5.27. 9 Certainly, the coefficient on the outside of the matrix is changed

from (∂µ/∂N)τ,V to h22 because of the variable shift, but this is due tothe nature of the problem since we have chosen to use ideal gas as aspecific example in the latter case.



5 Euler’s . . .Looking backThe theorem of Schwarz1 Repeated differentiation with respect to several variable sets putsus in the danger of committing a systematic error — if the variableswould prove to be interdependent.

2 This is not normally the case, but here it is.

3 The property A (n)V is calculated by differentiation with respect to V

in the first place.

4 When we afterwards replace the free variables V and N with Y1 =

Vk and Y2 = Nk we can no longer use Schwarz’s theorem on the crossterms.

5 That is, not across the variable sets X and Y .

6 A simple example illustrates the problem:( ∂2pıg

∂V∂(1/V)

)T ,N = −2NRT

V ,( ∂2pıg

∂(1/V)∂V)T ,N = 0

7 To our first surprise the cross derivatives are different.

8 But, partial differentiation with respect to V at constant 1/V doesnot really make any sense because constant 1/V also entails constantV .

9 Hence, the entire expression above is erroneous.

10 However, in the heat of the battle it is easy to overlook such details!233 / 1776

1 F(X1, . . . ,Xn, ξn+1, . . . , ξm) = λk f(x1, . . . , xn, ξn+1, . . . , ξm)

2 Scaling: Xi = λxi for all i ∈ [1,n]

3 A = π(τ,V ,n)V +n∑

i=1µi(τ,V ,n)Ni

4 G =n∑

i=1µi(τ, π,n)Ni

5 Gibbs–Duhem equation: S dτ+ V dπ+∑n

i=1 Ni dµi = 0

6 Intensive state: τ = τ(π, µ), π = π(τ, µ) or µ = µ(τ, π)

7 Singularity:

(∂2f∂x∂xT

)

ξ

x = 0

8 Note:(∂µ∂N

)τ,π

= 0 while(∂µ∂N

)τ,V , 0


News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds

Part 7

Equations of state




7 Equations . . .New concepts

1 Integration of Gibbs–Duhem’s equation: Ideal gas

2 Canonical potentials: Helmholtz, Gibbs, Grand canonical

3 Molecular degrees of freedom: Translation, vibration, rotation

4 Atomic gases: Sackur–Tetrode’s equation

5 Solids: Einstein’s equation

6 Quantum mechanics: Photons, phonons, electrons

7 Debye temperature

8 High temperature limit: Dulong–Petit’s rule

9 Statistical mechanics: Virial theorem

10 Empirism: Van der Waals equation of state



7 Equations . . .State description

1 2 Although thermodynamics builds on some basic, fundamental prin-ciples of physical science, which form part of a well-defined mathemat-ical theory, it is essentially an empirical science.3 This is particularly true of the many semi-empirical models that are

used in engineering research.4 That is how it was in the past, and that is how it will continue to be

in the future, because with the exception of a handful of simplificationsused in statistical mechanics[a] there are no precise models, and therewill always be the need for essentially physical models with adjustableparameters.5 A thermodynamic state description needs one or more functions of

state that define the asymptotes of the system (mixture) in a consistentmanner.6 Applying parameters to the functions gives engineers the flexibility

they need, while it also ensures thermodynamic consistency.

[a] Above all crystalline phases, ideal gases and dilute electrolytes. In this chapter we learn about a variety of useful concepts, such asideal gases, virial development, the Einstein–Debye phonon model andVan der Waals theory, but first of all we need to define the concept ofan equation of state.

7 Let us start with any canonical differential, e.g. of internal energy

dU = T dS − p dV +n∑

i=1µi dNi , (7.1)

defined using the canonical variables S, V og Ni .



7 Equations . . .State description (2)

8 If we know the fundamental state functions referred to as the equa-tions of state of the system

T = T(S ,V ,n) , (7.2)

p = p(S ,V ,n) , (7.3)

µi = µi(S ,V ,n) , (7.4)

then Eq. 7.1 can be integrated using Euler’s 1st theorem to give U =

TS − pV +∑n

i=1 µiNi , as shown in Theme 30 in Chapter 5. However,in practice this doesn’t work, because the relationships defined by 7.2–7.4 only exist as implicit functions Expanded footnote [a]:

[a] Only very rarely can entropy be written as an explicit (independent) variable in thestate functions. .



7 Equations . . .State description (3)

9 It is therefore more normal to start with

(dG)T ,p =n∑

i=1µi dNi (7.5)

or alternatively

(dA)T = −p dV +n∑

i=1µi dNi , (7.6)

10 which can be integrated in the same way if the state function(s)µi(T ,p,n) or p(T ,V ,n) and µi(T ,V ,n) are known,[a]

[a] Many textbooks refer to p(T ,V ,n) as if there were no other state functions, but thatis wrong! but it is still difficult to accurately describe the chemical potential of thecompounds in a multicomponent mixture.

11 The exceptions to this are ideal (gas) mixtures for which we canobtain exact solutions using statistical mechanics or, as we shall do ina moment, from integrating the Gibbs–Duhem’s equation using just oneadditional assumption.

12 By taking the ideal gas model as a reference, it is relatively easyto obtain the residual functions Gr,p(T ,p,n) = ∫ (V(T , p,n) − V ıg)dpor A r,v(T ,V ,n) = ∫ (p(T ,V ,n) − pıg)dV , which describe the differencebetween a real mixture and an equivalent ideal gas under the samephysical conditions.13 If we combine this with an expression for the energy of the refer-ence gas, we get the Gibbs or Helmholtz energy of the mixture, and afull set of state functions can then be derived using differentiation.14 This leaves us with just V(T ,p,n) or p(T ,V ,n) as the state func-tions.15 The alternative to the above approach is to derive the thermody-namic functions in non-canonical coordinates:

U(T ,V ,n) rather than U(S,V ,n)

H(T ,V ,n) rather than H(S,p,n)...

S(T ,p,n) rather than S(H,p,n)

This is particularly useful if you only need to obtain U, H or S, but notA or G.16 Typically, this is the case for physical state changes that occur at aconstant composition like for instance fluid flow in pipes, compressorsand valves.17 For systems of variable composition, where the chemical potentialis important, it is more suitable to use A and G because the expressionsµi = (∂A/∂Ni)T ,V ,Nj,i = (∂G/∂Ni)T ,p,Nj,i can then be obtained usingdifferentiation with respect to the canonical variables.18 If we choose to disregard canonical theory for a moment, thenthe chemical potential must be defined as µi = (∂H/∂Ni)T ,p,Nj,i −T (∂S/∂Ni)T ,p,Nj,i = hi − Tsi or a similar expression based on U, butthis approach normally involves more work.



7 Equations . . .Ideell (perfect) gas

1 2 The term ideal (perfect) gas refers to ideal, and from a physicalpoint of view hypothetical, descriptions of the conditions in a real gas. An ideal gas consists of individual particles (atoms, molecules or

radicals) that move freely and independently in space, without any kindof interaction.

7 3 This is the ideal situation because it involves ignoring the inter-molecular forces that exist between the particles in a real gas.4 This includes all of the interactions caused by induced dipoles(Van der Waals forces), dipole-dipole forces, dipole-quadrupole forces,etc.5 In spite of this massive simplification, the total energy of the gas isnot equal to zero, as the particles have their kinetic energy (translation),moment of inertia (rotation), internal degrees if freedom (vibration, in-ternal rotation, torsion, electronic excitation) and chemical binding en-ergy.6 Furthermore, electron and nuclear spin can also play a role. Ideal gases are only valid as a special case of the true thermody-

namic state when pressure approaches zero, and can never be verifieddirectly in a real (physical) system. 8 Nevertheless, we must be allowed to consider the ideal gas law

one of the great scientific discoveries.9 Perhaps it is just as important as Maxwell’s equations, which de-

scribe electromagnetism, although that is a very subjective judgement,as Maxwell’s equations are a coupled set of four complex partial differ-ential equations, whereas the ideal gas law is just a simple algebraicequation.10 What they have in common, then, is not their mathematical formu-lae, but rather their understanding of the underlying physics and of thepractical and theoretical implications.



7 Equations . . .Gibbs–Duhem IENGLISH text is missing.The ideal gas law1 It is not a new idea that the world is composed of more or less

distinct material points, but the experiments that underpin the ideal gaslaw as we know it today are more recent:

Name Observation YearBoyle ( PV )T ,N = c1 1662Charles ( V/T )p,N = c2 1787Dalton ( p/N )T ,V = c3 1801Gay-Lussac ( P/T )V ,N = c4 1809Avogadro (V/N )T ,p = c5 1811

Given what we know today, the above table accurately represents thescientific progress, but at the time the scientists lacked our historicalperspective, and we should be cautious about attributing entirely ratio-nal motivations for their observations.

2 It was only in 1834 that Émile Clapeyron successfully formulatedthe ideal gas law in the simple form:

pıg = NRTV . (7.7)

3 The question we are going to look at is to what extent, and if so how,this law can be used to derive expressions for the Helmholtz energy ofthe gas.

4 We can see that the unit of the product p · V is the same as forA , which means that there may be a deeper relationship between pres-sure–volume and the energy functions.

5 The argument that follows is a great example of how physical andmathematical analysis can be successfully combined with experimentalresults (thanks to the insight of the old scientists, not the author).

6 We will admittedly need to make some choices along the way,which mean that the results are not entirely conclusive, but this mustbe seen in the context of established practice for measuring and report-ing basic thermodynamic functions.

275 / 1776

1 The quickest way to obtain a useful result is to apply the Gibbs–Duhem Eq. 5.28 described in Chapter 5. For a thermodynamic systemwith only one chemical component, the folowing applies:

s dτ (s, v) + v dπ (s, v) + dµ (s, v) = 0 .

2 This differential has no general solution, as s and v are two inde-pendent co-ordinates of the material state, but for a change of state thatis assumed to be isothermal, it become possible to replace s, v by τ, vsuch that dτ = 0 and since v = V/N it follows that dπıg = Rτ/v2 dv,giving us Expanded footnote [a]:

[a] The Gibbs–Duhem equation includes the term v dπ where π is understood as theequation of state π(τ, v). The term v dπ can, therefore, not be given the interpretationv(τ, π) dπ. That would require a full variable transformation given by v dπ = dpv−πdv.In the case above, d(pv) = 0 because it is restricted to an ideal gas at constant tem-perature and mass — thus making v(τ, π) dπ and v dπ (τ, v) seemingly equivalent —but the condition of constant mass is not universal and will in fact break in the nextsection when we start talking about multicomponent systems.

(dµıg)T = − RTv dv ,



7 Equations . . .Gibbs–Duhem I (2)

or:

µıg(T ,p) − µ(T) = −RTv∫v

dνν = RT ln

(vv

).



7 Equations . . .Euler’s homogeneous function theorem

1 2 It is also important to remember that thermodynamic energy func-tions are first-order homogeneous functions to which Euler’s theoremapplies. The total Helmholtz energy of a single-component system can

therefore be written as the integral A = −Npv + µN.

6 3 Substituting in pıg and µıg from the above equations gives us:

A ıg = −NRT + Nµ(T) + NRT ln( vv

) .4 We have thus derived the Helmholtz energy from the ideal gas law,using the fact that Euler’s theorem applies to the extensive properties ofthe gas, and the Gibbs–Duhem equation, which is itself a direct result ofEuler’s theorem. The interpretation of the two parameters µ and v ispostponed till we start discussing multicomponent mixtures in the nextsection.5 Here, it is more important to stress the fact that knowlegde of Ameans knowlegde of all the energy functions. From Chapter 4 on Legendre transforms we know the general

equations S = − (∂A/∂T)V ,N and U = A + TS, which means that italso follows that:

Uıg = N

[µ − T

(∂µ∂T

)]− NRT = Nh − NRT .

7 In other words, the internal energy of the gas depends on the sys-tem’s size N and temperature T through the constant of integration Expanded footnote [a]:

[a] The internal energy U ıg is inevitably independent of both pressure and volume:Ideal gases are a collection of free material points without any form of interaction. Itis only the internal degrees of freedom of the particles that contribute to their energy,and these are also what determine the integration constant µ.

called µ. 8 The latter constant must be determined experimentally in each in-dividual case, or theoretically–experimentally if there are good simula-tion models for the chemical compound (based on spectrometric mea-surements).9 Our provisional conclusion is therefore that we can calculate theHelmholtz energy of an ideal gas with only one chemical componentand its internal energy. From this the other energy functions can beobtained in a similar way using Legendre transforms, apart from onetemperature dependent constant.



7 Equations . . .Gibbs–Duhem II

1 Our next challenge is to generalise this result into one that is sim-ilarly valid for a multicomponent system. In order to do that, we mustwrite the ideal gas law in its extended form

pıg =∑i

NiRTV (7.8)

so that we can identify the components that constitute the mixture.

5 2 This is a function of state in the form p(T ,V ,n), which gives us agood way to approach the question of Helmholtz energy A(T ,V ,n), butin order to determine A using the Euler method of integration as shownin Chapter 5, we also need to know the equation of state µıgi (T ,V ,n).3 This function cannot be obtained experimentally in the same way

as pıg, and must therefore be derived theoretically.4 One way to do this is to apply the Gibbs–Duhem equation again:

S dτ+ V dπ+∑i Ni dµi = 0 . The Gibbs–Duhem equation from Chapter 5 follows directly from

the fact that U is a homogeneous function to which Euler’s theoremapplies when expressed using the variables S, V and n.



7 Equations . . .Gibbs–Duhem II (2)

7 6 This means that there is an interdependence between the differ-entials of U which can be usefully exploited, and using the equation ofstate π = −pıg and the Gibbs–Duhem equation we will obtain a possi-ble solution for µi . Let us substitute for π = −pıg while keeping τ = T constant as

shown below:

−V∑i

NiRT−V2 dV − V

∑i

RTV dNi +

∑i

Ni(dµi)ıgT = 0 ,

or∑i(RT dV

V − RT dNiNi

+ (dµi)T ) = 0 .

8 The simplest solution for an expression with a sum equal to zero isto assume that each individual term is equal to zero:

(dµi)ıgT

RT = − dVV + dNi

Ni; ∀i ∈ [1,n] .



7 Equations . . .Gibbs–Duhem II (3)

13 9 Thermodynamics is a phenomenological science, and we cannotbe sure that the above expression correctly describes the gas mixture,but at least it is a possible solution.10 However, the result agrees with experimental observations of dilutegases, which in the limit case p → 0 behave like a chaotic collection ofparticles without any interaction, and where the macroscopic propertiesassociated with each particle are independent of the other particles.11 Interestingly, deriving the chemical potiential of a component of anideal gas using statistical mechanics gives us exactly the same an-swer[a].12 This is a convincing demonstration of the great power of formalthermodynamic analysis.

[a] Statistical mechanical theory is based on a microscopic description of the propertiesof particles, and involves calculating their average values based on a relatively smallnumber of axioms. Whereas at the macroscopic level we have to accept that ourresult is merely a possible result, the theoretician can assert that it is a correct resultfor a gas consisting of hypothetical particles without any physical extension or mutualinteraction. In other words, a perfect gas. Integrating the differential gives us

1RT (µıg

i (T ,V ,n) − µi (T)) = − ln V

V+ ln Ni

Ni.

Note that the constant of integration µi , which will hereafter be calledthe standard state of component i, is a function of temperature only.14 This means that for any given temperature T we must calculate or

estimate a new value for µi .15 If the function of state were available in the form p(S,V ,n), then µiwould become a genuine constant of integration.16 In practice there are very few systems that can be described in thisway, but radiation from black bodies, as described in Theme 46, is oneexample.



7 Equations . . .Standard potential

1 The normal convention is to assume a pure-component standardstate where V/Ni = RT/p and where p = 1 bar, or, in older referenceworks, where p = 1 atm.

3 2 The standard state µi should in that case be interpreted as rep-resenting µi (T ,p), or more properly µıgi (T ,p,N1 = 0, . . . ,Ni = 1 mol,. . . ,Nn = 0). Bearing in mind this fundamental understanding of what µi means,

the chemical potential of the component can be expressed as

µıgi (T ,V ,n) = µi (T ,p) + RT ln

(NiRTpV

). (7.9)

Here we should particularly note that (dµıgi )T ,V ,n = 0 must be true for all

values of p.



7 Equations . . .Standard potential (2)

10 4 This is of course the case by definition, since T , V and n completelydetermine the thermodynamic state, but at the same time p acts as aparameter in the expression for µıgi .5 It is therefore slightly unclear what happens if p changes.6 The passive role of the standard pressure becomes more obviousif we partially differentiate µıgi with respect to p.7 This gives us (∂µi /∂p)T − RT/p = 0, or put another way: If wechoose to change p then we must also correct µi to ensure that thetotal impact on µıgi is zero.8 It is important for our own self-esteem to acknowledge that in or-der to accurately determine the standard state one needs informationat a high level including precise calorimetric measurements, advancedspectroscopy and quantum physics, or a combination of these things,and that these activities lie outside our area of responsibility.9 We must therefore adapt our way of working to the available ther-modynamic databases, and not vice-versa, which means that it is im-portant to use the data properly. Essentially, standard states are presented in four different ways in

the literature:

µi (T ,p) = f(T) , (7.10)

where f is any suitable parameterized function like e.g. f(T) = a +bT +

cT ln T + · · · ,

µi (T ,p) = ∆fhi (T) − Tsi (T ,p) , (7.11)

µi (T ,p) = ∆fhi (T) + ( hi (T) − hi (T) ) − Tsi (T ,p) , (7.12)




11 which follow from the definition G = U + pV − TS = H − TS fromChapter 4 applied to the standard state as defined in the above para-graph, and finally:

µi (T ,p) =TTµi (T,p) + ∆fh

i (T)

(1 − T

T

)+

T∫

T

cp,i(τ)(1 − T

τ

)dτ .

(7.13)

12 The last relation is derived from the Gibbs–Helmholtz equation inChapter 4:4.6 which in this case says that (∂(µ/T)/∂(1/T))p = h.




15 13 Moreover, it is an experimental fact that all molecules, with the ex-ception of monoatomic ones, have internal energy due to the bondingwithin the molecule.14 Normally this energy is reported as the enthalpy of formation∆fhi (T) of the compound. All of the formation properties (enthalpy, volume, entropy, etc.) of

any given compound AaBb · · ·Zz in an ideal gas state are defined anal-ogously by

∆fhAaBb ···Zz

= hıgAaBb ···Zz

− ahαA − bhβB − · · · − zhζZ ,

where α, β and ζ indicate the phases (configurations) of the variouselements.

18 16 The phase can be liquid or solid, in which case the crystalline con-figuration must also be given, or a hypothetical ideal gas for gaseouselements.17 The chemical formula AaBb · · ·Zz should here be seen as the ele-mental composition, i.e. not the atomic composition, of the molecule. For example, hydrogen bromide would have the formula

(H2)0.5(Br2)0.5, giving the enthalpy of formation

∆fhHBr = hıg

HBr− 0.5hıg

H2− 0.5h lıq

Br2.

19 The reason for this discrepancy, with respect to the establishedpractice for chemical equilibrium reactions, is that according to the IU-PAC convention the enthalpy of a chemical element is defined as zerofor the stable configuration of the element[a]

[a] Such as H2(gas), Ar(gas), S(rhombıc), Br2(lıq), Al(sol), etc. at T = 298.15 K and p = 1 bar.




20 Hence, in the above equation hıgH2

= h lıqBr2

= 0 and the enthalpy offormation ∆fhHBr

≡ hıgHBr

is given relative to this arbitrary reference.

23 21 In particular, note that ∆fhBr2 , 0, as the enthalpy reference bydefinition relates to bromine in the liquid phase and not for bromine inthe (ideal) gas phase.22 This is the main principle that determines how the thermodynamicstandard state is reported, but there are, of course, exceptions to therule: At least one important database (JANAF) uses the more specificdefinition hi (T) = 0 for the elements at all temperatures T , while an-other database perhaps uses T = 0 K as its reference and yet anotherone may use gi (T,p) instead of si (T, p). There are therefore many ways of reporting the standard state, but

for all of them it is true that

hi (T) − hi (T) =T∫

T

cp,i(τ) dτ , (7.14)

si (T ,p) − si (T,p) =T∫

T

cp,i(τ) dln τ . (7.15)

24 One of these equations (it doesn’t matter which) can be viewed asa definition of cp i.e. either cp = (∂h/∂T)p,n or cp = T (∂s/∂T)p,n, whilethe other one follows from the partial differential (dh)p,n = T ds.



7 Equations . . .The superposition principle

1 2 The sections 3.1–3.4 explain a little bit about the fundamental phys-ical principles that, in their different ways, contribute to cv ,i . In practice, all thermodynamic reference works are based on a

large number of carefully recorded measurements supported by oneor more theoretical calculations, which together produce figures for∆fhi (T), si (T,p) and cp,i(T).

3 The enthalpy of formation is generally measured using calorimetry,whereas the heat capacity of an ideal gas component essentially ad-heres to the superposition principle Expanded footnote [a]:

[a] The gas constant R in the above equation is a result of the defintiion of H = U+pV ,which for an ideal gas gives us Hıg = Uıg +(pV)ıg = Uıg +NRT . If we then differentiatewith respect to temperature we get cp,i = cv,i + R.

cp,i = c transv ,i + crot

v ,i + cvıbv ,i + celec

v ,i + R

and can be calculated from spectrometric measurements. 4 The JANAF tables[a] are a good example of how far this develop-ment has gone.5 For relatively small molecules (fewer than 8–10 atoms) it is now

possible to estimate cp,i more precisely than using high-quality calori-metric measurements.

[a] JANAF thermochemical tables. 3rd edition. Part I,II. J. Phys. Chem. Ref. Data,Suppl., 14(1), 1985.



7 Equations . . .Euler integration

1 It is easy to lose one’s way in the details of these calculations.It is therefore worth reiterating that using Euler’s theorem to integrateeuqations 7.8 and 7.9 always works, and that the Helmholtz energy ofthe gas can be expressed in the following general form:

A ıg(T ,V ,n) = πıgV +∑i

Niµıgi

= −∑i

NiRT +∑i

Niµi (T ,p)

︸︷︷︸G(T ,p)

+∑i

NiRT ln(NiRT

pV).

2 Note that, in this expression, the standard state term is a conven-tional Gibbs energy function, because T and p have been selected asthe state variables of µi , rather than T and V. 3 In some strange manner, something has gone wrong here, but

what?4 It is clear that we no longer have a canonical description ofA(T ,V ,n), and that the unfortunate decision to combine T , V and n

from the original set of variables of pıg, with p from the standard stateof µıgi , is creating unintended confusion.



7 Equations . . .Helmholtz potential A(T ,V ,n)

12 The consequences become even clearer if we take A ıg as ourstarting point for differentiating Uıg = (∂(A ıg/T)/∂(1/T))V ,n or −S ıg =

(∂A ıg/∂T)V ,n.3 The partial derivative of A ıg is in both cases defined at constant

volume, but how is it possible to implement this for µi (T ,p)?4 Let us use Uıg as an example. We start with

A ıg

T = −∑i NiR +∑

i Ni µi (T ,p)T +∑

i NiR ln(NiRTpV )

followed by

d(A ıg

T)

V ,n = ∑i Ni

(∂(µi /T)

∂(1/T)

)

p

d(1T)+

(∂(µi /T)

∂p

)

Tdp

−∑i NiR

[T d(1T

)+ 1p dp] .

5 Finally we use the Gibbs–Helmholtz equations from Theme 28 onpage 160 together with (∂(µi /T)/∂p)T = R/p from the equivalentdiscussion on page 279.6 The differential can now be expressed in the form

Uıg(T ,n) =

(d(A ıg/T )

d(1/T)

)

V ,n= ∑

i Nihi (T) −∑i NiRT = Hıg − pıgV ,

which shows that Uıg is only a function of T and n because p and Vcancel out during the derivation.7 The total derivative now has one degree of freedom, and there-

fore takes the same value as the partial derivative (∂(A ıg/T)/∂(1/T ))

throughout the domain of definition.8 But why is it necessary to go via the total differential?9 Well, because otherwise we risk writing (∂(µi /T)/∂(1/T ))V ,n = ui .10 Firstly µi is a function of V and not of V , and secondly V varieswith T when p has been chosen as the standard state.11 It is easy to make a mistake in those circumstances, and it is there-fore safest to go via the total differential.12 The derivation also illustrates how risky it is to introduce non-canonical variables for the thermodynamic potentials. In summary, the following equations apply to ideal gas mixtures:

A ıg(T ,V ,n) =∑i

Niµi (T ,p) +

∑i

NiRT[ln

(NiRTpV

) − 1]

(7.16)

−S ıg(T ,V ,n) =∑i

Nisi (T ,p) +∑i

NiR ln(NiRT

pV)

(7.17)

Uıg(T ,n) = Hıg(T ,n) − NRT (7.18)

Hıg(T ,n) =∑i

Nihi (T) . (7.19)



7 Equations . . .Gibbs potential G(T , p,n)

1 2 Historically speaking, T ,p,n has long been the favoured set of vari-ables for modelling all chemical systems, probably because tempera-ture and pressure are particularly easy to control in a laboratory. In our case, it is possible to calculate Gibbs energy by using the

Euler method to integrate the chemical potential in Eq. 7.9, and substi-tuting in the ideal gas law from Eq. 7.7

µıgi (T ,p,n) = µi (T ,p) + RT ln

(NipNp

), (7.20)

which gives us:

Gıg(T ,p,n) =n∑

i=1µıg

i Ni =n∑

i=1µi Ni + RT

n∑i=1

Ni ln(Ni

N)+ NRT ln

( pp

).

(7.21)

3 Alternatively, we can calculate the Gibbs energy as a Legendretransform of Helmholtz energy (or vice versa).

7 4 This is undoubtedly possible, but since Gıg and A ıg have differentcanonical variables, we also need to invert p(T ,V ,n) to V(T ,p,n).5 If this is impossible, the result of the transformation will be an im-

plicit function [a].6 For most equations of state, with the exception of ideal gases, the

2nd virial equation and certain other models, an explicit transformationfrom T ,p,n to T ,V ,n is impossible.

[a] The specification V(p) has 3 solutions in the steam/liquid phases, and is not afunction in the ordinary sense of the word. On the other hand, p(V) is an unambiguousspecification, which means that there is no problem with doing a Legendre transformof Helmholtz energy to Gibbs energy. This significantly complicates the practical application of Legendre

transforms.Tore Haug-Warberg (Chem Eng, NTNU) Termodynamic methods TKP4175 02 September 2013 290 / 1776


7 Equations . . .Molecular contributions

ENGLISH text is missing.Grand canonical poten-

tial Ω(T ,V ,µ)1 The ideal gas law is a simple model, which allows us to choose any

system variables that we like.

2 For example, we can rewrite µ(T , p,n) or µ(T ,V ,n) as expressionsfor n(T ,V ,µ).

3 Very few models allow us to do this, and so we will make use of theopportunity now that we have it.

4 An alternative way of expressing Eq. 7.20 is:

Nıgi (T ,V ,µ) = pV

RT exp(µi−µi

RT). (7.22)

5 Substituted into the ideal gas law, Eq. 7.22 gives us an adequateequation of state for the grand canonical potential Ω(T ,V ,µ):

pıg(T ,µ) = RTV

n∑i=1

Nıgi = p

n∑i=1

exp(µi−µi

RT). (7.23)

6 The total differential of Ω for a one-component system was derivedin Chapter 4, see Theme 24 on page 144.

7 For an n-component system, we can state that

(dΩ)T ,µ = −p dV . (7.24)

8 The differential (dΩ)T ,µ can be integrated directly, as long as thefunction of state p(T ,V ,µ) is expressed explicitly. Eq. 7.23 is explicit,and if we integrate using Euler’s method at a given T and µ we obtain

Ωıg(T ,V ,µ) = −pıgV = −pVn∑

i=1exp

(µi−µiRT

). (7.25)

9 It may at first sight appear that p is a free variable in the aboveequations, but if we change p, then µi (T ,p) varies accordingly, alwayskeeping the value of Ωıg unchanged (show this).

10 Also note that pıg(T , µi) is a function of only two variables inEq. 7.23, whereas Ωıg(T ,V , µi) in Eq. 7.25 requires three variables.

11 The intensive state is evidently given when n + 1 intensive vari-ables have been determined, while the extensive state requires n + 2variables.

12 This is a general result, which is not limited to ideal gases[a].

13 From Theme 35 in Chapter 5 we know that the partial differentialsof functions with only one extensive variable fulfil the requirements forhomogeneity[a]:

∂T∂S

p,µ1,...,µn =

∂p∂V

T ,µ1,...,µn =

∂µ1∂N1

T ,p,µ2,...,µn

= . . . =∂µn

∂Nn

T ,p,µ1,...,µn−1

= 0 .

14 The intensive variables are clearly independent of the size of thesystem, and the state is therefore given when n + 1 intensive variableshave been determined, cf. Theme 39 on page 195.

15 However, in order to determine the size of the system, we alsoneed an extensive variable.

16 In other words, a thermodynamic system requires n + 2 state vari-ables, in spite of the fact that the intensive state can be described withonly n + 1 variables.

291 / 1776

[a] In contrast to e.g. Uıg = U(T ,n), which only applies to ideal gases.

[a] Zero is zero, most people probably think, but the partial differentials have differentunits in this case, and the similarity only applies to the measure (the absolute value) ofthese quantities.

§ 411 Derive the partial derivatives of Ωıg(T ,V , µj) = −NıgRT with re-

spect to T , V and µi . Show that the results can be interpreted as −S ıg,−pıg and −Nıg

j , without knowing the differentiation rules of the Legen-dre transform beforehand.

292 / 1776

§ 41 Partial differentiation of Ω(T ,V ,µ)1 It’s a good idea to start by establishing some shadow values before

starting on the differentiation process itself.

2 Instead of differentiating pıgV in Eq. 7.25 we will use the fact that−Ωıg = pıgV =

∑i Nıg

i RT = NıgRT . If we differentiate Nıgi (T ,V ,µ) in

Eq. 7.22 we get:(∂Nıg

i∂T

)V ,µ = Nıg

iRTsi −R(µi−µi )

(RT)2 − Nıgi

T =Nıg

iRT

(sıg

i − R),

(∂Nıgi

∂V)T ,µ =

Nıgi

V ,

(∂Nıgi

∂µj

)T ,V ,µk,j

=δij Nıg

iRT .

3 Here we have made use of the relationship −si = − (∂µi /∂T)p,n,which also means that −si = − (∂µi /∂T)V ,µ.

4 Think about this result for a moment.

5 Apart from this fundamental observation, the simplification ofRTsi − R(µi − µi ) in the first partial differential equation relies on anunderstanding of certain principles of thermodynamics.

6 In the general case, µi = hi − Tsi . In the special case of idealgases, hıg

i = hi , and hence Tsi − (µi − µi ) = Tsıgi .

7 We are now ready to derive the partial differentials of Ωıg

(∂Ωıg

∂T

)V ,µ = −RT

n∑i=1

(∂Nıgi

∂T

)V ,µ − NR = −

n∑i=1

Ni(sıgi − R) − NR = −S ıg ,

(∂Ωıg

∂V)T ,µ = −RT

n∑i=1

(∂Nıgi

∂V)T ,µ = −RT

n∑i=1

Nıgi

V = −pıg ,

(∂Ωıg

∂µj

)T ,V ,µk,j

= −RTn∑

i=1

(∂Nıgi

∂µj

)T ,V ,µk,j

= −n∑

i=1Nıg

i δij= −Nıg

j

293 / 1776

1 2 The ideal gas law was established in several stages over the courseof the 17th-19th centuries, as an experimental law for the so-called per-manent gases (air, N2, H2, etc.).3 We now know that there is a direct link between the macroscopic

characteristics of the gas and the microscopic properties of the individ-ual molecules.4 The related discipline is called statistical mechanics or statistical

thermodynamics.5 It would be far too big a job to discuss this theory in detail, but

we will take the time to look at some of the details that underpin ourunderstanding of ideal gases. Without further introduction, let us assume that the relationship

between the particles’ microstates and Helmholtz energy is

ANRT = − ln Q ,

where Q is the total partition function of the system Expanded footnote [a]:

[a] D. F. Lawden. Principles of Thermodynamics and Statistical Mechanics. John Wiley& Sons, 1987.


[a] Robert P. H. Gasser and W. Graham Richards. An Introduction to Statistical Ther-modynamics. World Scientific Publishing Co. Pte. Ltd., 2nd edition, 1995. .

7 6 For a system with n particles in defined locations, Q = qn, but ifthe particles are not in a fixed location (as in a gas), then Q = 1n!qn. The molecular partition function q approximates to the product of

q ≈ qtransqvıbqrot · · · over the (assumed) number of independent terms.

9 8 In principle, this is all we need to know about the gas, but in orderto calculate the necessary properties, we must first know the quantummechanical description of each individual input to the model. We can only barely scratch the surface of quantum mechanics, and

the descriptions given in this section have therefore been kept as simpleas possible.10 However, before we start looking at microstates, we will first estab-

lish the fundamental macroscopic thermodynamic equations that applyto ideal gases.



7 Equations . . .Free translation

1 A particle-in-a-box has three identical degrees of translational free-dom.

3 2 Even though the particle can in principle move freely within thecontrol volume, its kinetic energy is quantified (only standing waves arepermitted). In a cubic box, each direction will have its own quantum number jx ,

jy and jz , which is given the same unit of energy ε = h2/8ml2.

4 The associated partition function Expanded footnote [a]:

[a] Per Chr. Hemmer. Statistisk mekanikk. Tapir Forlag, 1970. In Norwegian. can be written

qtrans =∞∑

jx=1

∞∑jy=1

∞∑jz=1

exp(−(j2x+j2y+j2z )ε

kT).

5 For high quantum numbers j2 = j2x + j2y + j2z ≫ 0 is a virtuallycontinuous function over (an eighth of) the surface of a sphere with aradius of j.



7 Equations . . .Free translation (2)

6 The surface area of the octant of the sphere is πj2/2, which leadsus to the integral

limkT≫ε

qtrans = π2

∞∫

0

j2 exp(−j2ε

kT)dj

= π2(kTε

)3/2∞∫

0

x2e-x2dx

︸︷︷︸√π/4

, x2 =j2εkT ,

→ (2πmkT)3/2Vh3 .



7 Equations . . .Free translation (3)

7 When going from summation to integration, it has been assumedthat kT ≫ εx . Combining this with j2 ≫ 0 gives us j2ε/kT < 1 overmuch of the area of integration, which ensures that the integral is agood approximation of the sum.



7 Equations . . .Sackur–Tetrode S § 421 Calculate the typical translational temperature ε/k = h2/8ml2k

for the light gases hydrogen, helium and neon. Assume that the lengthsof the sides of the box are ≥ 1 nm. Comment on the answer.

298 / 1776

§ 42 Particle-in-a-box1 The table below gives the values for ε/k and the standard boil-

ingpoint of these three low-boiling-point compounds.

Mw[g/mol]ε/k[K]

Tb[K]

H2 2.0159 1.19 20.39He 4.0026 0.60 4.224Ne 20.179 0.12 27.09

2 The length of the sides of the box is l = 1 nm (which is right at thelimit of the range of validity of classical thermodynamics).

3 Except in the case of He, ε/k is reassuringly far below the boilingpoint of the compounds, and in practice all gases have a virtually con-tinuous spectrum of translational energies[a].

299 / 1776

[a] Note that ε ∝ l-2. If the length of the box is changed to e.g. 10 nm, then ε/k will bereduced by a factor of 100. This means that even He behaves in the classical mannerin a macroscopic control volume.

1 For an ensemble of n particles-in-a-box without fixed locations, thetotal partition function becomes Q trans = 1

n!(qtrans)n.

2 This results in a function for communal entropy that only dependson the system’s volume and the mass and number of particles, whilethe geometry and bonds are irrelevant:

Q trans =(qtrans)

n

n!n→∞→ en(qtrans)

n

nnε≪kT−→ (e

n)n [V(2πmkT)3/2

h3

]n.

3 Let us substitute in n = NA N, m = N-1A Mw , k = N-1

A R and ln e = 1.



7 Equations . . .Sackur–Tetrode S (2)

4 Differentiating gives us a set of canonical equations of state forHelmholtz energy:

ptransVNRT = 1 ,

µtrans

RT = 1 − ln[V(2πMwRT)3/2

NN4A h3

].

5 Two of thermodynamic properties that can be derived from this arec transv /R = 3/2 and standard entropy

strans(1 atm)R = −1.164855 · · ·+ 52 ln T[K] + 32 ln Mw[g/mol] , which is also known as the Sackur Expanded footnote [a]:

[a] O. Sackur. Ann. Physik, 36:398, 1911. –Tetrode Expanded footnote [a]:

[a] H. Tetrode. Ann. Physik, 38:434, 1912. Correction in 39 (1912) 255. equation.



7 Equations . . .§ 043 § 42 § 44

Use the Sackur–Tetrode equation to estimate s(298.15 K,1 bar)for all of the noble gases. Compare your answer with

experimental values.



7 Equations . . .§ 043 Sackur–Tetrode

1 A convincing comparison of measured and calculated values isshown in Table 7.1.

2 These calculations are more precise than experimental results, butthe theory cannot be tested on multi-atomic molecules, because theinternal degrees of freedom of the molecules also contribute energy.

3 This topic will be further discussed in the following sections onvibrational and rotational degrees of freedom.



Figure 7.1: Comparison of experimental standard entropies with the Sackur–Tetrode equation (the margin of error for R, NA , h and Mw is less than0.01 cal/mol K).

Mw[g/mol]

s[cal/K mol] [†] strans

[cal/K mol]Mw

[g/mol]s

[cal/K mol] [†] strans

[cal/K mol]

He 4.0026 - 30.11 Kr 83.80 39.17 ± 0.1 39.18Ne 20.179 35.01 ± 0.1 34.93 Xe 131.29 40.70 ± 0.3 40.52Ar 39.948 36.95 ± 0.2 36.97 Rn 222.0 - 42.08

[†] Gilbert Newton Lewis and Merle Randall. Thermodynamics. McGraw-Hill Book

Company, Inc., 2nd edition, 1961.



7 Equations . . .Free vibration

1 The atoms in a molecule are held together by forces which, whensubjected to small perturbations, allow the atoms to vibrate as if theircentres were connected by elastic springs.

2 The partition function for a harmonic oscillator of this kind, with acharacteristic frequency ν, quantum number j and energy factor ε =

hν = hω/2π = ~ω, is

qvıb =∞∑

j=0exp

[−(j+ 12)ε

kT]= exp

( −ε2kT

) ∞∑j=0

exp(−jε

kT).

3 The summation can be written as a geometric series qvıb =√

u∑∞j=0 uj =

√u/(1 − u) where u = e-x = exp(−ε/kT).



7 Equations . . .Free vibration (2)

4 If we divide the numerator and denominator by√

e-x, we get

qvıb =√

u1−u →

√e-x

1−e-x = 1√ex−√

e-x.

5 We can obtain a canonical equation of state if we substitute2 sinh(x/2) =

√ex −

√e-x.

6 When inserted into x = ε/kT this produces

µvıb

RT = − ln qvıb = ln[2 sinh

( hν2kT

)].

7 Differentiating twice with respect to temperature gives us

cvıbv

R =( hν

2kT )2

sinh2( hν2kT )

. (7.26)



7 Equations . . .§ 044 § 43 § 45

The heat capacity of a harmonic oscillator has anupper temperature limit of limT→∞ cvıb

v = R. Show this.



7 Equations . . .§ 044 Harmonic oscillator

1 The power series expansion of sinh(x) = x + 13!x

3 +O(x5) appliedto kT ≫ hν gives:

limkT≫hν

cvıbv

R = limx≪1

x2

(x+ 13!x

3··· )2 = 1 .

2 A harmonic oscillator has two equally important energy contribu-tions (kinetic and potential energy), which gives us cv → R/2+R/2 = Ras the upper temperature limit.

3 In a multi-atomic molecule with many vibration frequencies, thecontributions of each individual oscillator must be summed (linearly) inaccordance with the equipartition principle.



7 Equations . . .Einstein cv

1 2 The table below shows the characteristic vibrational temperaturesε/k = hν/k = hc/λk for a few di- and multi-atomic molecules.3 Unlike the contributions from translation and rotation, the vibra-tional state is only fully developed at high temperatures:

g1,2,3,... λ-11,2,3,...[cm-1] ε/2k[K]

H2 1 4162 ≈ 3000CH4 3,2,1,1 1306,1534,2916,3019 900–2200HCl 1 2885 ≈ 2100H2O 1,1,1 1595,3657,3756 1100–2700 One of the striking features of the oscillator equations is that the

ground state (with the quantum number j = 0) makes a positive contri-bution to the total energy.

4 If we ignore the ground state, the partition function simplifies to

qEinstein → 11−exp( −εkT )

= 11−e-x .

5 The heat capacity can then be expressed as Einstein’s Expanded footnote [a]:

[a] Albert Einstein, 1879–1955. German physicist. equation:

cEinsteinv = x2ex

(ex−1)2 . (7.27)

6 Our intuition tells us that the Eqs. 7.26 and 7.27 are equivalent, be-cause the ground state only makes a constant (temperature-indepen-dent) energy contribution. The explanation for this is by no means obvi-ous at first glance, but if we compare the equations, we can see that itis the case (show this).



7 Equations . . .Free rotation

1 A rotating system with a fixed distribution of mass is called a rigidrotor. The moment of inertia of the rotor determines how much kineticenergy it stores at any given frequency.

2 In quantum mechanics, the moment of inertia of isolated atoms iszero, whereas diatomic (linear) molecules have degenerate moments ofinertia about two of their axes of rotation and a zero moment of inertiaabout the third axis.

3 In multi-atomic (non-linear) molecules, the moments of inertiaabout each axis of rotation are different, unless the symmetry of themolecule dictates otherwise.



7 Equations . . .Free rotation (2)

4 The partition function for a linear molecule with a moment of inertiaI, quantum number j and energy factor ε = ~2/2I = h2/8π2I can beexpressed

qrot,2D = 1σ

∞∑j=0

(2j + 1) exp[−j(j+1)ε

kT],

where σ corrects for any rotational symmetry of the molecule (σH2O = 2,σNH3 = 3, σCH4 = 12).

5 There is no analytical limit to the above summation, but expandingthe power series Expanded footnote [a]:

[a] Gilbert Newton Lewis and Merle Randall. Thermodynamics. McGraw-Hill BookCompany, Inc., 2nd edition, 1961. gives us qrot = kT

ε (1 + ε3kT + 1

15(ε

kT )2 + 4

315(ε

kT )3 +

· · · ) provided that 0.3kT > ε.




7 6 The upper temperature limit can be calculated using integration toobtain an approximation of the partition sum, as shown below:

limkT≫ε qrot,2D = 1σ

∞∫

j=0

(2j + 1)exp [−j(j+1)εkT] dj

= − kTσε e-j(j+1) εkT ∣∣∣∞0→ 8π2IkTσh2 . The rotational behaviour of multi-atomic (non-linear) molecules is

more complex than that of linear molecules, and we will therefore simplystate the partition function without further explanation Expanded footnote [a]:

[a] Note the close analogy between qrot,3D and qtrans. Both of the terms include a kineticenergy contribution, and so there are clear similarties. :

limkT≫ε

qrot,3D =(8π2IkT)3/2

σh3 .

In this case, the moment of inertia is a much more complicated functionof the atomic configuration of the molecule, but it still turns out that theend result can be put on the simple form I = 3√λ1λ2λ3 =

3√detΛ where

λ1, λ2 and λ3 are the eigenvalues of the inertia tensor (a symmetrictensor→ real eigenvalues):

M =∑i

Mi

(yT

i yi I − yiyTi

)∼ QΛQT .




8 Here yi = xi − x represents the molecular coordinates of each ofthe atoms i measured in relation to the molecule’s centre of mass x =∑

i M-1w Mixi .



7 Equations . . .Debye cv § 451 Determine the moment of inertia of an H2O molecule if the bond

angle is 104.5 and the O–H distance is 0.958 Å.314 / 1776

§ 45 Moment of inertia1 Let us number the atoms H(1), H(2) and O(3).

2 Their respective atomic masses are M1 = 1.00794 g mol-1, M2 =

1.00794 g mol-1 and M3 = 15.9994 g mol-1.

3 The molecule can be oriented in an infinite number of ways, andthe coordinates given below are simply based on an arbitrary choice(origin at O(3) and H(1) along the x axis):

x1

[Å] =

−.95800

, x2

[Å] =

0.958 · sin(14.5·2π360 ) = 0.240

0.958 · cos(14.5·2π360 ) = 0.927

0

, x3

[Å] =

000

.

4 From this information, the inertia tensor follows

M

[g mol-1 Å2]=

0.818 −.262 0−.262 0.954 0

0 0 1.772

.

5 Diagonalising M y QΛQT gives us the eigenvalues λ1 = 0.615,λ2 = 1.156 and λ3 = 1.772, which have the same units as M itself.

6 The moment of inertia can now be calculated as I =3√0.615 · 1.156 · 1.772 = 1.080 g mol-1 Å

2= 1.794 · 10-47 kg m2.

315 / 1776

ENGLISH text is missing.Non-linear molecules1 Finally, the canonical equation of state describing the rotation of

non-linear molecules can be written

µrot,3D = −RT ln (8π2IRT)3/2

σN3A h3 ,

as long as we understand that I = 3√λ1λ2λ3 represents the molar mo-ment of inertia.

2 Most molecules, with the exception of hydrogen (isotopes), there-fore have a relatively high moment of inertia, which means that the up-per temperature limit is as low as T . 50 K:

Ref. [1] σ λi ·1047

[kg m2]I·1047

[kg m2]ε/k[K]

Tb[K]

H2 2 0.460 0.460 0 0.460 87.6 20.39CH4 12 5.313 5.313 5.313 5.313 7.6 111.66HCl 1 2.641 2.641 0 2.641 15.3 188.15H2O 2 1.022 1.920 2.942 1.794 22.5 373.15

3 This gives us the classical heat capacity values of R and 3/2R forlinear and non-linear molecules respectively.

316 / 1776

[a] G. M. Barrow. Physical Chemistry. McGraw-Hill, 3rd edition, 1979.

ENGLISH text is missing.Internal degrees of free-

dom1 Vibration and rotation are the “classical” contributions to the intra-

molecular energy function.

2 For simple molecules, these contributions can be added together,provided that the amplitude of vibration is small and doesn’t disturb themolecule’s centre of mass (harmonic oscillator).

3 However, the general description of the molecule is far more com-plex, and must take into account anharmonic vibration or internal rota-tional and torsional fluctuations.

4 In hydrocarbons, e.g. CH3, groups will rotate freely around thebond axis, whereas double bonds such as CH2 = CH2 only allow tor-sional fluctuations.

5 Internal rotation can also be partially prevented by sufficiently lowtemperatures (staggered rotation).

6 An important theoretical limit in this context is the equipartition prin-ciple, which states that a fully developed energy degree of freedom con-tributes 1

2R to cv (vibration results in two degrees of freedom, consist-ing of potential and kinetic energy).

317 / 1776

ENGLISH text is missing.Excitation & quantum

spin1 Electron excitation is a phenomenon that typically occurs at high

temperatures, and the outer electrons in a stable molecule will (gradu-ally) be excited at temperatures & 3000 K.

2 In radicals, excitation can occur at . 1000 K, because their elec-trons are more weakly bonded.

3 The spin of elementary particles can also contribute to the energystate of a molecule.

4 In the case of hydrogen, nuclear spin (cf. parahydrogen and ortho-hydrogen) plays a key role at temperatures . 400 K.

5 Similarly, electron spin is significant if the molecule has unpairedelectrons (cf. NO, O2 and NO2).

318 / 1776

ENGLISH text is missing.Photon gas1 Thermal radiation (in a closed control volume) can be treated as athermodynamic system, provided that we’re not interested in the spec-tral distribution[b] of the radiation.

2 In this case it is actually possible to set up the canonical equationsof state[b] for internal energy

prad =14

3√

34b

(SV)4 ,

T rad = 3√

3S4bV ,

hvor

b = 8π5k 4

15h3c3 = 7.56591 · · · × 10-16 Jm3 K4 .

3 Note that the number of photons present in the radiation space isnot included in the equations (determined by S and V ).

4 In this respect eletromagnetic radiation deviates from the classicalnotion that the particle number of a system can be treated as a freevariable because it is countable.

5 An unexpected consequence of this is that the chemical potentialof the photon must be set at zero (this follows from a Lagrange optimi-sation of the equilibrium state).

6 Using Euler’s method to do a simplified integration with respect toonly S and V gives us the internal energy of the radiation space:

Urad(S ,V) = T radS − pradV =34

3√

3S4

4bV .

7 By eliminating S we can obtain the Stefan[b]–Boltzmann[b] law ofradiation Urad = bVT4. Alternatively, we can eliminate T and keep p,which gives us Urad = 3pV .

319 / 1776

[b] An energy distribution function which has frequency as a free variable.

[b] Herbert Callen. Thermodynamics and an Introduction to Thermostatistics. Wiley,2nd edition, 1985.

[b] Jozef Stefan alias Joseph Stefan, 1835–1993. Slovenian physicist and poet.

[b] Ludwig Boltzmann, 1844–1906. Austrian physicist.

§ 461 One of the most important fission reactions in a uranium bomb

can be written 10 n+235

92 U→ 14256 Ba+91

36 Kr+10 n. The energy released is

3.5 · 10-11 J per fission event. Calculate T rad and prad if 1% of the atomstake part in the fission reaction.

320 / 1776

§ 46 Atomic bomb1 The density of the 235 uranium isotope is 19.1 g cm-3 and its molar

mass is 235.04 g mol-1.

2 If 1% of the mass is involved in the reaction, the energy density willbe u = V -1U = 0.01 · 6.0221 · 1023 · 3.5 · 10-11 · 19.1 · 106/235.04 =

1.71 · 1016 J m-3.

3 From u = bT4 = 3p we can obtain T rad = 6.9 · 107 K and prad =

5.7 · 1010 bar.

4 These figures are representative for the uranium bomb Little Boy,which was dropped on Hiroshima on 6 August 1945.

321 / 1776

ENGLISH text is missing.Debye A1 The Debye[b] model of vibration is based on a periodic crystal

lattice.

2 At each lattice point there is a particle (atom, group of atoms ormolecule) oscillating in the force field of the neighbouring particles.

3 The partition function adheres to the Einstein model on page 301,but in such a way that the frequency can vary for each individual particle.

4 Together all of the particles vibrate within a wide spectrum of fre-quencies (called phonons), right up to the critical frequency ωmax, whichis an indirect measure of the total number of degrees of freedom in thesystem:

ωmax∫

0

g(ω) dω = 3n .

5 The function g(ω) indicates the multiplicity for each frequency, andthe next important assumption in the Debye model is g(ω) ∝ ω2.

6 This is not, of course, entirely accurate, but it is nevertheless amuch better assumption than saying that all of the particles vibrate withthe same frequency, as is assumed in the Einstein model.

7 The explanation for this[b] is complicated, so we will not dwell on ithere, but the result is incredibly simple

g(ω) = 9nω3max

ω2 .

8 The partition function can now be expressed in a modified Einsteinmodel

QDebye =ωmax∏ω

1

[1−exp( −~ωkT )]g(ω),

where the ω represent all of the discrete frequencies in the crystal.

9 At the limit value n → ∞ there will be a continuum of frequencies,and the Helmholtz energy can be expressed as

limn→∞ADebye = 9NRTω3max

ωmax∫

0

ω2 ln[1 − exp

(−~ωkT

)]dω .

322 / 1776

[b] Petrus Josephus Wilhelmus Debije, alias Peter Joseph William Debye, 1884–1966.

Dutch–American physicist.

[b] Herbert Callen. Thermodynamics and an Introduction to Thermostatistics. Wiley,2nd edition, 1985.

1 2 The canonical equation of state can be obtained in the normal waythrough differentiation. When expressed in the dimensionless form it makes sense to in-

troduce a characteristic vibrational temperature θD = ~ωmax/k also re-ferred to as the Debye temperature. The chemical potential then be-comes

µDebye

RT = 9( TθD

)3θD/T∫

0

x2 ln (1 − e-x) dx ,

where x = ~ω/kT = θD/T .



7 Equations . . .Debye cv (2)

3 If we differentiate the partition function twice with respect to tem-perature, it gives us

cDebyev

R = 9( TθD

)3θD/T∫

0

x4ex

(ex−1)2 dx ,

but note that these expressions are two independent approximations ofthe partition function.

4 This introduces a thermodynamic inconsistency where cDebyev ,

−T ∂2µDebye/∂T∂T .



7 Equations . . .The Dulong–Petit law

1 There is no analytical solution to the above integral, but at the uppertemperature limit θD/T → 0 and expanding the power series ex = 1 +

x + O(x2) gives us

limT→∞

cDebyev

R = 9( TθD

)3θD/T∫

0

x4(1+x+··· )(x+··· )2 dx ≃ 9

( TθD

)3θD/T∫

0

x2 dx = 3 .

2 This result shows that each atom in the lattice oscillates in 3 inde-pendent directions if the temperature is sufficiently high.

3 Many metals (Pb, Sn, Ag, . . . ) are essentially in this state at roomtemperature. The limit value is also referred to as the Dulong Expanded footnote [b]:

[b] Pierre Louis Dulong, 1785–1838. French chemist. –Petit Expanded footnote [b]:

[b] Aléxis Thérèse Petit, 1791–1820. French physicist.

law.



7 Equations . . .The Dulong–Petit law (2)

4 The table below shows the original data measured by the two menin 1819. Recalculating their data to our present values for the atomicmasses and heat capacity of water gives the limit value cp/R = 3.020±0.004 Expanded footnote [b]:

[b] The value of cp ·Mw for each of the substances is multiplied with the conversion fac-tor Mw ,O ·cp,H2O /(R ·Mw ,H2O) where Mw ,O = 15.9994 g mol-1, cp,H2O = 75.327 J mol-1 K-1,R = 8.3145 J mol-1 K-1 and Mw ,H2O = 18.0153 g mol-1 before the average valueand the standard deviation is calculated in the usual way as x = N-1 ∑ xi and

σ =

√N-1 ∑ (xi − x)2. while the theoretical estimate stated above is 3. 5 This must said to be an extraordinary good fit— not least so when

considering that the measurements were done almost 200 years ago.



Figure 7.2: Experimental heat capacities of 12 metals plus sulphur measuredby Dulong and Petit Expanded footnote [b]:

[b] Aléxis T. Petit and Pierre L. Dulong. Ann. Chim. Phys., 10:395, 1819. . It is maybe a little weird that sulphur is on this list butaccording to the classic thoughts of the time the element was closely relatedto the metals.

cp [†] Mw [‡] cp ·Mw cp [†] Mw [‡] cp ·Mw

Bi 0.0288 13.30 0.3830 Pb 0.0293 12.95 0.3794Au 0.0298 12.43 0.3704 Pt 0.0314 11.16 0.3740Sn 0.0514 7.35 0.3779 Ag 0.0557 6.75 0.3759Zn 0.0927 4.03 0.3736 Te 0.0912 4.03 0.3675Cu 0.0949 3.957 0.3755 Ni 0.1035 3.69 0.3819Fe 0.1100 3.392 0.3731 Co 0.1498 2.46 0.3686S 0.1880 2.011 0.3780 - - - -

[†] Relative heat capacity. [‡] Relative atomic mass. The reference values are theheat capacity of liquid water and the atomic mass of oxygen respectively.



7 Equations . . .Low temperature limit

1 For the lower temperature limit, the ratio θD/T → ∞ and the inte-gral is a constant.

2 This result has a practical application in the proportional relation-ship

limT→0

cDebyev

R ∝ T3

when determining the entropy of a substance by calorimetry.

3 Instead of the highly time-consuming and difficult process of mea-suring near 0 K, all you need to do is measure the heat capacity downto 15–20 K. 4 Often this is sufficient to determine the factor of proportionality

in the above equation, allowing you to extrapolate the measurementsdown to 0 K.



7 Equations . . .Second virial coefficientENGLISH text is missing.Fermions in metals1 The bonding electrons in a metal belong to a group of particles

called fermions, named after the physicist Enrico Fermi[b].

2 One of the characteristics of fermions is that they cannot share thesame one-particle state (the Pauli [b] exclusion principle).

3 In a metal, the electron energy levels will therefore be filled up fromthe lowest level up to the Fermi level equivalent to the chemical potential(µ) of the electrons in the conduction band.

4 The internal energy of these electrons can be expressed

Uelec

NRT = uRT + π2

4 · kTµ + · · ·

5 The chemical potential of electrons is a relatively abstract concept,and an alternative to this value is the Fermi temperature TF = µ

k , whichfor normal metals is approximately 105 K.

6 Differentiating Uelec gives us the lower temperature limit for theelectronic contribution to the heat capacity of metals:

limT→0

celecv

R = π2T2TF ∝ T .

7 This is in contrast to crystalline phases with covalent bonds wherelimT→0(cv) ∝ T3 due to the phonon contribution.

8 The practical implication of this is that the heat capacity of met-als is dominated by the fermion contribution at sufficiently low tempera-tures (T . 5 K).

329 / 1776

[b] Enrico Fermi, 1901–1954. Italian–American physicist.

[b] Wolfgang Ernst Pauli, 1900–1958. Austrian physicist.

1 The virial equation relates to ideal gases. 2 There is an equivalent power series expansion for osmotic pressurein dilute solutions, but it is harder to derive. Expand the power series

p/RT in molar density ρ = N/V :

pvırVNRT =

∞∑k=1

Bk (T)(NV)k−1

= 1 + B2(T) NV + · · · (7.28)

3 This is the normal way to express the virial development of gases,but it is also possible to invert the series from p(ρ) to ρ(p) as shownbelow. 4 Here the series has been truncated after the second term, which

means that the index 2 can be left off the symbol B:

pV2.vır

NRT = 1 + BNV + · · · ≃ 1 +

BpRT + · · ·

5 Multiplying both sides by NRT/p gives the volume-explicit formV2.vır = NRT/p + BN, which reveals that B is a correction factor forthe molar volume of the gas.



7 Equations . . .Second virial coefficient (2)

6 Experimentally B = limp→0(V −RT/p) is used, but this limit tells usnothing about how B varies with temperature.

7 In practice B < 0 for T < TB ≈ 6Tc, where TB is known as theBoyle Expanded footnote [b]:

[b] Robert Boyle, 1627–1691. English chemist and philosopher. temperature.

8 Statistical mechanics explains the theoretical relationship betweenB and the associated pair potential φ(r) for the interaction between twomolecules in the gas phase:

B(T) = 2πNA

∞∫

0

(1 − e

−φ(r)kT

)r2 dr . (7.29)

9 This equation holds very well for non-polar molecules with vir-tually perfect spherical symmetry, but for rod-shaped molecules andmolecules with permanent dipole moments integration must be donewith respect to the (two) inter-molecular angles of orientation as well asthe inter-molecular distance.10 This results in an integral that looks like the one in 7.29, but whereφ(r) represents the mean pair potential of the molecules.



7 Equations . . .Hard spheres

1

σ r

φLet us assume that φ = 0 for r > σ and φ = ∞for r ≤ σ.

3 2 That means viewing the molecules as hard spheres, with no pair-wise interaction. Substituting this into the integral for B gives

us

Bhs = NA

σ∫

0

r2 dr = 2πNAσ3

3

4 The hard sphere potential produces a reasonable approximationof a real system at high temperatures, but it doesn’t explain why B istemperature-dependent.



7 Equations . . .Potential well

1

−ε σ r

φLet us assume that φ = 0 for r > ασ, φ = −ε forσ < r ≤ ασ and φ = ∞ for r ≤ σ.

2 That means viewing the molecules as hardspheres with pairwise interaction within the po-tential well.

3 Substituting this simple potential into the integral for B (whereα > 1), demonstrates that B(T) is a strongly temperature-dependentfunction with the hard-sphere potential as its upper temperature limit:

Bsw = Bhs + 2πNA

ασ∫σ

(1 − e

εkT

)r2 dr

= Bhs[1 + (α3 − 1)(1 − e

εkT )

].



7 Equations . . .Potential well (2)

4 Another connection between ε,TB and α is revealed if you solvethe equation Bsw(TB) = 0:

εkTB = ln α3

α3−1 ⇔ α3 = e εkTB

e εkTB −1



0 200 400 600 800−400

−300

−200

−100

0

100

BLJ[c

m3

mol

-1]

T [K]

TB

Bhs

Bsw

BLJ

Figure 7.3: The second virial coefficient of methane calculated using thehard sphere, square-well and Lennard-Jones 6:12-potential where σ = 3.85 Åand ε = 1.96 · 10-21 J. Both Bsw and BLJ satisfy the lower temperature limitlimT→0 B = −∞, whilst the upper temperature limit limT→∞ B = 0+ is only ful-filled for BLJ (this can only be seen once you reach T & 104 K).



7 Equations . . .Lennard-Jones[2]

1

−ε σ r

φExpanded footnote :

[a] John Edward Lennard-Jones, 1894–1954. English mathematician and physicist. Let us assume that φ = 4ε[(σr )

12 − (σr )6]

forr > 0.

3 2 This version is often referred to as the 6:12 potential. The exponent 6 has a theoretical basis inthe London Expanded footnote [b]:

[b] Fritz London, 1900–1954. German–american physicist. theory of dispersion forces involv-ing instantaneous dipoles, whilst the exponent12 is empirical.

4 BLJ cannot be determined analytically, and numerical integration istherefore required.

7 5 This is simple in theory, but not quite so straightforward in practice.6 In order to prevent numerical problems, the lower limit is estimated

using a hard-sphere contribution and the upper limit using a power se-ries expansion, see the Program 1.5 in Appendix 31. A typical temperature progression is shown in Figure 7.3 along with

the experimental data for methane Expanded footnote [b]:

[b] J. H. Dymond and E. B. Smith. The Virial Coefficients of Pure Gases and Mixtures.Oxford University Press, 1980. .



7 Equations . . .The Van der Waals equation

1 For a given vector x consisting of the mole fractions x1, x2, . . . , xn

the Van der Waals Expanded footnote [c]:

[c] Johannes Diderik van der Waals, 1837–1923. Dutch physicist. equation of state can be expressed as

pVdW(T , v ,x) = RTv−b(x) −

a(x)v2 , (7.30)

where a(x) =∑

i∑

j aijxixj tells us something about the attractive forcesbetween the molecules and b(x) =

∑i bixi represents the hard-sphere

volume.



7 Equations . . .The Van der Waals equation (2)

2 This means that there must be a relationship between theVan der Waals equation and the virial equation in Section 7.7. Wecan demonstrate this relationship by expanding the power seriespVdWV/NRT at the limit V →∞ (ideal gas):

pVdWVNRT = 1

1−b(NV )− a

RT(NV)

= 1 + b(NV)+ b2 (N

V)2

+ · · ·+ bn (NV)n − a

RT(NV).

3 If we compare the coefficients in this series with the ones inEq. 7.28, we find that Bvdw

2 = b − a/RT , Bvdw3 = b2, Bvdw

4 = b3 etc.



7 Equations . . .The Van der Waals equation (3)

6 4 Only Bvdw2 bears any resemblance to real systems; the other virialcoefficients are completely unrelated to reality.5 In spite of the massive simplification, the Van der Waals equation

of state is often used as a starting point for more complicated equationsof state, see Chapter 13. It is beyond the scope of this chapter to discuss all of the impli-

cations of the Van der Waals equation here, but we will at the veryleast note one important property: The parameters aii and bi can beobtained by measuring the critical point of each individual compoundexperimentally. 7 This does not mean that the equation is valid under near-critical

conditions (it isn’t), but these parametric values produce universal be-haviour which is very valuable.



7 Equations . . .§ 047 § 46 § 48

Show that the parameters b = RTc/8pc, vc = 3b anda = 27(RTc)

2/64pc meet the criteria for a mechanical–critical point defined by (∂p/∂v)T ,x = 0, (∂2p/∂v∂v)T ,x =

0 and p(Tc, vc) = pc.



7 Equations . . .Cubic equations of state

1 2 The values given can obviously be verified by direct substitution,but we will choose a different approach. If you multiply both sides of the equation by v2(v − b) you can

rewrite the equation of state 7.30 in the form p(v − b)v2 = RTv2 −a(v − b), or as

v3 − v2(b + RT

p

)+ v a

p − abp = 0 . (7.31)

4 3 This shows that (at constant pressure) the Van der Waals equationcan be treated as a cubic equation with volume as a free variable. Fromthe problem statement, we know that the pV isotherm has a degenerateextreme point[c] as its critical point.

[c] Both a maximum and a minimum (a horizontal point of inflection). This implies that the equation of state must have three coincidentroots (v − vc)

3 = 0, or in its expanded form:

v3 − v23vc + v3v2c − v3

c = 0 . (7.32)



7 Equations . . .§ 047 Critical points

1 The two Eqs. 7.31 and 7.32 have the same algebraic structure,which allows us to compare the coefficients term by term.

2 This gives rise to the following set of equations:

3vc = b + RTcpc

,

3v2c = a

pc,

v3c = ab

pc,

which must be solved with respect to a, b and vc.



7 Equations . . .§ 047 Critical points (2)

5 3 Alternatively we could have solved the equations with respect toe.g. a, b and R, but traditionally vc is favoured because it is subject toa relatively large degree of experimental uncertainty.4 However, this means that the Van der Waals equation of state hastoo few degrees of freedom to accurately reproduce all measurementsof pc,Tc and vc. The solution to the above set of equations can be expressed

38 =

vcpcRTc

, (7.33)

b = 13 vc , (7.34)

a = 3pcv2c , (7.35)

which can in turn be rewritten in the form used in the problem statement.

6 We should note that the critical compressibility factor zVdWc =

pcvc/RTc = 38 behaves as a universal constant because vc is a de-

pendent variable in the solution.

7 Experimentally this is no more than a rough approximation as0.25 < zc < 0.35 for most molecular compounds.



7 Equations . . .§ 048 § 47 § 49

Pressure is an intensive property (homogeneous functionof degree zero), which implies that p(T , v ,x) = p(T ,V ,n).

Rewrite the Van der Waals Eq. 7.30 in this form.



7 Equations . . .§ 048 VDW in the extensive form

1 Dimensional analysis shows that if v → V and x→ Nx in Eq. 7.30,then homogeneity will be preserved.

2 Consequently, the equation of state can be written in the form

pVdW(T ,V ,n) = NRTV−Nb(x) −

N2a(x)V2 (7.36)



7 Equations . . .§ 049 § 48 § 50

Show that, in its reduced form, the Van der Waalsequation can be written as pr = 8Tr/ (3vr − 1)− 3/v2

r .Draw the function for Tr ∈ 0.75,1,1.5. Identify all ofthe asymptotes to the graph. Specify the physical

range of the function.



7 Equations . . .§ 049 Corresponding state

1 Rewrite the Van der Waals equation in its reduced form by substi-tuting Eqs. 7.34 and 7.35 into Eq. 7.30:

pVdW = RT

v−13 vc

− 3pcv2c

v2 . (7.37)

2 Then divide by pc and define reduced pressure as pr = p/pc andreduced volume as vr = v/vc:

pVdWr =

(RT

pcvc

)

vr−13

− 3v2

r. (7.38)

3 Finally substitute in R/(pcvc) = 1/(zcTc) = 8/(3Tc) from Eq. 7.33and define reduced temperature as Tr = T/Tc.



7 Equations . . .§ 049 Corresponding state (2)

4 This gives us the Van der Waals equation as formulated in theproblem statement:

pVdWr = 8Tr

3vr−1 − 3v2

r. (7.39)

vr

pr

bc.p.

5 The physical range is defined by vr ∈⟨13 ,∞

⟩. Note that the equation contains no

arbitrary model parameters.

7 6 We call this a universal equation, and here we will take the oppor-tunity to mention the principle of corresponding states, which states thatrelated substances behave virtually alike at reduced states. For the lowest isotherm (Tr = 0.75)

there are up to three possible solutions forpr = pr(vr). 8 The fluid is thermodynamically unstable in the mid-range where

(∂pr/∂vr) > 0. This results in a phase transi-tion for the system as described in Chap-ter 14 on vapour–liquid equilibria.



7 Equations . . .Looking backENGLISH text is missing.Condensed phases1 According to Murnaghan[c]’s equation, isobaric expansivity α and

isothermal compressibility β are two independent state functions of tem-perature and pressure respectively.

2 The equation is most useful for pure solids and stoichiometric com-pounds.

3 To derive the chemical potential, we start with the total differentialof molar volume, for our current purposes expressed as (dln v)mur

N =

α(T) dT − β(p) dp, also see Section 8.7 on page 377.

4 Two commonly used functions for α and β are:

α(T) = a1 + a2T + a3T2 ,

β(p) = b11+b2p .

5 Let us start by integrating α(T) at a given reference pressure pbetween the reference temperature T and an arbitrary system temper-ature T . This gives us: ln[v(T , p)/v(T, p)] = a1(T − T) + 1

2a2(T2 −T2 ) − a3(T -1 − T -1 ) = (T − T)[a1 + 1

2a2(T + T) + a3(TT)-1].

6 Next we integrate β(p) at constant T between the reference pres-sure p and an arbitrary system pressure p. This gives us: ln[v(T ,p)/v(T ,p)] = −b -1

2 b1 ln[(1 + b2p)/(1 + b2p)].

7 We can now integrate (dµ)T ,N = v dp between the two states T ,pand T ,p. This produces µmur(T , p) = µ(T ,p) +

∫ pp v(T ,p) dp which

with a little bit of effort can be rewritten as

µmur(T ,p) = µ(T ,p) + v(T , p)p∫p

( 1+b2p1+b2p

)-b1/b2 dp

= µ(T ,p) + v(T , p)1+b2pb2−b1

[( 1+b2p1+b2p

)1−b1/b2 − 1]

= µ(T ,p) +b1

b2−b1

vβ

[ββ

vv − 1

],

where β = β(p) and v = v(T ,p).

8 It is now easy to produce a complete equation of state µmur(T ,p),provided that we have an expression for the compound’s standardchemical potential µ(T ,p).

9 For solid phases at high pressures and polymorphous compounds,it is normal to express the standard molar Gibbs energy as a (kind of)power series of temperature:

µRT = · · ·+ a

T2 + bT + c + d ln

(TT

)+ eT + fT2 + · · ·

10 A trivial integration of µmur using the Euler method gives us theGibbs energy function

Gmur(T ,p,N) = Nµmur(T ,p) .

349 / 1776

[c] F. D. Murnaghan. Proc. Natl. Acad. Sci. U. S. A., 30:244–247, 1944.

§ 501 Murnaghan’s equation is derived by integrating the expansivity

with respect to T at a given p, and then integrating the compressibilitywith respect to p at constant T , but this provides no guarantee that ourstarting point was thermodynamically consistent. Use differentiation toshow that α = f(T) and β = g(p) genuinely represent a consistent basisfor a model.

350 / 1776

§ 50 The Murnaghan model1 It is easy to make apparently independent assumptions without

checking whether they are mutually coherent.

2 For the model to be consistent, the following third-order Maxwellequation

∂2v∂T∂p = ∂2v

∂p∂T

must hold. Substituting in (∂v/∂T)p = vf(T) and (∂v/∂p)T = −vg(p)gives us

− (∂v∂T

)p g(p) ?=

(∂v∂p

)T f(T) ,

where f(T) and g(p) are functions of different state variables.

3 This avoids circular references.

4 If we rewrite the last equation slightly we get

vf(T)g(p) ?= vg(p)f(T) .

Here the right-hand side is identical to the left-hand side, so the test hasbeen satisfied.

351 / 1776

§ 511 Assume that thermal expansivity is constant, i.e. that a2 = 0 K-2

and a3 = 0 K. Find an expression for (∂cp/∂p)T ,N that is true for thissimplified Murnaghan model. What is the sign of the derivative?

352 / 1776

§ 51 Murnaghan CP1 The fact that cp is pressure-dependent can be derived as shown

below.

2 Note the Maxwell relationship used at the start of the equation:

(∂cmurp

∂p)T ,N = ∂

∂p

[T

(∂s∂T

)p,N

]T ,N = T ∂

∂T

[(∂s∂p

)T ,N

]

p,N

= − T ∂∂T

[(∂v∂T

)p,N

]p,N

= − T(∂(vα)∂T

)p,N

= − Tvα2

≤ 0

353 / 1776

§ 521 For a one-component system to be thermodynamically stable, itmust satisfy the conditions β > 0 and cp > 0. Are these conditions(always) met for Murnaghan’s equation?

354 / 1776

§ 52 Thermal instability1 The first condition is satisfied for all T and p if b1 > 0 and b2 > 0.

2 With respect to the second condition, in Theme 51 we showed that(∂cmur

p /∂p)T ,N

< 0 for all V > 0.

3 Since cp < ∞, there will always be a pressure p < ∞ at whichcmur

p = 0, unless

∞∫p

vmur dp?< ∞ .

4 It can be demonstrated that this is not the case, and that Mur-naghan’s model therefore does not satisfy the requirement for thermalstability at all T ,p.

355 / 1776

1 Helmholtz energy of an ideal gas:

A ıg(T ,V ,n) =∑i

Niµi (T ,p) +

∑i

NiRT[ln

(NiRTpV

) − 1]

−S ıg(T ,V ,n) =∑i

Nisi (T ,p) +∑i

NiR ln(NiRT

pV)

Uıg(T ,n) = Hıg(T ,n) − NRT

Hıg(T ,n) =∑i

Nihi (T)

2 Standard state contribution:

1) µi (T ,p) = µi (T ,p)

2) µi (T ,p) = ∆fhi (T) − Tsi (T ,p)

3) µi (T ,p) = ∆fhi (T) + (hi (T) − hi (T)) − Tsi (T ,p)



7 Equations . . .Looking back (2)

3 Temperature dependency:

hi (T) − hi (T) =T∫

T

cp,i(τ) dτ ,

si (T) − si (T) =T∫

T

cp,i(τ) dln τ

4 The superposition principle:

cp,i = c transv ,i + crot

v ,i + cvıbv ,i + celec

v ,i + R

5 The virial expansion theorem:

pvırVNRT =

∞∑k=1

Bk (T)(NV)k−1

= 1 + B2(T) NV + · · ·



7 Equations . . .Looking back (3)

6 The 2nd virial coefficient:

B(T) = 2πNA

∞∫

0

(1 − e

−φ(r)kT

)r2 dr

7 Corresponding state principle:

pVdW = RT

v−13 vc

− 3pcv2c

v2


News f(ξ) vs f(x) f(T ,V) f(T , p) Bridgman CP and CV α and β Ideal gas Waals Olds

Part 8

State changes at constant composition




8 State . . .New concepts

1 Non-canonical variables

2 The Bridgman table

3 Heat capacity

4 Isothermal expansivity

5 Isobaric compressibility

6 Critical point divergence



8 State . . .U, H and S

1 The most important quantities in process engineering include en-ergy, enthalpy and entropy, generally expressed as functions of temper-ature and pressure (or density), and composition.

4 2 These are particularly important to the energy balances for mediaflowing through valves, turbines, compressors and pipelines.3 It is therefore of great practical relevance to have a fundamental

understanding of the thermodynamic properties of a fluid (of constantcomposition). In many cases it is sufficient to know U, H and S as functions of

just one (common) set of variables, but from a theoretical point of viewthere is great value in being able to choose the variables that are mostsuited to the particular application.

5 In this context, T ,p,n and T ,V ,n are particularly relevant, as arethe functions’ canonical variables.



8 State . . .Canonical variables I § 531 Find the total differentials of U(S ,V ,n), H(S ,p,n) and S(U,V ,n)in the specified sets of variables. Assume that the composition is ineach case constant. How easy do you think it would be to integratethese differentials?

362 / 1776

§ 53 Revision1 As a warm-up exercise, we will start by revising the canonical

expressions for dU, dH and dS:

(dU)n = T dS − p dV , (8.1)

(dH)n = T dS + V dp , (8.2)

(dS)n = 1T dU + p

T dV . (8.3)

The structure of these differentials is clearly simple, but this makes themall the more difficult to integrate, as the equations of state are rarely ornever known for T(S ,V ,n) and p(S ,V ,n), or alternatively T(S , p,n)and V(S , p,n), or 1

T (U,V ,n) and pT (U,V ,n).

363 / 1776

1 Canonical means conforming to a pattern or rule, and in thermody-namics this refers to the structure of the differentials.

2 For example, S ,V ,n are canonical variables for U, whereas T ,V ,nand T ,p,n are not.

4 3 This becomes clear if we compare the above equations with 8.13and 8.24, which show that (dU)n becomes more complex if we strayaway from the canonical description. Other important functions (stated with their associated canonical

variables) include: H(S ,p,n), A(T ,V ,n), G(T ,p,n) and S(U,V ,n).The relationship between these functions can be derived from U, whichin this case is a fundamental energy function:

U(S ,V ,n) = TS − pV +n∑

i=1µiNi . (8.4)



8 State . . .Canonical variables I (2)

7 5 Alternatively we can view S as a new fundamental relationshipthat gives us the same information as U, in which case it is called aMassieu[a] function:

S(U,V ,n) = 1T U + pT V − n∑i=1

µiT Ni . (8.5)

6 The remaining functions in the list can be obtained by performingLegendre transformations on the relevant fundamental relationship[a].For instance, transforming U with respect to S gives rise to a new func-tion, which is called Helmholtz energy, which is defined by the equationA = U − TS.

[a] François Jacques Dominique Massieu, 1832–1896. French mathematician andphysicist.

[a] If you start out from U(S,V ,n), the coordinates will always be positive. If on theother hand you use S(U,V ,n), the coordinates make more sense in relation to thephysical system, in the sense that it is easier to accept energy as a free variable thanentropy. Thanks to our study of the theory behind Legendre transforms in

Chapter 4, we know that the new function is particularly useful if wereplace S with T as a free variable:

−S =(∂A∂T

)V ,n , (8.6)

−p =(∂A∂V

)T ,n , (8.7)

µi =(∂A∂Ni

)T ,V ,Nj,i

. (8.8)

8 This result is the reason behind the term canonical variables. 9 Also note that the differential of A with respect to T is −S, whereasthe differential of U with respect to S is T .10 This duality means that S and T are referred to as conjugate vari-ables.11 The differentials of A with respect to V and n are the same as thoseof U, but remember that T has to replace S as one of the variables thatis being held constant.



8 State . . .Canonical variables II

1 Assume for a change that we do not remember the rules of differ-entiation leading to Eqs. 8.6–8.8.

2 What alternatives exist for the differentiation in that case?

5 3 Let us apply Eq. 8.6 to a single component system and use that asan example.4 Substitution of respectively the Legendre transform A = U − TS

and the Euler form A = −pV + µN leads to:(∂A∂T

)V ,N = (∂(U−TS)∂T

)V ,N = (∂(−pV+µN)∂T

)V ,N . By inserting Eq. 8.6 on the left side and writing the partial deriva-

tives as differential quotients we get:

−S =(d(U−TS)

dT)V ,N =

(d(−pV+µN)dT

)V ,N ,

=(dU−TdS−SdT

dT)V ,N =

(−pdV−Vdp+µdN+NdµdT

)V ,N .



8 State . . .Canonical variables II (2)

6 Replacing the differential dU = T dS −p dV +µ dN in the middle ofthe expression, and the Gibbs–Duhem equation S dT−V dp+N dµ = 0in the last part of the expression, yields:

−S =(−pdV+µdN−SdT

dT)V ,N =

(−pdV+µdN−SdTdT

)V ,N ,

= −S = −S .

7 In the end, all the three derivatives were proved to be equal, al-though it is was not easy to be in charge of the operation and maintaina good overview all the time. 8 We shall therefore make use of the next subchapter to learn more

about the limitations that start to play when, or if, we want to deviatefrom the canonical formulations.



8 State . . .Non-canonical variablesENGLISH text is missing.Non-canonical variables I1 The canonical variables are of prime concern in the general de-

scription of thermodynamic states.

2 The question we are now going to ask ourselves is to what extentwe can choose among the co-ordinates (S,V ,n), (T ,V ,n), (T ,p,n),(U,V ,n), etc.

3 We know that these variables are the natural choices for describingU, A , G and S respectively, but could e.g. (T ,V ,n) serve as a commonset of variables for all of the functions?

4 That would at be immensely practical, if nothing else.

5 The answer proves to be ambiguous.

6 On the one side this choice is entirely possible but on the other sideit does complicate our formulas.

7 In order to get to the bottom of the matter, you need a thorough un-derstanding of function transformations in general and Legendre trans-forms in particular.

8 That discussion is not based so much on thermodynamics, and soit may be helpful to leave the thermodynamic equations out of it.

9 We will therefore study an anonymous function f(x) with an asso-ciated Legendre transform φ(ξ).

368 / 1776

1 2 But, anonymity is not synonymous with denial and we shall con-tinue to keep our thermodynamic cutlery sharp.3 In particular we must inculcate ourselves that a thermodynamicsstate is inextricably linked to the information contained in the graph(x, f(x)).4 From Chapter 4 we know that the Legendre transform φ = f − ξxpreserves all of the information about the thermodynamic state.5 Hence the graph (ξ, φ(ξ)) is equivalent to (x, f(x)).6 At the same time we argued that neither f(ξ) nor φ(x) possessesthe same property.7 In other words, the knowlegde of f(x) is sufficient for calculatingboth the function value and ξ whereas f(ξ) is less valuable becausex(ξ) cannot be derived from this function.8 The alternative is to use φ(ξ) which holds the same information asf(x).



8 State . . .Non-canonical variables III

1 To shed some light on this issue we shall nevertheless have acloser look at the possibility of using f(ξ) as an alternative to f(x) andφ(ξ). 2 The possibility of using φ(x) is also true but it is of less practical

interest. The following definition can serve us as an example:

f(x) = c + ln(x) .

3 The transformed variable is defined by

ξ(x) =(∂f∂x

)= x-1 ,

4 giving us x = ξ-1 and

φ(ξ) = f − ξx = c − ln(ξ) − 1

f(ξ) = f(x(ξ)) = c − ln(ξ)



8 State . . .Mappings Function ranges1 Note that the range of f(ξ) is exactly the same as the range of f(x).This assumes that ξ(x) is an invertible function, but provided that thisassumption holds, the two functions are identical until x is substitutedby x(ξ).

2 After substitution, the two functions are different.

3 The function f(ξ) should therefore be given a symbol different fromf(x), but in thermodynamic applications it is usual to use the argumentto distinguish between the functions.

4 This is not an ideal situation, but the same practice is used in e.g.computer programming[a].

371 / 1776

[a] One example is the so-called multiple dispatch methods where e.g. sqrt(x) in mostFORTRAN-like languages is meant to return the square root of x— be it a real or acomplex number— independently of whether the input value of x is a natural numberor a real number.

1 The link between the original two functions f(x) and ξ(x) on theone side and f(ξ), x(ξ), and φ(ξ) on the other side is illustrated below:



8 State . . .Mappings (2)

e-1 1 e c − 1 c c + 1

ξ(x) f(x)

f(ξ)

x(ξ)

φ(ξ)



8 State . . .Mappings (3)

2 In the figure x ∈ [1,e] has been chosen as a suitable domain ofdefinition for f .3 The mappings have the corresponding function domains:

f(x) : [1,e ]→ [c, c + 1]

ξ(x) : [1,e ]→ [1, e-1 ]

f(ξ) : [1,e-1]→ [c, c + 1]



8 State . . .Legendre transform

1 2 In the present case ξ(x) is invertible which means we can alsodetermine the inverse function x(ξ) = ξ-1(x).3 But, after substituting x(ξ) into f(ξ) = f(x(ξ)), it is no longer possi-

ble to reverse the transformation.4 Information about x(ξ) has been lost, and even if f(x) = f(ξ)

throughout the entire value domain, it is impossible to recreate x fromf(ξ). In order to recreate the original function graph we must know both

(ξ, f(ξ)) and (ξ, x(ξ)). Alternatively we can rely on the Legendre trans-form φ = f − ξx, which has the special property that [∂/∂]φξ = −x andthat f(ξ) = φ − (∂φ/∂ξ) ξ.

9 5 An equivalent description of the fucntion graph (x , f(x)) can thenbe stated in terms of (ξ, φ(ξ)).6 The new situation is also captured by the figure above.7 The Legendre transform is appearantly sufficient for recreating theentire function graph because x is equal to − (∂φ/∂ξ) and f(x) is equalto φ − (∂φ/∂ξ) ξ.8 In addition we learn a new function φ to know which can be ofpractical interest in itself. In other words, in a thermodynamic context we can recreate

the information contained in U(S ,V ,N) either from the two functionsU(T ,V ,N) and S(T ,V ,N), or from the Legendre transform A(T ,V ,N).

10 If T and p are chosen as free variables we must either know thethree functions: U(T ,p,N), S(T ,p,N) and V(T ,p,N), or the Legendretransform G(T ,−p,N).



8 State . . .§ 054 § 53 § 55

ENGLISH text is missing.U(T ,V ,N)1 Chemical process simulation software often uses p(T ,V ,n) equa-tions of state to predict the behaviour of fluids in phase transitions.

2 Based on what we have learnt so far, it seems wise to describethe system in terms of its Helmholtz energy, and to derive the internalenergy, enthalpy and entropy from A(T ,V ,n), but below we will look ata more direct approach to deriving U, H and S — partly because it isuseful practice and partly because it is a key topic in most textbooks.

3 The pivot of the derivations is the assumption that all real fluidseventually behave as ideals gases whence the pressure is sufficientlylow. In terms of Helmholtz energi we write:

limV→∞

A = A ıg .

4 How innocent it looks, this equation has far-reaching conse-quences.

5 Beacuse, if A approaches A ıg assymptotically when all its partialderivatives must do the same!

6 In particular, we would like to differentiate with respect to T and Vsince the composition is assumed constant. However, it would also beof interest to include the Legendre transform(s) of A in the study, andfor this reason we shall include the derivative with respect to 1/T (byapplying the Gibbs–Helmholtz equation):

limV→∞

U =(∂(A/T )

∂(1/T )

)V ,n

→ Uıg

S = − (∂( A )

∂( T )

)V ,n

→ S ıg

p = − (∂( A )

∂( V )

)T ,n

→ pıg

(8.9)

7 The discussion of the limits does not end here.

8 We can in principle continue to differentiate A for as long as wewish.

9 That is, however, not for very long because we in practice only needto know three of the second order derivatives:

limV→∞

(∂U∂V

)T ,n = ∂2(A/T )

∂(1/T ) ∂V = − (∂(p/T )

∂(1/T )

)V ,n

→ 0

(∂S∂T

)V ,n = − ∂2( A )

∂( T ) ∂T = CV

T →C ıg

V

T(∂S∂V

)T ,n = − ∂2( A )

∂( T ) ∂V =(∂( p )

∂( T )

)V ,n︸︷︷︸

Definition & Maxwell rules

→ NRV

(8.10)

10 These results make it possible to derive closed expressions forU and S of an arbitrary real fluid where the reference is an ideal gassharing the same macroscopic state variables T , V , and n.

376 / 1776

Set out the total differentials of U(T ,V), H(T ,V) andS(T ,V) in a form that can be integrated if C ıg

P (T) andp(T ,V) are known state functions.



8 State . . .§ 054 Differential of U(T ,V)

1 2 One general approach to problems of this type is to start out fromone of the canonical differentials 8.1–8.3 and to transform it by system-atically eliminating unwanted variables (in this case S). The total differential that we are after is

(dU)n =(∂U∂T

)V ,n dT +

(∂U∂V

)T ,n dV . (8.11)

If we compare this differential with the equivalent differential in Eq. 8.1we can see that S needs to be replaced with T in order for us to makeany progress.

3 One possible approach is to convert dS into a differential ex-pressed in terms of T and V :

(dS)n =(∂S∂T

)V ,n dT +

(∂S∂V

)T ,n dV

=CV

T dT +(∂p∂T

)V ,n dV . (8.12)



8 State . . .§ 054 Differential of U(T ,V) (2)

4 This eliminates the problem of U being a free variable, as it is inEq. 8.3, but the definition (∂S/∂T)V ,n = CV/T does not bring us muchcloser to our goal— we still have some work to do.

5 Also note that the Maxwell relation Expanded footnote [a]:

[a] For the derivation of Maxwell relations, see Theme 27 on page 155. (∂S/∂V)T ,n = (∂p/∂T)V ,n hasbeen used in the last line.

6 Let us combine Eqs. 8.1 and 8.12: Expanded footnote [a]:

[a] In order to derive the second line from the first line we need to use the mathematical

identity ∂(y/x)∂(1/x) ≡ y + 1x ∂y

∂(1/x) ≡ y − x ∂y∂x , also known as the Gibbs–Helmholtz’

equation, from Theme 28. :

(dU)n = CV dT +[T

(∂p∂T

)V ,n − p

]dV

= CV dT − (∂(p/T)∂(1/T)

)V ,n

dV (8.13)



8 State . . .§ 054 Differential of H(T , p) ENGLISH text is missing.U(T ,V)1 Eq. 8.13 can be integrated if we know CV (T ,V) and p(T ,V), butsince C ıg

V = C ıgP − NR and (∂(p/T)/∂(1/T))ıg

V ,n = 0 it is possible to

replace CV (T ,V) with C ıgP (T) by going via an ideal gas reference state.

This makes it much easier to integrate the equation.

2 Note that the canonical differential of U in Eq. 8.1 requires knowl-edge of altogether different and more complex equations of state.

3 Incidentally, the differential in Eq. 8.13 is in the same form as thedifferential in Eq. 8.11, from which it follows that

(∂U∂V

)T ,n = − (∂(p/T)

∂(1/T)

)V ,n , (8.14)

(∂U∂T

)V ,n = CV , (8.15)

where 8.15 represents an alternative definition of CV .

4 Our next task is to determine the total differential of H(T ,V). Theeasiest way to do that is to start from H = U + pV , but on this occasionwe will derive dH from first principles in order to demonstrate how it isdone.

380 / 1776

1 With reference to Eq. 8.2 we will here express dp as a differentialwith respect to T and V :

(dp)n =(∂p∂T

)V ,n dT +

(∂p∂V

)T ,n dV . (8.16)

2 If we substitute the above equation, along with Eq. 8.12, intoEq. 8.2, we get the following result:

(dH)n =[CV + V

(∂p∂T

)V ,n

]dT +

[V

(∂p∂V

)T ,n + T

(∂p∂T

)V ,n

]dV . (8.17)

4 3 Relationships equivalent to the ones that we found in Eqs. 8.14–8.15 can be derived for H, but they are of little practical value. Eqs. 8.12, 8.13 and 8.17 summarise dS, dU and dH as differentials

expressed in T ,V coordinates. All of them can be integrated, providedthat you know the functions C ıg

P (T) Expanded footnote [a]:

[a] Normally C ıgP is included in reference works on thermodynamics — rather than

C ıgV (T). and p(T ,V).



8 State . . .§ 055 § 54 § 56

Find the integral expressions for U, H and S by startingout from the differentials that were derived in Theme 54.

Use ∆fHıg(T) and S ıg(T,p) as the constants ofintegration.



0 V -1 V -1

T

T

Uıg(T,V)

∞∫

V

dUıg = 0

T∫

T

dUıg

V∫∞

dU

U(T ,V)

Figure 8.1: Calculation of the internal energy of a fluid at constant compo-sition. The most common standard state for fluids is an ideal gas at T =

298.15 K and V = RT/p where p = 1 bar, but because (dU)ıgT ,n < f(V) the

standard state pressure is not important in this context.



8 State . . .§ 055 Integration of U(T ,V) ENGLISH text is missing.Ideal gas U1 A selection of useful differentials were derived in Theme 54 on

page 347.

2 Since U, H and S are state functions, we can largely choose tointegrate the differentials in whatever way suits our purpose.

3 In the case of internal energy, the most sensible way to integrateis as shown in Figure 8.1 on the previous page: First you integrate dUfrom T to T assuming ideal gas behaviour, and then you integrate fromV = ∞ to V substituting in the relevant equation of state. It may appearwe are switching between an ideal gas and a real fluid as if by magic,but remember that

limV→∞

U = Uıg .

4 In other words, all real fluids approach ideal behaviour at the limitV →∞⇔ p → 0.

5 This is precisely where the switch from ideal to real takes place, sothere is no discontinuity in the method.

383 / 1776

1 Assuming that there is an equation of state that can be expressedin the form p = p(T ,V), and that the heat capacity is a known functionof temperature, it is possible to integrate dU(T ,V) in Eq. 8.13 to get

U(T ,V) = Uıg(T) +T∫

T

C ıgV dT −

V∫∞

(∂(p/T)∂(1/T)

)V ,n

dV

Residual: Ur,v

. (8.18)

2 The internal energy of an ideal gas is independent of the volumeof the gas. The contribution of the integral of dU from V to V = ∞ istherefore zero in the equation above Expanded footnote [a]:

[a] This means that the internal energy of the fluid can be given without specifying itspressure, if you assume ideal gas behaviour. .

3 This is evident in Eq. 8.13, where if you substitute in pıg/T =

NR/V , you get (∂(p/T)/∂(1/T))ıgV ,n = 0 and (dU)

ıgn = C ıg

V dT .



8 State . . .§ 055 Integration of U(T ,V) (2)

4 The residual term does on the other hand explain the departurein the internal energy of the real fluid from the internal energy of ahypothetical ideal gas assuming the two fluids are compared at thesame temperature, (volume) and chemical composition. 5 The residual concept is more thoroughly explained in Chapter 13.



8 State . . .§ 055 Integration of H(T ,V) ENGLISH text is missing.Aggregate state1 From the definition H = U + pV , the enthalpy can be determined

using Eq. 8.18, but here it is worth commenting on the standard state.

2 Normally the standard state is chosen to ensure that the elementshave zero enthalpy in the stable aggregate state at 298.15 K and 1 bar(ideal gas, liquid or crystal).

387 / 1776

1 The internal energy must therefore be expressed in terms of thestandard enthalpy, and not vice-versa:

Uıg(T) = Hıg(T) − (pV)ıg

= ∆fHıg(T) − NRT . (8.19)

2 The definition is trivial, but once the zero point has been chosen,there are certain consequences if it is changed. One of the conse-quences is that Expanded footnote [a]:

[a] With reference to Eq. 8.21, the molar heat of formation of a compound AaBb · · ·Zz

in the ideal gas state is defined as being: ∆fuıgAaBb ···Zz(T) = uıgAaBb ···Zz(T) − auıgA (T) −buıgB (T) − · · · − zuıgZ (T) = ∆fhıgAaBb ···Zz(T) − RT + aRT + bRT + · · · + zRT, wherethe coefficients a,b , . . . , z are the number of formula units of the elements A ,B , . . . ,Zin the compound. For example, water has the chemical formula (H2)1 (O2)1/2 because

H and O naturally occur as diatomic molecules. Only if a + b + · · · + z = 1 does∆fuıg = ∆fhıg.

Hıg(T) = ∆fHıg(T) , (8.20)

Uıg(T) , ∆fUıg(T) (8.21)



8 State . . .§ 055 Integration of S(T ,V) ENGLISH text is missing.Ideell gass Ideal gas S1 In principle, S is determined in the same way as U, but here we

have to bear in mind that dS diverges at the limit V →∞ [a].

2 In order to get around this problem, dS is integrated in three steps,as shown in Figure 8.2 on page 359.

3 First dS is integrated between V and V , and then it is integratedbetween T and T assuming ideal gas behaviour.

4 Finally, dS−dS ıg is integrated between V = ∞ and V in accordancewith the selected equation of state.

5 Once again, switching between an ideal gas and a real fluid doesnot imply discontinuity because

limV→∞

S = S ıg ⇔ limV→∞

(∂p∂T

)V ,n = NR

V .

389 / 1776

[a] Integrating dS in Eq. 8.12, assuming ideal gas behaviour, confirms that the expres-sion diverges.

1 If we substitute (∂p/∂T)ıgV ,n = NR/V into Eq. 8.12, the integral

of S is as shown below, but the limit V → ∞ remains undefined, andrequires closer analysis:

S(T ,V) = S ıg(T,V) +∞∫

V

NRV dV +

T∫

T

C ıgV

T dT +V∫∞

(∂p∂T

)V ,n dV

= S ıg(T,V) + NR ln(VV

)+

T∫

T

C ıgV

T dT +V∫∞

[(∂p∂T

)V ,n −

NRV

]dV .

(8.22)

2 Note the term∫ V∞ (NR/V) dV , which we have explicitly added and

taken away from the right-hand side in order to avoid divergence at thelimit V →∞.



8 State . . .§ 055 Integration of S(T ,V) (2)

3 All real gases approach ideal gas behaviour when p → 0, andthe volume integral in Eq. 8.22 is well-defined at this limit, since theintegrand tends to zero there.

7 4 In thermodynamics, an integral that expresses the difference be-tween a thermodynamic property of a real gas and an ideal gas in thesame state is often called a residual.5 Chapter 13 covers this topic in greater detail.6 Unlike in the case of U, the standard state of S is a function of

volume. However, the ideal gas entropy is generally tabulated at T =298.15 K and p = 1 bar, and in spite of the fact that the volume is a freevariable in the function, the standard volume V = NRT/p will have anumerical value. We will therefore rewrite Eq. 8.22 in a slightly different form using

p as the standard state: Let us first calculate the ratio

VV

=Vp

NRT=

VpNRT · T

T.

8 The result is substituted into Eq. 8.22. At the same time we may take advantage of the relationship C ıgP =

C ıgV + NR:

S(T ,V) = S ıg(T,p) + NR ln( pVNRT

)

+T∫

T

C ıgP

T dT +V∫∞

[(∂p∂T

)V ,n −

NRV

]dV

Residual: S r,v

(8.23)



8 State . . .§ 055 Integration of S(T ,V) (3)

9 Here, the residual term in the last equation explains the departurein the entropy of the real fluid from the entropy of a hypothetical idealgas when the two fluids are compared at the same temperature, volumeand chemical composition.10 The residual concept is more thoroughly explained in Chapter 13.



0 V -1 V -1

T

T

S ıg(T,V)

V∫

V

dS ıgT∫

T

dS ıg

V∫∞

dS −V∫∞

dS ıg

S(T ,V)

cancels

Figure 8.2: Calculating the entropy of a fluid at constant composition. Thehatched lines illustrate the nominal steps involved in integration. The stepsused in practise differ from the nominal ones because the integral divergesat the limit V → ∞. It is possible to switch between the two routes because(dS)

ıgT ,n < f(T) and (dS)V ,n < f(V). The most common standard state for fluids

is an ideal gas at T = 298.15 K and V = RT/p where p = 1 bar.



8 State . . .The Bridgman tableENGLISH text is missing.T , p versus T ,V1 Temperature and pressure occupy a unique place amongst the

thermodynamic state variables.

2 This is because we generally live under constant temperature andpressure conditions, and consequently a large number of experimentshave been performed at room temperature and atmospheric pressure.

3 It is therefore a useful exercise to derive the formulae discussed inthe previous section again, but this time using p as a free variable.

4 Without going into details, the total differentials of U(T , p), H(T ,p)and S(T , p) can be derived in an analogous way to the equivalent dif-ferentials of internal energy, enthalpy and entropy expressed in T ,V co-ordinates:

(dU)n =[CP − p

(∂V∂T

)p,n

]dT −

[p(∂V∂p

)T ,n + T

(∂V∂T

)p,n

]dp , (8.24)

(dH)n = CP dT +(∂V/T∂1/T

)p,n dp , (8.25)

(dS)n =CP

T dT − (∂V∂T

)p,n dp . (8.26)

5 It is entirely possible to integrate these expressions if we know theheat capacity C ıg

P (T) and the equation of state V(T , p).

6 In that case, we get

U(T , p) = H(T ,p) − pV , (8.27)

H(T ,p) = ∆fHıg(T) +T∫

T

C ıgP dT +

p∫

0

(∂V/T∂1/T

)p,n dp , (8.28)

S(T , p) = S ıg(T,p) − NR ln( pp

)+

T∫

T

C ıgP

T dT −p∫

0

[(∂V∂T

)p,n −

NRp

]dp ,

(8.29)

but in practice only the ideal gas law, the 2nd virial equation expressedas pV/NRT = 1 + Bp, and certain equations of state for solids areexplicit functions of V(T ,p,n).

7 This means that Eqs. 8.28 and 8.29 are used less than the equiva-lent Eqs. 8.18 and 8.23.

394 / 1776

1 In an arbitrarily chosen system of constant composition, there aretwo possible degrees of freedom.

2 For practical (and historic) reasons, these degrees of freedom areexpressed in terms of 9 experimentally determined properties that areall related to physical chemistry: T , p, s, v, u, h, α, β and cp.

3 The latter three quantities in this list are defined as

α = v -1 (∂v∂T

)p , β = −v -1 (∂v

∂p)T, cp = T

(∂s∂T

)p ,

4 cf. Eq. 8.37 on page 377 and Eq. 8.26.



8 State . . .The Bridgman table (2)

5 In view of the requirement for thermodynamic consistency, theremust also be 9− 3− 2 = 4 independent differentials that tell us what weneed to know about the behaviour of the system. 6 It is possible to imagine several alternative sets of differentials.

7 Here we will select the set that describes the change in the sys-tem’s energy in the simplest manner.8 This is a practical choice, in keeping with the philosophy that un-derlies the rest of this book:

(du)n = T ds − p dv ,

(dh)n = T ds + v dp ,

T(ds)n = cp dT − Tαv dp ,

p(dv)n = pvαdT − pβv dp .




9 The differentials are linearised functions (here as elsewhere), andthe information that they contain can (here as elsewhere) be sum-marised in matrix form. 10 Doing so will, in this case, make the next stages of our task signifi-

cantly easier.11 We will choose to express the coefficient matrix in dimensionlessform, with the right-hand side equal to 0 J mol-1:

−1 0 1 −1 0 0

0 −1 1 0 0 1

0 0 −1 0 1 −Tα

0 0 0 −1 pvαc-1p −pβ

du

dh

T ds

p dv

cp dT

v dp

=

0

0

0

0

. (8.30)




12 By performing a straightforward Gaussian elimination of the rowsin the coefficient matrix, taking the pivot elements from columns 1, 2, 3and 6, it is possible to derive the following set of coefficients:

−1 0 0 (−pβ+ Tα)/(pβ) (cpβ − Tvα2)/(cpβ) 0

0 −1 0 (−1 + Tα)/(pβ) (cpβ − Tvα2 + vα)/(cpβ) 0

0 0 −1 Tα/(pβ) (cpβ − Tvα2)/(cpβ) 0

0 0 0 −1/(pβ) vα/(cpβ) −1




13 Back substitution assuming p dv and cp dT to be the free variablesgives us:

(du)n = (cpβ − Tvα2)β-1 dT − (pβ − Tα)β-1 dv ,

(dh)n = (cpβ − Tvα2 + vα)β-1 dT − (1 − Tα)β-1 dv ,

(ds)n = (cpβ − Tvα2)T -1β-1 dT + αβ-1 dv ,

(dp)n = αβ-1 dT − v -1β-1 dv .




14 If we now bear in mind that

(∂p/∂T)v ,n = αβ-1 ,

v (∂p/∂v)T ,n = −β-1 ,cv = cp − Tvα2β-1 ,

then we see that the result of the matrix operation is identical to thecalculations performed in Eqs. 8.13, 8.17, 8.12 and 8.16.

16 15 This means that we have uncovered a general method for calculat-ing an arbitrarily chosen differential expressed in the six primary vari-ables: T , p, s, v, u og h. By systematically selecting the relevant pivot elements, we can now

calculate all of the 6×5×4 = 120 differentials that describe the system.

17 A Ruby program that calculates all of these differentials is de-scribed in Appendix 31:3.1



Table 8.1: A6 × 5 × 4 = 12phase at conssubstance (or

(δT)p = −(δp)T = 1

(δv )p = −

(δs )p = −

(δu )p = −

(δh )p = −



(δv )T = −

(δs )T = −

(δu )T = −

(δh )T = −

(δs )v = −

(δu )v = −

(δh )v = −

(δu )s = −

(δh )s = −

(δh )u = −



Table 8.2: A complete listing of all of the 6 × 5 × 4/2 = 60 total differentialsof u, h, s, v, T and p in a homogeneous phase at constant composition, i.e.either one kilogram or one mol of the substance (or alternatively the sum ofall substances in the mixture). The differentials are calculated by performingGaussian elimination on the system of Eqs. 8.30 on page 362 followed byalgebraic reduction and simplification in Maple R©.

dh (u , s ) =cp

p(cpβ−Tvα2)du +

(pβcp−pTvα2−cp+pvα)Tp(cpβ−Tvα2)

ds

dh (u , v ) =cpβ−Tvα2+vα

cpβ−Tvα2 du +pβcp−pTvα2−cp+pvα

cpβ−Tvα2 dv

dh (u , T ) = 1−Tαpβ−Tα du +

pβcp−pTvα2−cp+pvαpβ−Tα dT

dh (u , p ) =cp

cp−pvα du − (pβcp−pTvα2−cp+pvα)vcp−pvα dp

dh (s , v ) =(cpβ−Tvα2+vα)T

cpβ−Tvα2 ds − cp

cpβ−Tvα2 dv

dh (s , T ) = Tα−1α ds +

cp

Tα dT



dh (s , p ) = T ds + v dp

dh (v , T ) = Tα−1β dv +

cpβ−Tvα2+vαβ dT

dh (v , p ) =cp

vα dv +cpβ−Tvα2+vα

α dp

dh (T , p ) = cp dT + (1 − Tα)v dp

dp (u , h ) =cp

(pβcp−pTvα2−cp+pvα)v du − cp−pvα(pβcp−pTvα2−cp+pvα)v dh

dp (u , s ) =cp

p(cpβ−Tvα2)v du − (cp−pvα)Tp(cpβ−Tvα2)v ds

dp (u , v ) = αcpβ−Tvα2 du − cp−pvα

(cpβ−Tvα2)v dv

dp (u , T ) = 1(pβ−Tα)v du − cp−pvα

(pβ−Tα)v dT

dp (h , s ) = 1v dh − T

v ds

dp (h , v ) = αcpβ−Tvα2+vα dh − cp

(cpβ−Tvα2+vα)v dv

dp (h , T ) = 1(1−Tα)v dh +

cp

(Tα−1)v dT



dp (s , v ) = Tαcpβ−Tvα2 ds − cp

(cpβ−Tvα2)v dv

dp (s , T ) = −1vα ds +

cp

Tvα dT

dp (v , T ) = −1vβ dv + α

β dT

ds (u , h ) =−cp

(pβcp−pTvα2−cp+pvα)T du +p(cpβ−Tvα2)

(pβcp−pTvα2−cp+pvα)T dh

ds (u , v ) = 1T du +

pT dv

ds (u , T ) = −αpβ−Tα du +

p(cpβ−Tvα2)

(pβ−Tα)T dT

ds (u , p ) =cp

(cp−pvα)T du − p(cpβ−Tvα2)v(cp−pvα)T dp

ds (h , v ) =cpβ−Tvα2

(cpβ−Tvα2+vα)T dh +cp

(cpβ−Tvα2+vα)T dv

ds (h , T ) = αTα−1 dh +

cp

(1−Tα)T dT

ds (h , p ) = 1T dh − v

T dp

ds (v , T ) = αβ dv +

cpβ−Tvα2

βT dT



ds (v , p ) =cp

Tvα dv +cpβ−Tvα2

Tα dp

ds (T , p ) =cp

T dT − vαdp

dT (u , h ) = Tα−1pβcp−pTvα2−cp+pvα du +

pβ−Tαpβcp−pTvα2−cp+pvα dh

dT (u , s ) = Tαp(cpβ−Tvα2)

du +(pβ−Tα)T

p(cpβ−Tvα2)ds

dT (u , v ) = βcpβ−Tvα2 du +

pβ−Tαcpβ−Tvα2 dv

dT (u , p ) = 1cp−pvα du − (pβ−Tα)v

cp−pvα dp

dT (h , s ) = Tαcp

dh +(1−Tα)T

cpds

dT (h , v ) = βcpβ−Tvα2+vα dh + 1−Tα

cpβ−Tvα2+vα dv

dT (h , p ) = 1cp

dh +(Tα−1)v

cpdp

dT (s , v ) = βTcpβ−Tvα2 ds − Tα

cpβ−Tvα2 dv

dT (s , p ) = Tcp

ds + Tvαcp

dp



dT (v , p ) = 1vα dv +

βα dp

du (h , s ) =p(cpβ−Tvα2)

cpdh − (pβcp−pTvα2−cp+pvα)T

cpds

du (h , v ) =cpβ−Tvα2

cpβ−Tvα2+vα dh − pβcp−pTvα2−cp+pvαcpβ−Tvα2+vα dv

du (h , T ) =pβ−Tα1−Tα dh +

pβcp−pTvα2−cp+pvαTα−1 dT

du (h , p ) =cp−pvα

cpdh +

(pβcp−pTvα2−cp+pvα)vcp

dp

du (s , v ) = T ds − p dv

du (s , T ) =−(pβ−Tα)

α ds +p(cpβ−Tvα2)

Tα dT

du (s , p ) =(cp−pvα)T

cpds +

p(cpβ−Tvα2)vcp

dp

du (v , T ) =−(pβ−Tα)

β dv +cpβ−Tvα2

β dT

du (v , p ) =cp−pvα

vα dv +cpβ−Tvα2

α dp

du (T , p ) = (cp − pvα)dT + (pβ − Tα)v dp



dv (u , h ) =−(cpβ−Tvα2+vα)

pβcp−pTvα2−cp+pvα du +cpβ−Tvα2

pβcp−pTvα2−cp+pvα dh

dv (u , s ) = −1p du + T

p ds

dv (u , T ) =−β

pβ−Tα du +cpβ−Tvα2

pβ−Tα dT

dv (u , p ) = vαcp−pvα du − (cpβ−Tvα2)v

cp−pvα dp

dv (h , s ) =−(cpβ−Tvα2)

cpdh +

(cpβ−Tvα2+vα)Tcp

ds

dv (h , T ) =β

Tα−1 dh +cpβ−Tvα2+vα

1−Tα dT

dv (h , p ) = vαcp

dh − (cpβ−Tvα2+vα)vcp

dp

dv (s , T ) =βα ds − cpβ−Tvα2

Tα dT

dv (s , p ) = Tvαcp

ds − (cpβ−Tvα2)vcp

dp

dv (T , p ) = vαdT − vβdp



8 State . . .§ 056 § 55 § 57ENGLISH text is missing.Calorimetry1 Heat capacity is a measure of the thermal inertia of the system.

2 If this inertia is very high, a lot of energy is needed to raise thetemperature by one degree.

3 For instance, if you were to measure the heat capacity of waterat 1 atm without being aware of the vapour–liquid transition, you wouldconclude that cp →∞ at 100 C.

4 But the integral of cp dT over a small temperature range dT ∈〈100 − ǫ, 100 + ǫ〉 will be finite, and equal to what is defined as theheat of vaporization.

5 Analogously, substances with a low heat capacity, e.g. carbon, forwhich cp ≈ 3.5 J mol-1 K-1 in the temperature range T ∈ 〈0,300〉K, re-quire only a small amount of energy for a large temperature change.

6 Hence heating one gramme of carbon from 0 → 300 K will requireabout the same amount of energy as is needed to vaporize just 0.04 gof water.

7 For practical reasons it is normal to define the heat capacity of aclosed system in two different ways:

8 you measure how energy varies with temperature at either constantpressure, or at constant volume.

9 A third option, which is widely used in solid and fluid mechanics, isto assume adiabatic conditions and to look at how energy varies withtemperature along an isentrope.

10 The relationship between these heat capacities can be derivedfrom the volumetric properties of the fluid.

11 In the next section we will look at the relationship between the firsttwo of the abovementioned heat capacities.

409 / 1776

Entropy and heat capacity are closely linked, and bycomparing the differentials dS (T ,V)n and dS (T ,p)n

from Eqs. 8.12 on page 348 and 8.26 on page 360,it is possible to obtain two alternative relationshipsbetween CP and CV . Show that these relationships

can be expressed as:

i) CP − CV = −T(∂V∂T

)2p,n

(∂V∂p

)-1T ,n

, (8.31)

ii) CP − CV = −T(∂p∂T

)2

V ,n

(∂p∂V

)-1T ,n . (8.32)



8 State . . .§ 056 Heat capacity

1 A total differential contains the same information whether it is ex-pressed in T ,V or T ,p coordinates. Based on Eqs. 8.12 and 8.26, for(dS)n we can write

CV

T dT +(∂p∂T

)V ,n dV =

CP

T dT − (∂V∂T

)p,n dp . (8.33)

2 If we assume constant volume we get: (dp)V ,n = (∂p/∂T)V dT .

3 Similarly, at constant pressure: (dV)p,n = (∂V/∂T)p dT .

4 In both cases, the elimination of dT from the equation above gives:

CP − CV = T(∂p∂T

)V ,n

(∂V∂T

)p,n . (8.34)



8 State . . .§ 056 Heat capacity (2)

5 We can now choose to get rid of either p or V as a dependentvariable. If we substitute dp = 0 into the total differential of p, we canshow Expanded footnote [a]:

[a] The total differential of p at constant composition is: (dp)n = (∂p/∂T)V ,n dT +

(∂p/∂V)T ,n dV . Assume constant pressure i.e. dp = 0 and divide by dT . Then isolatethe derivative: (dV / dT)p,n = (∂V/∂T)p,n = . . . that

(∂V∂T

)p,n = − (∂p

∂T)V ,n

(∂p∂V

)-1T ,n . (8.35)

6 Similarly, by substituting dV = 0 into the total differential of V , wecan show that

(∂p∂T

)V ,n = − (∂V

∂T)p,n

(∂V∂p

)-1T ,n

. (8.36)

7 Substitution of the last two equations into Eq. 8.34 produces thetwo expressions that we set out to find Expanded footnote [a]:

[a] With a pressure-explicit equation of state, it may be sensible to work in reducedvariables. For your own benefit, show that (cp − cv)/R = zcTr (∂pr/∂Tr)2vr ,x (∂pr/∂vr)-1Tr ,xfor one mole of fluid. .



8 State . . .Definition of α and β

1 2 Heat capacity at constant pressure was one of the first thermody-namic properties to be investigated systematically. Other historically important properties include isobaric expansivity

and isothermal compressibility:

α = v -1 (∂v∂T

)p , β = −v -1 (∂v

∂p)T. (8.37)

3 For some applications it is useful to use these quantities as theygive a more compact set of formulae. In the case of molar volume, for instance, it gives us

(dv)n = vαdT − vβdp , (8.38)

while for molar entropy we get

(ds)n =cp

T dT − αv dp , (8.39)

(ds)n =cv

T dT + αβ dv . (8.40)



8 State . . .Definition of α and β (2)

6 4 These expressions are obtained by substituting 8.37 into the totaldifferentials of volume and entropy; see Theme 56 on page 374 for thenecessary details.5 Incidentally, Eqs. 8.31–8.32 in Theme 56 are expressed in terms of

extensive variables. In the literature on thermodynamics, intensive variables are gener-ally used, which makes the size of the system irrelevant to the problem,and one widely used alternative to Eq. 8.31 is the molar form

cp − cv = Tvα2

β . (8.41)



8 State . . .§ 057 § 56 § 58 ENGLISH text is missing.Applications1 Ideal gases are an important concept in thermodynamics, and it isreassuring to know that this simple model gives a good description ofmany real systems, e.g. in process engineering and meteorology.

2 This section is therefore one of the most important ones in thewhole book, at least in terms of its practical applications.

415 / 1776

What are the expansivity α and compressibility β of anideal gas mixture, expressed in terms of the variables

T and p? And what is C ıgP − C ıg

V for the mixture?



8 State . . .§ 057 Ideal gas I

1 If we combine the ideal gas law pıg = NRT/V with the definitionsin Eq. 8.37, it follows that

αıg = NRPV = T -1 and βıg = NRT

P2V = p-1 ,

which gives us the molar relationship c ıgp − c ıg

v = pv/T = R whensubstituted into Eq. 8.41. For an arbitrary system size, we can say that

C ıgP − C ıg

V = NR ,

also known as Mayer’s relation. 2 Consistency: Use the Eqs. 8.31 and 8.32 to verify this result.



0 500 1000

−10

0

10

20

30

0 500 10000

50

100

150

200

250

T [K]T [K]

Enthalpy

0

EntropyH[k

J]

S[J/K

mol]

Figure 8.3: Approximated (linearised) ideal gas enthalpy and entropy calcu-lated for Ar, N2 and CH4, compared to precise values taken from the JANAFtables.



8 State . . .§ 061 § 60 § 62§ 581 Rewrite the equations 8.18–8.23 and 8.27–8.29 assuming ideal

gas behaviour.418 / 1776

§ 58 Ideal gas II1 In general the results speak for themselves, but is worth noting

that Uıg and Hıg are only functions of T (and not of p or V):

Uıg(T) = ∆fHıg(T) − NRT +T∫

T

C ıgV dT

= ∆fHıg(T) − NRT +T∫

T

C ıgP dT , (8.42)

Hıg(T) = Uıg(T ,V) + (pV)ıg

= ∆fHıg(T) − NRT +T∫

T

C ıgV dT + NRT

= ∆fHıg(T) +T∫

T

C ıgP dT , (8.43)

S ıg(T ,V) = S ıg(T,p) +T∫

T

C ıgV

T dT + NR ln( pVNRT

)

= S ıg(T,p) +T∫

T

C ıgP

T dT − NR ln(NRT

pV), (8.44)

S ıg(T ,p) = S ıg(T,p) +T∫

T

C ıgP

T dT − NR ln( pp

)(8.45)

419 / 1776

§ 591 Use c ıg

p (298.15 K) to obtain an approximation for the enthalpyHıg(T) −Hıg(298.15 K) of argon, nitrogen and methane in the tempera-ture range 0–1000 K. Make a graphical illustration of the functions andcompare with the values from JANAF[a]. Comment on the discrepan-cies.

420 / 1776


§ 59 The enthalpy function1 The expression for enthalpy becomes ∆H(T) = c ıg

p (T −T), whereT = 298.15 K. The predictive accuracy of the function can be seen onthe left-hand side of Figure 8.3 on the previous page.

2 The discrepancy between the calculated and measured values isbarely noticeable for argon, slightly bigger for nitrogen and biggest formethane (see Program 1.3 in Appendix 31).

3 This difference between the elements is due to the internal degreesof freedom of the molecules.

4 Argon is a monatomic gas without any rotational and vibrationalenergy, nitrogen has one vibrational and two rotational degrees of free-dom, while methane has nine vibrational and three rotational degreesof freedom.

5 At 298.15 K it is essentially only the rotational energy that is rele-vant, whereas the vibrational energy only comes into play at higher tem-peratures.

6 This is why the enthalpy curves turn upwards, and why the devia-tion increases with the size of the molecule.

421 / 1776

§ 601 Use c ıg

p (298.15 K) to obtain an approximation for the entropyS ıg(T ,p) − S ıg(298.15 K, p) of argon, nitrogen and methane in thetemperature range 0–1000 K. Make a graphical illustration of the func-tions and compare the curves with the values from JANAF[0]. Com-ment on the discrepancies.

422 / 1776

§ 60 The entropy function1 The expression for entropy becomes ∆S(T ,p) = c ıg

p ln(T/T),where T = 298.15 K. The predictive accuracy of the function can beseen on the right-hand side of Figure 8.3 on the preceding page.

2 Unsurprisingly, the distance between the calculated values and theexperimental ones increases as T moves away from T.

3 An explanation of the behaviour at high temperatures follows fromTheme 59, but what happens at low temperatures?

4 You should particularly note that S(0,p) < 0 (see inset figure).

5 Negative entropy is theoretically possible and errors of prediction ofthis kind can have major consequences in conjunction with simulatingreal processes. Program 1.3 in Appendix 31 gives all of the details.

423 / 1776

Show that the following relations are true for an idealgas mixture if the heat capacity is constant over the

interval of integration:

(T2T1

)ıg

S ,n=

(p2p1

) γ−1γ

S ,n, (8.46)

(p2p1

)ıg

S ,n=

(V2V1

)-γS ,n

, (8.47)

(T2T1

)ıg

S ,n=

(V2V1

)1−γS ,n

. (8.48)

What will the exponent in Eq. 8.46 be for an isochoricprocess? And what is it for an isothermal process inEq. 8.47? And for an isobaric process in Eq. 8.48?



8 State . . .§ 061 Isentropic relations

1 Use the differential in Eq. 8.39 as a starting point and then assumeideal gas behaviour for α.

2 For a isentropic process we also know that ds = 0, so we can setup the following integral:

T2∫

T1

c ıgp (dln T)s,n =

p2∫p1

αıgv dp = Rp2∫p1

dln p .

3 Integrate between states 1 and 2, assuming that the heat capacityis constant over this range.



8 State . . .§ 061 Isentropic relations (2)

4 This assumption holds perfectly for monatomic gases, whereas forother gases it is necessary to use a mean value:

ln(T2T1

)ıg

s,n= R

c ıgp

ln(p2p1

)s,n

,

(T2T1

)ıg

s,n=

(p2p1

) γ−1γ

s,n.

6 5 Here the heat capacity ratio is defined as being γ = cp/cv . Note that R = c ıgp − c ıg

v , which has been used to obtain the finalline.



8 State . . .§ 061 Isentropic relations (3)

8 7 The other relations referred to in the question can be derived byanalogous methods starting from a combination of Eqs. 8.39 and 8.40. For an isothermal, isobaric or isochoric change of state, the ideal

gas law immediately tells us that:(T2T1

)ıg

V ,n=

(p2p1

)V ,n

,

(p2p1

)ıg

T ,n=

(V2V1

)-1T ,n

,

(T2T1

)ıg

p,n=

(V2V1

)p,n

9 These are evidently related to 8.46–8.48, but there is no value of γcapable of converting that system of equations into this one.



8 State . . .§ 062 § 61 § 63 ENGLISH text is missing.Van der Waals p(T ,V)1 The Van der Waals equation represents an important milestone inour understanding of fluids.

2 By itself, the equation is too inaccurate to be of any practical use,but it has laid the foundations for a whole series of cubic[a] equationsof state that do have practical applications.

3 A number of these equations (RK, SRK, PR) are very widely usedin relation to liquid–gas separation and gas treatment equipment.

429 / 1776

[a] Known as cubic because v(p) can be expressed as the solution to a cubic equation.

Substitute the Van der Waals equation of state (7.30)into Theme 55 on page 351. Go on to show that the

internal energy can be expressed

UVdW(T ,V) = Uıg(T)Reference

+∆Uıg(T)CV integral

− aN2

VResidual

.



8 State . . .§ 062 Van der Waals U

1 Let us first state the Van der Waals equation in the form p(T ,V ,n);cf. Eq. 7.36.

3 2 The form of the equation of state is important, because subse-quently it is integrated with respect to the total (rather than molar[a])volume V .

[a] Mixing up total and molar values is a common mistake in thermodynamics! Then we differentiate the equation of state with respect to T , whichgives us

(∂(p/T)∂(1/T)

)VdW

V ,n= − aN2

V2 .

4 If we substitute this into the expression for U(T ,V) in Theme 55 onpage 351 (Eq. 8.18), we get

UVdW(T ,V) = Uıg(T) +T∫

T

C ıgV dT +

V∫∞

aN2

V2 dV



8 State . . .§ 062 Van der Waals U (2)

= ∆fHıg(T) − NRT︸︷︷︸

Uıg(T)

+T∫

T

C ıgV dT − aN2

V . (8.49)

5 The physical model for the internal energy of a Van der Waals fluidinvolves an arbitrarily chosen reference state, a non-linear temperatureterm and a composition term that is proportional to the system’s molardensity.



8 State . . .§ 063 § 62 § 64

Find an expression for cVdWv based on Theme 62.

Determine the function of cVdWp − cVdW

v and draw iton a suitable diagram. What happens as the stateapproaches the mixture’s critical point? What is the

upper temperature limit and what happens at extremelyhigh pressure?



8 State . . .§ 063 Van der Waals CP

1 Differentiating UVdW with respect to T at constant V gives us

CVdWV =

(∂U∂T

)VdWV ,n = C ıg

V ,

provided that a is a constant parameter in the equation. This shows,quite unexpectedly, that heat capacity at constant volume is the sameas for an ideal gas.

4 2 Experimentally this is not true.3 In order to determine the heat capacity at constant pressure we

need to derive the Van der Waals Eq. 7.36 again:(∂pVdW

∂T

)V ,n = NRV−Nb , (8.50)

(∂pVdW

∂V)T ,n = −NRT(V−Nb)2 + 2aN2

V3 . (8.51) Now substitute the differentials into Eq. 8.32 on page 374 and ex-press the result in terms of molar variables:

(cp − cv)VdW =

−T(

Rv−b

)2

2av3 − RT

(v−b)2

.



8 State . . .§ 063 Van der Waals CP (2)

5 If we substitute in a = 9RTcvc/8 and b = vc/3, the equation canbe rewritten using reduced variables:

(cp−cv)VdW

R =1

1 − (3vr−1)2

4Trv3r

. (8.52)

8 6 According to the Van der Waals theory, the volume is defined in therange 〈b ,∞〉.7 This means that vr ∈ 〈1/3,∞〉, and moreover that cVdWp −cVdWv →∞

in Eq. 8.52 if both vr → 1 and Tr → 1. At temperatures Tr < 1, the gas and liquid states can exist in equi-librium, but our method is not designed to deal with that situation. Wewill therefore stick to supercritical temperatures Tr > 1.



100

101

0

2

4

6

8

10

1

1.2

1.5

25

vr

c p−c

v

R

Figure 8.4: Dimensionless heat capacities calculated from the Van der Waalsequation of state. The numbers in the figure refer to Tr = 1.0,1.2,1.5,2.0,5.0.



8 State . . .§ 064 § 63 § 65 ENGLISH text is missing.Van der Waals cp − cv1 A few selected isotherms have been drawn on Figure 8.4 on the

preceding page using the Program 31:1.6.

2 Note that cVdWp − cVdW

v → R due to ideal gas behaviour at hightemperatures Tr >> 1 and low pressures vr >> 1.

3 Funnily enough, the same limit also applies at extremely high pres-sures i.e. vr → 1/3.

4 In spite of the fact that at constant volume the heat capacity variesin the same way as that of an ideal gas, at constant pressure it deviatesgreatly.

5 It is particularly worth noting that cVdWp → ∞ at the critical point,

which is in line with experimental values.437 / 1776

The inversion temperature of a fluid is defined as thetemperature where (∂T/∂p)H,n = 0. What are the

practical implications of this? Determine the functionfor the Van der Waals equation of state and draw the

graph in a pr,Tr diagram.



8 State . . .§ 064 Joule–Thomson coefficient

1 2 This exercise would become incredibly unwieldly if we were to de-rive everything from first principles.3 For the sake of clarity we will therefore employ a schematic solu-tion with references to other chapters (this is not generally an advisableapproach, but every rule is made to be broken). Our starting point is a relevant expression for the Joule–Thomson

coefficient which we derived in Theme 93 on page 514:

κ =(∂T∂p

)H,n

=T

(∂p∂T

)V ,n + V

(∂p∂V

)T ,n

T(∂p∂T

)2

V ,n − CV(∂p∂V

)T ,n

.

4 Next we take the partial derivatives of the Van der Waals equationfrom Theme 63 on page 389

(∂p∂T

)VdW

V ,n = NRV−Nb ,

(∂p∂V

)VdW

T ,n = −NRT(V−Nb)2 + 2aN2

V3 ,



8 State . . .§ 064 Joule–Thomson coefficient (2)

8 5 and substitute this result into the expression for κ:

κVdW = −b + (v−b)2

RT 2av2

R + cv − cv (v−b)2

RT 2av3

.

6 In Theme 63 we showed that the heat capacity cv of aVan der Waals fluid is the same as that of an ideal gas.7 In addition we need the relationship between the parameters a, b

and Tc,pc as discussed in Theme 47 on page 322, and the relationshipbetween c ıgp and c ıgv from Theme 56 on page 374. In summary, we can state that:

cVdWv = c ıg

vvcpcRTc

= 38

c ıgp = c ıg

v + R b = 13 vc

c ıgp = γc ıg

v a = 3pcv2c = 9

8 RTcvc

9 Substituted into the expression for κVdW this gives us

κVdWRvc

=−1

3 +(3vr−1)2

4Trv2r

γγ−1 − 1

γ−1(3vr−1)2

4Trv3r



8 State . . .Inversion temperature

1 2 The inversion state is defined at κVdW = 0.3 In this state the enthalpy does not change with pressure, and hence

the temperature remains unchanged even if there is a slight fall in pres-sure. Assuming that the denominator is finite throughout the interval of

solution, the criterion for inversion will be fulfilled when the numerator iszero. If we solve with respect to vr we get

(vVdWr )κ=0 = 1

3−√

43 Tr

.

6 4 As a final trial the answer needs to be converted into reduced co-ordinates.5 We know the Van der Waals equation in reduced coordinates from

Theme 49 on page 328

pVdWr = 8Tr3vr−1 − 3v2r ,

and it is easy enough to substitute in (vVdWr )κ=0 from the equationabove. That gives us our final result:

(pVdWr )κ=0 = 24

√3Tr − 12Tr − 27 .

7 This expression can usefully be compared with the empirical corre-lation (pr)κ=0 = 24.21−18.54T -1

r −0.825T2r , which gives a relatively ac-

curate approximation of the experimental values for N2, CO2 and somesimple hydrocarbons Expanded footnote [a]:

[a] Donald G. Miller. Ind. Eng. Chem. Fundam., 9(4):585–589, 1970. . 8 In spite of the fact that the two equations are entirely different inform, the graphs that they produce are qualitatively similar; cf. Fig-ure 8.5[a].

[a] Theoretical expressions and empirical correlations are in some ways mathematicalhalf-brothers — they have the same origins and areas of application, but in terms oftheir content and form they are completely different from each other.



0 5 100

1

2

3

4

5

6

7

Miller

Van der Waals

pr

Tr

Figure 8.5: Inversion temperature calculated from the Van der Waals equationof state compared with an empirical correlation obtained by D. G. Miller (1970).



8 State . . .Looking back

1 Canonical: U(S ,V ,N), H(S ,−p,N), A(T ,V ,N), etc.

2 Non-canonical: U(T ,p,N) + S(T ,p,N) + V(T ,p,N), etc.

3 The Bridgman table:

−1 0 1 −1 0 0

0 −1 1 0 0 1

0 0 −1 0 1 −Tα

0 0 0 −1 pvαc-1p −pβ

du

dh

T ds

p dv

cp dT

v dp

=

0

0

0

0



8 State . . .Looking back (2)

4 Constant volume heat capacity:

(dS)n =(∂S∂T

)V ,n dT +

(∂S∂V

)T ,n dV =

CV

T dT +(∂p∂T

)V ,n dV

5 Constant pressure heat capacity:

(dS)n =(∂S∂T

)p,n dT +

(∂S∂p

)T ,n

dV =CP

T dT − (∂V∂T

)p,n dp

6 Isobaric expansivity: α = V -1

(∂V∂T

)

p,n

7 Isothermal compressibility: β = −V -1

(∂V∂p

)

T ,n

8 Heat capacity relation: CP − CV = NTvα2β-1

9 Van der Waals heat capacity: CVdWV = C ıg

V



8 State . . .Looking back (3)

10 Critical point divergence: limTr,pr→1,1

CVdWP = ∞


News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds

Part 9

Closed control volumes




9 Closed . . .New concepts

1 Closed control volume

2 The extended 1st law of thermodynamics

3 Compression work of ideal gas

4 Continuity of mass, energy and momentum

5 Thermodynamic speed of sound



9 Closed . . .Closed system

1 A closed control volume is synonymous with a closed system, andhence its energy balance reflects the 1st law of thermodynamics, andcan be expressed as dE = δQ − δW .

4 2 There is no transport term in this equation, but that does not meanthat the control volume must remain stationary.3 For example, later in this chapter we will study the changes of stateof a centre of mass moving along a streamline. The inexact differential δQ and − δW represent the transfer of heat

and work from the surroundings to the control volume. The symbol δhighlights the fact that Q and W are not state functions.

5 The extent to which the system reacts to external changes is givenby the exact differential of E = EK + EP + U.

9 6 In the absence of external fields and internal stress–strain condi-tions, the 1st law can be written:

Mυ dυ+ Mg dz + T dS − p dV + n∑1 µi dNi = δQ − δW ,

but it is rarely necessary to include all of the terms in order to obtain asatisfactory description of the system.7 Remember that any mathematical solution to a problem involvingphysical phenomena always simplifies what actually happens.8 In thermodynamics you have to be particularly careful in relation tothe assumption of equilibrium, and we should note that the compositionof the system can vary as a result of chemical reactions in the controlvolume. It is possible to choose values for the thermodynamic variables

that make them invariant Expanded footnote [a]:

[a] Except in the case of nuclear reactions, the number of moles of each atom will bepreserved in the control volume. One option is therefore to let Ni be the mole numberof each atom i. in response to chemical reactions, which

results in all dNi∈1,n = 0 Expanded footnote [a]:

[a] Admittedly there will be small, natural fluctuations in composition, but this will notaffect the energy, provided that the system is thermodynamically stable. We will notdiscuss the concept of stability in any greater detail here, but will instead refer toChapter 21 on page 1312. , although this requires the system to be in

equilibrium.



9 Closed . . .Closed system (2)

10 At thermodynamic equilibrium, for each independent Ni there willbe a chemical potential µi that is constant both in time and space.

14 11 Hence, if suitable values are selected for the thermodynamic vari-ables, all dNi = 0, but if there are any irreversible reactions, diffusionprocesses or phase transitions in the control volume, this no longer ap-plies.12 In such cases, the composition may vary both in space and in time,and the energy equation must be integrated over small partial volumeswith virtually uniform properties.13 This quickly leads us to partial differential equations (see Chap-ter 22 on page 1368). In order to avoid insuperable mathematical complications, we will

here limit ourselves to quasi-static systems, which can be describedusing ordinary differential equations.



9 Closed . . .Thermostatic state change

1 This chapter started by discussing the 1st law of thermodynamics,but a general analysis based on the principle of conservation of energyis not always required or practicable.

2 In many cases it is simpler to start with an explicit temperatureequation.

3 This is particularly true for temperature regulated systems and sys-tems that are in contact with thermal reservoirs.



9 Closed . . .Isochoric pressure change § 651 The temperature inside a refrigerator is 7 C. The door is openedand air at a temperature of 21 C fills the whole refrigerator. What isthe negative gauge pressure in the refrigerator after the door has beenclosed and the air has cooled back down to 7 C? How much force isneeded to open the door again if the refrigerator measures 50×140 cmon the inside? The atmospheric pressure can be assumed to be 1 bar.How do you solve the negative gauge pressure problem in practice?

451 / 1776

§ 65 Refrigerator1 Let T = 294 K and T1 = 280 K.

2 Assuming that the door seals tightly, the number of moles insidethe refrigerator will remain constant during cooling.

3 Applying the ideal gas law gives us N = pVRT = pV

RT1 , which implies

that the negative gauge pressure is ∆p = p − p1 = p(1 − T1T ) =

14294 · 105 = 4762 Pa.

4 The negative gauge pressure in the refrigerator means that an ex-ternal force will act inwards on the door.

5 The moment about the hinge is in this case M = ∆pL∫ B

0 x dx =12∆pLB2 = 1

2 4762 · 1.4 · (0.5)2 = 833.3 N m.

6 If we assume that the handle is right at the edge of the door, a forceequivalent to |F|B = 833.3 N m, i.e. |F| = 1666.7 N, must be applied inorder to open the door.

7 The volume of air in the refrigerator is not relevant to the problem,as it will not affect the result, provided that the air can flow freely whenthe door is open.

452 / 1776

§ 661 An apparatus for measuring diffusion at high pressures consists

of a test cell placed in a thermostatically controlled environment kept ata temperature of T = T. Initially the cell is filled with oil that can becompressed to high pressures using a piston. The piston is locked intoposition once the working pressure has been reached. The compres-sion is rapid, giving a virtually isentropic change of state. The subse-quent equalisation of the temperature occurs at constant volume. As-sume that all of the volume of oil is within the walls of the thermostati-cally controlled environment. Describe the fall in pressure as the oil ap-proaches thermal equilibrium, i.e. TCnH2n+2 → T. What compression isneeded if you want the final pressure in the cell to be p = 300 bar? Usedata from Table 9.1 on page 411.

453 / 1776

1

eplacem

TT ∆

T S

∆TV

t

2 The oil is practically incompressible, so the changes in both volumeand temperature are modest, in spite of the big change in pressure. It is therefore possible to obtain an approximation of ∆p by integrating

the differential

(dp)tot = (dp)s,n + (dp)v ,n

= γ-1(dp)s,n ,

using the constant heat capacity ratio γ = cp/cv .



9 Closed . . .Isochoric pressure change (2)

4 3 Instead of following the naive experimental procedure of repeatedlycompressing the oil to 300 bar, and then allowing the temperature tostabilise, the pressure can be raised higher than that, before levellingoff to 300 bar at t →∞. Obviously, the key to the problem is working out what compression

is required to achieve a final pressure in the cell of p = 300 bar.



9 Closed . . .§ 066 Pressure cell

1 2 The size of the system is irrelevant, and the solution can be ex-pressed in terms of moles. Take the total differentials of s and v from Chapter 8:

(ds)n =cp

T dT − vαdp ,

(dv)n = vα dT − vβdp .

3 In this case, the energy balance is replaced by the temperatureequation (dT)s + (dT)v = 0, because the system is in a thermostat-controlled environment; Expanded footnote [a]:

[a] In a thermostat-controlled environment we can control the temperature at all times,but we cannot normally control heat flow across the system boundary. , also see the Figure on page 406.

4 Let us combine the differentials of s and v and eliminate (dT)s,n

using the temperature equation.



9 Closed . . .§ 066 Pressure cell (2)

5 Finally we substitute in (dT)s,n = Tvα/cp(dp)s,n as the variable ofintegration for the change in pressure:

(dp)tot = (dp)s,n + (dp)v ,n

=cp

Tvα (dT)s,n + αβ-1(dT)v ,n

=( cp

Tvα − αβ-1)(dT)s,n

=(1 − Tvα2

cpβ

)(dp)s,n .

6 This expression can be significantly simplified by introducing the re-lationship cp−cv = Tvα2β-1, which is again from Chapter 8 on page 378.This gives us the final result

(dp)tot = γ-1(dp)s,n ,

where γ = cp/cv .Tore Haug-Warberg (Chem Eng, NTNU) Termodynamic methods TKP4175 02 September 2013 457 / 1776


9 Closed . . .§ 066 Pressure cell (3)

10 7 Before the differential equation can be integrated, we need to knowhow γ varies with temperature and pressure.8 Far away from the oil’s critical point, the heat capacity ratio will berelatively constant, at least across a moderate range of temperaturesand pressures.9 In this case it is sufficient to know that γ is constant ±10%. As we don’t have data for hydraulic oil, we will use the data for

benzene:∆p[bar] ≈

91.6134.3 (300 − 1) = 204 . (9.1)

12 11 The final pressure deviates significantly from the desired pressure,which has practical implications when setting the pressure prior to start-ing an experiment. An analogous calculation to the one shown in 9.1 reveals that the

initial pressure must be 439 bar in order to ensure that pt→∞ = 300 bar.



Figure 9.1: Physical properties of some common compounds. All values aregiven for T = 298.15 K and p = 1 bar.

Property[†] cp α β v cp − cv cv

Units JK mol

104

K1011

Pa106 m3

molJ

K molJ

K mol

Ideal gas − 33.54016 106 24790 8.3145 −H2O(lıq) 75.5 2.35 46 18 0.63 74.9

C(diamond) 6.07 0.035 0.68 3.4 0.0018 6.07SiO2(quartz) 44.5 0.35 2.65 22.6 0.31 44.2

NaCl 49.7 1.21 4.2 27 2.8 46.9H2O(sol) 36.6 2.55 12 19.6 3.2 33.4

Cu 24.5 0.48 0.72 7.1 0.68 23.8Ag 25.5 0.57 0.99 10.2 1.28 23.3Pb 26.8 0.86 2.37 18.7 1.74 25.1

[†] G. M. Barrow. Physical Chemistry. McGraw-Hill, 3rd edition, 1979. p. 151.



9 Closed . . .§ 067 § 66 § 68ENGLISH text is missing.Limits of thermodynam-

ics1 Heat can be transported by diffusion, convection and radiation, soan intelligent question to ask is: When can we no longer rely on ther-modynamic analysis?

2 When do we instead need to embark on more esoteric approachesbased on solving partial differential equations?

3 The answer to the latter question is relatively obvious: Thermody-namic functions have n + 2 variables, none of which describe the ge-ometry of the system.

4 Consequently, we can only solve problems where the geometry isirrelevant (i.e. where the system has uniform properties).

5 If there are gradients in e.g. T or p within the control volume, thenwe must use a distributed model, i.e. a partial differential equation.

6 In practice, convection and radiation can (often) be described interms of simple thermodynamic relationships, whereas diffusion re-quires real mathematics.

460 / 1776

A thermos flask is filled with 1 l of boiling water. It isthen buried in snow which has a temperature of 0 C.

Calculate how the temperature in the flask changes asa function of time. Does it make any difference whether

the flask maintains a constant volume or constantpressure? Does the geometry of the system affectthe calculation? Assume that the flask is insulated

with 2 cm thick expanded polystyrene. Find the timeconstant. Is the material used for the inner flask, either

2 mm thick glass or 0.5 mm thick stainless steel, ofany practical significance? What about the amount of

liquid in the flask?



9 Closed . . .Isochoric cooling

1 2 It will be assumed that the thermos is closed (in other words thereis no evaporation). For a closed system, the 1st law of thermodynamics can be written:

dU = δQ − δW . In this case, the system (the thermos flask) doesn’tperform any work, and the energy equation can be expressed as

(dU)V ,n = CV dT = δQ ,

or, if we want to emphasise the role of time: dU /dt = CV dT /dt .

3 A simple heat transfer model gives us

CV dT = − k∆l A(T − T) dt .

10 4 This equation assumes that the temperature in the flask is uniform(i.e. it is based on a lumped model).5 This gives a good approximation, as the flask is designed to pre-vent heat loss, while the water is a relatively good heat conductor (ap-proximately 20 times better than expanded polystyrene).6 However, if the thermos flask is filled with air, the assumption nolonger holds.7 Next we need to consider whether the flask is full with water, orwhether there is steam present.8 If we assume that there is steam in the flask, we must be carefulwhen calculating the CV term.9 Admittedly CV is defined as being the derivative of internal energy

with respect to temperature at constant volume, but in this context thedefinition refers to a two-phase system consisting of water and steam. The variation in pressure will be hugely different when there is

steam present compared to what is observed for a single phase liq-uid and we cannot simply use the value of CV for pure liquid water.11 On the other hand, if we assume that the flask is initially full, then

the contraction of the water as it cools will cause a significant reductionin pressure.12 If the negative gauge pressure is significant, then steam will in-evitably be produced, which takes us back to a two-phase system.



9 Closed . . .§ 067 Thermos

1 We are not at this point going to attempt to mathematically deriveCV for a two-phase water/steam system. Instead we will say that if theflask is reasonably full, then its heat capacity will be almost entirelydetermined by the water phase.

2 The change in temperature from 100 C to 0 C will reduce thewater vapour pressure from 1 atm to approximately 0.005 atm, but thishas little influence on the heat capacity of the water, and for all practicalpurposes we can assume that CV ≈ CP,H2O + CP,SiO2

.

3 For the moment we will ignore the heat capacity of the flask itself,and say that

T+θ∫

T+100

dT(T−T)

= −t∫

0

kACP,H2O ∆l dt ⇒ θ

[C]= 100 exp

(− tτ

).



9 Closed . . .§ 067 Thermos (2)

4 Some representative design parameters are k = 0.025 W m-1 K-1,A = 0.06 m2, CP,H2O = 75.5 · 1000

18 ≈ 4194 J K-1 and ∆l = 0.02 m, whichgives us a time constant of τ = 15.53 h.

5 The latter depends on the geometry of the system, in so far as thecontact area A affects the calculation, but the area isn’t very significantunless there are big changes to the shape.



0 5 10 15 200

20

40

60

80

100

1/1

1/3

1/8

Time [h]

Tem

pera

ture

[C]

Figure 9.2: Time–temperature profile for a thermos flask buried in snow. Solidlines show temperature change if the flask is ignored. Dotted lines take theflask into account. The flask is 1/1, 1/3 and 1/8 full respectively.



9 Closed . . .Time constant

1 Including the flask in the calculation affects the temperature changeover time, but it does not make much difference whether it is made of0.5 mm thick steel or 2 mm thick glass.

2 The volumetric heat capacity of steel is about a quarter of theequivalent value for glass.

4 3 Bearing in mind the thickness of the flask, the contribution of thesteel and glass to the overall heat capacity is the same. A rough but nevertheless realistic estimate is CP,SiO2

≈ 44 ·0.2·600·2.6

60.1 = 228 J K-1. This contribution will change the time constantfrom 15.5 h to 16.4 h.

5 Moreover, the initial temperature will fall from 100 C to 94.7 C, asthe flask is raised to the same temperature as the water.



9 Closed . . .Time constant (2)

9 6 This assumes that the temperature of the flask is 0 C when it isfilled.7 So, the initial temperature falls, while the time constant rises.8 At a given point in time, in this case after 16.0 h, the water temper-ature will be the same for both calculation methods. If the flask contains less water, both the initial temperature and the

time constant fall, and if the flask is an eighth full, the initial temperaturewill be 69.0 C, while the time constant will be only 2.8 h.10 This has a huge impact on the water temperature, as can be seen

from Figure 9.2 on page 416.11 The three temperature progressions were drawn using the Pro-gram 31:1.12.



U2 ≃ 0

U1 =−aN2

1V1

Figure 9.3: Gas leak in a closed room. The reference state is an ideal gas atthe original temperature of the system.



9 Closed . . .Mechanical energy balance § 681 A 40 l gas cylinder pressurised to 300 bar is placed in an insulatedroom of volume 50 m3. The room temperature is 20 C and the airpressure is 1 bar. The behaviour of the gas can be described usingthe Van der Waals equation of state p = RT/(v − b) − a/v2, wherea = 1.35 atm l2 mol-2 and b = 0.038 l mol-1. The valve fails and thegas flows out of the cylinder and into the room. Calculate how thetemperature and pressure of the gas have changed once equilibrium isreached. The room has a window of area 2 m2. How big is the forcethat acts on the window frame?

468 / 1776

§ 68 Closed room1 We can safely ignore the potential energy contribution.

2 The kinetic energy is significant in the release phase, but graduallyits contribution tends to zero.

3 A differential energy balance equation is unnecessary for the pur-poses of this problem, as the equations can be solved algebraically atlimt→∞.

4 This is because the system does not interact with its surroundings.

5 It is therefore sufficient to solve the energy and mass balance equa-tions without looking at the sequence of events.

6 The energy balance equation for the room, including the gas in theflask, can be expressed as

(∆U)V ,n = 0 .

7 We will start by finding the internal energy of the pressurised gas.

8 Let us convert a and b to SI units: a = 1.35 · 101325 · 10-6 =

0.13679 Pa m6 mol-2 and b = 3.8 · 10-5 m3 mol-1.

9 Then we can calculate the number of moles of gas in the cylinderusing the equation

300·105

[Pa] = N1R ·293.150.04−N1b −

N21a(0.04)2 ,

which gives us N1 = 443.6 mol, cf. Program 1.9 in Appendix 31.

10 Incidentally, the pressure in the cylinder is ≈ 10% lower than itwould be if the gas were ideal (which appears realistic).

11 In the rest of the room there are N2 = 2049.7 mol of air.

12 We will use an ideal gas at 293.15 K as our reference state for in-ternal energy— see explanation in Figure 9.3 on the preceding page—and will ignore non-ideal contributions in the expanded gas (verify thatthis is a reasonable assumption).

13 By using the results that we derived in Chapter 8, we can thereforesay that[a]

14 (N1 + N2)cıgv ∆T = −aN2

1/V1, or:

∆T[K] =

0.13679 (443.6)2

0.04 (443.6 + 2049.7) (29 − 8.3145)= −13.0 .

15 The change in temperature is modest, but the positive gauge pres-sure in the room is 0.1607 bar. On a 2 m2 window, this is equivalent to31237 N, which is far more than the frame can withstand.

469 / 1776

[a] Assuming that the heat capacity is independent of temperature— which is a goodapproximation when ∆T is small.

1 The thermos flask in Theme 67 on page 412 represents an extremecase, from a thermodynamic point of view, in that the energy balanceequation is reduced to simply a heat transfer problem.

2 An adiabatic system that only does pressure–volume work on itssurroundings is another extreme case.

6 3 This kind of system is more common than you might expect.4 If state changes are rapid, e.g. the air in a bicycle pump or the gas

in a gas turbine can be described in terms of adiabatic changes of state.5 For an ideal gas, the internal friction caused by flow resistance and

turbulence can be ignored. This is the same as saying that the entropy is conserved, whichmeans that the energy balance equation can be written

(dU)s,n = − δW = −p dV .

8 7 Systems that exchange both heat and work with their surroundingsare an intermediate case. Here we will compare adiabatic compression with a couple of other

types of compression. 9 We will return to the connection between heat and work in Chap-ter 18 on page 938.



9 Closed . . .§ 069 § 68 § 70

Determine the compression work done during a variablechange in volume in a closed system. The answerevidently depends on how the compression takes

place. Find comparable expressions for the isobaric,isothermal and isentropic compression of an idealgas. Assume a constant heat capacity of C ıg

P . Howmuch heat must be removed/added in each case?



9 Closed . . .§ 069 Compression work

1 For all of the parts of the question, the energy equation can be ex-pressed in the form (dU)

ıgn = δQ − δW , where the work term is δWrev =

pıg dV .

2 The internal energy of an ideal gas is only dependent on its tem-perature and composition, so we can say that C ıg

V dT = δQ − pıg dV .

5 3 Note that assuming that the gas is ideal means that all of the partsof the question have the same structure.4 The only question that remains is how to integrate p dV . The simplest thing to do is to start by looking at isobaric

compression Expanded footnote [a]:

[a] Perhaps compression is a misleading term given that the gas pressure is constant.An alternative way of looking at it is to think of a cooling process in which the gascontracts as a result of the fall in the temperature of the system. at a given system pressure p:

W ıgp =

V∫

V

p dν = pV (κ − 1) = NRT (κ − 1) .



9 Closed . . .§ 069 Compression work (2)

8 6 In the above equation, and in the rest of the section, κ refers to therelative reduction in volume V/V.7 The amount of heat that must be removed during compression isδQ = C ıgV dT + pıg dV = C ıgV dT + NR dT = C ıgP dT F (Q ıg)p = (∆Hıg)p . The next item on our agenda is to calculate the isothermal com-

pression work at the system temperature T:

W ıgT =

V∫

V

p(T, ν) dν =V∫

V

NRTν dν = NRT lnκ .

14 9 Here the amount of heat released is δQ = C ıgV dT + pıg dV =

pıg dV = δW i.e. (Q ıg)T = (W ıg)T .10 Now what remains to be done is to obtain an expression for theisentropic compression work.11 Normally we only know the entropy of the system indirectly throughT ,p or T ,V .12 The integral is therefore expressed in terms of these variables (so,for the initial state, T is used, rather than S).13 We have already derived the isentropic relations for an ideal gas;see Chapter 8:8.8. In order to save space, we will not do so again here. Instead, refer

to Eq. 8.47 on page 382:

W ıgS =

V∫

V

p(S, ν) dν =V∫

V

p( νV

)–γ dν = pVκ∫

1

κ–γ dκ

= NRT1−γ

(κ1−γ − 1

). (9.2)



9 Closed . . .§ 069 Compression work (3)

17 15 In this case δQrev = 0, which means that (Q ıg)S = 0.16 Show that this statement is true. What, then, is the difference between the three kinds of compres-

sion work? Well, for a given volume reduction of κ, (W)p < (W)T <

(W)S .18 The smaller κ is, the greater the difference, as shown in Figure 9.4on the following page.19 See Program 31:1.2 for further details.20 Note that W is defined as being positive when the system performswork on its surroundings.21 The compression work shown in Figure 9.4 is therefore negative.



0 0.2 0.4 0.6 0.8 110

2

103

104

1 − VV

W[J/m

ol]

Figure 9.4: Isentropic, isothermal and isobaric compression work for one moleof an ideal gas (γ = 1.40).



9 Closed . . .Heat capacity ratio γ § 701 Modern kitchen taps have ceramic valves that stop the water flowwithin a fraction of a second. This sudden blocking of the flow causeswater hammer, which can damage pipe connections and plumbing fix-tures. The water hammer originates where the valve is shut, and movesup the pipes at the speed of sound. The physics of the interface be-tween the water and the pipe is complex, and we will only look at whathappens in the water phase. The compression of this phase is ex-tremely rapid, and it is a good approximation to assume that it is isen-tropic. Derive an expression for the pressure build-up in the water whenthe flow of water is blocked. Calculate ∆pmax for a couple of represen-tative flow velocities and temperatures. Use the physical data from thepreceeding sections.

477 / 1776

ENGLISH text is missing.Kinetic energy conver-

sion1 In this case, the energy balance equation for the control volume

has an unusual form:

dEK + (dU)S ,n = 0 . (9.3)

2 It is tempting to include a pV term, because instinctively you thinkthat the control volume is compressed by the liquid upstream of it.

3 However, this is not the case, as the size of the control volume isarbitrary, and in theory it could encompass all of the liquid that is inmotion.

4 Hence there is no external mass that can perform pV work on thesystem.

478 / 1776

§ 70 Water hammer1 What remains to be done is to substitute in the differentials ofinternal energy at constant entropy (dU)S ,n = −p dV and of kineticenergy dEK = Mυdυ.

2 Note that the size of the system is irrelevant, so we can expressthe equations on an intensive form,

3 which gives us the following energy balance equation:

p(dv)s,n = Mwυdυ .

4 The differentials of s and v that we need can be found in Chapter 8:

(dv)n = vα dT − vβdp ,

(ds)n =cp

T dT − vα dp .

5 Eliminating dT at constant molar entropy and composition givesus:

(dv)s,n = vα Tcp vα dp − vβdp

=(Tv2α2

cp − vβ)dp . (9.4)

6 Combining the differential from Eq. 9.4 with the relation cp − cv =Tvα2

β , which is taken from Section 8.7 on page 377 gives us the finalresult

(dv)s,n = − vβγ dp . (9.5)

7 From the energy equation we know that p(dv)s,n = Mwυdυ. Com-bining this with Eq. 9.5 gives us γ-1vβp(dp)s,n = −Mwυdυ.

8 For isentropic changes of state in liquids, the molar volume and thespeed of sound are roughly constant, so the maximum pressure can beexpressed as

∆p ≈ υ√

Mwγvβ = υρ

√ γρβ = υρυ⋆

where υ⋆ is the speed of sound in the liquid; see Eq. 9.10 on page 443.For water flowing at a speed of 1 m s-1, the water hammer will have apressure of 14.5 bar.

9 This is at the limit of what pipes and plumbing fixtures can with-stand, so the moral of the story is that modern taps should be turnedoff with some care.

10 The best taps have slow-closing valves to prevent pipes from com-ing loose from their supports.

479 / 1776

1 2 The heat capacity ratio γ appears in a wide variety of contexts, andit is worth saying a bit about what values it can have. Based on statistical mechanics it can be shown that γ = 5/3 for a

monoatomic ideal gas, and that γ = 7/5 for a diatomic ideal gas with allof its vibrational and rotational levels occupied.

4 3 But what about the value of γ for liquids and solids? In their casesit is harder to make any generalisations, but for many substances thevalue lies between 1.2 and 1.5, as shown in Table 9.1 on page 411. However, it is worth noting a few exceptions: For solids with a high

modulus of elasticity, such as quartz and diamond, γ ≈ 1.

5 That is also the value for water at 4 C: This is the temperatureat which its density reaches a maximum, which means that α = v -1

(∂v/∂T)p,n = 0, and therefore that cp − cv = 0 in Eq. 8.41 on page 378.

6 From this it follows that γ = 1, which must also be considered theminimum Expanded footnote [a]:

[a] cp ≥ cv is a fundamental prerequisite for stability in thermodynamics, cf. Chapter 21on page 1312. value for γ.



xx +∆x

υx dtυx+∆x dt

Figure 9.5: Quasi one-dimensional flow in the horizontal plane. The impactof the y and z planes has been ignored. The arrows indicate the distancetravelled by the control volume during the time interval dt (the fluid acceleratesto the right). The fluid has been shown simultaneously expanding in order tofocus attention on compressible flow.



9 Closed . . .§ 071 § 70 § 72ENGLISH text is missing.Uniform flow1 Classic thermodynamics deals with equilibria and the properties ofequilibrium states.

2 In this chapter we are applying control volume theory to flowingmedia, which raises several fundamental questions about the nature ofa state.

3 It is far from obvious that results from thermodynamic equilibriumtheory apply to media in motion, but we will nevertheless assume thiswithout any further justification[a].

4 Next we must write Newton’s 2nd law in a form that relates velocityand time to thermodynamic state variables.

5 Velocity is distance over time, and multiplying it by the local cross-sectional flow area gives us the volume transported per time unit.

6 By considering a control volume that travels υ dt over a short spaceof time dt , it is possible to relate the velocity to familiar thermodynamicvariables such as mass, volume, energy etc, but it is not possible torigorously analyse the flow.

7 The theory in this section therefore only applies to the adiabatic,stationary and quasi-one-dimensional flow of a frictionless fluid withoutchemical reactions.

8 This may appear to be a hopelessly unrealistic special case, but inpractice there are many situations that fall within this category.

9 A number of important applications are discussed in Chapter 12(gas dynamics).

481 / 1776

[a] This topic is covered in detail in Chapter 22 on page 1368.

Figure 9.5 on the previous page shows a volumeelement moving with the fluid flow. There are no chemicalreactions in the control volume, and the flow is assumed

to be stationary. Show that the mass balance of theelement can be expressed as (dv

v )n = dυυ + dA

A whereυ is the flow velocity, v is the molar volume and A is

the cross-sectional flow area.



9 Closed . . .§ 071 Continuity of mass

1 Figure 9.5 shows the control volume at t = 0 and t = dt . 2 In the time period dt the system boundaries[a]

[a] In fluid dynamics, the coordinate system is either based on a fixed point in space(Euler), or it moves freely with the flow (Lagrange). The latter system has been chosenhere. move

from the shaded areas to the dotted lines in the drawing.

3 The left-hand system boundary moves υx dt in parallel with theright-hand boundary moving υx+∆x dt .

4 This creates two sub-volumes, one at the left and one at the rightof the figure, which both have the same mass Mx = Mx+∆x , givenstationary flow.

5 If we substitute in the flow variables, the mass balance can beexpressed as

(Aυdtv

)x =

(Aυdtv

)x+∆x . (9.6)



9 Closed . . .§ 071 Continuity of mass (2)

8 6 Since there are no chemical reactions in the control volume, wecan alternatively view v as a molar volume, but in that case the massbalance must be expressed in the form Nx = Nx+∆x .7 Regardless of whether Eq. 9.6 is expressed in molar or mass

terms, the differential dt will be a common factor to both sides of theequation. If we divide by dt , the equation simplifies to (Aυ

v )x = (Aυv )x+∆x

which means that Aυv is conserved in the direction of flow. In other

words ∆(Aυv ) = 0.

9 Provided that A , v and υ are continuous functions of the path lengthx, then d

(Aυv)n= 0, or expressed in logarithmic form:

∆(Aυv ) = 0

continuum⇒ (dln v)n = dln υ+ dln A . (9.7)

11 10 Note that the implication sign in Eq. 9.7 is pointing one way only. Even if the derivative doesn’t exist, meaning that it cannot be de-fined experimentally, it will still be possible to satisfy ∆(y) = 0 withoutdy = 0.

12 This kind of sudden change in the state variables is referred to asa normal shock in Chapter 12:12.5.



9 Closed . . .§ 072 § 71 § 73

Show that the energy balance of the (adiabatic) controlvolume in Figure 9.5 can be expressed as (dh)n = −Mwυ

dυ where Mw and h are respectively the molar mass andmolar enthalpy of the fluid. The remaining information

you need follows from Theme 71.



9 Closed . . .§ 072 Continuity of energy

1 The energy balance can be expressed (Me+pAυdt)x = (Me+pAυdt)x+∆x since the total energy content E = Me, plus the pV work re-quired to displace the fluid, must be the same for the two sub-volumes Expanded footnote [a]:

[a] These are still defined by the displacement of the control volume in the time perioddt

.

2 If we divide by Mx = Mx+∆x from the mass balance, we get (e +

pv)x = (e + pv)x+∆x , where v = M-1Aυdt is the specific volume of thefluid.

3 This shows us that e + pv ≈ u + ek + pv = h + ek is conserved inthe horizontal plane (ignoring the potential energy ep).

5 4 As there are no chemical reactions in the fluid, we can alternativelyuse molar values. If we substitute in molar kinetic energy Expanded footnote [a]:

[a] It has been assumed that υ2 = υ2 = υ2. Thtat allows us to use the same velocityvariable in the expressions for (mean) mass flow ρA υ and (mean) kinetic energy 12 ~mυ2.This simplification only applies in the case of fully developed turbulence. ek = 1

2Mwυ2 we end upwith

∆(h + ek ) = 0continuum⇒ (dh)n = −Mwυdυ (9.8)



9 Closed . . .§ 072 Continuity of energy (2)



9 Closed . . .§ 073 § 72 § 74

ENGLISH text is missing.Comments1 In deriving equations 9.7 and 9.8, the main emphasis has been ondescribing the physical properties of a flowing fluid, rather than on theunderlying mathematics.

2 In continuum mechanics, the same equations are derived in a moreuniversal and elegant way from the divergence theorem, but the moreabstract the derivation, the further removed it is from thermodynamics.

3 This difference will become even clearer when we now go on tolook at the last of the balance equations — the momentum balanceequation.

4 Unlike the continuity and energy equations, the momentum bal-ance equation is not related to thermodynamics.

5 A thermodynamic equilibrium describes a static state, and is unre-lated to the dynamics of the system, whereas the momentum equationis based on Newton[a]’s 2nd law.

6 One of the ways that it can be formulated involves the integral offorce with respect to time (impulses).

7 From Newton’s 2nd law it follows that∫

F dt = M∫

a dt = M∆υ. Inother words, an impulse changes the momentum of the system.

8 With reference to Figure 9.5, any pressure gradients in the controlvolume result in contact forces, which in turn affect the velocity of thefluid.

9 If we relate the pressure to the mechanical state, it tells us some-thing about the movement of the fluid.

10 But it is important to note that the force vector F has both a spatialdirection and a length (a measuring number combined with a unit).

11 This means that the momentum balance equation is vectorial,which is unhelpful to us, since the thermodynamic variables are as-sumed to vary only along the x axis.

12 If we are to use scalar equations, it is no longer sufficient for theflow to be quasi-one-dimensional, it must be genuinely one-dimensionaland have a constant cross-sectional flow area.

489 / 1776

[a] Sir Isaac Newton, 1642–1727 by the Julian calendar. English physicist and mathe-matician.

Show that in a frictionless control volume with a constantcross-sectional flow area, the momentum equationcan be expressed in the form v(dp)A ,n = −Mwυdυ.

The flow is only in the horizontal plane, it is genuinelyone-dimensional and frictionless, and there are no

chemical reactions in the fluid.



9 Closed . . .§ 073 Continuity of momentum

1 By expressing the momentum equation in scalar form, we areassuming that the flow is one-dimensional.

2 In that case, the pressure is isotropic, and Newton’s 2nd law canbe formulated as (Aυdt

v · υ)x+∆x − (Aυdtv · υ)x = (pA dt)x+∆x − (pA dt)x .

3 If we divide the balance equation by (A dt)x = (A dt)x+∆x we findthat p + v -1υ2 is conserved in the direction of flow.



9 Closed . . .§ 073 Continuity of momentum (2)

4 Since we know that d(υv)A ,n = 0 from the continuity Eq. 9.7 on

page 430, this means that the momentum equation can be expressedin the form

∆(p + υ2

v )A = 0continuum⇒ v(dp)A ,n = −Mwυdυ , (9.9)

where v represents the specific volume in the left-hand equation andthe molar volume in the right-hand one.



9 Closed . . .§ 074 § 73 § 75

For the purposes of the momentum Eq. 9.9, it is importantto assume that the flow is frictionless. Show that thisassumption also implies that the flow is isentropic.



9 Closed . . .§ 074 Frictionless vs isentropic

1 Combining the equations 9.9 and 9.8 gives us (dh)n = v(dp)A ,n.

2 Bearing in mind that the differential (dh)n = T ds + v dp, it followsthat T ds = 0.

3 In other words, the flow is isentropic.

4 Since the momentum Eq. 9.9 requires frictionless flow, whereasthe energy Eq. 9.8 does not, it is true to say that “frictionless” in thiscontext is the same as “isentropic”.



9 Closed . . .Speed of sound

1 It is perhaps surprising that a concept like the “speed of sound” isrelated to thermodynamics, but in fact it can be derived by looking at asimple control volume.

4 2 However, it should be stressed that the derivation is subtle, ar-guably even philosophical, and that in many ways it appears to be aretrospective[a] interpretation.3 Hence it doesn’t tell us a great deal about speed of sound as aphysical phenomenon, but it does still give us a useful perspective forlooking at the connection between thermodynamics and flow.

[a] Based on a historical perspective and with the benefit of hindsight. The equations for an adiabatic control volume can be summarisedas follows:

Energy equation; (dh)n =−Mwυdυ .

Continuity equation; (dln v)n = dln υ+ dln A .

1st law; (dh)n = T ds + v dp .

2nd law; T(ds)n ≥ 0 .



9 Closed . . .§ 075 § 74 § 76

According to the 2nd law of thermodynamics, ds /dt ≥ 0.This only applies to adiabatic systems. Combining

this law with the balance equations from Themes 71and 72, as well as with the differential (dh)n = T ds +

v dp, reveals that the flow velocity cannot exceedthe speed of sound of the fluid, which is given by

υ2⋆ = −M-1

w v2 (∂p/∂v)s,n. Show this.



9 Closed . . .§ 075 Choked flow

1 Let us first isolate the v dp term from the 1st law and then substi-tute in the energy balance, mass balance and 2nd law in that order:

v(dp)n = dh − T ds

= −Mwυdυ − T ds

= −Mwυ2v -1 dv − T ds

≤ −Mwυ2v -1 dv .

2 According to the above inequality, v(dp)n will reach a maximumunder isentropic conditions.

3 Under other conditions, −T ds will make a negative contribution tothe right-hand side.



9 Closed . . .§ 075 Choked flow (2)

4 Next we divide by dv assuming that s is constant, and isolate thevelocity term:

υ2 ≤ υ2⋆ = −M-1

w v2 (∂p∂v

)s,n .

5 This result is an expression for the speed of sound in the fluid.

6 The same definition is used in wave mechanics, although it is de-rived in a completely different way.



9 Closed . . .§ 076 § 75 § 77ENGLISH text is missing.Comments1 Here we must accept that the speed of sound and the maximum

flow velocity are two sides of the same coin, but it is fascinating to seehow well-known results from wave mechanics and gas dynamics canbe related to thermodynamics even when you have so little to go on.

2 Incidentally, the speed of sound is relevant far beyond the confinesof pipe flow, and the above equation correctly defines the speed ofsound in all homogeneous substances (but this cannot be shown usingthermodynamic principles).

3 This classical interpretation of the speed of sound is interesting inits own right, but the reverse application is equally important,

4 since the speed of sound of a fluid can be measured relativelyeasily and precisely.

5 This has made υ⋆ an important parameter for experimentally de-termining the equations of state of gases and liquids.

499 / 1776

As a general rule, there are many ways of expressingthermodynamic relationships, and the speed of sound isno exception. Show that υ⋆ in Theme 75 on page 440

can alternatively be expressed

υ2⋆ = −M-1

w v2 (∂p∂v

)s,n =

cpvMwcvβ

=γρβ . (9.10)



9 Closed . . .§ 076 Speed of sound I

1 Compare Eq. 9.10 with the equivalent expression in Theme 75 onpage 440. The exercise boils down to demonstrating that (∂p/∂v)s,n =

−cp/(cvvβ) is true.

2 The following three differentials from Chapter 8 on page 377 pro-vide a useful starting point:

(ds)n =cv

T dT +(∂p∂T

)v ,n dv ,

(ds)n =cp

T dT − vαdp ,

(dv)n = vαdT − vβdp .

3 Here we can eliminate dT from the first two equations, providedthat s is constant:

− Tcv

(∂p∂T

)v ,n (dv)s,n = T

cpvα(dp)s,n .



9 Closed . . .§ 076 Speed of sound I (2)

4 We still need to get rid of (∂p/∂T)v ,n.

5 At constant v, the volume differential can be rearranged to(∂p/∂T)v ,n = αβ-1, which if inserted into the above equation gives us−Tcv

-1αβ-1(dv)s,n = Tcp-1vα(dp)s,n.

6 After dividing by (dv)s,n, the final result is(dpdv

)s,n =

(∂p∂v

)s,n = − cp

cv vβ



9 Closed . . .§ 077 § 76 § 78

T [K] υ⋆ [ms ] T [K] υ⋆ [

ms ]

283.148 1447.290 323.137 1542.470293.145 1482.320 333.134 1550.880303.143 1509.040 347.131 1555.010313.140 1528.800 363.126 1550.320

Calculate the speed of sound of each of the fourteencompounds in Table 9.1 on page 411. Is there a

general pattern to which substances have a high orlow speed of sound? For water, compare the re-sult obtained by calculation with the experimental

values Expanded footnote [a]:

[a] Del A. J. Grosso. J. Acoust. Soc. Am., 45:1287, 1969.


[a] W. Kroebel and K. H. Mahrt. Acustica, 35:154, 1976. in the above table. All of the values relateto 1 [bar]. When doing calculations for ideal gases,



9 Closed . . .§ 077 § 76 § 78 (2)

you can either use Eq. 9.10 on page 443, or you canconsult Eq. 9.11 on page 451.



9 Closed . . .§ 077 Speed of sound II

1

υ⋆ [ms ] υ⋆ [

ms ] υ⋆ [

ms ]

He 1015 CCl4 931 NaCl 3412H2 1318 C6H6 1312 H2O(sol) 3152N2 352 CHCl3 1050 Cu 3998SF6 136 H2O(lıq) 1480 Ag 3153Hg 1599 Diamond 6455 Pb 2016CS2 1149 Quartz 3780

For gases, υ⋆ ∝ M-1/2w , which produces values in the range

100–1000 m s-1.

2 Solids with a high modulus of elasticity (particularly diamond) havethe highest speeds of sound, often in the range 3–4000 m s-1.



9 Closed . . .§ 077 Speed of sound II (2)

3 If you calculate υ⋆ for liquid water you get 1480 m s-1 at 25 C, whichis close to the value in the table. 4 The other liquids have similar values to water, albeit generally

slightly lower.



9 Closed . . .§ 078 § 77 § 79

In gas dynamics, the speed of sound in an ideal gas isa very important concept. Show that υıg

⋆ =√γRT/Mw

based on Theme 76.



9 Closed . . .§ 078 Ideal gas III

1 For an ideal gas βıg = −v -1 (∂v/∂p)T ,n = p-1.

2 Substituted into Eq. 9.10 this gives us: υ2⋆ =

cpRTMwcv

.

3 The equation below follows from the definition γ = cp/cv :

υıg⋆ =

√γRTMw

(9.11)



9 Closed . . .Looking back

1 The 1st law of thermodynamics: d(EK + EP + U) = δQ − δW

2 Isothermal compression: W ıgT =

∫ VV

p(T, ν) dν = NRT lnκ

3 Isentropic compression: W ıgS =

∫ VV

p(S, ν) dν = NRT1−γ

(κ1−γ − 1

)

4 Continuity of mass: ∆(Aυv ) = 0

5 Continuity of energy: ∆(h + ek ) = 0

6 Continuity of momentum: ∆(p + υ2

v ) = 0

7 Speed of sound: υ2⋆ = −M-1

w v2 (∂p∂v

)s,n =

cpvMwcvβ

=γρβ

8 Applied to ideal gas: υıg⋆ =

√γRTMw


News Emptying a gas cylinder Filling a gas cylinder Acoustic resonance Olds

Part 10

Dynamic systems




10 Dynamic . . .New concepts

1 Inlet and outlet control volumes

2 Emptying an ideal gas flask

3 Filling an ideal gas flask

4 Joule’s gas expansion experiment

5 Kvasistatic state

6 Helmholtz resonance

7 Eigenfrequency



10 Dynamic . . .Dynamic systems

1 This chapter will look at what happens when a closed system isopened, allowing mass transfer through an inlet or outlet.

2 The result is a dynamic system, which can be filled or emptied overa period of time.

3 Apart from the equilibrium state, which is reached at the limit t →∞, the system has no stationary state.

6 4 It is hard to conceive of this kind of system as an abstract controlvolume without physical limits, so it can be helpful to look at a specificobject such as a storage cylinder.5 However, although most people find it easier to get to grips witha physical entity than an abstract system, by using one we create anunintended problem for ourselves: Where should the system boundarygo? Figure 10.1 on the following page shows three identical cylinders

with the flow limiter — for instance a valve or a crack in the cylinderwall— in different locations.



a) b) c)

Figure 10.1: Outflow from cylinder. Three alternative locations for the flowlimiter.



10 Dynamic . . .§ 079 Outflow § 791 Where would you draw the system boundaries in Figure 10.1 if

you wanted to describe the state inside the cylinder? Then assume thatthe direction of flow is reversed. Where would you draw the systemboundaries now?

513 / 1776

1 The system boundary should be drawn inside the cylinder, as closeto the valve as possible.

2 It is important to know the state of the fluid as it leaves the system,so it is assumed that the control volume has uniform properties.

3 Heat exchange with the cylinder walls must be considered on acase-by-case basis.

4 In the figure on the left, you must also consider whether there is anyheat exchange between the small section of pipe inside the cylinder andthe fluid in the cylinder.



10 Dynamic . . .§ 079 Inflow

1 The system boundary should once again be drawn inside the cylin-der, but this time so as to include the valve.

2 This is because we need to know the state of the fluid flowing intothe system.

3 Heat exchange with the walls of the cylinder must once again beconsidered on a case-by-case basis.

4 In the figure on the right, you must also consider whether there isany heat exchange between the section of pipe outside the cylinder andthe flowing fluid.



10 Dynamic . . .§ 080 § 79 § 81

Derive an expression for the temperature change thattakes place when a cylinder containing an ideal gasis emptied. Assume that the gas has a constant heat

capacity in the relevant temperature range. Ignore anyheat gained from the surroundings, including from thewalls of the cylinder. What will the final temperature

be in a compressed air cylinder with an initial state of10 bar and 20 C? Does the size of the cylinder affect

the numerical result?



10 Dynamic . . .§ 080 Cylinder emptyingENGLISH text is missing.System boundary1 The system boundary must be drawn in such a way as to allow us

to describe the state of the fluid passing through the control surface.

2 In order to avoid having to digress into a discussion of adiabaticflow and gas dynamics, we will therefore aim to place the system bound-ary somewhere that allows us to disregard the flow of kinetic energythrough the control surface.

3 This means that the pressure and temperature of the flowing fluidwill be the same as inside the cylinder itself.

4 If we are dealing with a leak, e.g. through a hole in the wall, thecontrol surface may need to be inside the cylinder.

5 If the size of the leak is large in relation to the surface area of thewalls of the cylinder, we will need to put the control surface so far intothe cylinder that it will be at the expense of the control volume itself.

6 Then we lose track of a significant proportion of the contents of thecylinder, and the description breaks down[a].

518 / 1776

[a] If a pipe breaks in two, we cannot treat it as a lumped problem because the cross-sectional area of flow of the gas has the same dimensions as the control surface.

1 Provided that the cross-sectional area of flow is small in relation tothe typical cross-sectional area of the control volume, the energy andmass balances for the emptying process can be expressed as

dUc-vdt = −~Uout − pout~Vout = −~Hout ,

dNc-vdt = −~Nout .

4 2 We have tacitly assumed that there are no chemical reactions tak-ing place in the control volume.3 If that is not the case, then the mass M must replace the molenumber N as the free variable. For an ideal gas with constant heat capacity we can say that

Uıg = Nc ıgv T ,

Hıg = Nc ıgp T ,

6 5 where the reference energy has been defined as Uıg(0) = 0. The relationship Hıg = Nc ıgp T follows from the definition H = U +

pV , which can be restated as Hıg = Uıg+NRT = N(c ıgv +R)T = Nc ıg

p T .



10 Dynamic . . .§ 080 Cylinder emptying (2)

7 By combining the mass and energy balances, while simultaneouslysubstituting in molar values, we get

d(Nc ıgv

T)c-v

dt = −c ıgp Tout~Nout = c ıg

p ToutdNc-v

dt .

9 8 This brings us to the crux of the derivation. The temperature inside the cylinder is the same as the temperaturein the flowing fluid, 10 which is a direct consequence of our foresight when choosing the

location of the control surface. This allows us to eliminate the indices from thetemperature equation.

11 Expanding out the total differential on the left-hand side of theequation gives us

c ıgv T dN

dt + Nc ıgv

dTdt = c ıg

p T dNdt .



10 Dynamic . . .§ 080 Cylinder emptying (3)

12 Now we can substitute in γ = (cp/cv)ıg and rewrite the expression

in logarithmic form:

(dln Tdln N

)ıg= γ − 1 ≈

2/3 for He,Ne,Ar,Kr

2/5 for H2,N2,O2,etc.0 for ∞ degrees of freedom

15 13 Note that the differential dt has disappeared.14 Thermodynamics is quite literally a timeless[a] discipline, and thereare never any time variables in thermodynamic expressions unless theyare linked to a transport term[a].

[a] Thermodynamics does not involve time variables, and can also be considered atimeless discipline, as most of its principles are based on work carried out over 100years ago, by people like Ludwig Ferdinand von Helmholtz, Rudolf Julius EmanuelClausius, James Clerk Maxwell and Josiah Willard Gibbs.

[a] From fluid mechanics, heat transport, diffusion etc. Given an initial state T1,N1, the differential equation can be inte-grated to obtain

T ıg2 = T1

(N2N1

)γ−1.

16 Alternatively, by substituting in the values for an ideal gas we get

T ıg2 = T1

(p2VRT2

)γ−1 (RT1p1V

)γ−1= T1

(p2T1p1T2

)γ−1 ⇒ T ıg2 = T1

(p2p1

) γ−1γ



10 Dynamic . . .Temperature change

1 In other words, the temperature inside the cylinder changes in thesame way as it does when a closed system expands adiabatically.

2 If a compressed air cylinder at T1 = 293.15 K and p1 = 10 bar isemptied quickly until it reaches a final pressure of p2 = 1 bar, the finaltemperature will be T2 = 151.8 K (assuming γ = 1.40).

4 3 In other words, it is hardly surprising that condensation forms whenyou empty a compressed air cylinder. The size of the cylinder doesn’t affect the calculation, but that is

only because we have questionably assumed that the emptying processis adiabatic.

5 In practice this is not true if: i) the volume of gas is small, or ii) thecylinder is emptied slowly.



υ ≈ 0 υ⋆ υ

Figure 10.2: Control volume for rocket engine



a) b)

~W~Q~Q

∆~S = 0 ∆~H = 0

Figure 10.3: Alternative methods of emptying a gas cylinder to produce atemperature drop: a) is isentropic and b) is isenthalpic.



10 Dynamic . . .§ 083 § 82 § 84 § 811 Figure 10.2 is a schematic diagram of a solid-fuel rocket. Set out adifferential energy balance equation for the control volume shown, andintegrate it over the period that the rocket operates. Assume that thechemical reaction C+NH4NO3 → N2 +2 H2O+CO has a constantrate of reaction and that the state of the combustion chamber is station-ary. Derive an expression for the outflow velocity if all of the internal en-ergy of the reactants is converted into kinetic energy.

523 / 1776

ENGLISH text is missing.Rocket engines1 An ideal rocket engine is assumed to have the following charac-

teristics: Chemical equilibrium[a] in the combustion chamber; exhaustgases follow the ideal gas law; flow is frictionless and adiabatic; anduniform mass flow rate with a quasi-one-dimensional velocity field.

2 In spite of all of these assumptions, in many cases the numericalresults are within 10% of the observed behaviour.

525 / 1776

[a] It is normal to assume either that the composition of the gas remains unchangedthrough the nozzle (frozen equilibrium), or that the equilibrium is constantly changingthroughout the nozzle (dynamic equilibrium). In real life, the truth lies somewherebetween these two extremes.

§ 81 Rocket engine1 From a thermodynamic point of view, there is little difference be-

tween a rocket engine and a gas cylinder being emptied.

2 The basic principles set out in Section 80 therefore remain valid,but we must add a term for the kinetic energy of the exhaust gases tothe energy balance equation:

dUc-vdt = −~Hout − ~EK ,out .

3 Under stationary conditions, the energy equation can be integratedto obtain EK ,out = Uc-v − Hout.

4 This simplification is possible because all of the quantities are time-invariant, and in that sense the description of the thermodynamic stateof a rocket is simpler than the description of a pressure vessel.

5 From the energy equation we can see that the outflow velocitywould be zero if the enthalpy of the exhaust gases exactly matchedthe internal energy of the fuel.

6 The thermodynamic description of a similar system is discussed indetail in Chapter 17.

7 The lower the enthalpy of the exhaust gases (and thus the temper-ature), the greater the velocity.

8 Even supersonic flow is possible, if the nozzle is designed in aparticular way (see Chapter 12 on page 644 for a discussion of that).

9 In this exercise we will assume that the temperature of the gasflowing out of the rocket is the same as that of the unreacted fuel.

10 In that case, the outflow velocity will be

υ =

√2(Uc-v−Hout)

Mw ≃√−2∆rxH

Mw = 2229 m s-1 ,

where ∆rxH = −2×241826−110527+365560 = −228619 J[a] [a] andMw = 92.054 g (values given for the reaction equation, not per mole).

11 This calculation is based on the assumption that U = H − pV ≃ Hfor low-molecular-weight solid compounds, e.g. ammonium nitrate andcarbon[a].

526 / 1776

[a] The NBS tables of chemical thermodynamic properties. Selected values for inor-

ganic and C1 and C2 organic substances in SI units. J. Phys. Chem. Ref. Data, 11(2):1–392, 1982.


[a] Even at a pressure of 100 bar the pV term comes to less than 500 J.

§ 821 Figure 10.3 on the preceding page shows two identical gas cylin-ders being emptied in two different ways. Assume perfect heat ex-change between the gas that leaves the cylinder and the gas that re-mains, and that heat loss through the wall is negligible. Derive expres-sions for the temperatures in the two cylinders as functions of time.Which of the two methods of emptying results in the lowest final tem-perature?

527 / 1776

§ 82 Isenthalpic emptying1 It is easy to fail to see the wood for the trees, but what goes on

inside the control volume is actually irrelevant to the changes of state.

2 In practice, it is only the interaction between the control volume andits surroundings that differentiate this example from Section 80.

3 If the gas is ideal, even that difference no longer exists, as thetemperature is then the same on both sides of the valve.

4 Consequently, in an ideal gas there is no net heat exchange be-tween the contents of the cylinder and the gas flowing out of it.

5 However, the temperature of a real gas will fall as it passes throughthe valve (or rise if its state is outside the inversion curve shown inFigure 397).

6 This results in heat exchange between the gas and the contents ofthe cylinder, which in turn changes the outflow conditions.

7 Here it is worth noting that if the outflow pressure is sufficiently low,the behaviour of the gas will match that of an ideal gas as described inSection 80, because the impacts of the gas’s non-ideal properties willcancel each other out during the heat exchange.

529 / 1776

§ 82 Isentropic emptying1 This time the control volume does work that needs to be included

in the energy balance equation.

2 Apart from that, the principles are the same as in Section 80 onpage 459 (and Section 83 below):

dUc-vdt = −~Hout − ~Ws ,

dNc-vdt = −~Nout .

3 Let us take a formula for isentropic work of expansion from Sec-tion 91 on page 504, Eq. 11.7:

~W ıgS = γ

γ−1 RT

[1 − (p

p) γ−1

γ

]~Nout

= γγ−1 RT(1 − α)~Nout .

4 The parameter α ≤ 1 has been introduced purely out of laziness.

5 T ,p is the state inside the cylinder and p is the outflow pressure.

6 Let us now add the work of expansion ~W ıgS , as well as Uıg = Nc ıg

v T

and Hıg = Nc ıgp T , to the energy balance:

d(Nc ıgv T)

dt = −c ıgp T ~Nout − γ

γ−1 RT(1 − α)~Nout .

7 If we expand the left-hand side (using the chain rule) and substitutein the mass balance we get

c ıgv T dN

dt + Nc ıgv

dTdt = c ıg

p T dNdt + γ

γ−1 RT(1 − α) dNdt .

8 Let us substitute in c ıgp − c ıg

v = R , which gives us

Nc ıgv

dTdt = RT dN

dt + γγ−1 RT(1 − α) dN

dt ,

and then multiply the whole expression by dt /(c ıgv TN).

9 Rearranging the equation slightly gives us the following expression:

dln Tdln N = (γ − 1) + γ(1 − α) .

10 Compared with Section 80, this equation has the extra term γ(1 −α) > 0 on the right-hand side, which tells us that the temperature dropis greater in this case[a].

530 / 1776

[a] It is possible to obtain an exact answer, but to do so we first need to integrate thedifferential equation. Hint: From the definition of α it follows that dlnα = − γ−1

γ (dln N +

dln T). Use this to show that ln( α2(2−α1)α1(2−α2)

)= −2(γ − 1) ln

(N2N1

). Substituting in p2 = p =

p1/10, γ = 1.4 gives you T2 = 0.2318T1.

Repeat the operation from Section 80 on page 459,but this time changing the direction of flow, so thatthe cylinder is filled instead of being emptied. What

will be the final temperature in a helium cylinder if it isfilled using a high-pressure hose at 300 bar and 20 C?The problem becomes simpler if you assume that the

cylinder has been evacuated prior to being filled.



10 Dynamic . . .§ 083 FillingENGLISH text is missing.Kinetic energy1 This time we must assume that the kinetic energy of the fluid flow-

ing into the cylinder can be disregarded.

2 For this to be true, the valve must be located close to the cylinderand the cylinder must be filled slowly, to ensure that the velocity of thegas in the fill hose is low.

3 By analogy to Section 80 on page 459, it follows that the mass andenergy balances can be expressed

dUc-vdt = ~Hın ,

dNc-vdt = ~Nın .

4 Quite unexpectedly, the two balance equations have an algebraicsolution if the cylinder is evacuated prior to the filling.

5 Straighforward integration gives the result Uc-v = Hın and Nc-v =

Nın showing that the enthalpy from the supply line is converted to inter-nal energy when the gas enters the cylinder.

6 Because Hıg > Uıg we can conclude that the temperature of thecylinder gas must increase.

7 We can also deduce that this temperature increase will happenmomentarily because the U/H ratio is independent of the degree offilling.

532 / 1776

1 In the same way as in Section 80, we can substitute in Uıg = Nc ıgv

T and Hıg = Nc ıgp T , but this time Tc-v , Tın.

3 2 Let T be the temperature inside the cylinder and T be the temper-ature in the fill hose (which is assumed to be constant). Eliminating the mass balance and then differentiating gives us

c ıgv T dN

dt + Nc ıgv

dTdt = c ıg

p T dNdt .

4 Next substitute in γ = (cp/cv)ıg and then rearrange the differential

equation in two steps:

dNdt (c ıg

v T − c ıgp T) =−Nc ıg

vdTdt ,

N-1 dN =c ıg

v

c ıgp

T−c ıgv

TdT ,

dln N =−dln(c ıgp T − c ıg

v T) .



10 Dynamic . . .§ 083 Filling (2)

5 Now let us integrate between two states: N1,T1 and N2,T2. 6 Note that T1 describes the temperature in the cylinder at t = 0,while T is the temperature in the fill hose. The

result of the integration is

ln(N2N1

)ıg= ln

(c ıgp

T−c ıgv

T1

c ıgp

T−c ıgv

T2

).

In order to obtain as simple an expression as possible, with the fewestpossible parameters, we will assume that T = T1.

7 By substituting accordingly and taking the exponential we get

(N2N1

)ıg=

(c ıgp−c ıg

v)T1

c ıgp

T1−c ıgv

T2=

(γ−1)T1γT1−T2

.

8 If we use the definition α = (N2/N1)ıg, we can write the final ex-

pression as

T ıg2 = T1

γ(α−1)+1α



10 Dynamic . . .§ 083 Filling (3)



10 Dynamic . . .Limiting case

1 We should particularly note that T2 = γT1 when α→∞.

2 If the cylinder has been evacuated prior to filling, this limit valuewill be reached instantaneously at t → 0, and the temperature of thecylinder’s contents will remain constant throughout the filling process.

7 3 Also note that the pressure is not explicitly part of the problem.4 This follows from our decision to assume ideal gas behaviour,5 since for an ideal gas Hıg = H(T ,N), which means that the en-

thalpy of the gas flowing into the cylinder is only a function of tempera-ture.6 Having said that, it should be added that the pressure is implicitlypresent in the expression through the quantity α. In order to achieve α≫ 1, i.e. to fill the cylinder with a large quantity

of gas, the pressure of the fill hose must be much higher than the initialpressure of the cylinder.

8 Some numerical results are shown in Figure 10.4 on the followingpage.

9 To do the calculations, the temperature equation first has to berestated as T ıg

2 = T1γβ/(β+ γ − 1) where β = p2/p1.10 The figure was drawn using the Program 1.13 in Appendix 31.



0 100 200 300

50

100

150

200

250

p [bar]

T[

C]

Figure 10.4: Temperature increase when filling a helium cylinder with fourdifferent initial pressures: p1 = 1,5,25,125 bar. The upper line shows thetheoretical limit for p1 → 0.



10 Dynamic . . .§ 084 § 83 § 85

Two cylinders are connected in series as shown inFigure 10.5 on page 474 Expanded footnote [a]:

[a] The set-up is the same as in James Prescott Joule’s famous experiment of 1843. . Calculate the temperaturechanges in the cylinders if the flow rate through the

valve is ~N = K√∆p and the heat transfer between the

two cylinders is assumed to be ~Q = UA∆T , whereUA is the overall heat transfer coefficient [J/K s]. Thesmaller cylinder has an initial pressure of 10 bar anda temperature of 300 K, while the larger cylinder hasbeen evacuated. The heat capacity ratio of the gas isγ = 1.40. Use a total energy balance to show thatdUdt = 0. Assume ideal gas behaviour. What will the

final state be if UA = 0? And if UA > 0?



10 Dynamic . . .Quasi-steady state

1 Let us start by putting system boundaries on each side of the valve,giving us three control volumes: The smaller cylinder, the valve and thelarger cylinder.

9 2 Let us moreover assume that the gas is flowing slowly out of thesmaller cylinder, and that both cylinders are in a quasi-steady state atall times.3 In the case of the larger cylinder, this assumption is questionable,

as you would expect the gas to be flowing quickly downstream of thevalve.4 On the other hand, the kinetic energy will quickly be transformed

into thermal energy due to internal friction, so the assumption producesaccurate results over an extended period of time.5 Thus, if the small cylinder is emptied over a long time period in re-

lation to the typical delay time constant of the turbulence, the assump-tion will be valid.6 This also means that E2 = U2 can be used as an approximation for

the energy of the large cylinder.7 Finally we have to say something sensible about the valve.8 It is out of the question to describe it in detail, as the situation there

is far too disordered. However, if the small cylinder is much bigger than the control vol-ume in the valve, then the latter will be in a quasi-stationary state(in = out):

(dUdt

)1 = ~Q − ~H1 ,

(dEdt

)JT = ~H1 − (~H2 + ~EK ,2) = 0 ,

(dUdt

)2 = ~H2 + ~EK ,2 − ~Q .



10 Dynamic . . .Quasi-steady state (2)

V1 V2

~Q

Figure 10.5: An isolated system consisting of two gas cylinders.

10 Combining the three energy equations gives us(dU

dt)1 = ~Q − ~H1 = − (dU

dt)2 .

11 From this it follows that (dUdt )1 + (dU

dt )2 = 0, which means that thetotal energy of the system is conserved.



10 Dynamic . . .§ 084 Joule[1]’s experiment

1 Expanded footnote :

[] James Prescott Joule, 1818–1889. English experimentalist.

2 In order to solve this problem, we need to know what happens tothe state variables T , p and N in each of the cylinders, in other wordssix variables in total.3 The differential equations for p and N can be derived from the ideal

gas law and the equation for the valve given above, but the tempera-tures require a bit more thought. Based on the energy balance for the small cylinder, and the pro-

posed heat transfer model, we can say that

d(N1c ıgv T1)

dt = c ıgp T1

dN1dt + UA (T2 − T1) .

4 This makes use of the fact that the enthalpy rate can be expressedas ~H1 = cpT1

~N1 and that ~N1 = − dN1dt , which we know from the mass

balance equation for the cylinder.

5 Expanding the left-hand side and moving the dN1dt term to the right-

hand side gives us:

N1c ıgv

dT1dt = c ıg

p T1dN1dt + UA (T2 − T1) − c ıg

v T1dN1dt

= (c ıgp − c ıg

v )T1dN1dt + UA (T2 − T1) .



10 Dynamic . . .§ 084 Joule[2]’s experiment (2)

7 6 Dividing by N1c ıgv results in a pure temperature equation for thesmall cylinder. By analogy we can obtain a temperature equation for the large

cylinder, and the differential equations can be summarised asdp1dt = RT1

V1

dN1dt + N1R1

V1

dT1dt ,

dp2dt = RT2

V2

dN2dt + N2R2

V2

dT2dt ,

dN1dt = sign(p2 − p1)K

√|p1 − p2| ,

dN2dt = sign(p1 − p2)K

√|p1 − p2| ,

dT1dt = (γT1 − T1)N-1

1dN1dt + UA (N1cv)

-1(T2 − T1) ,dT2dt = (γT1 − T2)N-1

2dN2dt + UA (N2cv)

-1(T1 − T2) .

9 8 It is possible to reduce the number of equations from six to four, byeliminating e.g. dp2dt and dN2dt , but we will keep all six in order to empha-sise their symmetry[a].

[a] It’s a different matter if we have to solve them by hand. This set of equations constitutes a classic initial value problem thatcan be solved by any numerical method of integration, but it isn’t allplain sailing.



10 Dynamic . . .§ 084 Joule[3]’s experiment (3)

10 The initial values are as follows:

p1 = 10 bar p2 = 0 bar

N1 =p1V1RT1

mol N2 = 0 mol

T1 = 300 K T2 = γ · 300 K

11 Note that T2,t=0 = γ · T1t=0.

12 This condition is necessary if dT2dt is to have a finite solution Expanded footnote [a]:

[a] Also see Section 83 on page 466. .

13 Hence T2 = γT1 at t = 0.

14 If N2 → 0 then dT2 /dt → ∞, which makes the set of equationsstiff.15 In order to avoid numerical problems, an adjusted heat transfer co-

efficient has been used in the Program 31:1.14

U′A = UA min(p1p2 ,p2p1

) .16 This adjustment prevents artificially high heat transfer values whent → 0, but it also means that the time-axis is non-linear.17 The numerical results are shown in Figure 10.6 on the next page.



0 0.5 1150

200

250

300

0 0.5 1300

350

400

450Small cylinder Big cylinder

T[K]

UA = 0UA = 0UA = 0

UA > 0UA > 0UA > 0

UA ≫ 0UA ≫ 0UA ≫ 0

Figure 10.6: Simulation of Joule’s famous experiment to demonstrate thatthe internal energy of a gas is unrelated to its volume. The graphs showtemperature variations over time for three different heat transfer coefficients(time axis is not scaled).



300 400 500 600 7005

10

15

20

25

30

Sig

nals

tren

gth

[dB]

Frequency [Hz]

497 Hz

480 Hz

Figure 10.7: Helmholtz resonance in a flask with the following dimensions:∆V = 25 ml, V = 120 ml and L = 50 mm. The theoretical eigenfrequency(blue) is 497 Hz, while the graph peaks at 480 Hz. The discrepancy is dueto the fact that the effective mass of the oscillating gas plug is bigger than∆V = AL suggests. By increasing L to L ′ = 53.5 mm we get a more realisticresult (red).



10 Dynamic . . .Looking backENGLISH text is missing.Helmholtz resonance1 Leaving behind flow processes for a moment, let us look at a con-trol volume that is disturbed by acoustic pressure waves in the sur-rounding gas (typically air).

2 The term Helmholtz resonator refers to a physical control volume(container) that interacts with its surroundings through a small hole (orneck) through which the gas can move freely and uniformly in bothdirections.

3 In practice the resonator can be a beer bottle, a guitar, a flute, etc.

4 When surrounded by a spectrum of sound frequencies, the instru-ment will pick up a characteristic frequency that resonates with the “aircushion” inside the control volume[d].

5 The air cushion changes volume in response to the sound wavescoming from the outside, while the gas that oscillates back and forth inthe neck is assumed to act like a stiff plug (piston).

6 The resonant frequency is mainly determined by the mass ratiobetween the air cushion and the gas plug.

7 Sound propagates as longitudinal waves[d] consisting of alter-nately compressed and rarefied air, and for small amplitudes there isvery little dissipation of energy.

8 It is therefore important to assume that the density variations in thecontrol volume are isentropic:

(dp)s,n =(∂p∂v

)s,n dv ,

−N dv = A dx .

9 Here A is the cross-sectional area of the hole, N is the mole num-ber of the air cushion and dx is the displacement of the air plug in theneck (which is assumed to be incompressible).

10 For small density variations, the state variables are roughly con-stant, so using Eq. 9.10 the integral of dp over the amplitude can begiven as:

p − p ≈ −(∂p∂v

)s,n

AN x = υ2⋆ρA

vN x .

545 / 1776

[d] The control volume must be small in relation to the wavelength of the sound, andthe volume of the neck must not be much bigger than the control volume itself. Ifthese assumptions don’t hold, the theory becomes more complex. A didgeridoo is anexample of a control volume where the assumptions break down.

[d] Waves that oscillate in the same direction as they are travelling, as opposed totransverse (surface) waves, where the centre of mass oscillates at right angles to thewave’s direction of travel.

ENGLISH text is missing.Eigenvalue analysis1 A pragmatic interpretation of Newton’s 2nd law is that “force equalsmass times acceleration”.

2 In this case A(p − p) is a force that causes the mass ρAL of theair plug in the neck to move in the opposite direction of the pressuregradient:

ρAL d2xdt2 = −A(p − p) .

3 Combining the last two equations gives the following 2nd orderlinear differential equation:

− d2xdt2 = υ2⋆ρA

vNρL x =(υ⋆

L)2 ∆V

V x = cx .

4 ∆V and V represent the air plug and the control volume respec-tively.

5 The differential equation can be written in the standard form cx ′′ +x = 0. This solution to the equation involves restating it as

0 −c1 0

x′1x′2

=

x1

x2

where x2 = x′1.

6 Show that the coefficient matrix has the eigenvalues λ = ±i√

cand calculate the eigenvectors in S. The diagonal form can be formallyexpressed as:

SΛS-1x′ = x .

546 / 1776

ENGLISH text is missing.Eigenfrequency1 Substituting in z = S-1x gives us the system of equations Λz′ = z.

2 Without getting bogged down in the details, we can affirm that thesolutions are in the form zi ∝ exp(t/λi) = exp(±i t√c ).

3 The eigenvalues are imaginary (two complex conjugate eigenval-ues with a real component of zero), so the oscillations of the wave arenot damped.

4 This agrees with our initial assumption of isentropic changes indensity.

5 If we apply Euler’s formula e ix = cos x + i sin x we find that thecharacteristic period of oscillation is τ = 2π

√c.

6 Alternatively, this solution can be stated in terms of the eigenfre-quency

ν = τ-1 = 12π

υ⋆L

√∆VV .

In the above equation, we have substituted in c =(υ⋆

L)2 ∆V

V .

7 Note that we do not need an exact solution to the wave equation(besides which, the limits are unclear), and an eigenvalue analysis ofthe problem is sufficient to give us the resonant frequency of the controlvolume which dominates the audible sound.

8 Figure 10.7 shows the frequency diagram[d] for a homemadeHelmholtz resonator made of glass (specifically a 100 ml round-bottomflask from the organic chemistry laboratory).

9 The shape of the container is fairly irrelevant, but is important forthere not to be space for standing sound waves inside the control vol-ume.

10 In air, a 480 Hz sound wave has a wavelength of approximately0.7 m, and the typical length of the resonator must be much less thanthat.

11 If the length is increased to around a metre, the acoustics willchange dramatically (becoming more like those of a didgeridoo).

547 / 1776

[d] The audio file was recorded in mono (2.5 s) using GarageBand 2 on an iBook Macwith a built-in microphone. The file was then converted from AIFF to WAVE formatusing iTunes (again on a Mac), before being read and processed with Matlab runningin Windows XP; see program 31:1.15.

1 Depletion: T ıg2 = T1

(N2N1

)γ−1= T1

(p2p1

) γ−1γ

2 Filling: T ıg2 = T1

γ(α−1)+1α

3 Limit value: limα→∞

(T2) = γT1

4 Helmholtz resonator: cx′′ + x = 0

5 ODE form (eigenvalues λ = ±i√

c):

0 −c1 0

x′1x′2

=

x1

x2


News Incompressible flow Compressible flow Olds

Part 11

Open control volumes




11 Open . . .New concepts

1 Pipe flow control volume

2 Velocity profiles

3 Laminar versus turbulent flow

4 The steady state

5 Incompressible flow

6 Bernoulli’s equation

7 Ideal gas compression work

8 Joule–Thomson coefficient

9 Linde plant



11 Open . . .Open control volumes

1

~Eın

~Eout~Q

~W

dEdt

Figure 11.1: A simplified open control volume.

With only one inlet and one outlet, the control volume shown in Fig-ure 11.1 is of limited validity, but in spite of that the model is generalenough for many practical purposes.



11 Open . . .Open control volumes (2)

2 For example, it is possible to fill and empty the volume at the sametime, which will later enable us to study stationary processes.

4 3 In this chapter we will focus primarily on describing the control vol-ume itself, and less on properly understanding the flow at the inlets andoutlets. It will therefore be assumed that all inflows and outflows have uni-

form properties measured along (or at) their respective control surfaces.

5 For practical reasons we will also assume that the control surfacesare at right angles to the direction of flow. 6 Incidentally, these assumptions are so fundamental, and above all

so hard to get around, that in problem formulations they are often onlyalluded to.7 Before going further, we will therefore spend a moment to reflecton them and to think about their practical consequences.8 In particular, we will see how, in the absence of the assumption ofuniformity, the rates of energy and mass transport would diverge.



11 Open . . .§ 085 § 84 § 86

What is the ratio between the mean velocity υ and themaximum velocity υ of laminar flow? The velocityprofile is defined by υ(R) = υ(1 − (R/R)2); see

Figure 11.2 on the next page. What is the equivalentratio between the square of the mean velocity υ2 and

the mean square velocity υ2?



υ(R)

RR

Figure 11.2: Parabolic velocity profile for laminar pipe flow



11 Open . . .§ 085 Velocity profile

1 The mean velocity can be obtained by calculating the first momentof the axial velocity component about R = 0:

υ =

∫ R0 υ2πRdR∫ R0 2πRdR

= 2πυπR2

R∫

0

[1 −

(RR

)2]R dR = υ

R2

R2∫

0

[1 −

(RR

)2]dR2 .

2 Integrating between the limits gives us the ratio

υ = υR2−

R4

2R2

R2

= υ2 .

3 For laminar pipe flow, the mean velocity is half the maximum veloc-ity component.



11 Open . . .§ 085 Velocity profile (2)

4 The mean square velocity can be calculated using the same prin-ciples, which gives us the following result:

υ2 =

∫ R0 υ22πRdR∫ R0 2πRdR

=υ2

R2

R2∫

0

[1 −

(RR

)2]2

dR2 =υ2

3 .

Thus, for the square velocities, υ2 = 34 υ

2 = 14υ

2.

6 5 This result is relevant to energy balances that include kinetic en-ergy transport. While we use the mean velocity to calculate the transport of in-

ternal energy and potential energy, we cannot use it for kinetic energytransport.

7 In fact, if we were to use it for laminar flows, the kinetic energywould be underestimated by as much as 25%. 8 In the rest of this chapter we will therefore stick to uniform (i.e.

turbulent) velocity profiles where υ = υ and υ2 = υ2.9 This means that we can leave out the line above the variable and

just use the symbol υ to mean velocity.



11 Open . . .§ 086 § 85 § 87

Set out the energy balance for the control volumeshown in Figure 11.1 on page 483. The balance should

include internal energy, kinetic energy and potentialenergy. Assume that the inflows and outflows have

uniform properties over the cross-sectional area of flow.What happens if the latter assumption does not hold?



11 Open . . .§ 086 Energy balance

1 In process engineering, a control volume will normally have virtu-ally constant kinetic and potential energy. The total energy of the vol-ume will therefore be E = U.

4 2 This requires the control volume to be stationary, without any ki-netic energy bound up as eddies or turbulence in the flow medium.3 Moreover, the centre of gravity of the control volume must not moveup or down. For a control volume that satisfies these conditions, the energy

balance can be expressed as

dUdt = (~E + p~V)ın − (~E + p~V)out + ~Q − ~W

= (~U + ~EK + ~EP + p~V)ın − (~U + ~EK + ~EP + p~V)out + ~Q − ~W= ( ~EK + ~EP + ~H)ın − ( ~EK + ~EP + ~H)out + ~Q − ~W . (11.1)

5 Here ~E, ~H, ~Q , ~W are energy rates [J/s] associated with the trans-port of mass, heat and work.



11 Open . . .§ 086 Energy balance (2)

8 6 If the intensive properties of the fluid vary over the cross-sectionalarea of flow Ω, they must be considered average energy rates,7 which can be determined by integrating ~H = ∫

Ω hTn dΩ, where n isthe unit normal vector of the control surface and h [J/m2 s] is the fluid’senthalpy flux vector. If we introduce the specific variables e = M-1E and h = M-1H,

Eq. 11.1 can be expressed in the formd(Mu)

dt = ~Mın(ek + ep + h)ın − ~Mout(ek + ep + h)out + ~Q − ~W= ~Mın(

12 υ

2 + gz + h)ın − ~Mout(12 υ

2 + gz + h)out + ~Q − ~W .

(11.2)

9 The mass flows in and out of the system are by definition ~M =

Aρυ, but remember the assumption of a uniform velocity profile fromTheme 85.

10 In practice this assumption means that υ = υ and υ2 = υ2, where

υ is the net flow velocity.



11 Open . . .§ 087 § 86 § 88

Set out the energy balance for a stationary controlvolume. Can the control volume be said to be stationaryif the states of the inflows and outflows vary as functions

of time? If so, what conditions must be fulfilled?



11 Open . . .§ 087 Stationary state ENGLISH text is missing.Stationary processes1 For stationary processes the state variables of the control volume

are constant over a time period ∆t ≫ τ, where τ is a characteristic timeconstant for the process.

2 In other words, the change in the state variables shouldn’t reflectany transient conditions in the control volume, cf. a quasi-static process.

3 For example, it is a good approximation to assume that a heatexchanger is stationary, even if the temperature of the cooling watervaries with the seasons.

562 / 1776

1 For a stationary process, dU /dt = 0 and dM /dt = 0, whichmeans that the mass balance can be expressed ~Mın = ~Mout.

2 Eliminating the flow rates ~Mın and ~Mout in Eq. 11.2 gives us

(12 υ

2 + gz + h)ın + q = (12 υ

2 + gz + h)out + w .

3 The values of e, h, q and w are specific, i.e. they are stated in[J/kg].

4 Conventionally the final result is expressed in ∆ notation:

12 ∆υ2 + g∆z +∆h = q − w . (11.3)

5 If there are no chemical reactions in the volume, the energy bal-ance can alternatively be expressed in molar ∆ notation:

12 Mw∆υ2 + Mwg∆z +∆h = q −w . (11.4)

6 It probably makes sense to stick to the symbols h, q and w, al-though here the unit is [J/mol] rather than [J/kg] as in the previousequation.



11 Open . . .§ 088 § 87 § 89

Assume that the flow through the control volume 11.1on page 483 is steady, incompressible and lossless Expanded footnote [a]:

[a] Lossless means that the flow is frictionless (inviscid), and that there are no temper-ature gradients in the fluid, although heat can be transferred between the control vol-ume and its surroundings. .

Show that the energy balance is an expanded Bernoulliequation. Use this equation to look at how h, ek and ep

vary along the pipeline in Figure 11.3 on the followingpage. A qualitative analysis is sufficient. How do T , p

and µ vary along the same pipeline?



~Q ~W

υz

x

14 υ

Figure 11.3: Simplified diagram of Bernoulli flow.



11 Open . . .§ 088 Bernoulli[1]’s equation


[] Daniel Bernoulli, 1700–1782. Dutch mathematician (son of Johann Bernoulli). For lossless flow, δqrev = T ds.

2 Substituting this into (dh)n = T ds + v dp gives us (dh)n,rev = δq +

v dp.

3 For a compressible fluid, i.e. a fluid with a constant specific volumethat is independent of the pressure, the differential can be integrated togive (∆h)n,rev = q + v∆p. If we substitute this into Eq. 11.3 and divideby v [m3/kg], it gives us the expanded form of Bernoulli’s equation. 4 The expansion is trivial and only involves the work term w:

12 ρ∆υ

2 + ρg∆z +∆p = −w . (11.5)



11 Open . . .§ 088 Bernoulli[2]’s equation (2)

8 5 Here the unit of w is [J/m3], but that shouldn’t cause any majorconfusion.6 In the absence of chemical reactions, Bernoulli’s equation can also

be expressed in molar form:

12 Mw∆υ2 + Mwg∆z + v∆p = −w . (11.6)

7 Bernoulli’s equation expresses a direct relationship between theflow profile of p and the equivalent profiles of υ and z. Provided that we know the state of the fluid at a given point in the

streamline, it is possible to calculate the pressure along the rest of thestreamline as a function of the cross-sectional area of flow (a simplemass balance gives the velocity) and the elevation.

9 The associated temperature profile follows indirectly from the as-sumption δqrev = T ds.

11 10 For example, in the case of isobaric flow there is the simple rela-tionship cp dT = δqrev. If we know both T and p, we can work out the profile of µlıq(T ,p).

14 12 It is also possible to calculate the difference between this potentialand the associated potential for a hypothetical ideal gas (steam) at thesame temperature and pressure as the liquid.13 The steam is not physically present, as that would contradict theassumption of incompressibility, but the difference ∆µ = µlıq − µıg hasan important function: It tells us where in the system there is a risk ofboiling, i.e. where ∆µ < 0; see Figure 11.4. The calculation involves using the simplifications µlıq ≈ µsat(T) +

v(p − psat(T)) and µıg = µ(T ,p) + RT ln(p/p).

16 15 If we apply them to the equilibrium state µlıq = µıg, we get µsat =

RT ln(psat/p). Hence ∆µ = µlıq − µıg = RT ln(psat/p) + v(p − psat).17 Chapter 15 on page 791 discusses this topic in greater detail.



−1

−0.5

0

0.5

0

5

10

15

−3

−2

−1

0

−5

0

5

10

∆T∆p

∆µlıq µvap − µlıq

[kJ/

mol]

[J/m

ol]

[bar]

[mK]

a)b)

c) d)

Figure 11.4: T ,p, µ profiles for the Bernoulli flow shown in Figure 11.3 (adetailed explanation is given in the body of the text).



11 Open . . .Cavitation

1 2 Using Bernoulli’s equation to interpret variation in chemical poten-tial isn’t a common approach, but Figure 11.4 confirms the usefulnessof the exercise.3 The first three sub-figures show the p, T and µ profiles for flowingwater with the inlet conditions Tın = 50 C, pın = 1 bar and υın = 8 m s-1. The total elevation is z = 10 m, the heat added is 55 kJ m-3 and

the work done is 75 kJ m-3 (see Program 1.20 in Appendix 31 for furtherdetails).

6 4 This results in an interesting µlıq profile, but more importantly itgives us the graph shown in Figure 11.4d, which shows the differencein chemical potential between liquid water and a hypothetical steamphase at the same temperature and pressure.5 In the liquid phase it is weakly dependent on pressure and strongly

dependent on temperature, whereas in the steam phase it is stronglydependent on both temperature and pressure. This means that ∆µ varies considerably along the streamline, and

we should particularly note the coloured area where the difference isnegative (this represents the top of the pipeline just before it gets wider).

7 Here the steam phase is more stable Expanded footnote [c]:

[c] As an aside, it should be noted that there is an equivalent test for multicomponentsystems, but it involves a more advanced criterion called the minimum tangent planedistance; see Chapter 21 on page 1312. than the liquid phase, so

both phases will be present in this area. 8 This undermines the foundations for Bernoulli’s equation, and inpractice the steam bubbles can cause cavitation problems.



11 Open . . .§ 089 § 88 § 90

You forget to shut the damper of the fireplace on theground floor. The temperature difference between

the house and the outside air produces a convectioncurrent in the chimney which means that you have to

use more heating to maintain your desired indoortemperature. Express the extra heating required as afunction of the outdoor temperature. Assume that thetemperature of the air passing through the chimney is

constant from the inlet to the outlet, and that Bernoulli’sprinciple ∆p + 1

2ρ∆υ2 + ρg∆z = 0 gives a reasonable

estimate of the flow velocity. The indoor temperatureis 20 C. The chimney has a standard 9” flue and

runs through two full stories plus a cold loft.



11 Open . . .§ 089 Boussinesq flowENGLISH text is missing.Chimneys1

Dz

Figure 11.5: Natural convection through a chimney flue.

A cross-section of the chimney is shown in Figure 11.5.

2 The pressure differential between the air inside and outside thehouse is taken to be equal to the hydrostatic pressure gradient in theatmosphere multiplied by the height ∆z.

3 This assumes that the convection current has not resulted in sig-nificant negative gauge pressure in the building (this is no doubt debat-able).

4 The velocity at the inlet of the chimney can be safely assumed tobe zero. The velocity right by the inlet is of course non-zero, but theassumption is valid because somewhere or other in space at the sameheight as the inlet the velocity is (more than) close enough to zero forour purposes.

5 Remember that Bernoulli’s equation applies along a streamline,which means that this assumption is consistent with the theory thatunderpins the equation.

6 Also assume that the density of the air depends on temperature,but not on pressure.

7 This may appear to be a strange assumption, but the differencein air density due to the height difference ∆z is small compared withthe difference in density between the cold air outside and the warm airinside the chimney[c].

571 / 1776

[c] In fluid mechanics, assuming incompressible gas flow and constant thermal expan-sivity is called the Boussinesq approximation. Named after the French mathematicianValentin Joseph Boussinesq (1842–1929).

1 We know that the chimney runs through 2 stories.

2 In addition there is the chimney stack and the section runningthrough the loft.

3 In total this gives us approximately 8 m of chimney.

4 The cross-sectional area of flow is 9 × 9” i.e. A = 0.052 m2.

5 The various pressure components of Bernoulli’s equation aretherefore:

Kinetic: 12 ρ∆υ

2 ⇒ − 12ρınsıde υ

2out .

Potential: ρg ∆z ⇒ ρınsıde g∆z .

Mechanical: ∆p ⇒ − ρoutsıdeg∆z .



11 Open . . .§ 089 Boussinesq flow (2)

6 Bernoulli’s equation can therefore be expressed 0 = ρınsıdeg∆z −ρoutsıdeg∆z + 1

2ρınsıdeυ2out, or since ρıg =

pMwRT :

υout =

√2g∆z

(TT − 1

).

7 Now we can calculate the amount of extra energy needed for heat-ing:

~Q = CP~N(T − T) = CP

~VpRT

(T − T) = CPυoutAp

RT(T − T) .

9 8 The heat lost from the room due to the convection current in thechimney is thus proportional to the inside/outside temperature differen-tial, and to the flow velocity in the chimney. We can eliminate the velocity υout by combining the last two equa-

tions:

~Q = CPApRT

√2g∆z

T (T − T)3/2



−20 −10 0 10 20 300

2

4

6

8

10

Q[k

W]

Outdoor temperature [C]

Figure 11.6: Convection losses through an open chimney at two differentindoor temperatures (18 C and 25 C) and for two different chimney lengths:4 m (dotted) and 8 m (solid).



11 Open . . .§ 091 § 90 § 92 ENGLISH text is missing.Fireplaces1 The extra heating required has been drawn using the Pro-

gram 31:1.21 and is shown in Figure 11.6 on the previous page.

2 As can be seen, the impact is frightening at low outdoor tempera-tures, and the moral of the story should be clear: Seal your house, butnot to the extent that you damage your health by having a poor indoorclimate (this is a difficult balancing act).

3 Note that the impact of the chimney height is also evident fromFigure 11.6.

575 / 1776

§ 901 In the mainstream press, it is occasionally claimed that hy-

dropower stations cause thermal pollution of the surroundings. Is itpossible that the imbalance is caused by the work done by the turbines,or could there be other factors at play? Hint: First set out the energy bal-ance for the waterfall. Then calculate the temperature change acrossthe turbine relative to the temperature change in free-falling water.

576 / 1776

§ 90 Hydro-power1 Let us start by defining an open control volume around the water-

fall; see Figure 11.7.

2 The water does not do any mechanical work apart from the pVwork related to the inflows and outflows.

3 Provided that there is little evaporation, as is the case in Norway, itdoesn’t exchange any heat with its surroundings either.

4 Under these conditions, the energy balance 11.4 simplifies to

12 Mw∆υ2 + Mwg∆z +∆h = 0 .

5 We haven’t said anything about the size of the control volume, andthe equation

Figure 11.7: Control volume over a water

is just as applicable to the whole waterfall as to a small section of it.

6 Assume that the vertical drop is 300 m and that the river’s flowvelocity is the same above and below the waterfall:

∆h = −Mwg∆z .

7 The process takes place at constant pressure and the heat capac-ity remains virtually constant, i.e. CP∆T ≈ −Mwg∆z.

8 If we insert numerical values we get ∆T ≈ 0.018 · 300 · 9.82/75 =

0.7 K.

9 The work done by the turbines can be calculated from the energybalance 11.6, assuming that qrev =

∫T ds and that the water can be

treated as an incompressible liquid.

10 There is little change in the pressure of the surroundings betweenthe reservoir and the turbine outlet, and the pressure drop in the pipelinecan be largely ignored (less than 5% loss of pressure):

w = −Mwg∆z .

11 An ideal turbine captures all of the available kinetic energy in thewater, and the temperature of the water therefore remains unchangedas it travels through the power plant.

12 As a result, the temperature is 0.7 K lower with the turbine installedthan without it.

13 However, this temperature reduction is so small that any thermalpollution is probably due to other factors.

14 Probably a much bigger problem is the release of cold water fromthe bottom of the reservoir in the summer because the water enters abiotope that is adapted to a higher water temperature.

15 In winter, conversely, the water from the bottom of the reservoir willbe too warm.

577 / 1776

The behaviour of a steadily flowing gas when it iscompressed or allowed to expand is of particular interest

in process engineering. Replace the incompressibleliquid in Theme 88 on page 494 with an ideal gas. Once

again assume that the flow is frictionless, and ignoreany heat transfer between the control volume and its

surroundings (adiabatic flow). Derive an expression forthe shaft power needed to compress the gas from its

state at the inlet Tın,pın to its state at the outlet Tout,pout.Assume that the composition and heat capacity of

the gas are constant. The molar flow rate is ~N.



11 Open . . .Work of steady flow

1 2 The basic equations describing the isentropic compression of anideal gas in a closed system were derived in Theme 69 on page 421.Here we will extend our analysis to include the compression of a gasunder stationary flow conditions. Let us hold onto the idea of a closed control volume, which in this

case is a hypothetical volume element moving with the fluid flow, butnow in addition to the work of compression, the control volume per-forms flow work by displacing the gas downstream of it, while the gasupstream of it performs flow work on the control volume itself.

3 These two extra contributions are shown in Figure 11.8 as −(pv)out

and +(pv)ın, which represent the flow work done by and on the systemrespectively.



11 Open . . .Work of steady flow (2)

vout vın

pın

pout

−(pv)out

+(pv)ın

Figure 11.8: Isentropic compression of a steadily flowing gas: ws =∫ out

ınp dv − (pv)out + (pv)ın or alternatively ws = −

∫ out

ınv dp.

5 4 The net amount of work involved in displacing the contents of thecontrol volume followed by compressing and moving the gas— from theinlet to the outlet— is equal to the sum of the three components. Thus, ws =

∫ out

ınp dv − (pv)out + (pv)ın. Partial integration also

enables us to state that ws = −∫ out

ınv dp.



11 Open . . .Work of steady flow (3)

7 6 A graphical interpretation of the three components is shown in thefigure[c].

[c] The explanation for this is as follows: Due to adiabatic compression work theinternal energy of the closed control volume increases with (∆u)s,n = −∫ p(dv)s,n.Partial integration gives us ∫ p(dv)s,n = ∆(pv)s,n − ∫ v(dp)s,n. Combining the twoequations produces (∆u)s,n + ∆(pv)s,n = ∫ v(dp)s,n, and because h = u + pv itfollows that (∆h)s,n = ∫ v(dp)s,n, as stated in the body of the text. The tempera-ture increase that takes place during compression means that ∆(pv)s,n is positive.Hence (∆h)s,n is greater than (∆u)s,n. In other words, more energy is requiredto compress a fluid flowing through an open system than a fluid in a closed sys-tem. As an aside it should be noted that the ratio between the two work terms is(∆h)s,n/(∆u)s,n = −∫ v(dp)s,n/∫ v(dp)s,n, which for an ideal gas with constant heat ca-pacity becomes (∆h)s,n/(∆u)s,n = cp∆T/cv∆T = γ. This means that the ratio be-tween the two integrals −∫ p(dv)s,n and ∫ v(dp)s,n in Figure 11.8 is also γ. The result is mathematically correct, but by assuming the existence

of a closed control volume moving with the fluid flow in the velocityfield we have gone slightly against the terms in which the problem wasoriginally formulated. 8 We will therefore derive the stationary work term again based on

an independent analysis of an open system.



11 Open . . .§ 091 Isentropic compression

1 From the energy Eq. 11.1, it follows directly that 0 = ~Hın− ~Hout− ~W .Or, expressed in molar terms, ∆h = −w; also see Eq. 11.4. 2 The difference in enthalpy at a given composition is (dh)n = T ds+

v dp, which for frictionless, adiabatic flow can be simplified to (dh)s,n =

v dp. Assumingideal gas behaviour, this becomes (dh)ıg

s,n = (RT/p) dp.

3 From Theme 61 on page 382 we are familiar with the adiabaticrelationship (T/Tın)s,n = (p/pın)

(γ−1)/γ, which gives us the followingequation:

(dh)ıgs,n = RT

p (dp)ıgs,n = RTın

p( ppın

) γ−1γ dp = RTın

( ppın

)−1γ d

( ppın

).

4 If we integrate this we get

−w ıgs = (∆h)ıg

s,n = RTın

pout/pın∫

1

( ppın

) −1γ d

( ppın

)=

γγ−1 RTın

( ppın

) γ−1γ

∣∣∣∣∣pout/pın

1.



11 Open . . .§ 091 Isentropic compression (2)

6 5 If we set the upper and lower limits of integration, −w ıgs expressesthe work required to compress one mole of a gas from its inlet conditionsTın,pın to its outlet pressure pout and outlet temperature Tout > Tın. At any given (stationary) molar flow rate ~N the total power required

is

~Ws-s,ıgS = ~Nw ıg

s =γγ−1

~NRTın

[1 − (pout

pın

) γ−1γ

]. (11.7)

7 but note carefully: Since pout > pın, then ~W ıgS < 0. We can see

that the delivered work done is negative, which means that the suppliedwork of compression must be positive, as you would expect.

8 Eq. 9.2 was an equivalent expression for the work required to com-press a closed system.

9 That requires less work, since the gas remains in the control vol-ume, and doesn’t need to be moved in the direction of flow.

12 10 The work of compression in Eq. 11.7 is therefore greater than theequivalent work in Eq. 9.2.11 We can determine the ratio between the two terms by substitutingan isentropic relationship from Theme 61 into Eq. 9.2 and then compar-ing our result with eqution 11.7. What we find is that (ws-s

s /wc-ss )ıg = γ.13 Incidentally, the same result is derived using a slightly different

method in footnote on page 507.14 In the reverse case, i.e. when pout < pın, work of expansion will bedone by the open system.15 We can calculate this work by swapping the integration limits in theabove equation.16 After a bit of manipulation we can obtain the same function, but asthe limits have been changed ~W ıgS > 0.



11 Open . . .Reversible vs frictionless

1 2 In fluid engineering, the terms totally reversible (externally and in-ternally reversible, frictionless) flow and adiabatic flow are used. In the case of totally reversible flow, the heat transfer is reversible,

i.e. δQrev = T dS, whereas for adiabatic flow δQ = 0.

3 In isentropic flow, on the other hand, δQ = 0 and dS = 0:

Adiabatic:Isentropic:Loss-free:

δQ = 0dS = 0δQ = T dS

4 If the temperature of the fluid changes due to friction, the flow can-not be totally reversible, since the entropy changes without any simulta-neous heat transfer between the fluid and its surroundings.

5 From this it follows that T dSırr > 0.



11 Open . . .Reversible vs frictionless (2)

6 However, the flow can still be adiabatic, i.e. not involve any ex-change of heat with the surroundings.

10 7 There is often a great deal of confusion over the distinction betweentotally reversible, adiabatic and isentropic processes, but as usual inthermodynamics, the terms may be vague but the maths isn’t.8 Once we have these definitions in place, we can derive most ofwhat we need from the following two equations:

1st law of thermodynamics:Fundamental energy equation:

(dU)n = δQ − δW(dU)n = T dS − p dV

9 Combining these equations gives us (T dS)n = δQ + p dV − δW . For an isentropic process, (δQ)n = δW − p dV , and if the processis also adiabatic, the work done on the system must be exactly equal tothe pressure–volume work: (p dV)n = δW .

13 11 Hence a control volume that moves with the fluid flow cannot doany work against friction, and the process must be reversible.12 On the other hand, if the process is totally reversible, p(dV)n = δWeven if the entropy of the fluid element doesn’t remain constant. In other words, isentropic-adiabatic flow is a special case of totally

reversible flow.



11 Open . . .§ 092 § 91 § 93

In Themes 88 and 91, the flow was totally reversible.We are now going to look at the opposite extreme,

which is completely turbulent flow. Here the entropyof the fluid changes, regardless of whether there isany heat exchange between the control volume andits surroundings. This kind of flow model gives a

good description of e.g. junctions, valves and burners.Provided that the flow is stationary and adiabatic, andno mechanical work is done, the energy balance canbe expressed in very simple terms. What are they?



11 Open . . .§ 092 Isenthalpic flow

1 From Eq. 11.1 it follows that ~Hın = ~Hout.

2 If we restate this in terms of the specific enthalpy, ∆h = 0,; alsosee Eq. 11.3.

3 This equation holds for turbulent and adiabatic flow in any fluid,provided that the acceleration and change in potential energy can beignored.

4 It is important to express the equation in terms of mass, and notin terms of moles. This is because the problem formulation does notmake any assumptions about the composition.

5 We will come back to this question in Chapter 17:17.7 in conjuctionwith the calculation of adiabatic flame temperatures.



11 Open . . .§ 093 § 92 § 94

Derive an expression for the temperature change thattakes place when a flowing fluid is forced through avalve, if the fluid can be described by an equation ofstate in the form p = (T , v). What result does this

give you if the fluid is an ideal gas? What about for agas that obeys the Van der Waals equation?



11 Open . . .§ 093 Joule[3]–Thomson[4]


[a] James Prescott Joule, 1818–1889. English experimentalist.

Expanded footnote :

[b] William Thomson, 1st Baron (Lord) Kelvin, 1824–1907. Irish–Scottish physicist andengineer. The question requires us to find a suitable expression for

(∂T/∂p)H.

2 From Chapter 8, and more specifically equations 8.1, 8.12and 8.16, we know that

(dH)n = T dS + V dp ,

(dS)n =CV

T dT +(∂p∂T

)V ,n dV ,

(dp)n =(∂p∂T

)V ,n dT +

(∂p∂V

)T ,n dV .

3 If we eliminate dS and dV from the equations we get

(dH)n = CV dT + T(∂p∂T

)V ,n

(∂p∂V

)-1T ,n

[dp − (∂p

∂T)V ,n dT

]+ V dp

=[CV − T

(∂p∂T

)2

V ,n

(∂p∂V

)-1T ,n

]dT +

[T

(∂p∂T

)V ,n

(∂p∂V

)-1T ,n + V

]dp .



11 Open . . .§ 093 Joule[5]–Thomson[6] (2)

4 For an isenthalpic process (dH)n = 0, so the Joule–Thomson co-efficient is

κ(T ,V ,n) =(∂T∂p

)H,n

=T

(∂p∂T

)V ,n + V

(∂p∂V

)T ,n

T(∂p∂T

)2


)T ,n

. (11.8)

5 From Eq. 8.32 we can see that the denominator can be expressedas CP(T ,V ,n), but that is not of any relevance to us here and now.

6 For an ideal gas (∂p/∂V)T ,n = −NRT/V2 and (∂p/∂T)V ,n = NR/V ,which means that κıg = 0.

7 In other words, the enthalpy of an ideal gas is independent of itspressure.



11 Open . . .Pressure–temperature drop

1 Assume that the depressurised gas in Theme 93 is heat exchangedwith the gas entering the system as shown in Figure 11.9 on the follow-ing page.

2 If the Joule–Thomson coefficient is positive, the exchange of heatwill cause the valve inlet temperature to fall.

3 Eventually the temperature will become so low that some of thegas will condense and in order to achieve a steady state the liquid mustbe tapped.

6 4 This is the principle that allows a Linde plant to produce liquid air.5 To improve efficiency, it is common to use a turbine to precool theair before forcing it through the valve. Note that in the case of light gases such as hydrogen, helium and

neon, the Joule–Thomson coefficient is negative at room temperature. 7 Under those circumstances, depressurisation leads to a rise intemperature, which means that it is essential to precool the gas be-fore passing it through the valve.



~Q

~Q

Figure 11.9: Schematic diagram of a Linde plant for producing liquid air.



11 Open . . .§ 095 § 94 § 96 § 941 Use an energy balance to show that the system shown in Fig-ure 11.9 cannot reach a steady state unless: i) the fluid is an idealgas, or ii) the Joule–Thomson coefficient is positive and liquid is tappedstraight after the valve. Assume perfect heat exchange between the twofluids and zero heat loss from the process as a whole.

593 / 1776

§ 94 Linde[7] plant1 In the absence of heat loss, the process becomes an isenthalpic

expansion.

2 Perfect heat exchange has also been assumed, which means thatthe inflow and outflow temperatures are the same.

3 Under steady state conditions, the inflow and outflow rates will alsobe the same.

4 This allows us to set out the following four relationships:

~Hın = ~Hout , pın > pout ,

Tın = Tout , ~Nın= ~Nout .

5 Overall the flow is isenthalpic, which means thatdh =

(∂h∂T

)p dT +

(∂h∂p

)T dp = 0, where h is the molar enthalpy as

seen in ~H = ~Nh.

6 Since the fluid temperature is the same at the inlet and outlet, itfollows that dT = 0.

7 We also know that fluids always flow through valves down a pres-sure gradient, and hence that dp < 0.

8 That allows us to formulate the necessary condition for the processto be stationary:

(∂h∂p

)T = 0.

9 This condition is only fulfilled in an ideal gas.

10 With a real gas, the system will get colder and colder if the Joule–Thomson coefficient is positive, and conversely it will get hotter if

(∂T∂p

)h <

0.

11 In a Linde plant, this cooling is counteracted by removing liquidstraight after the valve.

594 / 1776

[g] Carl Paul Gottfried von Linde, 1842–1934. German engineer.

p [bar] 77.347 K 240 K 290 K

1.01325−3405.3 (lıq)+1547.6 (vap)

6967.1 8425.0

100 - 6070.4 7837.9300 - 5232.8 7254.9

Calculate how much liquid nitrogen is produced if theinlet conditions are 290 K and 100 bar and the outletpressure is 1 atm. Assume perfect heat exchangebetween the two fluids. In order to improve the effi-ciency of the system, it is important to cool the gasbefore it reaches the inlet. How much liquid nitro-gen would be produced if the inlet conditions were

changed to 300 bar and 240 K? What happens if the



11 Open . . .§ 095 § 94 § 96 (2)

inlet pressure is reduced to 1 atm? Use the enthalpyvalues Expanded footnote [h]:

[h] Richard T. Jacobsen and Richard B. Steward. J. Phys. Chem. Ref. Data, 2(4):757–922, 1973. [J mol-1] in the above table.



11 Open . . .§ 095 Production of liquid air

1 Based on the analysis that we did in Theme 94, the energy balancecan be expressed in the form

~Hgasın = ~Hvap

out +~Hlıq

out =~N

[hvap

out (1 − α) + h lıqoutα

],

or hgasın = hvap

out (1 − α) + h lıqoutα , (11.9)

where α is the molar liquid fraction under stationary conditions. For ni-trogen at an inlet temperature of 290 K and pressure of 100 bar, hgas

ın =

7837.9 J mol-1.

2 The liquid fraction α can be calculated from Eq. 11.9: 7837.9 =

8425.0(1 − α) − 3405.3α, and thus α = 0.0496.



11 Open . . .§ 095 Production of liquid air (2)

4 3 Theoretically liquid will be produced at the valve if the inlet temper-ature is lower than the inversion temperature TJT of the gas. The inversion temperature, which is defined at κ(TJT) = 0, is a

function of the pressure (see Theme 64 on page 393), but TJT ≈ 6Tc isa reasonable approximation at low pressures.

5 For nitrogen this means that TJT,N2 ≈ 750 K.

6 In our process Tın = 290 K, which is comfortably below the inver-sion temperature.

7 The amount of liquid produced tends to zero when pın → pout,and at an inlet pressure of 2 bar the liquid fraction falls to a measlyα = 5.58e − 4.

8 If the inlet conditions are changed to 240 K and 300 bar, on theother hand, then α = 0.1672.



11 Open . . .Looking back

1 Laminar: υ = υ2 and υ2 =

υ2

3

2 Turbulent: υ = υ and υ2 = υ2

3 Steady state: 12 ∆υ2 + g∆z +∆h = q − w

4 Bernoulli’s equation: 12 ρ∆υ

2 + ρg∆z +∆p = −w

5 Ideal gas compression work:

~Ws-s,ıgS = ~Nw ıg

s =γγ−1

~NRTın

[1 − (pout

pın

) γ−1γ

]

6 Adiabatic: δQ = 0

7 Isentropic: dS = 0

8 Lossless: δQ = T dS



11 Open . . .Looking back (2)

9 Joule–Thomson coefficient:

κ(T ,V ,n) =(∂T∂p

)H,n

=T

(∂p∂T

)V ,n + V

(∂p∂V

)T ,n

T(∂p∂T

)2


)T ,n


News Gr,p(T , p,n) A r,p(T , p,n) A r,v(T ,V ,n) Olds

Part 13

Departure Functions




13 Departure . . .New concepts

1 Gibbs residual

2 Helmholtz residual

3 Fugacity coefficient

4 Redlich–Kwong’s equation of state



13 Departure . . .Multicomponent thermodynamics

1 From a thermodynamic point of view, it makes little differencewhether a system is composed of one, two or many components, pro-vided that their relative proportions are constant in time and space.

3 2 Such systems will only have three degrees of freedom[a] andwill have relatively simple descriptions, as discussed in Chapter 8 onpage 335.

[a] Chemical reactions change the mole numbers of the compounds in a system with-out affecting the components that only vary with mass, i.e. the atoms or atom groupsthat are reaction invariant. The internal degrees of freedom are hidden from the out-side world, so the composition of the system is perceived to be constant, although thesize of the system may well vary with e.g. temperature and pressure. Isotopic mix-tures can be particularly intricate in this respect. Natural water contains e.g. nine equi-librated isotopic compounds without being paid any attention by us under normal cir-cumstances. The situation becomes much more complicated, however, if the

components that vary with mass are included in the system description.

4 The number of degrees of freedom increases, and the thermody-namic functions describe abstract hyperplanes in more than three spa-tial dimensions.

5 This makes it difficult to visualize the mathematical description, andas we do not want to restrict the theory to any specific chemical system,we shall hereafter use generic indices i ∈ [1,n] instead of componentnames.



13 Departure . . .Multicomponent thermodynamics (2)

8 6 The thermodynamics of multicomponent mixtures therefore has amore formal structure than that of its single-component counterpart.7 The virial equation will be used repeatedly as an example in this

chapter, but let us start by looking at how we can extend the resultsfrom Chapter 8 on page 335 to systems with variable compositions. In ideal gases, the composition dependence is relatively simple,

and by referring to the Eqs. 8.42–8.44 on 382 we can write:

∆fHıg(T,n) =

n∑i=1

Ni∆fhi (T) , (13.1)

C ıgV (T ,n) =

n∑i=1

Nicv ,i(T) , (13.2)

C ıgP (T ,n) =

n∑i=1

Nicp,i(T) , (13.3)

S ıg(T,p,n) =n∑

i=1Nisi (T,p) − R

n∑i=1

Ni ln(Ni

N). (13.4)

9 Note that ∆fhi , cv ,i , c

p,i and si are contributions from the standard

state of the same kind as µi in Eq. 7.20 on page 287.10 For systems with constant composition, U, H and S are the keyfunctions, but in mixtures the chemical potential becomes increasinglyimportant.11 However, this quantity cannot be obtained by differentiatingU or H with respect to their canonical variables because µi =

(∂U/∂Ni)S,V ,Nj,i = (∂H/∂Ni)S,p,Nj,i requires S to be a free variable,which is only possible in very special cases.12 The solution to the problem is to take either the Gibbs[a] or theHelmholtz[a] energy as our starting point, as µi = (∂G/∂Ni)T ,p,Nj,i =

(∂A/∂Ni)T ,V ,Nj,i can then be determined through explicit differentia-tion[a].13 Whether we use G or A depends on the equation of state beingused (more on this later).

[a] Josiah Willard Gibbs, 1839–1903. American physicist.

[a] Hermann Ludwig Ferdinand von Helmholtz, 1821–1894. German physician andphysicist.

[a] The great majority of thermodynamic models, including both equations of state andactivity coefficient models, are explicit in T but not in S .



13 Departure . . .§ 104 § 103 § 105

Express the Gibbs energy of an ideal gas based onEqs. 13.1–13.4. Calculate the standard state contribution

of Gıg by using si , ∆fhi and cp,i(T). Show by differentiation

that the chemical potential can be expressed as µıgi (T ,

p,n) = µi (T ,p) + RT ln[Nip/(Np)]. Identify thefunction µi .



13 Departure . . .§ 104 Ideal gas Gibbs energy

1 The Gibbs energy G = H − TS of an ideal gas mixture can bederived from Eqs. 8.43 and 8.44:

Gıg(T ,p,n) = ∆fHıg(T,n) +

T∫

T

C ıgP (τ,n) dτ − T

S ıg(T,p,n)

+T∫

T

C ıgP (τ,n)

τ dτ − NR ln( pp

). (13.5)



13 Departure . . .§ 104 Ideal gas Gibbs energy (2)

2 Let us substitute Eqs. 13.1, 13.3 and 13.4 into Eq. 13.5 and differ-entiate with respect to Ni :

µıgi (T ,p,n) =

(∂Gıg

∂Ni

)T ,p,Nj,i

= ∆fhi (T) +

T∫

T

cp,i(τ) dτ − Tsi (T,p)

− R ln(Ni

N) −

n∑k=1

Nk R NNk

δik N−NkN2

+T∫

T

cp,i(τ)

τ dτ − R ln( pp

). (13.6)

3 The above equation contains Kronecker’s delta, which has thevalue 1 if i = k and 0 in all other cases.



13 Departure . . .§ 104 Ideal gas Gibbs energy (3)

4 By simplifying Eq. 13.6 we can state the chemical potential as

µıgi (T ,p,n) = µi (T ,p) + RT ln

(NipNp

), (13.7)

where the standard chemical potential µi is defined as:

µi (T) = ∆fhi (T) +

T∫

T

cp,i(τ) dτ − Tsi (T,p) − TT∫

T

cp,i(τ)

τ dτ (13.8)



13 Departure . . .Gibbs residual Gr,p§ 1051 Use a specific example to show that Uıg − TS ıg + pV ıg =

∑ni=1 µ

ıgi

Ni.719 / 1776

§ 105 Ideal gas internal energy1 It provides valuable experience to know that the general relationswe meet in thermodynamics are valid under special circumstances aswell.

2 This paragraph underpins such an experience.

3 First, Eq. 13.7 is substituted into the right-hand side of the equationgiven in the paragraph text.

4 The left-hand side then follows by inserting Hıg = Uıg + NRT .720 / 1776

ENGLISH text is missing.The thermodynamic

residual1 In the coming sections we are going to examine these ideas further

and use our knowledge about ideal gas mixtures to describe the devia-tion between them and the real fluid.

2 The interpretation of the word “real” is a little vague, however, sincewe most of the time will be dealing with different state models and notwith experimental data.

3 For us the Van der Waals equation will be as real as the virialequation— except for yielding different function expressions but that isalso all.

4 The theory behind is the same and that is what matters in thiscontext.

5 The deviation between the ideal gas mixture and the real fluid iscalled a residual.

6 Traditionally, the residual is calculated at given a temperature, pres-sure and composition but we will also be using the volume as a free vari-able whenever that suits our needs better.

7 It is important to understand that the residual concept is connectedto a change in the model space and not the state space.

8 The thermodynamic coordinates T , p,n (or T ,V ,n) are frozen whilethe model description is changed from the real behaviour to the hypo-thetically ideal gas.

9 By integrating from a reference point with known properties we canreach all other thermodynamic states we would like to investigate.

721 / 1776

1 2 The canonical variables of Gibbs energy include temperature andpressure.3 This makes it possible to compare the chemical potentials of the

components of real fluids with those of the same components in theideal gas state, at the same temperature and pressure. Using what we already know about Gıg allows us to define the

residual Gibbs energy as:

Gr,p(T ,p,n) = G(T ,p,n) −Gıg(T ,p,n) . (13.9)

4 As p → 0, all fluids have virtually ideal properties in the senseof the pressure of the fluid approaching NRT/V when the volume in-creases indefinitely. 5 This does not mean that the volume of the fluid is the same as that

of an ideal gas! This approximation means that

limp→0

Gr,p = 0,

(∂Gr,p

∂p)T ,n

= B′2, . . . ,(∂nGr,p

∂p···∂p)T ,n

= (n − 1)!B′n+1

(13.10)

is meaningful and also defined such that the parameters B′i are notfunctions of the pressure (but they can still be strong functions of tem-perature and composition).



13 Departure . . .Gibbs residual Gr,p (2)

8 6 In particular, we shall notice that B ′2 , 0. This quantity describesthe low-pressure difference in the volume of the fluid and the ideal gaswhen kept at the same total pressure and is very important for our phys-ical understanding of the matter.7 The disposition given here is consistent with the virial theorem inChapter 7 on page 268, and with the Legendre transform in Chapter 4,from which it follows that (∂G/∂p)T ,n is the system volume regardlessof whether the fluid is ideal or real. By taking the limits from Eq. 13.10 it is possible to rewrite Eq. 13.9

as Expanded footnote [a]:

[a] Note that π is used as the integration variable in order to avoid confusion with p,which appears in the upper limit of the integral. :

Gr,p(T ,p,n) =p∫

0

(V − V ıg) dπ =p∫

0

(V(π) − NRT

π

)dπ . (13.11)



13 Departure . . .Residual chemical potential

1 An equivalent expression for the chemical potential µi =

(∂G/∂Ni)T ,p,Nj,iis true for all conceivable Gibbs energy functions, in-

cluding Gr,p.

2 This allows us to define the residual chemical potential as µr,pi =

µi − µıgi , or

µr,pi (T ,p,n) =

(∂Gr,p

∂Ni

)T ,p,Nj,i

=p∫

0

[(∂V∂Ni

)T ,π,Nj,i

− RTπ

]dπ =

p∫

0

(vi − RT

π

)dπ , (13.12)

where vi = (∂V/∂Ni)T ,π,Nj,iis defined as the partial molar volume of the

component i.



13 Departure . . .The Schwarz theorem

1 2 To obtain Eq. 13.12 from Eq. 13.11, Gr,p must be differentiated atconstant pressure[a].

[a] Note that to calculate the derivative of Gr,p at a given pressure p we must know thederivative of V − V ıg for all pressures π ∈ [0,p] in the integration domain. This goesagainst usual practice when performing partial differentiation, but Gr,p is a functionalthat is defined for all pressures p > 0, and not merely for one arbitrarily chosen systempressure p. The integrand, on the other hand, only depends on the local pressure π, soit will not change in any way if p changes in the upper limit of the integral. The integral inquestion is thus uniquely defined in terms of the area under a specific curve, integratedfrom the lower limit p = 0 up to the system pressure p. Admittedly the curve dependson the specified isotherm T and specified composition n, but it does not depend on thesystem pressure p. Consequently, the curve must be defined by differentiating V −V ıg

at every value of π such that we can choose to stop the integration at any value ofp > 0. Since the system pressure is only present in the limit of the integral,

we need only worry about the derivative of the kernel, but there thepressure does vary (locally) throughout the integration!

5 3 This is an unusual situation, and it is important to realise that thecorresponding differential of V(π) is defined locally as

(dV)T ,π,Nj,i =(∂V∂Ni

)T ,π,Nj,i dNi

for all pressures π ∈ [0, p].4 What is unusual in this respect is that there are two different phys-ical interpretions of pressure — one being the system pressure in theupper limit of the integral and the other being the integration variable π. However, there is a given local pressure π at each stage of the in-

tegration, so the only remaining degree of freedom comes from varia-tion in the composition of component i.

6 Hence, the chemical potential must be integrated over the volumederivative as shown in Eq. 13.12.



13 Departure . . .The Schwarz theorem (2)

7 For historical reasons it is customary to rewrite the residual poten-tial as

RT lnϕi = µr,pi (T ,p,n) , (13.13)

where ϕi(T ,p,n) is the fugacity coefficient of component i. 8 This is a measure of the difference between the thermodynamicstate of a component in a real fluid and of the same component in idealgas under the system conditions T ,p,n.9 The fugacity coefficient has traditionally been given a great deal ofattention in the American literature, but really it is a vestige of the past.10 For gases at low pressures and phase equilibrium calculations us-ing the K -value method it has certain benefits, but in general it causesmore problems than it solves — it neither increases our basic under-standing nor improves the numerical performance of thermodynamicalgorithms.



13 Departure . . .§ 106 § 105 § 107

Show that µi = µıgi + µr,p

i = µi +RT ln[Niϕip/(Np)].Verify that the lower limit limp→0 ϕi = 1. What is ϕi for

a component in the ideal gas state?



13 Departure . . .§ 106 The fugacity coefficient

1 The chemical potential can be derived by combining Eqs. 13.7and 13.12.

2 The latter equation gives us limp→0 µr,pi = 0 and hence limp→0 ϕi =

1.

3 For an ideal gas v ıgi = RT/π, which means that µr,p,ıg

i = 0 regard-less of the pressure.

4 Hence ϕıgi = 1.



13 Departure . . .§ 107 § 106 § 108

Derive Gr,p for the 2nd virial equation if pV2.vır = NRT +Bp.Differentiate to find RT lnϕ2.vır

k . Use the quadraticmixing rule for B = N

∑i

∑j

xixjBij where Bij = Bji .



13 Departure . . .§ 107 Virial Gibbs energy I

1 The residual Gibbs energy can be determined by directly substi-tuting the 2nd virial equation into Eq. 13.11:

Gr,p,2.vır =p∫

0

(NRTπ + B − NRT

π

)dπ = Bp . (13.14)

2 In order to find RT lnϕk we must differentiate Gr,p, and hence B,with respect to Nk , but differentiating with respect to (mole) fractionshas an unfortunate tendency to cause what is known as “code bloat” incomputing.



13 Departure . . .§ 107 Virial Gibbs energy I (2)

3 We can considerably reduce the amount of writing that we needto do by using NB =

∑i∑

j NiNjBij , which, even though it is an implicitexpression, is much better suited for differentiation:

(∂NB∂Nk

)T ,Nl,k

=∑i

∑j

(∂NiNj

∂Nk

)Nl,k

Bij .

4 Differentiate both sides and substitute in Kronecker’s delta, whereδii = 1 if i = j and δij = 0 if i , j:

B + N( ∂B∂Nk

)T ,Nl,k

=∑i

∑j

(δik

Nj + Ni δjk

)Bij =

∑j

NjBkj +∑i

NiBik .

5 We can simplify the right-hand side considerably by substitutingBkj = Bjk and simultaneously changing one of the summation indicesfrom j to i.



13 Departure . . .§ 107 Virial Gibbs energy I (3)

6 This gives us two identical terms on the right-hand side, so thepartial derivative of B can now be written

bk =( ∂B∂Nk

)T ,Nl,k

= 2∑i

NiN Bik − B

N .

7 The residual potential can be calculated from Eqs. 13.13 and 13.14on page 724, as shown below:

RT lnϕ2.vırk =

(∂Gr,p,2.vır

∂Nk

)T ,p,Nl,k

= p( ∂B∂Nk

)T ,Nl,k

= pbk .

8 Alternatively, we can achieve the same result by substituting thepartial derivative of B into Eq. 13.12 on page 718 and integrating Expanded footnote [a]:

[a] It is worth noting that bk is the (pressure-independent) departure function of thepartial molar volume from the 2nd virial equation: v2.vırk − RT/π = bk (T). .



13 Departure . . .Helmholtz residual A r,p IENGLISH text is missing.Helmholtz residual1 Residual Gibbs energy has the canonical variables temperature

and pressure.

2 We must therefore use an equation of state in the form V = V(T ,p,n), which in practice means that the virial equation V2.vır = NRT/p + Bis of limited use.

3 However, there has been rapid development of equations of statein the form p = p(T ,V ,n), which changes the canonical variables totemperature and volume rather than pressure.

4 This makes it attractive to take Helmholtz energy rather than Gibbsenergy as our starting point.

5 Traditionally, the definition of the fugacity coefficient in Eq. 13.13 isnot changed in the transposition from G to A .

6 This causes a problem with the chemical potential, because ϕi

describes the difference between the real fluid and ideal gas at a givenpressure— not a given volume.

7 As a result, you end up with a non-canonical description, which isdiscussed here mainly because of its historical interest.

733 / 1776

1 Our goal is to describe the difference in Helmholtz energy betweena real fluid and an ideal gas at a given temperature and pressure.

3 2 To do this, we have to find the difference between two state func-tions applied to two different states. We can simplify our task by splitting the expression into two terms:

The first term compares the two functions in the same state, while thesecond term compares the ideal gas in two different states:

A r,p(T ,V(p),n) = A (T ,V(p),n) − A ıg(T ,V ıg(p),n)

≡ A (T ,V(p),n) − A ıg(T ,V (p),n) (13.15)

+ A ıg(T ,V(p),n) − A ıg(T ,V ıg(p),n) .

4 It should be emphasised that volume is a function of pressure p. 5 The pressure is by definition the same in the ideal gas and the realfluid, i.e. pıg = p.

It implies, however, that the two fluids have different volumes, sinceV , V ıg at a given p.



13 Departure . . .Helmholtz residual A r,p I (2)

6 From Chapter 4 we know that (∂A/∂V)T ,n = −p. 7 This is true for any Helmholtz energy function, regardless ofwhether it refers to an ideal gas or a real fluid. Analogously to

Eq. 13.11 on page 717 we can therefore write Expanded footnote [a]:

[a] This follows from limV→∞

A − A ıg = 0, (∂A−A ıg∂V )T ,n = 0, . . . , (∂nA−A ıg∂V ···∂V )T ,n = 0

in accor-

dance with the virial theorem.

A r,p(T ,V(p),n) =V(p)∫∞

(πıg − π) dν −V(p)∫

V ıg(p)

πıg dν

=V(p)∫∞

(NRTν − π

)dν − NRT ln z(p) , (13.16)

where the ideal gas volume is given by V ıg = NRT/p, and the com-pressibility factor z is defined as z = V/V ıg = pV/(NRT).



13 Departure . . .§ 108 § 107 § 109

ENGLISH text is missing.Problems1 Eq. 13.16 may look trivial, but please note that the volume is an

implicit variable in the integral.

2 The implicit form of the function A r,p(T ,V(p), p,n) prevents us fromcalculating the chemical potential by differentiation with respect to thecanonical variables T ,V ,n [a].

3 so we must instead use

µi =(∂A∂Ni

)T ,V ,Nj,i

⇒ µıgi =

(∂A ıg

∂Ni

)T ,V ıg,Nj,i

,

4 If we substitute this into the residual chemical potential µr,pi = µi −

µıgi we get

µr,pi =

(∂A∂Ni

)T ,V(p),Nj,i

− (∂A ıg

∂Ni

)T ,V ıg(p),Nj,i

=(∂A r,p

∂Ni

)T ,V(p),V ıg(p),Nj,i

.

5 Note that V(p) is constant in one of the partial derivatives, whileV ıg(p) is constant in the other one.

6 This means that both V(p) and V ıg(p) must be kept constant whendifferentiating A r,p, which in turn implies that z(p) = V(p)/V ıg(p) is aconstant factor.

7 The residual chemical potential can therefore be expressed

µr,pi (T ,V(p),n) =

V(p)∫∞

[RTν −

( ∂π∂Ni

)T ,ν,Nj,i

]dν − RT ln z(T ,V(p),n) .

(13.17)

8 Although this derivation is mathematically correct, it is hard to ex-plain logically, as it relies on there being two different volumes (both ofwhich remain constant, in line with the definition) during the same dif-ferentiation.

736 / 1776

[a] If we redefine the fugacity coefficient to mean the difference between an ideal gasand a real fluid at a given volume rather than a given pressure, it becomes easier toderive the chemical potential; see Section 13.4 on page 731.

Find A r,p for the 2nd virial equation if pV2.vır = NRT +Bp.



13 Departure . . .§ 108 Virial Helmholtz energy

1 The residual Helmholtz energy is calculated by inserting the2nd virial equation into Eq. 13.16, which gives us this surprising re-sult:

A r,p,2.vır =V∫∞

(NRTν − NRT

ν−B

)dν − NRT ln

( VV−B

)= 0 . (13.18)

2 So there is no difference between Helmholtz energy for the2nd virial equation and that for an ideal gas at the same pressure.

3 In other words: The volume term in A r,p,2.vır happens to be equal toRT ln z2.vır.

8 4 This does not mean that all residual properties cancel.5 For example, Gr,p,2.vır was found to be Bp in Eq. 13.14 on page 724,and you should verify that if the 2nd virial equation is substituted intoEq. 13.17, it gives the same answer as in Eq. 13.12 on page 718.6 Moreover, even though A r,p,2.vır is zero, this does not imply thatµr,p,2.vıri is also zero.

7 The fact is that Gr,p,2.vır, and hence µr,p,2.vıri , are direct measures ofthe non-ideality of the gas, whereas A r,p,2.vır is not. It is important to remember that A r,p is a non-canonical function.



13 Departure . . .(T , p,n) vs (T ,V ,n)ENGLISH text is missing.Helmholtz residual A r,p II1 At this point it may be helpful to offer an alternative, more intuitive

derivation of A r,p.

2 Let us first differentiate Eq. 13.15 with respect to the mole numberNi :

(∂A r,p

∂Ni

)T ,V ,Nj,i

= µi −(∂A ıg

∂Ni

)T ,V ,Nj,i

. (13.19)

3 This time we face the problem that constant V does not necessarilyimply constant V ıg.

4 In fact, in order to differentiate A ıg with respect to Ni, we need toknow how V ıg varies with Ni when V is kept constant.

5 At a given temperature and at fixed amounts of all components j,except i, the differential of A ıg is

(dA ıg)T ,Nj,i =(∂A ıg

∂Ni

)T ,V ıg,Nj,i

dNi +(∂A ıg

∂V ıg

)T ,n dV ıg = µıg

i dNi − pıg dV ıg ,

(13.20)where ıg is stated explicitly in order to make the next step a little bitclearer.

6 If we bear in mind the condition pıg = p, we can simplify Eq. 13.20to

(∂A ıg

∂Ni

)T ,V ,Nj,i

= µıgi − p

(∂V ıg

∂Ni

)T ,V ,Nj,i

. (13.21)

7 Combined with Eq. 13.19 this gives us:

µi − µıgi =

(∂A r,p

∂Ni

)T ,V ,Nj,i

− p(∂V ıg

∂Ni

)T ,V ,Nj,i

. (13.22)

8 Finally, Eq. 13.16 is differentiated:

(∂A r,p

∂Ni

)T ,V ,Nj,i

=V∫∞

[RTν −

( ∂p∂Ni

)T ,ν,Nj,i

]dν+ NRT

V ıg

(∂V ıg

∂Ni

)T ,V ,Nj,i

− RT ln z .

(13.23)

9 You will recognise NRT/V ıg = pıg = p in Eq. 13.23.

10 This means that the term (∂V ıg/∂Ni)T ,V ,Nj,i disappears when theEqs. 13.22 and 13.23 are combined.

11 The residual chemical potential µr,pi = µi − µıg

i is thus the same asin Eq. 13.17 on page 729.

739 / 1776

ENGLISH text is missing.Helmholtz residual A r,p III1 A third option is to use Gibbs energy as our starting point, which isa natural thing to do, since the residual chemical potential has the samecanonical variables as Gibbs energy.

2 Substituting G = A + pV into Gr,p = G −Gıg gives us:

Gr,p = A(T ,V(p),n) + pV(p) − [A ıg(T ,V ıg(p),n) + pV ıg(p)] . (13.24)

3 Furthermore, pV = zNRT and pV ıg = NRT . When we substitutethese into the above equation, and combine it with Eqs. 13.15 and 13.16on page 727, we get:

Gr,p = A r,p + NRT(z − 1)

=V(p)∫∞

(NRTν − π

)dν − NRT ln z(p) − NRT [z(p) − 1] . (13.25)

4 Our one remaining task is to differentiate this expression with re-spect to the mole number of component i.

5 It is evident that we will need to obtain the partial derivative ofz = pV/NRT , so let us start by doing that:

zi =( ∂z∂Ni

)T ,p,Nj,i

= pviNRT −

pVN(NRT) =

pviNRT − z

N .

6 This is where things start to get tricky: How can we differentiate theintegral?

7 The upper limit clearly depends on the composition at fixed pres-sure, while the differential of π at a given volume ν is:

(dπ)T ,ν,Nj,i =( ∂π∂Ni

)T ,ν,Nj,i

dNi .

8 It is important to note that ν is an integration variable in Eq. 13.25,and that it remains constant during the above differentiation.

9 Partial differentiation therefore gives us:

µr,pi =

V(p)∫∞

[RTν −

( ∂π∂Ni

)T ,ν,Nj,i

]dν+

(NRTV(p) − p

)vi

− RT ln z − ziNRTz + RT(z − 1) + ziNRT .

10 If we substitute in zi , most of the terms cancel, and the final resultis identical to Eq. 13.17 on page 729.

740 / 1776

§ 1091 Show that Gr,p,2.vır = NRT(z −1) = Bp when pV2.vır = NRT +Bp.

741 / 1776

§ 109 Virial Gibbs energy II1 From Eq. 13.25 we know the general expression Gr,p = A r,p +

NRT(z − 1).

2 Combined with A r,p,2.vır = 0 from Eq. 13.18 on the previous pagethis gives us Gr,p,2.vır = NRT(z − 1).

3 Substituting in z2.vır = 1 + Bp/NRT gives Gr,p,2.vır = Bp in accor-dance with Eq. 13.14 on page 724.

4 The results from the various sections are thus internally consistent.742 / 1776

1 2 The problems we experienced in the previous section were causedby the disastrous decision to use pressure as a free variable in theHelmholtz function. If we redefine the residual Helmholtz energy as

A r,v(T ,V ,n) = A(T ,V ,n) − A ıg(T ,V ,n)

we are back into the world of canonical variables, and we can use thesame approach as the one used for residual Gibbs energy on page 716.

3 The alternative departure function can be written

A r,v =V∫∞

(NRTν − p(ν)

)dν ,

while the associated chemical potential is

µr,vi =

(∂A r,v

∂Ni

)T ,V ,Nj,i

=V∫∞

[RTν −

( ∂p∂Ni

)T ,ν,Nj,i

]dν .



13 Departure . . .§ 110 § 109 § 111

Derive an expression for A r,v and the first derivatives∂A r,v/∂(T ,V ,n) for a fluid that conforms to the Redlich–

Kwong equation of state pRK = RT/(v−b)−a/√

Tv (v + b).Use the mixing rules b(x) =

∑i bixi and a(x) =

∑i∑

j√

aiaj

xixj where bi and ai are component-specific parameters.Finally, state bi and ai as functions of Tc and pc; also

see Theme 47 on page 322.



13 Departure . . .§ 110 Redlich–Kwong[1]


[] Otto Redlich and J. N. S. Kwong. Chem. Rev. (Washington, D. C.), 44:233–244,1949. Let us first estimate a and b for a fluid composed of a single

component where we supposingly have a measurement of the criticaltemperature and pressure of this component. For this purpose we willrewrite the equation of state as a cubic polynomial of volume:

v3 − RTp v2 +

(a

pT1/2 − b2 − RTbp

)v − ab

pT1/2 = 0 .

2 At the critical point this equation must fulfil (v − vc)3 = 0, or v3 −

3vcv2 + 3v2c v − v3

c = 0.

3 If compared term-by-term (see Theme 47 in Chapter 7), the twopolynomials can be used to produce three equations, which can thenbe solved for a, b and vc:

3vc = RTcpc

, 3v2c = a

pcT1/2c− b2 − RTcb

pc, v3

c = abpcT1/2

c.



13 Departure . . .§ 110 Redlich–Kwong[2] (2)

4 Combining the middle equation with the other two yields 2v3c −(vc+

b)3 = 0, which can be solved for the positive root b = (21/3 − 1)vc.

5 If expressed in terms of the critical temperature and pressure (forany component i), this is equivalent to:

bi = ΩbRTc,ipc,i

ai = ΩaR2T5/2

c,ipc,i

Ωb = 21/3−13 Ωa = 1

9(21/3−1)

6 We can now choose whether to express the rest of the derivationin terms of extensive properties (mole number and volume) or intensiveproperties (mole fraction and molar volume).

7 Here, the extensive properties have been chosen, in order to em-phasise the homogeneity of the thermodynamic functions.




8 The total Helmholtz energy of the fluid is thus

A r,v,RK =V∫∞

(NRTν − pRK

)dν = NRT ln

( VV−B

)+ A

B ln( VV+B

),

and the associated first derivatives are:(∂A r,v,RK

∂T)V ,n =NR ln

( VV−B

)+ AT

B ln( VV+B

),

(∂A r,v,RK

∂V)T ,n =NRT −B

V(V−B) +A

V(V+B) ,(∂A r,v,RK

∂Ni

)T ,V ,Nj,k

=RT ln( VV−B

)+ NRT bi

V−B + 1B

(Ai − Abi

B

)ln

( VV+B

) − AbiB(V+B)

.




9 The coefficients of the equation are defined below (note that αi hasthe same units as aVdW

i ):

B =∑i

biNi ,

A =∑i

∑j

(aiaj

T)1/2

NiNj =(∑

iα1/2

i Ni

)2,

AT = − A2T ,

Ai = 2α1/2i

∑jα1/2

j Nj = 2(αiA)1/2 ,

αi = aiT1/2



13 Departure . . .Looking back

1 Proper Gibbs residual:

Gr,p(T ,p,n) = G(T ,p,n) −Gıg(T ,p,n)

=p∫

0

(V(π) − NRT

π

)dπ

2 Proper Helmholtz residual:

A r,v(T ,V ,n) = A(T ,V ,n) − A ıg(T ,V ,n)

=V∫∞

(−p(ν) + NRT

ν

)dν



13 Departure . . .Looking back (2)

3 Helmholtz residual at given pressure:

A r,p(T ,V(p),n) = A(T ,V(p),n) − A ıg(T ,V ıg(p),n)

=V(p)∫∞

(−p(ν) + NRT

ν

)dν − NRT ln z(p)

4 Gr,p,2.vır = NRT(z − 1) = Bp

5 A r,p,2.vır = 0

6 A r,v,RK = NRT ln( VV−B

)+ A

B ln( VV+B

)

7 Chemical potential from Gibbs residual:

µr,pi (T ,p,n) =

(∂Gr,p

∂Ni

)T ,p,Nj,i

=p∫

0

(vi − RT

π

)dπ



13 Departure . . .Looking back (3)

8 Gibbs inspired chemical potential from Helmholtz residual:

µr,pi (T ,V(p),n) =

V(p)∫∞

[RTν −

( ∂π∂Ni

)T ,ν,Nj,i

]dν − RT ln z(T ,V(p),n)

9 Fugacity coefficient: RT lnϕi = µr,pi (T ,p,n)


News Clausius–Clapeyron Waals min A Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds

Part 14

Simple vapour–liquid equilibrium




14 Simple . . .New concepts

1 Saturation state

2 Clausius–Clapeyron’s vapour pressure equation

3 Van der Waals equation of state

4 Critical point behaviour

5 Maxwell’s area rule



14 Simple . . .Phase boundary ENGLISH text is missing.Perspective1 Particularly in the chemical process industry, vapour–liquid equilib-ria are essential to the design and operation of all kinds of distillationand absorption columns, oil and gas separators, evaporators and othersimilar process equipment.

2 In our everyday lives, the same physical phenomena cause theevaporation of solvents in paints and varnishes, allow us to enjoy thearomas of wines, cheeses and coffees, make it possible to use propaneand butane in camping gas stoves, etc.

3 Moreover, at a global level, all life on Earth is inextricably linked tothe transport of sea water (as vapour) from warmer to cooler parts ofthe globe, where it then condenses as rain or snow before it eventuallyfinds its way back to the ocean.

4 Although this cycle, which is driven by the sun, occurs under near-equilibrium conditions, it is responsible for the vast majority of masstransport at or near the Earth’s surface.

5 It is therefore not true to say that near-equilibrium processes are“slower” or in any way less important than irreversible processes.

754 / 1776

1 2 The physical explanation of most of the abovementioned processesrequire complex descriptions of multicomponent systems involving var-ious phases.3 The theory is too involved to be covered in one single chapter andwe shall rather learn from studying the characteristics of vapour–liquidequilibria based on the behaviour of a realistic, but much simpler, one-component system.4 This makes the formulation part of the theory far simpler becauseit does not need to take into account the variation in chemical compo-sition, and to some extent you would expect there to be analytical solu-tions to some of the selected problems. How to go about that is precisely what this chapter is about. Ac-

cording to our working hypothesis the conditions for thermodynamicequilibrium along a continuous phase boundary curve in the T ,p, µspace are:

Tvap = T lıq dTvap = dT lıq

pvap = p lıq ⇒ dpvap = dp lıq

µvap = µlıq −svap dTvap + vvap dpvap = −s lıq dT lıq + v lıq dp lıq



14 Simple . . .Clapeyron equation

1 2 The criteria on the left-hand side are the classical equilibrium con-ditions being derived in Section 14.4.3 They must be fulfilled at all thermodynamic equilibrium points, but

from Theme 39 on page 195 we know that the intensive state[a] of thesystem can be determinied when two of the three state variables T ,p, µare known.4 Consequently, the equilibrium state can be parametrised by either

T ,p or T , µ or p, µ, the first of which is the most common choice.5 For a small change in the variables, the conditions on the right-

hand side then apply.

[a] We have not formulated a mass balance for the equilibrium, so the phase distribu-tion is unknown, but that is irrelevant in this context because (∂µ/∂N)T ,p = 0 for allone-component systems, i.e. the equilibrium is the same regardless of the quantity ofvapour and liquid. It is normal to combine these three differentials into a single equa-

tion, which is referred to as the Clapeyron Expanded footnote [a]:

[a] Émile Clapeyron, 1799–1864. French engineer and mathematician. equation:(dpdT

)∆µ=0 = svap−s lıq

vvap−v lıq = ∆vs∆vv = ∆vh

T∆vv .

7 6 This equation brings together four important quantities in phasetheory: Vapour pressure, temperature, enthalpy of vaporization and sat-uration volume. Note that the relationship between the entropy and enthalpy of va-

porization follows from the equilibrium condition ∆vµ = 0, which substi-tuted into µ = h − Ts gives us ∆vs = ∆vh/T .



14 Simple . . .Clausius–Clapeyron equationENGLISH text is missing.Reduced Clapeyron

equation1 We could have stopped the derivation after the last section but

traditionally the expression is written on a dimensionless form.

2 Finally, the equation will be integrated using a couple of physicallybased assumptions. Reducing the right-hand side

(dpdT

)∆µ=0 = ∆vh

T∆vv = RTcTcvc ·

∆vhrTr∆vvr =

Rvc ·

∆vhrTr∆vvr

3 and then the left-hand side:

− pcpr

TcT2r· ( dln pr

d(1/Tr)

)∆µ=0 = R

vc ·∆vhr

Tr∆vvr .

4 The answer is:( dln prd(1/Tr)

)∆µ=0 = − RTc

pcvc ·Trpr ·

∆vhr∆vvr . (14.1)

5 To come any further it is required to use a mix of measurements,assumptions and models.

6 The ideal gas is one obvious candidate but how should the criticalconstants be interpreted in that case?

7 There exist no ideal critical state and we must formally eliminate allTc, pc and vc from the equation (reversing all the previous operationsyou can say).

8 At low pressures the denominator is taken over by the vapour vol-ume such that ∆vv ≈ vvap.

9 We then get the limit case:

limp→0

( dln pd(1/T)

)ıg

∆µ=0 = − ∆vhR . (14.2)

757 / 1776

1 If we integrate we get the Clausius Expanded footnote [a]:

[a] Rudolf Julius Emanuel Clausius, 1822–1888. German physicist. –Clapeyron vapour pressureequation, where b = ∆vh/R is assumed to be constant over the rele-vant temperature range:

ln( pp

)C-C= a − b

T . (14.3)

2 The enthalpy of vaporization is an independent quantity in this caseand must be measured experimentally.

4 3 In order to attain a consistent description of the vaporization pro-cess vi have to consider a model that can describe both phases with asingle formula. Being omnipresent throughout the rest of this chapter, the

Van der Waals equation 14.5 is a natural choice.

6 5 The following is then true (see also Theme 47 on page 322):

(pcvcRTc

)VdW = 38 , limTr≪1 pVdWr = 8Tr3vvapr , limTr≪1∆vvr = vvapr . Substitution into Eq. 14.1 gives the final result:

limTr≪1

( dln prd(1/Tr)

)VdW

∆µ=0= −∆vhr . (14.4)

7 The two Eqs. 14.2 and 14.4 are similar notwithstanding the latterequation is written on reduced form.8 The conclusion is that the Van der Waals equation is consistent

with the ideal gas assumption at low pressures.



14 Simple . . .Measured vapour pressure

1 If Tr → 1 then we cannot assume ideal gas behaviour and thevolume of liquid is no longer negligible either. This undermines thetheoretical basis for



14 Simple . . .Measured vapour pressure (2)

1 1.2 1.4−3

−2

−1

0

1

1/Tr = Tc/T

ln(p

r)=

ln( p/p

c)

Figure 14.1: Vapour pressure of CO2, where is the measured value and ×is the percentage deviation.



14 Simple . . .Measured vapour pressure (3)

the Clausius–Clapeyron equation, but the non-linear contributionslargely cancel each other out, so Eq. 14.3 still gives a reasonable ap-proximation of the vapour pressure.

4 2 This relationship is illustrated in Figure 14.1, which shows a vapourpressure curve that is surprisingly rectilinear, even close to the criticalpoint (cf. Program 1.11 in Appendix 31).3 The percentage difference between the estimated and measured

vapour pressure[a] is given in the figure by × and there are no cluesthat the linear estimate is fundamentally wrong.

[a] W. Duschek, R. Kleinrahm, and W. Wagner. J. Chem. Thermodynamics, 22:841–864, 1990. To begin to understand why the vapour pressure can be plotted

as a straight line even at high pressures, and to start learning aboutvapour–liquid equilibria in general, let us look at the Van der Waalsequation of state in greater detail. 5 The state description is not nearly precise enough for practical uses

in process engineering studies but it is the simplest equation we can useto illustrate the peculiarities we are after.



14 Simple . . .Van der Waals p and µ

1 2 The Van der Waals equation has been presented before in Chap-ters 7:7.8 and 8:8.9 but only in the context of single phase gases.3 The focus is now shifted towards two-phase vapour–liquid equilib-

ria.4 However, we do not want to undertake the full discussion of mul-

ticomponent mixtures and for the present the parameters b and a areconsidered to be true constants:

pVdW = NRTV−Nb − aN2

V2 (14.5)

5 Let us next calculate the chemical potential of the Van der Waalsfluid using the p(T ,V)-relation above. From the discussion of residuals in Chapter 13 the chemical poten-

tial can be expressed as µVdW = µıg + µr,v, where the residual term isdefined as being

µr,v =V∫∞

(RTν −

(∂p∂N

)T ,ν

)dν

=V∫∞

(RTν − RT

ν−Nb − NRTb(ν−Nb)2 + 2aN

ν2

)dν

= RT ln( VV−Nb

)+ NRTb

V−Nb − 2aNV ,

and the ideal gas contribution is

µıg = µ(T ,p) + RT ln(NRT

Vp

).



14 Simple . . .Van der Waals p and µ (2)

6 Combining the two equations above gives us the following expres-sion for the chemical potential:

µVdW = µ(T ,p) + RT ln( NRTp(V−Nb)

)+ NRTb

V−Nb − 2aNV .

7 If we use the generalised parameters 3b = vc, a = 3pcv2c and

8pcvc = 3RTc from Theme 47 on page 322, we can express the pres-sure and chemical potential in dimensionless form:

pVdW

pc= pVdW

r = 8Tr3vr−1 − 3

v2r, (14.6)

µVdW

RTc= µVdW

r = µr − Tr ln pr + Tr ln( 8Tr3vr−1

)+ Tr

3vr−1 − 94vr

. (14.7)



14 Simple . . .Van der Waals A

1 It is worth noting that these two equations represent a complete setof equations of state for Helmholtz energy.

2 Integration using the Euler method gives us A = −pV + µN, whichexpressed in dimensionless form becomes A

NRTc= − pV

NRTc+

µRTc

. Rewrit-ing this in reduced variables:

aVdWr = AVdW

NRTc= − 3

8 pVdWr vr + µVdW

r . (14.8)

3 Note that the critical compressibility factor zc drops out as the con-stant number pcvc/RTc = 38 when the Van der Waals equation of stateis used to solve for the critical state of the fluid. The universal behaviourindicated is what we call a corresponding state. A more thorough ex-planation is given in Themes 47 and 49 on page 328.



14 Simple . . .Van der Waals S and H

1 Given the expression of Helmholtz energy above it is easy to derivethe entropy of the fluid by differentiation with respect to T . 2 On a dimensionless form we have the general expression

−sr = −SNR = (NR)-1 (∂A∂T

)V ,N = (∂(A/NRTc)∂(T/Tc)

)V ,N = (∂ar∂Tr

)vr , which in the

current case becomes:

−sVdWr = − 3

8 · 83vr−1 · vr − sr − ln pr + ln

( 8Tr3vr−1

)+ 1 + 1

3vr−1

= −sr − ln pr + ln( 8Tr3vr−1

). (14.9)

3 We can thereafter do Euler integration of H = TS + µN to find theenthalpy:

hr =H

NRTc=

TS+µNNRTc

= Trsr + µr .

Since µ = h − Ts the latter expression becomes:

hVdWr = hr + Tr

3vr−1 − 94vr

. (14.10)



14 Simple . . .Van der Waals S and H (2)

4 The standard state contribution cancels out when calculating theenthalpy of vaporization:

∆vhVdWr = Tr

3vvapr −1

− 94vvap

r− Tr

3v lıqr −1

+ 94v lıq

r. (14.11)

5 The Eqs. 14.6–14.11 contain everything we need to know in orderto discuss the fundamental issues of vapour–liquid equilibria.6 Theoretically, the expressions are not particularly difficult but theirmathematical handling depend on quite a few physical details.7 The knowledge necessary is gained only with practical training anda one single example may contribute more than a thousand words.



14 Simple . . .Phase equilibrium ENGLISH text is missing.Brief comment1 Theme 112 in Section 14.4 takes you into that direction where

Figure 14.3 is a practical example used to illustrate the properties ofVan der Waals equation.

2 The equations that are necessary for calculating such phase dia-grams in general are treated in the next two sections.

767 / 1776

1 In this section we are going to define the thermodynamic equilib-rium of a closed two-phase system as a state of minimum Helmholtzenergy.

3 2 In principle this task sounds quite easy but in practise it is a littlemore complex. Helmholtz energy is a function of the standard state of the fluid,

and since standard state properties do not have absolute values, theminimum of the function won’t be an absolute value either.

5 4 Admittedly we are going to seek the minimum of the Helmholtzenergy function, but we will do that by distributing the total mass acrosstwo different phases. The equilibrium state is therefore characterised by having two

points on a single isotherm that can be connected by a line that is en-tirely below the rest of the function surface (the graph).

6 From a geometrical point of view, this condition will be met if thetwo points have a common tangent. 7 It can be shown that the axis intercepts in the diagram are propor-

tional to p and µ respectively, which means that the tangent require-ment is also equivalent to pvap = p lıq and µvap = µlıq.8 This can be considered the result of a differential condition for equi-

librium, whereas the Maxwell equal area rule in Theme 113 on page 762gives an integral formulation of the same condition.



14 Simple . . .§ 111 § 110 § 112

The necessary conditions for thermodynamic equilibriumin a closed system with only one chemical component,at a given temperature T , total volume V = Vvap +V lıq

and total composition N = Nvap + Nlıq, are

Aeq = minv,n

(A)T ,

ev = V ,

en = N .

Here e = ( 1 1 ), vT = ( Vvap V lıq ) and nT = ( Nvap Nlıq ).Show that pvap = p lıq and µvap = µlıq are necessary

conditions for equilibrium. Illustrate the energy surfaceusing a graph drawn in (N/V ,A/V) or (V/N,A/N)



14 Simple . . .§ 111 § 110 § 112 (2)

co-ordinates. Use reduced variables ρr, vr and ar

where ρr ∈ 〈0,3〉 when you draw the diagram.



0 1 2 30

0.5

1

1.5

2

2.5

3

ρr = vc/v

a r·ρ

r=

Av c/V

RT

c

c.p.

Figure 14.2: Helmholtz energy for the isotherms in Figure 14.4 on page 767.The equilibrium state (the tangent plane at minimum energy) is given for thesecond lowest isotherm. The standard state has been chosen for the pur-poses of clarity, and varies for each isotherm. The bell shaped curve indi-cates the system’s phase boundaries.



14 Simple . . .§ 111 Minimum of Helmholtz energy

1 2 In optimization theory there are necessary and sufficient conditionsfor the minimum.3 The necessary conditions, being of first order, lead to the equilib-

rium relations.4 The sufficient conditions, being of second order, lead to the stability

criteria. Here, the first order conditions will do fine.

5 A necessary condition for thermodynamic equilibrium is that(dA)T = 0 for variations in the extensive variables Vvap,Nvap andV lıq,Nlıq.

6 Taking the total differential for both phases gives us

(dA)T = (dA)vapT + (dA)

lıqT

= −pvap dVvap + µvap dNvap − p lıq dV lıq + µlıq dNlıq

= 0 .



14 Simple . . .§ 111 Minimum of Helmholtz energy (2)

7 The limits on the volume and mole number of the closed systemmean that dV lıq = −dVvap and dNlıq = −dNvap. Substituted into thedifferential of A that gives us

(dA)T = −(pvap − p lıq) dVvap + (µvap − µlıq) dNvap = 0 .

8 Since dVvap and dNvap are independent variables, then it must bethe case that pvap = p lıq and µvap = µlıq, which is what we needed toshow. 9 The next question is how we best should illustrate the minimum.

10 A minimum in two variables would normally belong to a surface inthree dimensions but for Helmholtz energy it is sufficient to map thefunction value along a straight line in the (V ,N) plane.11 The line must actually be traversing the entire (V ,N) plane.12 It is not sufficient that the line departs from the origin.13 This would imply a simpel scaling of the extensive variables V andN.14 A scaling of this kind does not influence on the intensive propertieslike p and µ and it has because of this no effect on the equilibrium.15 In order to traverse the entire (V ,N) plane we can e.g. keep Vconstant while changing N or we can do the opposite.16 Here, the first option is chosen.17 Figure 14.2 shows selected isotherms in the requested (N/V ,A/V)

diagram, along with the saturation curve of the fluid (see Program 1.10in Appendix 31).



100

101

0

0.5

1

1.5

2

2.5

3

c.p.

vr = v/vc

p r=

p/p

c

Figure 14.3: The Van der Waals equation of state expressed in pr, vr coordi-nates for Tr = 0.8,1.0,1.2. The horizontal line shows the experimentally es-tablished two-phase region for nitrogen at Tr = 0.8.



14 Simple . . .Maxwell equal area rule IENGLISH text is missing.The tangent plane1 In order to understand the details of the figure we need to know theslope of the graph and the intersection of the tangent bundle and theordinate axis.

2 The differential of A/V in reduced co-ordinates is:

d(ar · ρr)V = d( A

NRTc ·NvcV

)V = µvc

VRTc dN

3 The corresponding differential of N/V is:

(dρr)V = d(Nvc

V)V = vc

V dN

4 The elimination of dN gives the slope of the graph:(d(ar · ρr)

dρr

)

V= µ

RTc = µr

5 The intersection of the tangent and the ordinate is now easily cal-culated:

a⋆ = ar · ρr −(d(ar · ρr)

dρr

)

V· ρr

6 As a small curiosity we note that a⋆ is identical to the Legendretransform of ar · ρr with respect to ρr.

7 After re-calculating the right-hand side we get:

a⋆ =( A

NRTc −µ

RTc

)· ρr =

(−pV+µNNRTc −

µRTc

)· ρr =

pVNRTc · ρr = −pr · pcvc

RTc

8 In other words, the slope of the graph is equal to µr while the inter-section of the tangent with the ordinate is proportional to pr.

9 Furthermore, the solution of the energy minimization problem weposed earlier tells us that the pressure and the chemical potential mustbalance in the two phases.

10 The only way this can happen is that two points along the sameisotherm has the same slope and the same intersection.

11 This implies that the two points must actually share the same tan-gent.

12 Figure 14.2 shows what the construction looks like for the secondlowest isotherm.

13 The minimum principle was illustrated using a (N/V ,A/V) diagramabove but we could equally well have been using a (V/N,A/N) dia-gram.

14 This would, however, change the definition domain of the abscissafrom ρr ∈ 〈0, 3〉 to vr ∈ 〈13 ,∞〉 which is a little impractical.

15 At the same time the values of the slope and the intersection wouldswap (consider it a bonus task to show this) without having any negativeside effects.

774 / 1776

§ 1121 At 101 K nitrogen has the following saturation state: p = 8.344bar, ρlıq = 24.360 mol dm-3 and ρvap = 1.222 mol dm-3. The criticalpoint exists at pc = 34.000 bar, ρc = 11.210 mol dm-3 and Tc = 126.2 K.Calculate the saturation state expressed in reduced coordinates anddraw your result onto the graph pVdW

r = f(Tr, vr); also see Theme 49on page 328. Comment on the discrepancy between the calculatedvolumes of vapour and liquid and the experimental values.

775 / 1776

§ 112 Corresponding state1 The results calculated from the Van der Waals equation have beendrawn in reduced coordinates in Figure 14.3 on the preceding pageusing the program in Appendix 31:1.8.

2 Theoretically, the two experiments should fit the lowest isothermbut they are in this example located outside the calculated two-phasedomain.

3 This means that the priciple of corresponding states is not an exactscience.

4 Note in particular that the Van der Waals equation overestimatesthe volume of N2 in the liquid phase.

5 This is a weakness that is common to all cubic equations of state.

6 In the vapour phase the discrepancy is relatively speaking smaller,because there the molar volume is greater than in the liquid phase.

7 Also note that the areas above and below the hatched phaseboundary line at Tr = 0.8 based on experimental values are roughlyequal.

8 This is not a coincidence, and for a correctly calculated phase equi-librium the Maxwell equal area rule states that the two areas will beidentical.

776 / 1776

1 2 Maxwell’s area rule gives an elegant depiction of the thermody-namic criterion for phase equilibrium. Assume there exist a liquid phase brought to its boiling point by

external means.

5 3 We are next expanding the volume of this liquid (at constant tem-perature) thereby gradually changing its character from liquid to vapour.4 In practise the liquid would evaporate through boiling but we shalldo this experiment in our minds only and with a mathematical model toour disposal. The isothermal state change is then described by: (dµ)T ,N = v dp.

6 By integrating the state from the saturated liquid to a saturatedvapour we get

µvap∫

µlıq

(dµ)T ,N =pvap∫

p lıq

v dp = 0 ,

assuming that the isothermal equilibrium state is governed by pvap =

p lıq and µvap = µlıq.



14 Simple . . .Maxwell equal area rule I (2)

7 This being the case proves that the net area of the v(p) graphwill be equal to zero when integrating the volume from one saturationpressure to another. 8 The construction is conceptually simple but not very practical[a].

In this chapter we have been dealing exclusively with Helmholtz energyand it would be better if the representation is based on a p(v) graph.9 That variable change is easily undertaken using integration by

parts like it is discussed in the paragraph below.

[a] If you rotate Figure 14.3 ninety degrees in the mathematical sense, turns the pageand holds it up against the light, you will understand why. The v(p) graph forms apretty manifold which unfortunately is hard to exploit mathematically.



14 Simple . . .§ 113 § 112 § 114

The Maxwell equal area rule can be expressed as∫ Vvap

V lıq (p − psat)(dV)T ,N = 0 and be plotted on a figureequivalent to 14.3, but with the ordinary volume along

the abscissa. Show that the proposed equation actuallyholds for a one-component phase equilibrium.



14 Simple . . .§ 113 Maxwell equal area rule II

1 There are at least two alternative ways of deriving this relationship.

2 Integration by parts is one. Applied to the Maxwell area rule weget:

0 =pvap∫

p lıq

v dp = (vp)∣∣∣vap

lıq−

vvap∫

v lıq

p dv =(vvappvap − v lıqp lıq

)−

vvap∫

v lıq

p dv .

3 Introducing the equilibrium condition p lıq = pvap = psat and writingthe difference in the phase volumes as an integral:

pvap∫

p lıq

v dp = psat

vvap∫

v lıq

dv −vvap∫

v lıq

p dv = −vvap∫

v lıq

(p − psat) dv = 0 .

4 At constant composition dV = N dv which means we can also write(the negative sign is not important here):

Nvvap∫

v lıq (p − psat)dv =vvap∫

v lıq (p − psat)dV = 0 .



14 Simple . . .Vapour–liquid equilibriumENGLISH text is missing.Maxwell equal area rule

III1 Alternatively we may recognise −p as the partial derivative

of Helmholtz energy with respect to the volume along the chosenisotherm.

2 The integral along the isotherm can then be expressed with thefollowing:

Vvap∫

V lıq

(p − psat) (dV)Tsat,N = −Vvap∫

V lıq

[(∂A∂V

)T ,n + psat

]dV

= A lıq − Avap − psat

(Vvap − V lıq)

?= 0 . (14.12)

3 Since the system only has one chemical component, it must alsobe true that A = −pV + µN. Moreover, when the vapour and liq-uid phases are in equilibrium, plıq = pvap = psat. Hence the equalityin 14.12 can be simplified to

(µN)lıq ?= (µN)vap . (14.13)

4 The integration has been performed at constant temperature Tsat

and total composition N. Hence, as in 14.13, the mole number canbe considered a constant (common) factor for the vapour and liquidphases.

5 It then follows that

µlıq = µvap ,

which is identical to the phase equilibrium criterion for a one-componentsystem.

782 / 1776

1 2 At thermodynamic equilibrium, T , p and µ have uniform values overthe entire phase ensemble, which in this case is made up of the vapourand liquid phases.3 Since the Van der Waals equation is expressed on a pressure ex-

plicit form in terms of pr(Tr, vr), the equilibrium equations have to besolved simultaneously in vvapr , v lıqr coordinates.4 We can’t expect there to be an analytical solution of this (non-

linear) 2 × 2 equation system, but there is no harm in setting out theequations. The fact that the pressure is the same in both phases means that

8Tr

3v lıqr −1− 3

(v lıqr )2

= 8Tr

3vvapr −1

− 3(vvap

r )2 ,

which can be solved for temperature:

8Tr = 6 x2−y2

2x3−x −

2y3−y

. (14.14)

5 Here x ∈ [1,3〉 and y ∈ 〈0,1] are reduced (molar) densities definedby x = (v lıq

r )-1 and y = (vvapr )-1.

6 Similarly, the fact that the chemical potential is the same in bothphases means that

Tr ln( 8Tr

3v lıqr −1

)+ Tr

3v lıqr −1− 9

4v lıqr

= Tr ln( 8Tr

3vvapr −1

)+ Tr

3vvapr −1

− 94vvap

r.



14 Simple . . .Vapour–liquid equilibrium (2)

7 If we substitute in the temperature from Eq. 14.14 and rewrite theequation in terms of molar density, the chemical potential equation canbe expressed as

ln(x(3−y)y(3−x)

)+

(1 − 6

x+y

)3(x−y)

(3−x)(3−y) = 0 . (14.15)



14 Simple . . .Pressure–potential diagram

1 In order to determine the saturation state of the fluid, we need tosolve Eq. 14.15 for y = y(x) or x = x(y).

3 2 The latter was implemented in the Function 2.5 shown in Ap-pendix 31, which calculates x ∈ [1,3〉 for a given value of y ∈ 〈0,1]. The solution is shown graphically in Figure 14.4 on the next page

along with a corresponding (pr, µr)-diagram Expanded footnote [a]:

[a] Drawn as the parametrised graph pr(vr), µr(vr) where the volume increases fromthe top right corner (liquid) down to the bottom left corner (gas) of the figure. .

6 4 In the last figure the simultaneous solution of p lıqr = pvapr and µlıqr =

µvapr appears as the triple point of a manifold with three regions: Vapour,liquid and an unphysical fluid phase for which (∂p/∂v)T > 0.5 The figures were drawn using the Program 31:1.10. Note that the slope of the graph asymptotically approaches −1 as

it nears the critical point, where xc = yc = 1. 7 This means that the liquid density decreases at the same rate asthe vapour density increases at the far right in the left sub-figure. The average density of

the two phases therefore has the nearly constant value 12 (x + y) ≃ 1. 8 Experimentally it has been demonstrated that the value of x + y is

not quite constant, but nevertheless it is virtually a linear function (henceit is also referred to as the rectilinear density) of Tr over a wide range oftemperatures.



0 0.5 11

1.5

2

2.5

3

−1 0 1 2−1

0

1

2

3Molar density

y = ρvapr = vc/vvap

x=ρ

lıq r=

v c/ v

lıq

Pressure–chem.pot.

pr = p/pc

µr=µ/R

Tc

c.p.

Figure 14.4: ρvapr , ρlıq

r saturation graph, with the associated isotherms, ex-pressed in pr, µr coordinates (drawn with volume as the parameter).



14 Simple . . .Lower temperature limit ENGLISH text is missing.Enthalpy of vaporization1 The last piece of information needed is the enthalpy of vaporization

from Eq. 14.11 expressed on density form.

2 Not because it is required right now but in order to make thevapour–liquid formulae complete:

∆vhVdWr = (y − x)

( 3Tr(3−x)(3−y) −

94

). (14.16)

3 This is as far as we can get by looking at the general case, so wewill now move on to a discussion of the asymptotes of the left graph inFigure 14.4.

4 The discussion will go on for next two sections but there are manydetails and in order to appease the impatient reader we refer immedi-ately to the final results in Table 14.7.

5 The table summarises the main results from the coming discussionof the lower and upper temperature limits Tr ≪ 1 and Tr → 1.

786 / 1776

1 2 At temperatures far below the normal boiling point of the liquidphase, the molar density approaches a constant value of limTr≪1(x) = 3while the corresponding molar vapour density y tends to zero. At sufficiently low temperatures the reduced densities can therefore

be expressed as x = 3 − α and y = β, where α and β are small positivenumbers. When substituted into the equilibrium Eq. 14.15 this gives us

0 = ln((3−α)(3−β)

βα

)+

(1 − 6

3−α+β) 3(3−α−β)

α(3−β) ≃ ln( 9βα

) − 3α

if only the dominant terms are included.

3 The asymptotic solution of β is

limTr≪1

β = 9

α·exp(−3α )

, (14.17)



14 Simple . . .Lower temperature limit (2)

6 4 which shows that the vapour density declines exponentially withfalling temperature.5 In that limit case, the temperature Eq. 14.14 can be simplified as

limTr≪1(8TVdWr ) = 6 (3−α)2−(β)2

2(3−α)α − 2β3−β ≃ 9α , (14.18)

and if we substitute that into the Van der Waals equation 14.5 we get

limTr≪1 pVdWr = 8Tr3β−1 − 3β2 ≃ 9αα3 exp( 3α)−1 ≃ 27exp( 3α) . (14.19) Combining these latter two equations allows us to eliminate α, giv-ing us the temperature-explicit form:

limTr≪1

ln(pVdW

r

)= ln(27) − 27

8Tr

which conforms to the Clausius–Clapeyron vapour pressure Eq. 14.3,with a slope of −b = −27

8 = −∆vhr.

7 This result is confirmed by explicitly calculating the heat of vapor-ization due to Eqs. 14.16 and 14.18:

limTr≪1

∆vhVdWr = (β − 3 + α)

(3Tr

α(3−β) −94

)

≈ −(3 − α)(

Trα − 9

4

)

≈ 278 .



14 Simple . . .Lower temperature limit (3)

8 In practice, the low-temperature limit can never be reached, be-cause most liquids freeze at a temperature of Tr ≈ 0.7. Table 14.5 onpage 774 therefore gives a comparison with the heat of vaporization fora few simple substances Expanded footnote [a]:

[a] Robert C. Reid, John M. Prausnitz, and Thomas K. Sherwood. The Properties ofGases and Liquids. McGraw-Hill, 3rd edition, 1977. measured at the normal boiling point. 9 As can be seen from the table, the correlation between the two is

no more than reasonable.10 Confer also Table 14.7 to get an overview of the main results ob-tained in this section.



14 Simple . . .§ 114 § 113 § 115

Use the information available to calculate limTr≪1(pr ·∆vv).Afterwards, calculate the derivative in Eq. 14.1 at the

beginning of this chapter. Compare the result withEqs. 14.2 and 14.4.



14 Simple . . .§ 114 Clapeyron’s equation I

1 As previously mentioned, when the temperature approaches zero,the specific volume of the liquid tends to a ratio of limTr≪1(vVdW

r ) = 13 .

2 In the same limit, the volume of the vapour phase will completelydominate the vaporization volume:

limTr≪1

∆vvr = vvapr − v lıq

r = 1β − 1

3−α ≈ 1β .

Combined with the Van der Waals Eq. 14.6 and the temperaturelimit 14.18 we get:

limTr≪1

(pr ·∆vvr) ≈ prβ = 8Tr

3−β − 3β ≈ 8Tr3 = 3α .

3 Combined further with Eq. 14.1:

limTr≪1

( dln prd(1/Tr)

)VdW

∆µ=0= − RTc

pcvc· Tr

pr· ∆vhr∆vvr≈ − 8

3 ·Tr3α · 27

8 = − 278 .



14 Simple . . .§ 114 Clapeyron’s equation I (2)

4 If we combine this with the Clausius–Clapeyron vapour pressureEq. 14.2–14.4 we get the slope limTr≪1 b = −∆vhr = − 27

8 .



Figure 14.5: Enthalpy of vaporization measured at the normal boiling point ofa few selected substances compared with ∆vhVdW

r = 278 = 3.375.

Tb[K]

Tc[K]

∆vh[cal/mol]

∆vhRTc

∆vhVdW

[cal/mol]

N2 77.35 126.2 1333 5.32 846Ar 87.27 150.8 1560 5.21 1011O2 90.17 154.6 1630 5.31 1037

CH4 111.66 190.4 1955 5.17 1277Kr 119.74 209.4 2309 5.55 1404

H2O 373.15 647.3 9717 7.55 4341



14 Simple . . .Upper temperature limit

1 2 At low temperatures the liquid molar density approaches a constantvalue of limTr≪1(x) = 3 while the corresponding molar vapour density ytends to zero. As the temperature increases, the liquid phase gradually becomes

less significant, until it ceases to exist entirely at the critical point wherelimTr→1(x) = limTr→1(y) = 1.

5 3 In this state there is no fundamental difference between a vapourand a liquid.4 This idea is supported by Figure 14.6 on page 782, which showsthat the density of the two equilibrium phases is symmetrical about xc =

yc = 1 close to the critical point. Near the critical point, the reduced densities can be expressed asx = 1 + α and y = 1 − β, where α and β are small positive numbers. Ifwe substitute these values into the equilibrium Eq. 14.15 we get

ln((1+α)(2+β)(1−β)(2−α)

)+

(1 − 6

2+α−β) 3(α+β)(2−α)(2+β) = 0 .



14 Simple . . .Upper temperature limit (2)

7 6 Now let us expand ln(1+x) = x +O(x2) and multiply out the termsin brackets. We can ignore all higher order terms of the kind αβ, α2 and β2:

0 = α+ 12 β − (−β − 1

2 α) +(1 − 6

2+α−β) 3(α+β)

4−2α+2β−αβ

≃ 1 +(1 − 6

2+α−β)

12+β−α

= 4 − (β − α)2 + 2 + (β − α) − 6

≃ β − α . (14.20)

8 The conclusion is that β ≈ α, which shows that the vapour andliquid densities are symmetrical about the point xc = yc = 1.




10 9 This is an important result, which has also been confirmed exper-imentally[a] [a]. Confer also the previous Figure 14.4 on page 767 andthe next Figure 14.6 (cf. Program 1.11 in Appendix 31).

[a] The critical volume of a fluid can be estimated using straight-line extrapolation ofthe sub-critical vapour and liquid densities, while the critical pressure and temperaturecan be measured directly.

[a] Molar entropy displays a similar symmetry; see Figure 18.9 on page 975. By substituting into the temperature Eq. 14.14 we can obtain thelimit value

limTr→1

(8TVdWr ) = 6 (1+α)2−(1−α)2

2(1+α)2−α −

2(1−α)2+α

= 2(4 − α2) , (14.21)

which can in turn be substituted into the Van der Waals equation 14.5,to give us

limTr→1

(pVdWr ) = 8Tr

3x −1− 3x2 =

2(4−α2)3

1+α−1− 3(1 + α)2 = 1 − α2 . (14.22)

11 Here we should note that both TVdWr and pVdWr are simple functionsof α2.12 In other words, the phase envelope psatr (ρr) calculated in Fig-ure 14.6 is parabolic with the critical point as its origin. If α2 is eliminated from the last two equations, the saturation pressurecan instead be written as a function with temperature as its free variable:

limTr→1

(pVdWr ) = 4Tr − 3 .

13 The enthalpy of vaporization has a similar representation.




14 Substituting x = 1 + α and y = 1 − β where limTr→1(β) = α intoEq. 14.16 yields

limTr→1

∆vhVdWr = −(α+ β)

(3Tr

(2−α)(2+β) −94

)= −2α

(3Tr

4−α2 − 94

),

15 which due to limTr→1(8Tr) = 2(4 − α2) simplifies to:

∆vhVdWr = −2α

(3Tr4Tr− 9

4

)= 3α .

16 Or, we can write the enthalpy expression on a temperature explicitform:

limTr→1

∆vhVdWr = 3α = 3

√4 − 4Tr = 3

√1 − Tr . (14.23)

17 The well-known correlation due to Watson[a] states that∆vhWatson = a(1 − Tr)b where b ≈ 0.38 for ordinary nonpolar organiccompounds, goes a long way towards confirming this result.

[a] K. M. Watson. Ind. Eng. Chem., 35:398, 1943.



14 Simple . . .§ 115 § 114 § 116

Use the information available to calculate limTr→1(pr ·∆vv).Afterwards, calculate the derivative in Eq. 14.1 at the

beginning of this chapter. Compare the result withEqs. 14.2 and 14.4.



14 Simple . . .§ 115 Clapeyron’s equation II

1 Close to the critical point, the reduced densities of the two phasesare x = 1 + α and y = 1 − β:

4 2 From here, the derivation closely follows that given in the main textabove:

limTr→1(∆vvVdWr ) = vvapr − v lıqr = 11−β − 11−α ≈ 2α(1−α)2 = 2αpr = 23 · ∆vhrpr

3 See Eqs. 14.22 and 14.23 for more details. In particular you should note the expression to the far right whichtells us that the vaporization volume is proportional to ∆vhr/pr.

7 5 The denominator cancels thereby with the multiplication of pr

limTr→1(pr ·∆vvVdWr ) = 2∆vhr3 ,

6 which is the direct cause for that the vapour pressure curve in Fig-ure 14.1 looks almost straight. The central issue is that the right-hand side of Eq. 14.1 approaches

a constant value even though the volume of vaporization and the en-thalpy of vaporization are both tending towards zero:

limTr→1

( dln prd(1/Tr)

)VdW

∆µ=0= − RTc

pcvc· Tr

pr· ∆vhr∆vvr≈ − 8

3 · Tr · 32 = −4Tr = −(4 − α2) .

8 If we combine this with the Clausius–Clapeyron vapour pressureEq. 14.2–14.4 we get the slope limTr→1 b = −4T2

r ≥ −4. 9 From Section 14.6 we know that limTr≪1 b = −∆vhr = −27/8. Thedifference is surprisingly small.



14 Simple . . .Slope of vapour pressure

1 In other words, the predicted slope of the vapour pressure curvevaries in the range −4 < −b < − 27

8 .

3 2 This is what the theory predicts but at the small scale of Figure 14.1the variation in the slope is hardly noticeable for the experimental datapoints either, and the graph appears to be a straight line. The Van der Waals equation therefore gives us a qualitatively ac-

curate representation of the vapour pressure curve, although from ex-perience we know that the model is insufficiently accurate for quantita-tive purposes.

4 As seen in Figure 14.6 the one-phase region is not adequatelyrepresented either, but once again it is qualitatively accurate Expanded footnote [a]:

[a] W. Duschek, R. Kleinrahm, and W. Wagner. J. Chem. Thermodynamics, 22:841–864, 1990. .



0 0.5 1 1.5 2 2.50

1

2

3

ρr = vc/v

p r=

p/p

c

Figure 14.6: The saturation pressure of CO2 along with selected isothermsin the one-phase region (280 K, 313 K and 330 K respectively). Note the sym-metry of the saturation pressure psat

r around the critical density ρr = 1.



14 Simple . . .Near-critical densityENGLISH text is missing.Cubic equations of state1 As such, only small changes are needed to improve the predictive

accuracy.

2 In 1949, Redlich and Kwong[a] introduced the first improvement bymaking the a term temperature-dependent.

3 There are now around a hundred closely related models, with theRedlich–Kwong–Soave[a] (SRK) and Peng–Robinson[a] (PR) modelsbeing two of the best-known ones.

4 They are all cubic functions of the volume (and thereby of density),and their qualitative properties are the same.

5 This means that the Van der Waals equation of state still remainsan important conceptual model— 140 years after it was formulated thefirst time.

803 / 1776

[a] Otto Redlich and J. N. S. Kwong. Chem. Rev. (Washington, D. C.), 44:233–244,1949.

[a] Giorgio Soave. Fluid Phase Equilib., pages 1197–1203, 1972.

[a] Ding-Yu Peng and Donald B. Robinson. Ind. Eng. Chem. Fundam., 15(1):59–64,1976.

1 The volumes of the saturated phases can be expressed as x =

(vr)-1 = 1 ± α where α is a small number, and from Eq. 14.21 we can

then derive the asymptotic temperature function:

limTr→1

(vVdWr ) = 1

1±α = 11±√

4−4Tr≃ 1 ∓ 2

√1 − Tr .

2 This makes the logarithm of the saturation volume of the liquidphase

limTr→1

(ln(v lıqr )VdW) ≃ −2

√1 − Tr ,

while the Rackett Expanded footnote [a]:

[a] Harold G. Rackett. J. Chem. Eng. Data., 15(4):514–518, 1970. equation, which forms the basis for several semi-empirical correlations, states that ln(v lıq

r )Rackett = ln(zc) · (1 − Tr)2/7

where ln(zc) ≈ −1.2 for nonpolar substances. 3 The parameters in the two equations are different, but qualitativelythere is still a real correspondence between theory and practice[a].4 Confer also Table 14.7 to get an overview of the main results ob-tained in this section.

[a] The Van der Waals equation indicates that the function should take the form a(1 −Tr)b and also gives us good initial values that can be adjusted to reflect experimentaldata (both the proportionality constant ln(zc) and the exponent 2/7 in the Rackettequation are empirical).



Figure 14.7: Asymptotic solutions of the vapour–liquid phase equilibrium.

Tr ≪ 1 Tr → 1

Property Definition Approximation Definition Approximation

ρlıq

r 3 − α 3 − α 1 + α 1 + α

ρvapr β 9

α · exp (−3α ) 1 − β 1 − α

Tr − 98 α − 1

4 (4 − α2)

pr − 27 · exp (−3α ) − 1 − α2

v lıqr

13−α

13

(1 + α

3

)1

1−α 1 + α

vvapr

1β

α9 · exp (3

α)1

1−α 1 + α

pr ·∆vvr − 3α − 2α

∆vhr − 278

(1 − α

3

)− 3α

∂ ln pr

∂(1/Tr)− − 27

8 − −(4 − α2)



14 Simple . . .2nd order phase transitionENGLISH text is missing.Near-critical states1 In Theme 49 on page 328 we showed that the Van der Waals equa-tion can be written in its reduced form without the use of any arbitrarymodel parameters.

2 The underlying theory is based on the critical properties of the fluidunder the conditions (∂p/∂v)T = 0 and (∂2p/∂v∂v)T = 0, calculated atT = Tc and v = vc, and of course also at p = pc.

3 This produces two non-linear equations that contain the model pa-rameters a and b .

4 Conventionally, the equations are solved with respect to a and bas functions of Tc and pc.

5 This results in a useful form of the equation of state, which wehave used in this chapter to analyse the fluid’s behaviour in a numberof interesting limit cases.

6 Unfortunately, Figure 14.6 shows that the theory is by no meansuniversally applicable.

7 In this section we will look in greater detail at what happens close tothe critical point, and quantify to what extent the Van der Waals equationfails to predict experimental results.

806 / 1776

ENGLISH text is missing.Order parameter1 It has been theoretically proven— and experimentally confirmed—that the energy surface close to the critical point of a vapour–liquidequilibrium is of a universal form, which is parametrised by what we callan order parameter.

2 The order parameter reflects the characteristic behaviour of thefluid in phase transitions.

3 By definition it is zero at the critical point and in the surroundingone-phase region.

4 In two-phase regions it gradually increases with the distance fromthe critical point.

5 For reasons of symmetry, we wouldn’t expect the order parameterto differ between vapour–liquid and liquid–vapour transitions.

6 There is absolutely no reason for it to do so, since there is nothermodynamic distinction between a vapour phase and a liquid phase.

7 They are both fluids and nothing more.

8 For vapour–liquid equilibria, it is natural to use∣∣∣ρr − 1

∣∣∣ as an orderparameter[a], provided that it produces symmetrical phase transitions,which it indeed does.

9 In fact, that order parameter is precisely what explains the rectilin-ear density graph mentioned in conjunction with Figure 14.4.

807 / 1776

[a] The theory applies to all phase equilibria that have a single (scalar) order parameter.For example, ferromagnetic materials display relatively analogous behaviour, whichis described by the order parameter |Mr − 1|, where M is the magnetisation of thesubstance.

1 2 Let us assume we can do measurements along the fluid’s criticalisochore moving towards the critical point.3 In the direction of that point, but still outside the two-phase region,

it is possible to observe a gradual phase transition. The closer we measure towards the critical point, the more anoma-lous the p(T ,V) behaviour of the fluid.

4 This behaviour is called a λ transition or a 2nd order phase transi-tion.

6 5 The name λ transition comes from the variation in heat capacity atconstant volume as a function of the distance to the critical point: WhenTr − 1 is used as the measure of distance from the critical point, theshape of the graph is reminiscent of the Greek letter lambda. In the literature for this field, which incidentally covers a large and

exciting range of research Expanded footnote [a]:

[a] Kenneth G. Wilson received the Nobel prize in 1982 for his contribution to thedevelopment of renormalisation groups applied to, amongst other things, critical phasetransitions. , the critical anomalies of the fluid are ex-

pressed using a clever choice of 4 asymptotically divergent correlations:

One phase, ρr = 1 : cv ∝ (Tr − 1)-α , (14.24)

One phase, ρr = 1 :(∂ρr

∂pr

)Tr∝ (Tr − 1)-γ , (14.25)

Critical, Tr = 1 : |pr − 1| ∝ |ρr − 1| +δ , (14.26)



14 Simple . . .2nd order phase transition (2)

Two phases, Tr < 1 : |ρr − 1| ∝ (1 − Tr)+β . (14.27)



Figure 14.8: Critical exponents measured experimentally for selected com-pounds compared with the classical values calculated from the Van der Waalsequation of state.

System α γ δ β

Van der Waals 0 1 3 0.5Ising 3D [†] 0.110(1) 1.237(2) 4.789(2) 0.326(5)Heisenberg 3D [†] −.1336(2) 1.3960(9) 4.783(3) 0.3689(3)Ni, Fe, Gd2BrC, Gd2IC,Tl2Mn2O7, . . . [†]

−.1336(2) 1.3960(9) 4.783(3) 0.3689(3)

3 He, 4 He, Xe, CO2,H2O, O2 [†]

0.10(0) 1.19(0) 4.35(0) 0.35(5)

[†] J. M. H. Levelt Sengers and J. V. Sengers. Phys. Rev. A, 12(6):2622–2627, 1975.



14 Simple . . .Looking backENGLISH text is missing.Critical exponents1 In the next four sections we will derive the classical values of the

exponents α, γ, δ and β, based on the properties of the Van der Waalsfluid.

2 However, the results that we will obtain are not only true for theVan der Waals equation, they are also valid for all explicit equations ofstate, regardless of whether or not they are cubic.

3 This is because we are only interested in the exponents in theabove correlations. The proportionality factors are of no significance inthis respect.

4 It turns out that the exponents are the same, whether you start outfrom the Van der Waals equation or from a more advanced (modern)equation of state.

5 The reason is that a Taylor series expansion of the critical proper-ties gives the same classic exponents independent of the exact form ofthe equation of state.

6 This classification is also known as the Landau[a] theory.810 / 1776

[a] Lev Davidovich Landau, 1908–1968. Russian theoretical physicist.

ENGLISH text is missing.Isochoric heat capacity

(α)1 From our knowledge of thermodynamic theory, and particularly

from chapters 4 and 8 on Legendre transformations and on the thermo-dynamics of systems of constant composition, we know the general re-lationships (∂A/∂T)V ,n = −S and (∂S/∂T)V ,n = CV/T .

2 These equations can also be expressed in reduced coordinates as(∂ar/∂Tr)vr = −s/R and (∂s/∂Tr)vr = cv/Tr.

3 Eliminating s gives us cv = −RTr (∂2ar/∂Tr∂Tr)vr . In other words,the heat capacity of the fluid gives us the second derivative of Helmholtzenergy.

4 For a Van der Waals fluid, we can obtain the Helmholtz energy bycombining Eq. 14.8 with equations 14.6 and 14.7.

5 Double differentiation of ar with respect to Tr gives us

cVdWv = −RTr

(( ∂2µr∂Tr∂Tr

)+ 1

Tr

)= cp − R = cv (T) .

6 Note that the temperature derivative of µ gives us cp rather thancv .

7 This is because the standard state is defined at constant p, whichimplies that v(T , p) is a variable with respect to temperature.

8 This topic is discussed more thoroughly in the context of Eq. 7.9 onpage 279.

9 We can therefore conclude that αVdW = 0 in Eq. 14.24.

10 This assumes that the heat capacity in the standard state behaves“normally” close to the critical temperature of the fluid, but this is verylikely to be the case, given that cv refers to an ideal gas state.

11 After all, the critical state is a result of the non-ideal nature of thefluid, and the standard state doesn’t tell us anything about non-idealconditions.

12 Although cv is a function of temperature, it does never display anytemperature divergence of the type (Tr − 1)-α.

13 cv can therefore be considered a constant in this context.811 / 1776

ENGLISH text is missing.Bulk modulus (γ)1 The Van der Waals equation expressed in reduced Tr, vr coordi-

nates was discussed in Theme 49. Here we need the same equation,but this time as a function of Tr and ρr:

pVdWr = 8Tr

3vr−1 − 3v2r

= 8Trρr

3−ρr − 3ρ2r

2 Let us differentiate the pressure with respect to density, and rear-range the resulting equation in such a way as to isolate Tr − 1 in one ofthe numerators:

limρr→1

(∂pr∂ρr

)VdW

Tr= 24

( Tr−1(3−ρr)2 + 1

(3−ρr)2 −ρr

4︸︷︷︸

→0

)= 6(Tr − 1)

3 The exponent of the expression on the right side of the equation is1 which means that γVdW = 1 (not −1) in Eq. 14.25.

4 The expression above is written for the bulk modulus whileEq. 14.25 is valid for the isothermal compressibility. These are mu-tually resiprocal quantities which explains the sign shift of the exponent.

812 / 1776

ENGLISH text is missing.Critical isotherm (δ)1 We will start out from the Van der Waals equation expressed as a

function of Tr and ρr, just as we did for the bulk modulus.

2 Along the critical isotherm, the equation of state can be factored asfollows:

limTr→1

(pVdWr − 1) = 8Trρr

3−ρr − 3ρ2r − 1

= 33−ρr (ρ

3r − 3ρ2

r + 3ρr − 1)

= 33−ρr (ρr − 1)3

3 That gives us δVdW = 3 in Eq. 14.26. In other words, the criticalisotherm describes a third-degree curve with its origin at the criticalpoint.

4 This is why the Van der Waals equation and other equations of thesame familiy are sometimes called “cubic equations of state”.

813 / 1776

ENGLISH text is missing.Binodal curve (β)1 Based on our discussion of phase equilibrium problems in Sec-

tion 14.5, and with particular reference to Eq. 14.22, it follows that:

2 (ρr − 1) = ±(1 − Tr)0.5 .

2 The vapour pressure curve describes a parabola with the order pa-rameter ρr−1, which means that the exponent βVdW = 0.5 in Eq. 14.27.

814 / 1776

ENGLISH text is missing.Critical point measure-

ments1 Table 14.8 gives a summary of what we have learned so far about

phase transitions and how they relate to the Van der Waals equation.

2 As can clearly be seen from the table, the equation of state doesnot accurately describe what happens in real life, but we knew thatalready.

3 What is surprising is that the experimental values do not diverge;instead they suggest that physical quantities exhibit certain universalbehaviour during phase transitions.

4 Here we should note that all the exponents α, γ, δ and β varysurprisingly little, both in response to changes in the chemistry of thesystem and to the physical interpretation of the order parameter.

5 Intuitively you would expect there to be a big difference betweenvapour–liquid phase transitions and ferromagnetism, but it turns outthat the critical exponents of these phenomena are roughly the same.

6 We are clearly standing at the threshold of a new understanding ofthe world.

7 However, the theory that underpins it is unfortunately beyond ourreach, and we will therefore have to limit ourselves to a verbal discus-sion of the topic.

816 / 1776

ENGLISH text is missing.Nature is non-analytical1 What we can say for sure is that there is no equation with an ana-

lytical solution that accurately describes critical phenomena in nature.

2 In other words, the critical exponents are neither integers (z ∈ Z)nor simple fractions (q ∈ Q). That is what they would need to be forthere to be any hope of finding an equation of state with an analyticalsolution capable of describing the critical phase transition behaviour.

3 This is because a series expansion of such an equation wouldalways produce exponents that are either rational numbers or integers.

4 Nature is not like that, and we must therefore give up on tryingto find an all-encompassing universal equation to describe the laws ofnature.

817 / 1776

ENGLISH text is missing.Correlation length1 Our next observation is loosely related to statistical thermodynam-

ics and to the knowledge that some models are easier to deal with thanothers.

2 The simplest special case is a gas in which the molecules are con-stantly moving around without colliding or interacting with each other inany way. That is what we mean by an ideal gas.

3 We will assume that the correlation length of the gas is zero, evenat the risk of claiming to know something that we haven’t verified.

4 The second special case is a system in which the molecules (oratoms) are arranged in a repeating lattice with long-range order. Thatis what we mean by a crystalline structure.

5 In one sense, the correlation length of the crystal is infinite. We donot say that the oscillations of the particles are strongly correlated butthat the position of each single particle is known once the unit cell ofthe crystal is known.

6 The third special case only arises in conjunction with phase transi-tions.

7 Here the correlation length varies over several orders of magnitude,which means that this phenomenon is much more significant than thecontributions from the intra-molecular states of the particles.

8 At the critical point, the correlation length increases to a macro-scopic level and to thereby interfere with the wavelength of visible light.

9 This phenomenon is called critical opalescence.

10 It induces a characteristic light scattering that results in an array ofbeatifully tinted colours when white light hits a sample that is in a near-critical state.

818 / 1776

ENGLISH text is missing.Universal scaling1 In the critical state, there is a balance between the thermal energyand the potential (inter-molecular) energy of the fluid.

2 This results in the energy surface stretching out and losing its cur-vature. Both the first and second order contributions vanishes thereby.

3 Consequently, it is impossible for there to be a classic static equi-librium, as there appears to be at other pressures and temperatures,judging by observations.

4 The compressibility diverges (tends to infinity) and the fluid appearsto be constantly moving, without ever quite coming to rest.

5 One practical consequence of these conditions is that the force ofgravity becomes relatively more important, leading to relatively largevariations in density that can distort experimental results.

6 Measurements have therefore been taken under zero-gravity con-ditions in order to validate current theories relating to the field.

7 All observations seem to indicate that theory and practice agree,but that the Van der Waals equation (like other equations of state withanalytical solutions) produces inaccurate estimates of the critical expo-nents.

8 However, the principle that the critical phenomena display univer-sal scaling behaviour holds true, and there is in fact a general rule thatapplies to critical phase transitions in all substances, as stated in equa-tions 14.24–14.27. Moreover, we also know that

α = 2−β(δ+ 1)

γ = β(δ − 1)

The exponents summarised in Table 14.8 are judiciously adjusted (us-ing the least square principle) to fulfill these relations.

819 / 1776

1 Saturation state:(dpdT

)∆µ=0 = svap−s lıq

vvap−v lıq = ∆vs∆vv = ∆vh

T∆vv

2 Clausius–Clapeyron: limp→0

( ∂ ln p∂(1/T)

)∆µ=0

= − ∆vhR

3 Rectilinear density: ρlıqr + ρvap

r ≈ 1

4 Maxwell’s area rule:∫ Vvap

V lıq (p − psat)(dV)T ,N = 0

5 Van der Waals equation of state:pVdW

pc= pVdW

r = 8Tr3vr−1 − 3

v2r

µVdW

RTc= µVdW

r = µr − Tr ln pr + Tr ln( 8Tr3vr−1

)+ Tr

3vr−1 − 94vr



14 Simple . . .Looking back (2)

6 Critical point behaviour:

One phase, ρr = 1 : cv ∝ (Tr − 1)-α

One phase, ρr = 1 :(∂ρr

∂pr

)Tr∝ (Tr − 1)-γ

Critical, Tr = 1 : |pr − 1| ∝ |ρr − 1| +δ

Two phases, Tr < 1 : |ρr − 1| ∝ (1 − Tr)+β

7 Classical values for the exponents: α = 0, γ = 1, δ = 3 and β = 0.5


News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi Convergence orde

Part 15

Multicomponent phase equilibrium




15 Multicomponent . . .New concepts

1 Thermodynamic phases

2 Minimum (Gibbs) energy

3 Fugacity versus activity coefficient

4 Direct substitution

5 Newton–Raphson iteration

6 K values

7 Fugacity coefficients

8 Activity coefficients

9 Order of convergence



15 Multicomponent . . .Minimum Gibbs energy IENGLISH text is missing.Minimum energy princi-

ple1 At thermodynamic equilibrium the system has reached a state of

minimum energy.

2 This far-reaching postulate is of tremendous importance for ourunderstanding of matter and energy, but it remains to discuss which ofthe energy functions that are being minimised.

3 In Chapter 16 on page 851 it is for example proved that Gibbsenergy of a closed system decreases (due to chemical reactions) tillthe minimum value minn G(n)T ,p is reached.

4 This is generally true when the temperature and pressure are con-stant.

5 If the volume is kept constant and the pressure is varied theHelmholtz energy will be minimised.

6 Chapter 14 on page 741 illustrates this for the case of a simplevapour–liquid equilibrium.

7 The other energy functions U, H, etc. are minimised at constantvalues of their respective canonical variables.

824 / 1776

1 In general we shall investigate a closed system consisting of π ∈ α,β, . . . , ψ, ω phases Expanded footnote [a]:

[a] Phase α is usually the low temperature phase and β, . . . , ψ, ω are phases that be-come stable at successively increasing temperatures. and a set of i ∈ 1,2, . . . ,n components which are

common to all the phases Expanded footnote [a]:

[a] Reacting systems with (maybe) disjoint sets of the components in each of thephases is taken up in Chapter Phase Reactions. .

2 For the special case of given temperature and pressure

(dG)T ,p,nα+nβ+··· =ω∑π=α

n∑i=1µπi dNπ

i = 0 ,

is a necessary condition for thermodynamic equilibrium.

3 We note that dNπi are not independent variables in the differential

of G, but that they are interconnected by the n component balances ofthe system.



15 Multicomponent . . .Minimum Gibbs energy I (2)

4 From the component balances it follows that∑ωπ=α Nπ

i = Ni andeven

∑ωπ=α dNπ

i = 0 because the total moles Ni are constant in thesystem as a whole.

5 When combined, this information is sufficient for deriving an in-vertable set of equations for calculating the equilibrium state.

7 6 To ensure that the problem size does not escalate unreasonablythe main concern of this chapter is a two-phase system where α andβ may include vapour–liquid, liquid–liquid, solid–liquid and solid–solidequilibrium. The thermodynamic equilibrium criterion is then simplified to:

(dG)T ,p =n∑

i=1µαi dNα

i +n∑

i=1µβi dNβ

i = 0 ,

dNαi + dNβ

i = 0 .



15 Multicomponent . . .Minimum Gibbs energy II

1 Elimination of dNβi makes (dG)T ,p,n =

∑ni=1(µ

αi − µ

βi ) dNα

i = 0.

2 In the vicinity of an equilibrium point all the differentials dNαi are

freely varying quantities Expanded footnote [a]:

[a] At equilibrium dNαi are fluctuations beyond our control. These fluctuations are quiteinevitable from the laws of quantum mechanics and will be observed in every physicallyacceptable system. Thermodynamic equilibria are in other words dynamic, not static. which means the equilibrium relationship is

equivalent to:

µα1 = µβ1

...

µαn = µβn

(15.1)

ENGLISH text is missing.



15 Multicomponent . . .§ 116 § 115 § 117

Show that the necessary criteria for multiphaseequilibrium at given total entropy S, total volume V and

total composition Ni for each component i, are Tα =

Tβ = · · · = Tω, pα = pβ = · · · = pω and µα = µβ = · · · =µω. It is assumed that all the components i ∈ [1,n] arepresent in all the phases α, β, . . . , ω. Hint: Start from

Ueq = mins,v,ni

U(s,v,n1, . . . ,nn) ,

es = S ,

ev = V ,

eni = Ni , ∀i ∈ [1,n] ,



15 Multicomponent . . .§ 116 § 115 § 117 (2)

where e = ( 1 1 · · · 1 ) has as many elementsas there are phases in the system, and where the

other vectors

sT = ( Sα Sβ · · · Sω ) ,

vT = ( Vα Vβ · · · Vω ) ,

nTi = ( Nα

i Nβi · · · Nω

i ) ,

define the extensive co-ordinates of the phases.



15 Multicomponent . . .§ 116 Minimum internal energy

1 2 The internal energy of the system is minimised in the equilibriumstate. In this (stationary) state the differential of U must be zero for all

feasible variations in Sπ, Vπ and Nπi :

(dU)S ,V ,n =ω∑π=α

dUπ =ω∑π=α

(Tπ dSπ − pπ dVπ +

n∑i=1µπi dNπ

i

)= 0 .

5 3 Note that U has no absolute minimum.4 Due to its extensive properties the energy will diverge to ±∞ whenthe system size goes towards ∞. The variations in Sπ, Vπ and Nπ

i must therefore be constrainedsuch that the total entropy, volume and mole numbers (of each compo-nent) are conserved:

ω∑π=α

dSπ = 0 ,

ω∑π=α

dVπ = 0 ,

ω∑π=α

dNπi = 0 , ∀i ∈ [1,n] .



15 Multicomponent . . .§ 116 Minimum internal energy (2)

7 6 From these n + 2 balance equations we can eliminate dSω, dVω

and dNωi for all the components i ∈ [1,n]. Substituted into the differential of U:

(dU)S ,V ,n =ψ∑π=α

[(Tπ−Tω) dSπ− (pπ−pω) dVπ+

n∑i=1

(µπi −µωi ) dNπ

i

]= 0 .

9 8 In the neighbourhood of an equilibrium point the quantitiesdSα, . . . , dSψ and dVα, . . . , dVψ and dN1α, . . . , dNψn are truly inde-pendent variables. If (dU)S ,V ,n = 0 then it must be true that:

Tα = Tβ = · · · = Tψ = Tω , (15.2)

pα = pβ = · · · = pψ = pω , (15.3)

µα1 = µβ1 = · · · = µ

ψ1 = µω1 ,

... (15.4)

µαn = µβn = · · · = µ

ψn = µωn .



15 Multicomponent . . .Activity–activityENGLISH text is missing.Gibbs’ phase rule1 The equations above show that temperature, pressure and chem-ical potentials are uniform in all phases at thermodynamic equilibrium.This does not imply that we can specify these variables directly.

2 The Gibbs–Duhem’s equation

s dT − v dp + ( n1 n2 · · · nn ) dµ = 0 .

removes one degree of freedom per phase.

3 The number of independent intensive variables in a system istherefore F = dim(n) + 2 − dim(e), or in other words F = N + 2 − P,also known as the Gibbs phase rule for un-reacting systems.

4 The practical consequence of this law is that it we are not allowedto calculate the equilibrium of a system with F < 0 degrees of freedom.Doing so will lead to an over-specified system which eventually will leadto a singular set of equations to be solved.

5 Systems with non-negative degrees of fredom are in principle solv-able, but it is nevertheless not so easy to find a good solution algorithmin each case.

6 By and large, all we can do is to write algorithms that treat certainsubclasses of the problems we encounter. Vapour–liquid equilibria atlow to medium high pressure— maybe up to a few hundred bar— is onesuch subclass. In the next two sections we will look at two competingmethods, called direct substitution and Newton–Raphson iteration, tosolve this kind of equilibria. It should be stressed that the methods dowork for other types of phase equilibria as well, but the vapour–liquidequilibria are very common in the chemical process industry and it doesnot hurt to have these in mind.

832 / 1776

1 ENGLISH text is missing.

µαi (T ,p,Nα1 ,N

α2 , . . . ,N

αn ) = µ

βi (T ,p,N

β1,N

β2, . . . ,N

βn) ,

Nαi + Nβ

i = Ni ,(15.5)


6.1 Én felles fugasitetsmodell for damp og væske,eventuelt for to fluider med et kritisk punkt foravblanding, som vist i figur 15.1.6.2 Én felles aktivitetsmodell for to faste faser,

eventuelt for to væskefaser, med et kritisk punkt foravblanding som vist i figur 15.2.



15 Multicomponent . . .Activity–activity (2)

6.3 Blandet bruk av fugasitets- og aktivitetsmodell fordamp og væske ved forholdsvis lave trykk som vist ifigur 15.3.6.4 To ulike aktivitetsmodeller for to ulike faste faser

eventuelt fast fase og væske som vist i figur 15.4.




15 Multicomponent . . .Fugacity–Fugacity


µi = µi (T ,p) + RT ln(ϕixip

p

)(15.6)


µi (T ,p) + RT ln(ϕαi xαi p

p

)= µi (T ,p) + RT ln

(ϕβi xβi pp

). (15.7)


ϕαi xαi = ϕβi xβi .


xαi = 1 ∧ xβi = 1 ⇒ ϕαi = ϕβi ,




15 Multicomponent . . .Fugacity–Activity

1 If the chemical potentials of the pure components are the same inboth phases, at the temperature and pressure of the system, it can beconcluded that the phases are of the same nature.

3 2 If this is the case it is no longer necessary to integrate all the wayfrom the ideal gas in Eq. 13.12 but rather refer to the properties of purecomponent -i- as a reference. In practise this means that the fugacity coefficient ϕi can be re-

placed by the activity coefficient γi defined by:

γi(T ,p,n) =ϕi(T ,p,n)ϕ⋆i (T ,p)

. (15.8)


µi(T ,p,n) = µi (T ,p) + RT ln(ϕ⋆i p

p

)︸︷︷︸

Reference potential

+ RT ln(ϕixi

ϕ⋆i

), (15.9)

= µ⋆i (T ,p) + RT ln [γi(T ,p,n)xi] . (15.10)

4 because the pure component reference ϕ⋆i is the same in bothphases.

RT lnγi(T ,p,n) = RT lnϕi − RT lnϕ⋆i

= RT lnγi(T ,p⋆,n) +p∫p⋆

(vi(T , π,n) − v⋆i (T , π)) dπ .

(15.11)


T ≪ 100 K

RT | lnγi | ≪ 1000 J mol-1

p ≫ 10 MPa


γi(T ,p,n) ≈ γi(T ,n) ,

ENGLISH text is missing. ENGLISH text is missing.



15 Multicomponent . . .Ideal gas


µi (T ,p) = ∆fhi (T) + ( hi (T) − hi (T) ) − Tsi (T ,p) , (7.12)


hi (T) − hi (T) =T∫

T

cp,i(τ) dτ , (7.14)

si (T ,p) − si (T,p) =T∫

T

cp,i(τ) dln τ . (7.15)




15 Multicomponent . . .Condensed phase


µ⋆i (T ,p) = µ⋆i (T ,p⋆) +p∫p⋆

v⋆i (T , π) dπ . (15.12)


µ⋆i (T ,p⋆) = µi (T ,p) + RT ln(ϕ⋆i p⋆

p

),

= µi (T ,p) +V(p⋆)∫∞

[RTν −

( ∂π∂Ni

)T ,ν,Nj,i

]dν − RT ln

(pV(p⋆)NRT

).

(15.13)



15 Multicomponent . . .Condensed phase (2)


µi(T ,p,n) = µ⋆i (T ,p⋆) + RT ln [γi(T ,p⋆,n)xi] +p∫p⋆

vi(T , π,n) dπ .

(15.14)ENGLISH text is missing.



15 Multicomponent . . .Saturation state


µβ,sati (T ,psat

i ) = µα,sati (T ,psat

i ) ,

µi (T ,p) + RT ln(ϕsat

i psati

p

)= µ⋆i (T ,p⋆) +

psati∫

p⋆

v⋆i (T , π) dπ .


µi(T ,p,n) = µi (T ,p)

+ RT ln(ϕsat

i psati

p

)+

p⋆∫

psati

v⋆i (T , π) dπ

+ RT ln [γi(T ,p⋆,n)xi] +p∫p⋆

vi(T , π,n) dπ .



15 Multicomponent . . .Saturation state (2)


ϕiyip ≈ ϕsati psat

i γi(T ,p⋆,n)xi exp

1RT

p∫

psati

v⋆i (T , π) dπ

.



i γi(T ,p⋆,n)xi exp[

vi(p−psati )

RT

], (15.15)


yi = 1 ∧ xi = 1 ⇒ ϕip = ϕsati psat

i .




15 Multicomponent . . .Phase transition point


d(∆trµ⋆i

T)p = ∆trh⋆i d

(1T)


∆trµ⋆iT

∣∣∣∣T

T⋆i=

∆trh⋆iT

∣∣∣∣T

T⋆i−

T∫

T⋆i

d∆trh⋆iτ


∆trµ⋆i (T)

T − ∆trµ⋆i (T⋆i )

T⋆i︸︷︷︸0

=∆trh⋆i (T)

T − ∆trh⋆i (T⋆i )

T⋆i−

T∫

T⋆i

∆trc⋆p,iτ dτ



15 Multicomponent . . .Phase transition point (2)


∆trµ⋆i (T) = ∆trh⋆i (T

⋆i )+

T∫

T⋆i

∆trc⋆p,i(τ) dτ− TT⋆i·∆trh⋆i (T

⋆i )−T

T∫

T⋆i

∆trc⋆p,i(τ)

τ dτ


∆trµ⋆i (T) = ∆trh⋆i (T

⋆i )

(1 − T

T⋆i

)+

T∫

T⋆i

∆trc⋆p,i(τ)(1 − T

τ

)dτ . (15.16)



15 Multicomponent . . .K -value

1 Vapour–liquid equilibria are frequently calculated by an iterativemethod which is due to Rachford and Rice Expanded footnote [a]:

[a] H. H. Rachford and J. D. Rice. Trans. Am. Inst. Min., Metall. Pet. Eng., 195:327–328, 1952. .

2 This is also known as the K -value method because the equilibriumrelations are solved as a set of K -value problems on the form

xβi = Kixαi , ∀i ∈ [1,n] , (15.17)

where xαi and xβi are the mole fractions of component i in α (liquid) andβ (vapour) respectively.

3 We shall later learn that there is a one-to-one relationship betweenKi and the chemical potentials of component i in the two phases.

5 4 This tells us that the K -value method can be applied even whenα and β are two arbitrary phases, but prior to the discussion of exoticphase equilibria we shall derive a calculation scheme (algorithm) forEq. 15.17. The total component balance is Nα + Nβ = N and the individual

component balances can be written xαi Nα + xβi Nβ = Ni .



15 Multicomponent . . .K -value (2)

6 Elimination of Nβ produces xαi Nα + xβi (N − Nα) = Ni and a subse-quent substitution of xβi from Eq. 15.17 yields:

xαi = NiNα+Ki(N−Nα)

= zizα+Kizβ

. (15.18)



15 Multicomponent . . .Direct substitution

1 The corresponding expression for xβi is

xβi = KiNiNα+Ki(N−Nα)

= Kizizα+Kizβ

, (15.19)

where zα = Nα/(Nα+Nβ) and zβ = Nβ/(Nα+Nβ) are the phase fractionsand zi = Ni/N is the total (feed) mole fraction of component i.

2 Note that the right hand sides of Eqs. 15.18 and 15.19 contain asingle unknown variable zα, or equivalently zβ = 1 − zα.



15 Multicomponent . . .Direct substitution (2)

4 3 Definition-wise ∑ni xi = 1 in both phases and it is possible to solvefor zα from Eq. 15.18 or 15.19, but practical experience proves thatthe algorithm is more stable when the symmetric condition f(zα) =∑ni=1(xαi − xβi ) = 0 is being used. Summing up xαi and xβi from Eqs. 15.18 and 15.19 yields a closed

expression for f(zα)

f(zα) =n∑

i=1

(1−Ki)zi

zα+Kizβ=

n∑i=1

fizi = 0 ,

fi =1−Ki

zα+Kizβ,

which is easily solved with respect to zα using a Newton–Raphson iter-ation:

zα,k+1 = zα,k − ( ∂f∂zα

)-1f = zα,k +

( n∑i=1

f2i zi

)-1f . (15.20)

5 The recursion formula makes a new zα,k+1 available, which onback-substitution into Eqs. 15.18 and 15.19 produces new values ofxαi and xβi . 6 These updates (hopefully) improve the Ki-values, and subsequent

iterates in Eq. 15.20 converge finally to the equilibrium state.



15 Multicomponent . . .K -value algorithm § 1171 Computerise the function [xα,xβ, zα] = TpKvalue(n,k) using

Eq. 15.20 as your reference algorithm.848 / 1776

§ 117 Rachford–Rice1 See Function 2.6 in Appendix 31.

2 Note that the function interface is more elaborate than has beenasked for in the text.

3 The syntax has notably been extended to TpKvalue(n,k,u,U, v ,V)where U and V are data structures transmitting model specific parame-ters used by u and v .

4 This makes it possible to iterate on non-ideal equilibrium states bysupplying two fugacity (activity) coefficient models u and v.

5 The update of Ki assumes for each iteration that ln kk+1 = ln k +

u(xα,U) − v(xβ,V). Constant Ki values will be assumed if the functionsu and v are expelled from the argument list.

849 / 1776

1 2 Note that ∂f/∂zα < 0 in Eq. 15.20 except for the degenerated caseKi = 1∀i ∈ [1,n] then ∂f/∂zα = 0. A fixed derivative sign means that there is at most one solution in

the domain zα ∈ [0,1].

5 3 Similar to the Newton–Rapshon method, which is derived in Sec-tion 15.5, the number of floating point operations in the update of xα isproportional to n, i.e. the computation time increases linearly with thenumber of components in the system.4 These are the strengths of the K -value method, but it has also an

inherit weakness in that non-ideal equilibria must be solved by nestediteration. First, Eq. 15.20 is solved with respect to zα at given or estimated

Ki (inner loop).

6 The phase compositions are then updated by the Eqs. 15.18and 15.19 before new Ki-values can be calculated (outer loop).

7 The double iteration procedure is repeated until the phase compo-sitions xαi and xβi have converged. 8 The Newton–Raphson method in Section 15.5 is without this flaw

because ∆µi (equivalent to Ki) is updated in every iteration avoiding theouter loop[a].

[a] Also applies to the K -value problem if disregarding the monotonic form of f(zα).



15 Multicomponent . . .Linearization of chemical potential

1 Contrary to the K -value method of Rachford–Rice, the Newton Expanded footnote [a]:

[a] Sir Isaac Newton, 1642–1727 by the Julian calendar. English physicist and mathe-matician. –

Raphson Expanded footnote [a]:

[a] Joseph Raphson, ca. 1648–1715. English mathematician. s method fulfills the component balance in every iteration.

3 2 This requires a feasible starting point where nα+nβ = n is fulfilled,and which of course is as close to the solution point as possible. The Newton–Raphson iteration of Eq. 15.1 is then,

µαi +n∑

j=1

(∂µαi∂Nα

j

)T ,p,Nl,j

∆Nαj = µβi +

n∑j=1

(∂µβi∂Nβ

j

)T ,p,Nl,j

∆Nβj ,

−∆Nβj = ∆Nα

j ,

or on matrix form:

µα + Gα∆nα = µβ + Gβ∆nβ ,

−∆nβ = ∆nα .



15 Multicomponent . . .Inversion

1 The matrix G = (∂2G/∂Ni∂Nj)T ,p is the Hessian Expanded footnote [a]:

[a] Ludwig Otto Hesse, 1811–1874. German mathematician. of Gibbs energyand ∆n = nk+1 −nk is the composition difference between two (subse-quent) iterations k + 1 and k .

2 The Newton–Raphson equations can be combined into

(Gα + Gβ)∆nα = −(µα − µβ) ,

or even better: ∆nα = −H-1∆µ where H = Gα + Gβ and ∆µ = µα − µβ.

4 3 It must be realised, however, that the algorithm has to be guided. In particular the mole number updates must be checked fornα,k+1 = nα,k +∆nα > 0 and nβ,k+1 = nβ,k −∆nα > 0.



15 Multicomponent . . .Inversion (2)

5 If the relations are violated it is mandatory to shorten the step sizeaccording to

∆nα = −τH-1∆µ , (15.21)

6 where τ ∈ 〈0,1] is calculated such that all the updated mole num-bers are positive.7 In this way we can ensure that the component balance is fulfilled inevery iteration.8 The Newton–Raphson method converges to a state where ∆µ = 0,but note that the phase models have not been taken into account yet.9 It must therefore be anticipated that ∆µ is calculated from an equa-tion of state, or an activity coefficient model, that describes the systemwith sufficient accuracy.



15 Multicomponent . . .Update formulaENGLISH text is missing.Ideal mixture Hessian1 The calculation of H may be approximated by the Hessian of an

ideal mixture.

2 The convergence properties will deteriorate close to the solutioncompared to the rigorous implementation of H = Gα + Gβ, but the sim-plicity of the model makes it interesting in its own right.

3 In summary it means that ideal mixing is assumed in the calculationof H while ∆µ is calculated rigorously.

4 From the definition of an ideal mixture we can write,

Gıdij =

(∂µıdi∂Nj

)T ,p,Nk,j

= ∂∂Nj

(µ⋆i + RT ln Ni

N

)T ,p,Nk,j

= NRTNi

(∂(Ni/N)∂Nj

)Nk,j

= RT( δij

Ni −1N

),

where the Kronecker[a] delta δij = 1 if i = j and δij = 0 else.

5 On matrix form this formula is written — see also Theme 12 onpage 86 in Chapter 3 on page 57:

Gıd =RT(n-D − N-1eeT) ,

e =( 1 1 . . . 1 )T ,

6 Here, n-D is an (inverted) diagonal matrix having 1/Ni along themain diagonal.

7 The simple structure of this matrix makes it possible to calculateH-1 with an analytical formula as shown below.

8 From the definition H = Gα + Gβ it follows:

Hıd = RT[(nα)-D + (nβ)-D −

((Nα)-1 + (Nβ)-1

)eeT] .

854 / 1776

[a] Leopold Kronecker, 1823–1891. German mathematician.

ENGLISH text is missing.Sherman–Morrison in-

verse1 The matrix (nα)-D+(nβ)-D has diagonal elements (Nα

i )-1+(Nβ

i )-1 =

Ni/(Nαi Nβ

i ) and the factor (Nα)-1 + (Nβ)-1 can be rewritten to (Nα +

Nβ)/(NαNβ).

2 This makes the following factorisation of Hıd possible:

Hıd = RT Nα+Nβ

NαNβ

[diag

(NαNβ

Nαi Nβi

Ni

(Nα+Nβ)

)− eeT

],

or,

Nα+Nβ

RT Hıd = 1zαzβ (D

-1 − eeT) ,

D-1 = diag( zi

xαi xβi

), (15.22)

where zα and zβ are phase fractions, xαi and

3 As usual xβi are component mole fractions, and zi is the feed molefraction of component i.

4 Eq. 15.22 indicates that H is a rank one update of D.

5 The inverse of H can then be calculated from the Sherman–Morri-

son formula(D-1 − eeT)-1 = D + (1 − eTDe)-1DeeTD, or, equivalently, if

we define d = De:

RTNα+Nβ

(Hıd)-1 = zαzβ

(D + (1 − eTd)-1ddT) . (15.23)

855 / 1776

§ 1181 The Sherman–Morrison formula is widely applicable and not at alllimited to thermodynamic problems. Verify the formula by proving thatH-1H = I. Attempt to find a more general formula valid for (A − uvT)-1.

856 / 1776

§ 118 Sherman–Morrison1 Straightforward substitution.

2 The general formula for a rank one update of the matrix is: (A −uvT)-1 = A-1 + A-1u(1 − vTA-1u)-1vTA-1.

857 / 1776

1 From Eqs. 15.21 and 15.23 it is possible to express the simplifiedtwo-phase Newton–Raphson iteration as:

∆nα

Nα+Nβ = −zαzβ(D + 1

1−dTeddT

)∆µ

RT . (15.24)

2 Note that zα, zβ, d, D and ∆µ/RT are dimensionless variables.

3 This makes the iteration sequence independent of the system sizefor a given total composition z.

4 We may therefore scale the system Expanded footnote [a]:

[a] This is just another example on the special properties of extensive functions. such that Nα+Nβ = 1 withoutaffecting the mole fractions in the phases α and β during the iteration.



15 Multicomponent . . .Ki versus µi § 1191 Computerise the function [xα, xβ, zα] = TpNewton(n, k). Use

Eq. 15.24 as your reference algorithm.859 / 1776

§ 119 Newton–Raphson1 See Function 2.7 in Appendix 31.

2 Note that the function interface is more elaborate than has beenasked for in the text.

3 The syntax has notably been extended to TpNewton(n, k,u,U, v ,V)where U and V are data structures transmitting model specific parame-ters used by u and v.

4 This makes it possible to iterate on non-ideal equilibrium states bysupplying two fugacity (activity) coefficient models u and v.

5 The update of ∆µ assumes that for each iteration ∆µ = ∆µıd +

∆µr,p where ∆µıd = ln k + ln xα − ln xβ and ∆µr,p = u(xα,U) − v(xβ,V).

6 Constant ∆µr,p = 0 will be assumed if the functions u and v areexpelled from the argument list.

860 / 1776

ENGLISH text is missing.Computation time1 It should be mentioned that the number of floating point operations

in the update is proportional to n.

2 Hence, the computation time increases linearly with the number ofcomponents in the mixture.

3 However, if we had not spent time on the Sherman–Morrison for-mula, but rather used numerical inversion of H in each iteration, thecomputation time would be proportional to n3.

861 / 1776

1 The Rachford–Rice iteration spends time on calculating K -valueswhile the Newton–Raphson method deals directly with chemical poten-tials.

2 However, because the two methods aim at solving the same equi-librium problem there must be a rational connection between them.

3 The purpose of this section is to show that (and how) Eq. 15.1 onpage 794 can be used to derive some useful relations between Ki and∆µi .

µβi = µαi (15.1)

⇓

µi (T ,p) + RT ln(ϕβi xβi p

p

)= µi (T ,p) + RT ln

(ϕαi xαi pp

), (15.7)



15 Multicomponent . . .Ki versus µi (2)

where the left side stands for µβi and the right side stands for µαi . FromEq. 15.17 there is the alternativ formulation of the same equilibriumcriterion:

xβi = Kixαi . (15.17)

4 But it should be stressed that in the context of being an equilibriumconstant the K -value is quite misleading. A thermodynamic equilibrium constant depends on temperature, andnothing but temperature, whereas Ki is in fact a non-linear function intemperature, pressure and the compositions of both phases (and de-pends therefore on two thermodynamic states simultaneously). If not,the two alternative formulations would not be equivalent.



15 Multicomponent . . .Fugacity–fugacity

1 On the basis of the discussion above, the quantity ∆µi/RT whichappears in Eq. 15.24 can be defined as

∆µiRT =

µαi −µβi

RT = ln(Keos

i xαixβi

), (15.25)

where Keosi is defined as the ratio between the fugacity coefficients in

the phases α and β:

Keosi (T ,p,nα,nβ) =

ϕαi (T ,p,nα)

ϕβi (T ,p,nβ). (15.26)

2 It is used an excessively obvious notation in this case to emphasizethat Ki is not a thermodynamic equilibrium constant.



15 Multicomponent . . .§ 120 § 119 § 121

Find experimental data for a typical hydrocarbon vapour–liquid system and see how close the RK equationof state matches the measurements. Calculate the

phase diagram using TpKvalue or TpNewton.



15 Multicomponent . . .§ 120 Natural gas (T , p)

1 The phase diagram of a typical, albeit synthetic, natural gas Expanded footnote [a]:

[a] Mario H. Gonzalez and Anthony L. Lee. J. Chem. Eng. Data, 13(2):172–176, April1968. is

illustrated in Figure 15.1 on the next page.

2 The calculations have been using TpNewton as shown in Pro-gram 1.22 in Appendix 31.

3 The plot is given a high colour density where the liquid and vapourphases coexist in approximately equal amounts, and a light colour nearthe single-phase region (vapour or liquid). 4 The dense ridge running along the left edge of the diagram com-

bined with the shallow basin to the right is typical for methane-rich nat-ural gases.5 The bubble point is dominated by the large methane content while

the dew point is determined by the trace components comprising the“heavy tail” of the gas.6 The Newton–Raphson method does an excellent job in this

case[a], but note that the dew-point calculations are sensitive to thestart estimate.7 Increasing the liquid composition of the medium components by

less than one tenth of a mole fraction causes the rightmost part of thephase boundary to be left out.8 The K -value method has even larger problems in this part of thediagram, and needs some type of guidance which is not implementedhere.

[a] Far better than anticipated. In fact, it does a better job than for a rigorous imple-mentation of the Hessian. The reason for this somewhat strange behaviour is that theeigenvalues of the simplified Hessian are strictly positive while the eigenvalues of therigorous Hessian change sign wherever thermodynamic instability occurs, see Chap-ter 21. This is known to cause numerical problems in Newton–Raphson methods.



−200 −160 −120 −800

100

200

300

400

500

600

700

800

900

Temperature [F]

Pre

ssur

e[p

sia]

c.p.

N = 2913

Figure 15.1: Phase diagram of nitrogen–methane–ethane–propane–n-bu-tane put together from a total of 6093 triangular patches using a recursivedivide-and-conquer algorithm (each triangle is split into four smaller ones un-til all the vertices are in the same phase domain).



15 Multicomponent . . .Activity–activity ENGLISH text is missing.Some physical chemistry1 In the ages before the modern computer era it was mandatory to

make physical simplifications before any numerical calculation was at-tempted.

2 One of these simplifications is known as the Lewis mixing rule.

3 Under certain circumstances the fugacity coefficient ϕi will be al-most independent of the mixture composition.

4 This is always true for an ideal gas mixture and it is valid to ahigh degree of approximation for real gases at low pressures or hightemperatures. The approximation also holds at high pressures if Vex ≈0 in the entire pressure and composition region.

5 In the case of ϕi(T , p, x) ≈ ϕ⋆i (T ,p) then Ki will be constant inEq. 15.26 (at a given temperature and pressure).

6 The K -value method is then superior to the Newton–Raphson iter-ation because it solves the entire equilibrium problem in R and not inRn.

868 / 1776

§ 1211 Show that the Lewis mixing rule is exact if Leduc??–Amagat[a]slaw Vex = 0 is valid over the entire composition and pressure [0, p]range.

869 / 1776

[a] Émile Hilaire Amagat, 1841–1915. French chemist.

§ 121 Lewis[1] mixing rule1 Vex = 0 means that V =

∑ni=1 vi(T ,p)Ni .

2 The fugacity coefficient is defined by the Eqs. 13.12 and 13.13, seepage 718. Substituted for the Lewis rule we get

RT lnϕLewısi =

p∫

0

(vi(T , π) − RT

π

)dπ = f(T ,p) ,

which shows that ϕi depends on temperature and pressure only be-cause the partial molar volume is independent of composition.

870 / 1776

[a] Gilbert Newton Lewis, 1875–1946. American chemist.

1 2 In Eq. 15.26 it is common that α is the liquid phase and β the vapourphase, but this is just one out of many possibilities.3 We could alternatively assume that α and β are two near-critical

fluids being something in between a vapour and a liquid, or two well-defined liquid phases.4 There are also examples on supercritical gas–gas equilibria in the

literature.5 In a thermodynamic sense there is no fundamental difference be-

tween a vapour phase and a liquid phase[b], but it is still possible to de-cide whether the phases are of the same type by looking at the purecomponent states in the system.

[b] It is intuitive that the density of the vapour phase has to be lower than the densityof the liquid phase, but this is wrong. At high pressures the system H2–He will in factinvert because the vapour phase is dominated by He which has a higher molecularmass than H2. If the chemical potentials of the pure components are the same in

both phases, at the temperature and pressure of the system, it can beconcluded that the phases are of the same nature, but we cannot usethe terms vapour and liquid in a strict manner.

6 In practise this means that the fugacity coefficient ϕi can be re-placed by the activity coefficient γi from Eq. 15.8:

Keosi =

ϕαiϕβi≡ ϕαi

ϕα,⋆i· ϕ

β,⋆i

ϕβi· ϕ

α,⋆iϕβ,⋆i

=γαi1· 1γβi· ϕ

α,⋆iϕβ,⋆i

=γαiγβi· ϕ

α,⋆iϕβ,⋆i

.

7 Provided the same activity model is used to describe both phasesthe Ki-values are

ϕα,⋆i = ϕβ,⋆i ⇒ Keosi → Kex

i =γαiγβi,

8 because the pure component reference ϕ⋆i is the same in bothphases.



15 Multicomponent . . .§ 122 § 121 § 123

Find experimental data for a ternary liquid–liquidsystem with known NRTL, Van Laar or Margules model

parameters. Select a system which has a fairly largesolubility region and a critical end-point inside the ternary

region. Use TpNewton to calculate the phase diagram(TpKvalue will experience convergence problems).



15 Multicomponent . . .§ 122 Cycloalkanes–methanol (x1, x2)

1 The phase diagram of cyclohexane–cyclopentane–methanol Expanded footnote [b]:

[b] Ternary systems. In Liquid–Liquid Equilibrium Data Collection, volume V, part 2,.DECHEMA, Frankfurt/Main, 1980. p. 115. is

illustrated in Figure 15.2 on the following page.

3 2 It is a classic liquid–liquid diagram — quite symmetric and with acritical end-point inside the triangle. The diagram was successfully calculated using TpNewton as Pro-

gram 1.23 in Appendix 31 shows.

6 4 In this case the measurements were used as start values and thecalculations converged without difficulty even in the vicinity of the criticalpoint of the mixture.5 The agreement between the measured and the calculated valuesis otherwise extremely good. Still, the K -value method experiences serious problems close to

the critical point.



0 0.2 0.4 0.6 0.8 1

c.p.

Mole fraction of methanol

Mol

efra

ctio

nof

cycl

open

tane

Figure 15.2: Phase diagram of cyclohexane–cyclopentane–methanol at298.15 K and 760 mmHg. The iterations are started at interpolated pointsin the diagram (ensure safe convergence).



15 Multicomponent . . .Fugacity–activity

1 For mixtures of polar substances it may be appropriate to employan equation of state for the vapour phase and an activity model for thecondensed phase.

3 2 The phase diagram will in general not be “closed” at the criticalpoint but the hybrid models have greater flexibility and are favourable inthe modelling of complex liquids at low pressures and crystal phases ingeneral. The starting point is this time Eq. 15.15 on page 807 on page807

which gives the approximate formula:


i γi(T ,p⋆,n)xi exp[

vsati (T)(p−psat

i )

RT

], (15.15)



15 Multicomponent . . .Poynting factor

1 ∆µi/RT from the Newton–Raphson algorithm in Eq. 15.24 maynow be written on the same form as in Eq. 15.25 provided Ki is calcu-lated as

K vlei ≈ ϕsat

i (T) psati (T)γi(T ,p⋆,nα)

ϕi(T ,p⋆,nβ) p exp(vsat

i (T)·(p−psati (T))

RT), (15.27)

The exponential term is known as the Poynting Expanded footnote [b]:

[b] John Henry Poynting, 1852–1914. English physicist. factor of Ki . 2 This factor is often neglected at low pressures, but only if Ki 1.3 If this is not true the Poynting-factor may dominate Ki down to 5–

10 bar.4 Confer Section 2.2 for a more detailed discussion of this topic.



15 Multicomponent . . .§ 123 § 122 § 124

Find experimental data for a close to ideal binary vapour–liquid system with known NRTL, Wilson, Van Laaror Margules model parameters. Use TpKvalue or

TpNewton to calculate the phase diagram. Commenton the practical applicability of Raoult’s law versus

the accuracy of pure component data.



15 Multicomponent . . .§ 123 Hexane–toluene (T , x)

1 The phase diagram of hexane–toluene Expanded footnote [b]:

[b] Aliphatic hydrocarbons: C4–C6. In Vapor–Liquid Equilibrium Data Collection, volumeI, part 6a,. DECHEMA, Frankfurt/Main, 1980. p. 593. is illustrated in Figure 15.3

on the following page.

2 The calculations have been using TpNewton as shown in Pro-gram 1.24 in Appendix 31.

4 3 The system is almost ideal, but there are nevertheless some no-table deviations between the experimental and the calculated values. Correcting for non-ideality makes some improvement (compare σ1

and σ2 in the left subfigure), but the largest error is actually hidden inthe vapour pressure of toluene.

5 By increasing the vapour pressure 2% the agreement is improvedeven further (compare σ3 and σ4 in the right subfigure), while Raoult’slaw still shows a significant deviation from the measurements. 6 This shows that Raoult’s law is of limited value even for nearly ideal

systems.7 In fact, the pure component parameters are maybe more importantfor the mixture properties than the phase model itself.



0 0.5 160

80

100

120

ss1=0.356ss2=0.157

0 0.5 160

80

100

120

ss3=0.265ss4=0.0953

Mole fraction of hexaneMole fraction of hexane

Tem

pera

ture

[C]

Antoine vapour pressure Corrected vapour pressure

Figure 15.3: Phase diagram of hexane–toluene at 760 mmHg. Calculatedcurves show Raoult’s law (stipled) and Van Laar activity (solid). The rightsubfigure shows the effect of increasing the vapour pressure of toluene by2%. The sum-of-squares are denoted σ1, σ2, σ3 and σ4.



15 Multicomponent . . .Raoult[2]s law


[a] François-Marie Raoult, 1830–1901. French chemist. For an approximately pure liquid component i it is reasonable toset ϕi ≃ ϕsat

i , γi ≃ 1 and p − psati ≃ 0.

2 The mixture is by definition ideal and Ki appears to be a simplefunction of temperature and pressure:

KRaoulti = lim

xi→1K vle

i =psat

i (T)

p



15 Multicomponent . . .Henry[3]s law


[b] William Henry, 1775–1836. English chemist. In the same mixture it may also be appropriate to use a hypothet-ical vapour pressure Hij for the component i infinitly diluted in the sol-vent j, the value of which is chosen such that the phase equilibrium isconventionally reproduced by:

KHenryi = lim

xi→0xj→1

K vlei =

Hij(T ,p)p



15 Multicomponent . . .Acitivity–activity

1 When α and β are condensed phases it is often appropriate toneglect the influence of pressure on the system properties.

3 2 This is quite typical for metallurgical melts and refractory systemswhere the process conditions are close to atmospheric. In general, pressure explicit equations of state are not conceivable

for these systems and activity models are often used for both phases.

4 Eq. 15.14 yields for the phase equilibrium between β and α:

µβ,⋆i (T) + RT ln (γ

βi xβi ) ≈ µ

α,⋆i (T) + RT ln (γαi xαi ) . (15.28)


∆trµ⋆i (T) = µα,⋆i (T ,p⋆) − µβ,⋆i (T ,p⋆) ≈ ∆trh⋆i (T

⋆i )

(1 − T

T⋆i

). (15.29)



15 Multicomponent . . .Acitivity–activity (2)

5 ∆µi/RT required by the Newton–Raphson algorithm in Eq. 15.24can finally be written as in Eq. 15.25 provided Ki is calculated as:

K slei ≈ γαi (T ,n

α)

γβi (T ,nβ)

exp[∆trh⋆i (T

⋆i )

R

(1T − 1

T⋆i

)]. (15.30)



15 Multicomponent . . .§ 124 § 123 § 125

Find experimental data for a binary alloy (or a binary salt)with full solubility in the solid phase. Choose a system

with fitted Margules or Redlich–Kister model parameters.Calculate the phase diagram using TpNewton or TpKvalue.

How important is the temperature dependency of ∆trs⋆iin this context?



15 Multicomponent . . .§ 124 Gold–copper (T , x)

1 Figure 15.4 on page 841 illustrates the high-temperature portionof the gold–copper Expanded footnote [d]:

[d] H. Okamoto, D. J. Chakrabarti, D. E. Laughlin, and T. B. Massalski. Bull. Alloy PhaseDiagrams, 8(5):454–473, 1987. phase diagram, see Program 1.25 in Appendix 31.

2 The agreement between the measured and calculated values isgenerally good even though it is not possible to calculate the entirediagram— neither with TpNewton nor with TpKvalue.

3 The problem is quite persistent, and although the starting valuesare chosen judiciously there is a substantial region around the congru-ence point where things go wrong.

8 4 The flaw is due to the (large) negative deviations from ideality whichis observed in the two phases.5 This does in turn cause an oscillatory iteration sequence.6 Quite surprisingly, maybe, because the phase calculations in Fig-

ure 15.2 on page 828 converges nicely over the entire compositionrange.7 However, the latter system benefits from having a positive deviation

from ideality. The Hessian matrix for the assumed ideal mixture will then over-estimate the curvature of the energy surface and therefore yield a con-servative step size in TpNewton (and a safe, albeit very slow, conver-gence).



15 Multicomponent . . .§ 124 Gold–copper (T , x) (2)

9 With negative deviations from ideality the curvature of the enrgysurface is under-estimated and the step size may eventually grow solarge that the iterations start oscillating around the solution point Expanded footnote [d]:

[d] An exactly calculated Hessian would stabilise the Newton–Raphson method. .



0 0.2 0.4 0.6 0.8 1900

950

1000

1050

1100

Tem

pera

ture

[C]

Mole fraction of copper

Figure 15.4: Phase diagram of gold–copper showing a negative deviationfrom ideality. Both TpNewton and TpKvalue has convergence problems closeto the azeotrope.



15 Multicomponent . . .First order convergence ENGLISH text is missing.Consistency1 Concerning the accuracy of the calculations it is not easy to argue

that ∆trs⋆i varies over the temperature domain.

2 The simplification which is part of the pure component referenceis dominated by the activity model, and the calculated phase diagramlooks very promising.

3 But, a single phase diagram is not sufficient to claim any kind ofthermodynamic consistency. To do this we have to validate all types ofcalorimetric quantities, vapour pressures, cryoscopic measurements,etc.

4 A systematic study at this level would indeed reveal the noted in-consistency, as well as several minor ones.

888 / 1776

ENGLISH text is missing.Phase diagram morphol-

ogy1 While azeotropes are common for vapour–liquid phase diagrams,

the analogous phenomena of congruent melting is quite rare among thealloys—at least when disconnected from the formation of stoichiometricsolid compounds, or the homogenous phase splitting for one of thephases[d].

2 It has nevertheless been chosen to present an exotic system hereto stress the fact that phase diagrams are classified according to theirmorphology, and that the calculation method is quite independent of thephase models.

3 The thermodynamics of phase equilibria does not make any differ-ence between vapour, liquid and solid phases in this regard[d], but inpractise the number of empirical model parameters is often higher inthe description of crystalline phases as compared to the same descrip-tion of vapour and liquid phases of simple organic cmpounds.

4 In the gold–copper case there are e.g. six binary parametersneeded for the solid phases, while only three is needed for the liquid.

889 / 1776

[d] The gold–copper system exhibits solid phase separation at T < 400 C.

[d] Even though the equilibrium conditions 15.26, 15.27 and 15.30 look quite differently.

1 2 A sequence of iterative calculations has (convergence) order m and(convergence) factor h if the sequence approximates limk→∞ log ‖xk+1−x∞‖ = log(h) + m log ‖xk − x∞‖. The K -value method is typically of first order, but close to the critical

point the convergence factor h → 1 which means xk+1 ≃ xk .

3 Thus, convergence is severely hampered and it may be difficult todecide whether the sequence converges or not.

4 To broaden the applicability of first order methods it has been sug-gested to accelerate the iteration step along the strongest eigendirec-tion of the Jacobi Expanded footnote [d]:

[d] Carl Gustav Jacob Jacobi, 1804–1851. German mathematician. matrix. This is known as the Dominant EigenvalueMethod, see Appendix 28.



−6 −4 −2 0−6

−5

−4

−3

−2

−1

0

log10

∥∥∥∆µ∥∥∥k

log 1

0

∥ ∥ ∥ ∆µ∥ ∥ ∥k

+1

h1 = 0.675each 4th iterationhn = 0.965each 24th iteration

Figure 15.5: Convergence properties of the quasi-Newton–Raphson methodwhen using an ideal Hessian in the calculation of the bottom (h1) and top (hn)tie-lines in Figure 15.2. The latter is close to the critical point. Solid linesillustrate 1st order convergence. Note that h → 1 for the near-critical point.



15 Multicomponent . . .Second order convergence

1 The Newton–Raphson method is of second order — details con-cerning the convergence factor is discussed in Appendix 27.

2 This means that the number of significant digits will double in eachiteration provided that iteration k is sufficiently close to the solutionpoint.

4 3 The local convergence properties are irreproachable, but themethod needs quite good start estimates. But, the simplifications in Section 15.5 makes m = 1 unless for

ideal mixtures where the Hessian is exact.

5 Figure 15.5 illustrates what influence the simplification has on thecalculation of the bottom and top tie-lines of Figure 15.2.



15 Multicomponent . . .Second order convergence (2)

6 The iteration sequence approaches a line with slope 1, which istypical for first order methods, and for the near-critical point the conver-gence factor also approaches 1, in the sense that hn = 0.965 is prettyclose to unity in this context.



15 Multicomponent . . .Looking back

1 The general phase equilibrium criterion:

Tα = Tβ = · · · = Tψ = Tω ,

pα = pβ = · · · = pψ = pω ,

µα1 = µβ1 = · · · = µ

ψ1 = µω1 ,

...

µαn = µβn = · · · = µ

ψn = µωn

2 The Gibbs–Duhem equation on vector form:

s dT − v dp + ( n1 n2 · · · nn ) dµ = 0

3 K -value: xβi = Kixαi , ∀i ∈ [1,n]



15 Multicomponent . . .Looking back (2)

4 Chemical potential: ∆µiRT =

µαi −µβi

RT = ln(Keos

i xαixβi

)

5 Fugacity coefficient: Keosi =

ϕαiϕβi

6 Activity coefficient: Keosi =

ϕαiϕβi

=γαiγβi

ϕα,⋆iϕβ,⋆i

7 Pure component reference: ϕα,⋆i = ϕβ,⋆i ⇒ Keosi = Kex

i =γαiγβi

8 Poynting factor: K vlei =

ϕsati psat

i γi

ϕipexp

(vsati (p−psat

i )

RT)

9 Raoult’s law: KRaoulti = lim

xi→1K vle

i =psat

ip

10 Henry’s law: KHenryi = lim

xi→0K vle

i =Hij

p

11 Phase transition enthalpy: K slei =

γαiγβi

exp[∆trh⋆i (T

⋆i )

R

(1Ti− 1

T

)]




12 Two-phase equilibrium solved by direct substitution:

f(zα) =n∑

i=1

(1−Ki)zi

zα+Kizβ=

n∑i=1

fizi = 0 ,

fi =1−Ki

zα+Kizβ,

zα,k+1 = zα,k − ( ∂f∂zα

)-1f = zα,k +

( n∑i=1

f2i zi

)-1f

13 Two-phase equilibrium solved by Newton–Raphson iteration:

µα + Gα∆nα = µβ + Gβ∆nβ ,

−∆nβ = ∆nα




14 Assume ideal mixture when calculating G:

∆nα

Nα+Nβ = −zαzβ(D + 1

1−dTeddT

)∆µ

RT

15 Convergence: limk→∞ log ‖xk+1 − x∞‖ = log(h) + m log ‖xk − x∞‖


News The elements Reactions min G Calculation Le Chateliér Olds

Part 16

Chemical equilibrium




16 Chemical . . .New concepts

1 The periodic table

2 Stable elements

3 Chemical reactions

4 Reaction equilibria

5 Minimum Gibbs energy

6 Le Chateliér’s principle



16 Chemical . . .The elementsENGLISH text is missing.Chemistry1 Chemical equilibrium theory introduces the concept of chemical

bonds into thermodynamics, but without considering the physical forcesthat hold the compounds together. Neither is the geometry of the chem-ical bond an issue in this context.

2 Instead, it takes as its starting point a thermodynamic interpretationof the nature of chemical compounds, and looks at what net reactionsare possible between them.

3 Practical experiments have thereby shown that all chemical com-pounds are built up of one or more unique elements (atoms)[a]eachwith their own characteristic physical properties.

4 These elements are the unchangeable components of life on Earth,which manifest themselves through millions of different chemical com-pounds and an equally large number of chemical reactions.

5 They are what we call conserved quantities in accordance with theprinciple of conservation of mass[a].

6 The exceptions to the conservation principle are the nuclear re-actions that take place in nuclear power stations, in the Earth’s crust(granite rocks) and in the core of the sun.

7 These are governed by Einstein’s equation e = mc2, which statesthat ultimately mass and energy are equivalent concepts.

8 In this exceptional case, new elements can arise to replace oldones that disappear.

9 Approximately 112 elements have so far been identified, 80 ofwhich are considered stable.

10 The stable elements are all of the ones from hydrogen (1) tolead (82), with the exception of technetium (43) and promethium (61).

11 If we include all of the isotopes[a] of the various elements, thenumber of stable nuclei (species of nucleus) rises to 256.

12 This nuclear palette provides the basis for all life on Earth, and forall of the minerals, natural products and chemicals that we know fromeveryday life.

900 / 1776

[a] From the Greek átomos, which means indivisible.

[a] Demonstrated by Antoine-Laurent de Lavoisier in 1789. This paved the way forstoichiometry, which in turn led to the final rupture between chemistry and alchemy.

[a] Two isotopes of the same element always have the same number of electrons andprotons in their nuclei, but different numbers of neutrons.

1 2 The physical properties of the elements vary greatly— from beinggases at room temperature (air) and cryogenic liquids down to 0 K (he-lium) to being solids that are virtually impossible to melt (carbon). The big span in physical properties is clearly illustrated in the table

below. It shows the melting points of the stable elements rounded tothe nearest degree Celsius:

He nil Hg −39 In 157 Mg 650 Pr 931 Tb 1357 Tm 1545 Ru 2250H2 −259 Br2 −7 Li 181 Al 660 Ge 938 Dy 1407 Pd 1552 B 2300Ne −248 Cs 29 Se 221 Ba 729 Ag 961 Si 1410 Ti 1660 Ir 2443O2 −223 Ga 30 Sn 232 Sr 769 Nd 1016 Ni 1453 Lu 1663 Nb 2468F2 −220 Rb 40 Tl 304 Ce 798 Au 1065 Ho 1470 Pt 1772 Mo 2617N2 −210 P 44 Cd 321 As 817 Sm 1072 Co 1495 Zr 1852 Ta 2996Ar −189 K 63 Pb 328 Eu 822 Cu 1085 Er 1522 Cr 1857 Os 3027Kr −157 Na 98 Zn 420 Yb 824 Mn 1246 Y 1526 V 1902 Re 3180Xe −112 I2 114 Te 450 Ca 839 Be 1287 Fe 1535 Rh 1966 W 3422Cl2 −101 S 115 Sb 631 La 920 Gd 1312 Sc 1539 Hf 2227 C 3675



16 Chemical . . .StoichiometryENGLISH text is missing.Chemical compounds1 In nature, the elements are mainly found in various chemical com-

pounds, with the most obvious examples being water, dissolved salts inthe oceans and minerals in the Earth’s crust.

2 In some places there are native deposits of metals such as copper,silver and gold, and elsewhere you can find a few elements like sulphur,graphite and diamond in their free states, but these are the exceptionsthat prove the rule.

3 The discovery of the elements and their classification in the peri-odic table [a] therefore played a key role in the emergence of modernchemistry.

4 However, although the classification system itself may be a rela-tively recent innovation, a surprising number of methods for producingchemical reactions have existed for thousands of years.

5 The depth of knowledge in classical antiquity is amongst otherthings evidenced by the fact that all of the known metals of that time—silver Á, copper Ã, mercury Â, iron Ä, tin Å and lead Æ— with the excep-tion of gold À, were mined from ore and chemically processed ratherthan excavated from native deposits.

6 Our modern description of the associated metallurgical processesis so complex and extensive that you have to wonder whether trial anderror were the only tools they had at their disposal.

902 / 1776

[a] The Russian chemist Dmitri Ivanovich Mendeleev (1834–1907) is credited with thediscovery of the periodic table. He was’nt the only person working on the problem, buthe was the first person able to predict the properties of elements that had not yet beendiscovered (1869).

ENGLISH text is missing.The recipients1 Although it is an environmental sink in the same way as large bod-

ies of water and bedrock, the atmosphere occupies a unique position.

2 The total mass of the atmosphere constitutes no more than1 kg cm-2, which is approximately equivalent to a 10 m layer of con-densed material, but its composition is all the more interesting, as in ad-dition to small amounts of water, methane, ammonia and carbon diox-ide, it contains of a number of elements in their free states: 78% ni-trogen, 20.95% oxygen, 0.9% argon and trace amounts of other noblegases.

3 The noble gases are constantly being produced by radioactive pro-cesses in the Earth’s crust and leak slowly into the atmosphere.

4 Helium has a comparatively low mass and will eventually escapeinto outer space, while the other gases take part in the big geologicalcycles on Earth.

5 However, there is no equivalent inorganic explanation for why nitro-gen and oxygen dominate the atmosphere to the extent that they do.

6 The thermodynamic conditions on Earth are such that these ele-ments should only be found as nitrates dissolved in water and as ironoxide in various ores.

7 Actually, the relative abundance of both of them in their free statesin the atmosphere is entirely due to biological activity on Earth[a].

8 The atmosphere is in other words dominated by the waste productsfrom living organisms.

9 Life involves many other elements, such as hydrogen, carbon, iron,sulphur and phosphorous, but only nitrogen, oxygen (and under certainconditions sulphur) are produced in their free states in large quantities.

903 / 1776

[a] Nitrogen is formed by denitrifying bacteria under anaerobic conditions in the soil:2 NO−3 +C6H12O6 = N2 +3 O2, whereas oxygen is a waste product of photosynthesisin green plants: CO2 +H2O = CH2O+O2. Similarly, sulphur and copper can also beformed under anaerobic conditions in ??

ENGLISH text is missing.Chemical reactions1 In addition to the substances we are familiar with from nature, hu-

mans have managed to produce a number of new compounds in labo-ratory conditions.

2 The process that converts chemical compounds into other (new)compounds is called a chemical reaction.

3 This transformation involves breaking the existing bonds betweenthe atoms in the original compounds and forming new bonds underdifferent conditions.

4 The compounds that are transformed are called reactants, whilethe ones that are formed are called products.

5 The number of atoms of each element is the same before and af-ter the chemical reaction, but the properties of the products may never-theless be completely different from those of the reactants.

6 For example, if solid carbon reacts with chlorine gas, it producescarbon tetrachloride[a], a colourless liquid at ordinary conditions.

904 / 1776

[a] J. D. Blackwood and B. D. Cullis. Kinetics of the direct chlorination of carbon. Aust.J. Chem., 23(4):725–735, 1970.

ENGLISH text is missing.Extent of reaction1 Chemical reactions allow us to synthesise new compounds and toharness the bond energy of the reactants as thermal energy (throughcombustion) or as electrical energy (in a battery).

2 There are countless variations on the theme, but all chemical reac-tions can nevertheless be defined in terms of two basic properties: Theextent and rate of reaction.

3 The extent tells us how far the reaction has proceeded, while therate tells us how quickly it is progressing.

4 The above description sets out the fundamental principles that un-derpin chemical reactions, and although the theoretical study of ratesof reaction is perhaps unpredictable, all reactions will eventually reacha point at which the rate of reaction is zero.

5 Once equilibrium is reached, the extent of reaction has attained athermodynamic limit value, which we will learn to calculate.

905 / 1776

ENGLISH text is missing.Heterogenous reactions1 A chemical reaction always involves at least two compounds (reac-tant and product) in their gas, liquid or solid phases.

2 If there is only one macroscopic phase in the system, the reactionis referred to as being homogeneous.

3 A heterogeneous reaction, on the other hand, takes place acrossthe phase boundaries within the system, but the system as a wholemust be closed, or in some other way be consistent with the principlesof mass and energy conservation.

4 That is essential if we wish to formulate the problem in thermo-dynamic terms, and that is precisely our goal: To establish a physico-mathematical description of chemical reaction equilibria based on thelaws of conservation of mass and energy, as well as on the additionalassumption that equilibrium will be reached when the system is at thelowest possible energy state.

906 / 1776

1 The mathematical description of chemical reactions is closelylinked to the concept of stoichiometry and to the related changes inthe system’s composition. Stoichiometry is the study of the composi-tion of chemical compounds.

4 2 Based on the evidence of countless experiments, it has been foundthat in many compounds (albeit not all), elements combine in simpleratios to form stable structures, such as CH4, H2O, etc.3 This observation is incredibly important, as it gives us an accuratemathematical basis for predicting what will happen in a given situation. For example, when methane burns in air, the equation

CH4(gas) + 2 O2(gas)⇔ CO2(gas) + 2 H2O(gas) (16.1)

reliably predicts what will happen.

5 It is so accurate that in-situ measurements of the flame geometry,gas velocity and methane-to-air ratio are superfluous Expanded footnote [a]:

[a] Except in relation to small, unburned particles and pollutants, which form in varyingamounts depending on the physical conditions in the flame. . 6 However, it is by no means true that chemical conversions always

follow the same pattern and always yield the same products.7 In practice there may be competing reactions and alternative reac-tion pathways that affect the outcome, but for equilibrium reactions wecan predict what products will be formed and how far the reaction willproceed.8 We can do that by performing a thermodynamic analysis of the finalstate — the mechanism that brings about that state is irrelevant to theanalysis.



16 Chemical . . .Chemical equilibriumENGLISH text is missing.Thermodynamic phases1 When two or more compounds are combined in a homogeneous

mixture (as in Eq. 16.1) they constitute a thermodynamic phase.

2 The compounds in the mixture (the phase) are often referred to ascomponents.

3 For chemical reactions in a closed system, the composition of themixture is determined by the stoichiometry of the individual compo-nents.

4 We say that the reaction stoichiometry of the mixture is a functionof the observed component stoichiometries.

5 If we know the stoichiometry of the components we can predict thepossible changes in the composition of the mixture without needing toperform any further experiments.

6 This is a very powerful principle, which is illustrated by the combus-tion reaction shown in Eq. 16.1.

7 Here we can immediately conclude that NCH4+NCO2 = NC, 4NCH4+

2NH2O = NH and 2NO2 + 2NCO2 + NH2O = NO based on the principle ofconservation of mass.

8 The fixed mole numbers NC, NH and NO indicate that the amountsof carbon, hydrogen and oxygen in the system remain constant— theyare what we call reaction invariants.

908 / 1776

ENGLISH text is missing.Degrees of freedom1 Meanwhile, it is also the case that the reaction stoichiometry of the

system is determined by several atomic properties beyond our control.

2 The system thus loses an external degree of freedom for eachreaction that reaches equilibrium.

3 This means we cannot alter all of the composition variables inde-pendently.

4 One consequence of this is that the Gibbs’ phase rule must bechanged from F = C + 2 − P to F = C + 2 − P − R, where R is thenumber of chemical equilibrium reactions in the system.

5 On the other hand, the system gains an internal degree of freedom,because the associated extent of reaction both can and will vary withtemperature and pressure.

6 This makes the analysis of the system more challenging than theresults from Chapter 8, which deals with closed systems with a constantcomposition.

909 / 1776

1 The postulate that the equilibrium state is equivalent to a state ofminimum energy will be a key concept when we now look at chemicalreaction equilibria.

3 2 Amongst other things, we will see that there is a graphic link be-tween the concept of equilibrium and minimum Gibbs energy, beforeleaving the graphic analysis behind in Chapter 17 in favour of an analy-sis involving multivariate functions. Let the reaction equation

aA + bB ⇔ cC + dD (16.2)

serve as an example.

4 For a closed system at a given temperature and pressure, the equi-librium distribution of A , B, C and D will be such that

aµA + bµB = cµC + dµD . (16.3)



16 Chemical . . .Chemical potential

1 The symbol µi refers to the chemical potential of component i, aname that gives us a clear hint that it is a quantity related to work.

4 2 Put simply, you can say that the change in the chemical potential ofa component (mixed with several other components in the same phase)indicates the work required to transfer the chemical compound from onethermodynamic state to another.3 The change of state in question can be described in several differ-

ent ways, and there is a significant amount of freedom in the choice ofsuitable system variables, cf. the alternative definitions of chemical po-tential

µi = (∂U∂Ni

)S,V ,Nj,i =

(∂A∂Ni

)T ,V ,Nj,i =

(∂H∂Ni

)S ,p,Nj,i =

(∂G∂Ni

)T ,p,Nj,i

from Chapter 4. Here we will use the definition µi = (∂G/∂Ni)T ,p,Nj,i, and show

how it leads us to a standard formulation of chemical equilibria in idealgases.

5 This is the right choice for our purposes here, but there are alter-native formulations derived from the Legendre transformations in Chap-ter 4.

6 Imagine, for example, a chemical equilibrium in a closed reactorwith a constant volume. In that case µi = (∂A/∂Ni)T ,V ,Nj,i

would be amore natural choice.



16 Chemical . . .Minimum Gibbs energy

1 Eq. 16.3 agrees with the assumption that the point of minimumGibbs energy corresponds to the equilibrium state, i.e. Geq = min(G)T ,p

or (dG)T ,p = 0.

2 The differential equation gives us a necessary (but insufficient) con-dition for Gibbs energy having a possible minimum value.

4 3 The word possible complicates matters in the general case, but forideal gases both expressions produce the same answer[a].

[a] This is related to the fact that the Gibbs energy of an ideal gas mixture is a convexfunction of the number of moles at a given T and p. Using the components involved in the reaction Eq. 16.2, the sys-

tem’s total Gibbs energy can be expressed as:

G = U − TS + pV =D∑AµiNi . (16.4)

5 The challenge is to determine Geq = min(G)T ,p from the Eqs. 16.2and 16.4, either by minimising G, or indirectly by finding a state at which(dG)T ,p = 0. Here we will choose the second approach.



16 Chemical . . .§ 125 § 124 § 126

Eq. 16.3 is an example of a chemical equilibriumequation. Derive this equation by applying the principleof minimum Gibbs energy to Eq. 16.4 at a given T andp, and using the reaction in 16.2 as a constraint. Dothe phases of the components affect your calculation

in any way?



16 Chemical . . .§ 125 Reaction equilibrium

1 We have a closed system with four mole fraction variables and onereaction equation. Let us use NB , NC and ND to describe the overallcomposition of the system before the reaction takes place.

2 This leaves NA as a free variable, which tells us something aboutthe extent of reaction of the system. We can express the minimisationproblem using the following differential

0 = (dG)T ,p =D∑Aµi dNi , (16.5)

and similarly for the constraints:

dNB = + ba dNA ,

dNC = − ca dNA ,

dND = − da dNA .



16 Chemical . . .§ 125 Reaction equilibrium (2)

8 3 We can also say that aNB − bNA , aNC + cNA and aND + dNA arereaction invariants, i.e. they are conserved during the chemical reaction.4 Eq. 16.2 thus results in a principle of conservation quite similar to

the underlying principle for Eq. 16.1.5 This indicates that mass balance invariants and chemical reactions

are equivalent concepts, and that it is sufficient to know either the com-ponent stoichiometry or the reaction stoichiometry of the system.6 The truth isn’t quite that simple, and Chapter 17 gives a more de-

tailed analysis of this relationship using methods taken from linear al-gebra.7 For the moment it will suffice to say that combining the above con-

straints and 16.5 gives us

0 = µA dNA + ba µB dNA − ca µC dNA − da µD dNA

= (cµC + dµD − aµA − bµB) dNAa . (16.6) Since dNA is a free variable, and a is an arbitrary coefficient, theabove equation is equivalent to

cµC + dµD = aµA + bµB , (16.7)

cf. Eq. 16.3.

9 We have derived this equilibrium condition solely using generalthermodynamic relationships, and no assumptions have been madeabout behaviour specific to a particular state.10 This relationship therefore applies to all homogeneous and hetero-

geneous reactions in the gas, liquid and solid phases.



16 Chemical . . .Ideal gas

1 2 In order to calculate the equilibrium state we must find the pointat which the chemical potentials balance in accordance with the stoi-chiometry of the system (cf. Eq. 16.7), taking into consideration any ex-ternal constraints such as temperature, pressure, volume, energy, etc.3 In general this requires iteration involving N + 2 state variables, i.e.T , p and n or T , V and n.4 It is only possible to solve the equilibrium analytically in a few

cases: For example if there is one reaction in an ideal gas mixture at agiven T and p.5 But even in a simple case like that, you need to have a complete

understanding of the standard chemical potential µi . This is based on the chemical potential of a component of an idealgas mixture as expressed by the formula µıg

i = µi + RT ln(Nip/Np)where µi is the standard chemical potential of the component in itspure form at temperature T and standard pressure p.

6 In practice, µi is purely a function of temperature because the cur-rent convention is that p is equal to 1 bar (1 atm prior to 1982).



16 Chemical . . .Standard state

1 For most practical purposes µi is determined by consulting a ta-ble or using an explicit function of temperature. A common calculationmethod is µi (T ,p) = hi (T) − Tsi (T ,p) where hi (T) = ∆fhi (T) +∫

cp,i dT .

2 The formation enthalpy ∆fhi plays a particularly important role inthis calculation.

4 3 It is a measure of the energy level of component i relative to that ofthe elements that make up the compound. Under the IUPAC convention, hi = 0 for the chemical elements in

their stable configurations at T = 298.15 K and p = 1 bar. 5 The difference between the formation enthalpy of the products andreactants determines how far a chemical reaction can proceed beforereaching equilibrium.6 This contrasts with the criterion for phase equilibrium discussed inChapter 14, which makes no reference to the enthalpy level.



16 Chemical . . .§ 126 § 125 § 127 ENGLISH text is missing.Numerical solution1 If you support Eq. 16.3 with an explanation of chemical potential theresult is a simple, but nonetheless general, description of the criterionfor chemical equilibrium.

2 That leaves us with the question of how to solve this equilibriumproblem in practice.

3 Naturally there are several answers to such an open question, andwe must therefore be willing to make use of various numerical tech-niques.

4 In this chapter it is natural to solve polynomial expressions,whereas in Chapter 17 on simultaneous reaction equilibria it will bemore appropriate to use matrix-based methods.

918 / 1776

For reactions in ideal gas mixtures, the equilibrium 16.3can be expressed in the format Q(y) · f(p) = K(T).

Here we refer to Q as the reaction quotient and K asthe equilibrium constant of the reaction. The molefraction vector y includes all the components in the

systema— including the ones not participating in thereaction. It is a function y(ξ) of the extent of reaction

ξ. Determine the functionals Q, f and K .



16 Chemical . . .§ 126 Reactions in ideal gas mixtures

1 We will now show that the equilibrium equation can be separatedinto three constituent parts that are functions of composition, pressureand temperature.

2 We start by substituting the values for an ideal gas into Eq. 16.7:

0 = c[µC + RT ln

(pNCNp

)]+ d

[µD + RT ln

(pNDNp

)]

−a[µA + RT ln

(pNANp

)] − b[µB + RT ln

(pNBNp

)]. (16.8)

3 Then we multiply out the brackets and substitute mole fractions intothe equation:

−(cµC + dµD − aµA − bµB

)= cRT ln

(pyCp

)+ dRT ln

(pyDp

)

− aRT ln(pyA

p

) − bRT ln(pyB

p

). (16.9)



16 Chemical . . .§ 126 Reactions in ideal gas mixtures (2)

4 Finally we simplify and take the exponential of both sides:

ycC yd

D

yaA yb

B︸︷︷︸Q(y)

· ( pp

)c+d−a−b

︸︷︷︸f(p)

= exp[−(cµC+dµD−aµA−bµB)

RT

]

= exp(−∆rxG(T)

RT)

= K(T) (16.10)



16 Chemical . . .Extent of reaction ENGLISH text is missing.Reaction quotient1 Eq. 16.10 is the classic equilibrium equation for a single reaction

in an ideal gas mixture. The problem is thereby reduced to calculatingy(ξ) such that the given equation is fulfilled.

2 More often than not, it will require numerical iteration.

3 A closely related equation describes equilibria in other ideal sys-tems such as dilute solid in liquid solutions, but in that case the contri-bution of pressure is negligible.

4 For permanent gases at constant pressure and for condensedphases in general, pressure has little impact on the result and so wecan use the approximation Qeq ≈ Kc(T), which states that the reactionquotient Q must balance the constant Kc at the equilibrium point.

5 Note, however, that Kc has dimensional units and should not bethought of as a true thermodynamic equilibrium constant!

922 / 1776

1 Eq. 16.10 also explains what happens if we choose to start anexperiment with a composition where Q , Kc .

2 As the reaction only has one degree of freedom, it must either goto the right or left to reach equilibrium.

3 If we look slightly more closely at the fraction Q(y(ξ)) we can seethat the reaction in fact must go to the right if Q < Kc and to the left ifQ > Kc .

4 This result is known as the Guldberg Expanded footnote [a]:

[a] Cato Maximilian Guldberg, 1836–1902. Norwegian chemist and mathematician. –Waage Expanded footnote [a]:

[a] Peter Waage, 1833–1900. Norwegian chemist. law of mass action.



16 Chemical . . .§ 127 § 126 § 128 ENGLISH text is missing.Polynomial solution1 In order to solve Eq. 16.10 as a mathematical problem we must

know the exact stoichiometry of the system, as expressed by the coef-ficients a, b, c and d. We must also know the values for the equilibriumconstant K (T), the pressure p and last but not least: The initial compo-sition expressed by the numbers of moles NA , NB , NC and ND .

2 Next we must determine the roots of a polynomial of the ordern = max(a · b , c · d).

3 That is because Q(yi) is a rational function of the mole fractionsyA , yB , yC and yD , expressed such that when the function is rewritten toa polynomial form n becomes the highest exponent for any yi that takespart in the fraction.

924 / 1776

T [K] ln(K) T [K] ln(K)

298.15 −11.55 1000 −0.363400 −7.348 1800 1.336600 −3.345 3400 2.073

Let A = CO2, B = H2, C = CO and D = H2O inEq. 16.2. Calculate the reaction equilibrium at 1000 K.Assume that the molar ratio of unreacted H2O to CO

is 3 : 1. Equilibrium data are given in the table on theright. What happens in this case if you change the

pressure, or temperature?



16 Chemical . . .§ 127 Shift reaction

1 Substitute a = b = c = d = 1 into Eq. 16.10 together with thechemical symbols from the above question, i.e. yA = NCO2/N, yB =

NH2/N etc:(NCO

N

)1(NH2O

N

)1

(NCO2

N

)1(NH2N

)1 ·( pp

)1+1−1−1= e-0.363 = 0.696 .

2 Note that the total number of moles N of all of the componentsremains unchanged by the reaction, since c + d − a − b = 0. Thisimplies that f(p) = 1.0, which in turn means that the equilibrium is thesame at all pressures.

4 3 The next step is to substitute the mole balances from above intothe equation. The initial conditions are N

CO= 1 mol and N

H2O= 3 mol.



16 Chemical . . .§ 127 Shift reaction (2)

5 Let ξ ∈ [0,1] be the dimensionless extent of reaction of the system.

6 For an arbitrary value of ξ we have NCO = NCO− ξmol, NH2O =

NH2O− ξmol, NH2 = ξmol and NCO2 = ξmol, which gives rise to the

equation

( 1−ξ4 )( 3−ξ

4 )( ξ4)(

ξ4)

=(1−ξ)(3−ξ)

ξ2 = 0.696 .

7 The physical solution of the second degree equation gives us ξ =

0.798, or

NCO2 = 0.798 mol , NH2 = 0.798 mol ,

NCO = 0.202 mol , NH2O = 2.202 mol .



16 Chemical . . .Changes in pressure

1 A change in the total pressure does not affect the water-gas equilib-rium, since the reaction has the same number of moles of compoundson each side of the reaction arrow, but this result only applies to idealgases.

2 The key factor in a real system is whether the products occupy avolume that is bigger than, equal to or smaller than the volume of thereactants.

3 Depending on the change in volume, any increase in pressure willresult in the reaction going to the left, remaining unchanged or going tothe right respectively.

4 This is a result of Le Chateliér’s principle, which is discussed inSection 16.6.



16 Chemical . . .Changes in temperature

1 A change in temperature causes the equilibrium constant K(T) tochange, and hence the equilibrium composition will also vary.

2 Gibbs–Helmholtz’ equation 4.46 on page 161, applied to ln K =

−∆rxG/(RT), gives us the slope (∂ ln K/∂(1/T)) = −∆rxH/R.

3 In other words, you would expect ln K as a function of 1/T to pro-duce a virtually linear graph, provided that ∆rxH does not change signwithin the domain of definition.

4 Figure 16.1 illustrates this for the water gas shift reaction. Thegraph is clearly curved, but not so curved that there is any doubt aboutthe relationship between changes in ln K and changes in the reciprocaltemperature 1/T . 5 This relationship is known as the van’t Hoff[a] equation— originally

expressed as (∂ ln K/∂T) = ∆rxH/(RT2).

[a] Jacobus Henricus van’t Hoff, 1852–1911. Dutch chemist.



1 2 3−10

−8

−6

−4

−2

0

2

ln(K

)

1000T [K]

−∆rxH/R

Figure 16.1: van’t Hoff’s equation (∂ ln K/∂(1/T)) = −∆rxH/R; also see thederivation of Gibbs–Helmholtz’ equation in Theme 28, applied to the systemCO2–H2–CO–H2O at 1000 K. The heat of reaction is a function of tempera-ture, and the slope of the graph therefore varies over the domain of definition.



16 Chemical . . .§ 128 § 127 § 129 ENGLISH text is missing.Changes in composition1 Any change in the overall composition will also potentially affect theequilibrium, but only if the reaction quotient Q is changed.

2 This means that it is possible to change the size of the system atconstant T and p, i.e. by multiplying all of the numbers of moles by thesame factor, without disturbing the equilibrium.

3 What happens if a single component is added is slightly more com-plicated.

4 One specific example is discussed in Section 16.6 in conjunctionwith the discussion of Le Chateliér’s principle.

931 / 1776

Sketch the graph of G(NCO,NH2O,NCO2 ,NH2) as afunction of ξ ∈ [0,1] for Ni = Ni(ξ), where ξ = 0

represents the unreacted system and ξ = 1 is wherethe limiting reactant has been used up. Show that theminimum value of the function agrees with the result

from Theme 127.



16 Chemical . . .§ 128 Minimum Gibbs energy I

1 For the water gas shift reaction in an ideal gas mixture, equilibriumwill assert itself independently of pressure. This is due to the fact thatthe total number of moles of components is preserved during the reac-tion.

2 Bearing in mind that the Gibbs free energy should reach a mini-mum, we are able to write

G = C1

+ ξ(µCO2+ RT ln ξ) + ξ(µH2

+ RT ln ξ)

+ (1 − ξ)(µCO + RT ln(1 − ξ)) + (3 − ξ)(µH2O + RT ln(3 − ξ)) ,



16 Chemical . . .§ 128 Minimum Gibbs energy I (2)

5 3 where C1(p,N) = NRT ln(p/(pN)) = −4RT ln 4 is a constantthat does not affect the result.4 Admittedly the total number of moles (and the equilibrium pressure)does affect the vertical position of the graph, but shifting the graph ver-tically will not alter where the minimum value of G(ξ) is on the ξ axis. By simplyfying the function further we obtain

G = C1 + µCO + 3µH2O︸︷︷︸constant

+ ξ(µCO2+ µH2

− µCO − µH2O)︸︷︷︸

−∆rxG=RT ln K

+ RT ln[ξξξξ(1 − ξ)1−ξ(3 − ξ)3−ξ

],

which, if we introduce the new constant C2(T) can be written

G = C2 − 0.363RTξ+ RT ln[ξξξξ(1 − ξ)1−ξ(3 − ξ)3−ξ

].

7 6 Here C2 results in a further vertical shift of G(ξ), while RT ln K =

−0.363RT is incorporated in combination with ξ, altering the slope ofthe graph. This means that the position of the minimum value of G(ξ) will

move with temperature. G(ξ) is illustrated in Figure 16.2 using Pro-gram 31:1.26. 8 We notice that the function has a minimum at ξ = 0.798, which is

in agreement with Theme 127.



0 0.5 15

10

15

20

25

30

xi

kjo

ule

minvalue

Figure 16.2: Gibbs energy for the system CO2–H2–CO–H2O as a function ofthe extent of reaction ξ. The short branch of the graph shows the Taylor seriesto the second order. Note that the function’s minimum value agrees with thesolution of the classical equilibrium equation in Theme 127.



16 Chemical . . .Le Chateliér principle ENGLISH text is missing.Graphical solution1 Figure 16.2 illustrates the equilibrium principle in an easy-to-grasp

way, but it is only possible to illustrate systems with one degree of free-dom like this.

2 Here we should remember that our system has a total of four chem-ical composition variables, which means that the graph is an exampleof a one-dimensional manifold in four-dimensional space.

3 In other words, describing chemical equilibria geometrically is com-plicated.

936 / 1776

§ 1291 Find an analytical expression for min G(n(ξ))T ,p , based on thesame function as the one used in Theme 128. Show that the resultagrees with Theme 127.

937 / 1776

§ 129 Minimum Gibbs energy II1 The necessary condition for a minimum value of G is that the

gradient ∂G/∂ξ = 0.

2 Alternatively we can choose to divide by RT in order to obtain adimensionless expression. This does not alter the necessary conditionfor a minimum, as the temperature of the system remains constant.

3 If we subsequently differentiate with respect to ξ, we get

0 = − 0.363 + 2 ln ξ+ 2ξξ − ln(1 − ξ) − 1−ξ

1−ξ − ln(3 − ξ) − 3−ξ3−ξ

= − 0.363 + ln ξ2

(1−ξ)(3−ξ) .

It is now easy to rearrange the expression to show that it is identical tothe equilibrium equation solved in Theme 127. Clearly, the tangent ofthe graph has a zero slope in the minimum point. Besides, the graphitself is convex in the same point. These are the necessary and suffi-cient conditions of a function minimum.

938 / 1776

ENGLISH text is missing.Disturbances1 By taking as our starting point the reaction aA + bB ⇔ cC + dDin Sections 16.4 and 16.5 we have seen that the equilibrium principlemin G(n(ξ))T ,p results in the condition aµA + bµB = cµC + dµD .

2 The assumption is that ξ [mol] refers to the extent of reaction of thesystem— selected such that any increase in ξ moves the reaction fromthe left to the right. The composition of the reacting system can then atall times be expressed as

NA = NA − aξ , NC = NC + cξ ,

NB = NB − bξ , ND = ND + dξ .

3 By varying ξ it is possible to find a minimum value of G(n(ξ)) whichsatisfies the equilibrium condition, cf. Figure 16.2.

4 But what happens to the equilibrium if the external conditionschange, i.e. if the temperature, pressure or compositions of NA , NB ,NC and ND are altered?

5 The famous Le Chateliér’s principle[a] [a] states that: if you imposean external change on a system at equilibrium, then the equilibriumshifts to reduce or eliminate the effect of the imposed external change.

939 / 1776

[a] Henri Louis Le Chatelier, 1850–1936. French chemist.

[a] Aschehougs konversasjonsleksikon. Aschehoug & co., 5th edition, 1974. In Norwe-gian.

ENGLISH text is missing.Le Chateliers principle1 For example, any rapid increase in pressure will cause the reac-

tion, which has now been driven away from equilibrium, to move in a di-rection that results in the system taking up less space, i.e. the pressuregradually decreases as the equilibrium is reasserted.

2 This simple principle is incredibly elegant, but it can easily be mis-interpreted.

3 History also tells us that Le Chateliér presented his ideas througha number of works in 1884, 1888, 1908 and 1933, which also suggeststhat the principle is not quite perfect after all.

940 / 1776

ENGLISH text is missing.Adding mass1 The following example reveals part of the problem: Assume that

the equilibrium reaction AB2 ⇔ A + 2B is described by Q = yA y2B/yAB2

and f(p) = 1, where Qeq = Kc cf. Eq. 16.10.

2 The definition of the reaction quotient Q shows us that Q < Kc

moves the reaction to the right, whereas Q > Kc moves the reaction tothe left.

3 If we add e.g. component A to the system the equilibrium will bedisturbed, but in which direction will the reaction then move?

4 Based on Le Chaterlier’s principle we might draw the overly hastyconclusion that the added component A must always be consumed, asthis would counteract the addition of the component.

5 The requirement would then be that (∂Q/∂NA ) > 0 for all values ofKc , but this is not true.

6 In fact, if we differentiate the reaction quotient we get( ∂Q∂NA

)= N-2 N2B

NAB2− 2N-3 NA N2B

NAB2

= N-1A (1 − 2yA )Q ,

which shows that the change in Q is governed by the stoichiometry ofthe system (the derivative changes sign at yA = 0.5) and not by theequilibrium condition.

7 Consequently, Le Chateliér’s principle does not apply if you addmore mass of a component to an equilibrium system.

8 The direction of the reaction is in this case determined by 1−2yA ,eq,as shown in the table below:

Case yA ,eq yB ,eq yAB2,eq Kc ( ∂Q∂NA) Direction

i 2/3 1/6 1/6 1/9 −1/18 →ii 1/2 1/4 1/4 1/8 0 −iii 1/3 1/3 1/3 1/9 +1/9 ←

941 / 1776

ENGLISH text is missing.Closed system changes1 In other words, adding a component to the mixture is not equivalent

to imposing an external change on the system.

2 It now remains to be seen what happens in a closed system.

3 Our task is to give the principle a precise mathematical definition,so that we can remove all doubt as to what conditions apply.

4 We will take as our starting point the dimensionless Gibbs energy

Γ = (RT)-1∆rxG = (RT)-1∑iνiµi ,

where νi represents the stoichiometric coefficients a, b, etc.

5 At equilibrium Γeq = 0 and for changes in the state of equilibriumdΓeq = 0.

6 We will first derive the differential dΓ using appropriate coordinatesand then study the effect of changing the temperature, pressure andcomposition with one reservation: In this problem it is practical to used(1/T) and dln p as free variables rather than dT and dp.

7 Our point of departure is the chemical potential of an ideal gas,which is the same as in Theme 126 on page 862.

8 In a dimensionless form it can be written (RT)-1µıgi = (RT)-1µi +

ln(Nip/Np).

9 The total differential is:

d(µıg

iRT

)=

(∂µi /RT∂(1/T)

)d(1T)+ dln p + dln Ni −

∑j

(∂ ln N∂ ln Nj

)dln Nj

=hiR d

(1T)+ dln p + dln Ni −

∑j

yj dln Nj .

942 / 1776

ENGLISH text is missing.Mathematical analysis1 In the above equation, Gibbs–Helmholtz’ equation 4.46 on

page 161 has been applied to the first of the derivatives on the rightside, and (∂ ln N/∂ ln Nj) = (Nj/N) (∂N/∂Nj) = yj has been applied tothe other derivative.

2 In general terms, the changes to the composition of the system canexpressed as

dln Nj = N-1j dNj = N-1

j (dNj + νj dξ) ,

but since the system is closed dNj = 0 ∀j.

3 This means that dln Nj = νj N-1j dξ, from which it follows in the

normal way that the differential of Γ is:

dΓ =∑iνi d

( µiRT

)

=∑iνi

hiR d

(1T)+

∑iνi dln p +

∑iνi

νiNi dξ −∑

iνi

∑j

yjνj

Nj dξ ,

but for the sake of the subsequent explanation we will rewrite the ex-pression in vectorial [a] form

dΓ = ∆rxH

R d(1T)+ (

∑iνi) dln p + [qTq − (qT√y)2] dξ , (16.11)

where qi = νi/√

Ni og yi = Ni/N.

4 The first and second terms in this equation have clear physicalinterpretations, whereas the meaning of the third term is less clear.

5 But if we apply the chain rule to Γ we get the result(dΓdξ

)T ,p =

∑j

( ∂Γ∂Nj

)T ,p

(dNj

dξ)=

∑j

∑iνi

(∂µi/RT∂Nj

)T ,pνj =

∑j

∑iνi

(∂2G/RT∂Ni∂Nj

)T ,pνj

=(d2G/RT

dξdξ)T ,p = νTGν ,

and by inserting the values for an ideal gas we find that νTGν = qTq −(qT√y)2 where G is the matrix of second order derivatives of the Gibbsfree energy with respect to the mole numbers.

6 It is also known as the Hessian of G.

7 In other words, the third term in the Eq. 16.11 corresponds to thesecond order derivative of G/RT with respect to the extent of reactionξ.

8 Figure 16.2 demonstrates that this is in fact true in practice.

9 Here the truncated graph shows a second order Taylor series[a] forG(ξ) around the minimum, i.e. G = Geq + (νTGeqν)(ξ − ξeq)2 + O(ξ3).

10 Around the equilibrium point, the graph clearly bends upwards, andwe can actually apply this result to all reactions in ideal gas mixturesover the whole domain of definition.

11 The Cauchy[a]–Schwarz[a] inequality states that |qT√y| ≤ ||q|| ·||√y||, but as ||√y|| = (

∑i |yi |)0.5 = 1, then |qT√y| ≤ ||q||. From this it

follows that

qTq − (qT√y)2 ≥ qTq − qTq = 0 ,

which proves that νTGν ≥ 0 across the entire composition range.

12 For random changes in the state of equilibrium dΓeq = 0, and if wesubstitute the above result into Eq. 16.11 it is easy to calculate the shiftin the equilibrium composition as a result of changes in 1/T and ln p:

( dξd(1/T)

)eq = −(νTGν)-1 ∆rxH

R ,

( dξdln p

)eq = −(νTGν)-1

∑iνi .

943 / 1776

[a] The summation ∑i νi νiNi −

∑i νi

∑j yj

νjNj can be expressed as ∑

i(νi/√

Ni)2 −∑

i(νi/√

Ni)√

Ni/N ·∑

j(νj/√

Nj)√

Nj/N or ∑i q2i −(∑

i qi√yi)·(

∑j qj√yj) where qi = νi/

√Ni .

In vector form the expression reads qTq − (qT√y)2.

[a] Simplifies to a constant and a second order term because the first order term iszero, in accordance with the equilibrium criterion dG /dξ = 0.

[a] Augustin Louis Cauchy, 1789–1857. French matematician.

[a] Hermann Amandus Schwarz, 1843–1921. German mathematician.

1 These equations conclude Le Chateliér’s principle for a chemicallyequilibrated, closed system.

2 A summary of the principle, as it applies to ideal gas mixtures, isset out below:

Criterion Description dT dp Direction∆rxH > 0 endothermic + 0 →∆rxH < 0 exothermic + 0 ←∑

i νi > 0 expansion 0 + ←∑i νi < 0 contraction 0 + →



16 Chemical . . .Looking back

1 256 non-radioactive nuclei (80 different elements altogether)

2 Standard states: He(gas), O2(gas), C(sol), Br2(lıq), etc.

3 Chemical reactions: νAA + νBB + · · · = 0

4 Ideal gas: Q(y) · f(p) = K(T)

5 Reaction equilibrium:n∑

i=1νiµi = 0

6 van’t Hoffs slope:( ∂ ln K∂(1/T)

)= − ∆rxH

R

7 Minimum Gibbs energy: Geq = min G(n(ξ)))T ,p



16 Chemical . . .Looking back (2)

8 Le Chateliér’s principle:( dξd(1/T)

)eq

= −(νTGν)-1∆rxH

R( dξdln p

)eq

= −(νTGν)-1∑iνi


News Stoichiometry min G Lagrange Null space Combustion Flame temperature Olds

Part 17

Simultaneous reactions




17 Simultaneous . . .New concepts

1 Atom balances

2 Formula matrix

3 Null space

4 Independent reactions

5 Lagrange multipliers

6 Extent of reaction

7 Newton–Raphson iteration

8 Adiabatic (flame) temperature



17 Simultaneous . . .Linear algebra

1 From Chapter 16 we know that one single reaction equation can besolved using a polynomial function of the mole fractions, but in order tosolve several simultaneous reactions there is a need for more abstractmethods.

4 2 It is indeed quite possible to write down explicit equations for all ofthe concurrent reactions using the same format as explained in Chap-ter 16, but the challenge is to select the correct (physical) root of eachpolynomial (there is one for each independent reaction).3 Linear algebra turns out to be a good option in this respect, as itprovides us with an opportunity to structure the equations in a goodway, and at the same time is capable of solving the general equilibriumproblem without having to make arbitrary choices between several pos-sible solutions. Below we shall investigate the combustion of methane (CH4) in air

(0.78N2, 0.21O2, 0.01Ar) which gives the reaction products CO, CO2,H2O, H2, OH, H and NO.



17 Simultaneous . . .Linear algebra (2)

5 Altogether there are 11 compounds involved, but matrix notationsimplifies the mathematical representation considerably. 6 It is of course possible to choose a more traditional approach in-

volving stoichiometric coefficients and chemical symbols, but it is notclear that doing so would make it any easier for us.7 If the matrix notation is unfamiliar, it is advisable to use both ap-

proaches side-by-side until you are sufficiently experienced with it.

CH4

N2,

O2,

Ar

CO

2,

H2O,

CO,e

tc.

Figure 17.1: Sketch of an adiabatic gas burner



17 Simultaneous . . .§ 130 § 129 § 131

Write down the atom balances for the burner in Figure 17.1on the preceding page. Assume stationary conditions.Is there any connection between the atom balances

and the mass balance Expanded footnote [a]:

[a] Hint: Assume that the symbols C and H are assigned the correct atomic masses inthe chemical formula CH4. By inserting ∗ and + in the appropriate order you will seethat C + H ∗ 4 is an algebraic expression for the molecular mass of methane. of the system?



17 Simultaneous . . .§ 130 Atom balances

1 At stationary conditions there are exactly the same number ofmoles of C, H, N, O and Ar leaving the burner as there are enteringthe burner.

2 The atom balance for the apparatus is hence

~bın = ~bout ,

3 where ~b is the molar flow rate of all of the various atoms involvedin the problem:

~b = ( ~NC~NH

~NN~NO

~NAr )T .



17 Simultaneous . . .§ 130 Atom balances (2)

4 Making use of the system’s atom matrix (atoms × components)(also called the stoichiometric matrix)

A =

CH4 N2 O2 Ar CO CO2H2O H2 OH H NO

1 0 0 0 1 1 0 0 0 0 04 0 0 0 0 0 2 2 1 1 00 2 0 0 0 0 0 0 0 0 10 0 2 0 1 2 1 0 1 0 10 0 0 1 0 0 0 0 0 0 0

C

H

N

O

Ar

(17.1)

5 and the molar flow rate of all the components in the mixture

~n =(~NCH4

~NN2~NO2

~NAr~NCO

~NCO2~NH2O

~NH2~NOH

~NH~NNO

)T, (17.2)

it is possible to express the molar flow rate of the atoms as ~b = A~n.



17 Simultaneous . . .§ 130 Atom balances (3)

6 The atom balance is thus

A~nın = A~nout . (17.3)

9 7 The relationship with the mass balance is simple.8 Let mw be the molar mass of the atoms. The mass flow rate is then ~M = mT

w~b, which when substituted into

~bout = ~bın gives ~Mın = ~Mout.

10 From this we can conclude that the mass balance is taken care ofby the principle of atom conservation.

11 This is a straightforward consequence of the mass balance beinga scalar (inner) product over the (vectorial) atom balance.12 Hence the principle does not work the other way around (the atom

balance is not given by the mass balance).



17 Simultaneous . . .§ 131 § 130 § 132 ENGLISH text is missing.Atom balance1 An atom balance represents a general principle that can be ap-plied to all chemical reaction problems with the exception of nuclearprocesses in atomic reactors (fission) and inside the sun (fusion).

2 There must the atom balance principle give way to the more gen-eral principle of conservation of the nucleon number in the reactor.

3 At even higher energy densities we have to reformulate this princi-ple too, until we finally end up with Einstein’s equation e = mc2.

4 At that level, the mass and energy balances do not apply separatelyanymore.

955 / 1776

Set out the component balance for the burner inFigure 17.1 on page 882. Assume stationary conditions.

Is there any link between the system’s componentand mass balances?



17 Simultaneous . . .Mass balance principle

1 From Theme 130 we know that any arbitrary change in the numberof moles of the components must fulfil the conditions

~bout − ~bın = A(~nout − ~nın) = A∆~n = 0 . (17.4)

2 The atom matrix in Eq. 17.1 has 5 rows and 11 columns.

4 3 This means that Eq. 17.4 has 5 specified values and 11 unknowns,giving 11 − 5 = 6 degrees of freedom. We can therefore solve the set of equations with respect to 5 molar

flow rates, while the remaining 6 must be considered independent vari-ables. First we partition Expanded footnote [a]:

[a] The partitioning of A = ( A1 A2 ) and ∆~n = ( ∆~nT1 ∆~nT2)T )T requires A1 tobe quadratic, and the number of elements in ∆~n1 must also be equal to the numberof elements in ~b. Moreover, A1 must be invertible, but we don’t have space to gointo the details of that here. Suffice to say that it will always be possible to performthe necessary operations, albeit through a more laborious process involving partialpivoting of the coefficient matrix. the atom matrix and flow rate vector, and then

we left-multiply by A-11 :

( A1 A2 )(∆~n1

∆~n2

)∼ ( I A-1

1 A2 )(∆~n1

∆~n2

)= 0 . (17.5)



17 Simultaneous . . .Mass balance principle (2)

5 Finally we solve the set of equations with respect to ∆~n1 =

−A-11 A2∆~n2, which gives us the solution vector:

∆~n =( −A-1

1 A2

I

)∆~n2 = Nξ . (17.6)

7 6 Note that the first I matrix has dimension dim ~n1 × dim ~n1 whereasthe second I matrix has dimension dim ~n2 × dim ~n2 in the equationsabove. N = νij is the system’s stoichiometric matrix, where νij is the stoi-

chiometric coefficient of component i in reaction j.

8 The elements of vector ξ = ∆~n2 are the rates of reaction of theindependent reactions.



17 Simultaneous . . .§ 131 Component balances

1 If we combine equations 17.5 and 17.6, it is clear that AN = 0. Wecan say that N lies in the null space of A.

3 2 This is an important result, which amongst other things means thatthe mass and atom balances are part of the component balance. The reverse is only true to a limited extent: Given a matrix N we

can always calculate a matrix A′ with a rank equal to the number ofreaction invariants in the system.

4 But A′ will not necessarily be of the same shape as the atom ma-trix, as for the latter dim A > rank A if some of the atoms are alwayspresent in specific ratios.

5 For example, this is true of the equilibrium 2 NO2 ⇔ N2O4. Herethe ratio between nitrogen and oxygen is 1:2 on both sides of the reac-tion arrow, which means that (dim A = 2) > (rank A = rank A′ = 1).



17 Simultaneous . . .§ 131 Component balances (2)

7 6 The atom matrix is surjective, whereas the stoichiometric matrix isonly injective:

A surinj N If we substitute in the figures from Theme 130, the final result weget is (check that AN = 0):

∆~n = 18

0 −4 −4 −2 −2 00 0 0 0 0 −4−4 −6 −2 −5 −1 −4

0 0 0 0 0 0−8 4 4 2 2 0

8 0 0 0 0 00 8 0 0 0 00 0 8 0 0 00 0 0 8 0 00 0 0 0 8 00 0 0 0 0 8

ξ . (17.7)



17 Simultaneous . . .§ 131 Component balances (3)

8 The factor 1/8 has been used in order to give us integers in thematrix.

9 Note that the system’s rates of reaction are wholly determined bythe components included in the ∆~n2 vector, which in this case meansCO2, H2O, H2, OH, H and NO.10 It is possible to choose a different basis by changing the order in

the n vector, but the columns in A must be in the same order as therows in n.



17 Simultaneous . . .§ 132 § 131 § 133 ENGLISH text is missing.Inert compounds1 As previously mentioned, A1 must be invertible.

2 This means that the rows and columns in the matrix must be lin-early independent.

3 Isn’t it odd, then, that the fourth row of N is 0?

4 What is the physical explanation for this? Hint: Which compoundis fourth in the atomic number vector, and how reactive is it?

5 compound Inert .962 / 1776

Find a (mathematical) set of independent reactionequations for the system in Theme 131. What are

your options?



17 Simultaneous . . .§ 132 Reaction equations

1 In this case there are 11 − 5 = 6 independent reactions. Moregenerally, this number is given by dim(n) − rang(A).

2 A set of independent reaction equations can be inferred directlyfrom matrix N in Eq. 17.7:

4 O2 + 8 CO = 8 CO2

4 CH4 + 6 O2 = 4 CO+8 H2O

4 CH4 + 2 O2 = 4 CO+8 H2

2 CH4 + 5 O2 = 2 CO+8 OH

2 CH4 + 1 O2 = 2 CO+8 H

4 N2 + 4 O2 = 8 NO



17 Simultaneous . . .§ 132 Reaction equations (2)

3 This is just one of the many options, and there are countless com-binations of these reaction equations — we say that they constitute abasis of the vector space.



17 Simultaneous . . .Minimum Gibbs energy

1 Finding the energy minima of systems is essential to all thermody-namic equilibrium calculations.

2 This also applies to chemical reactions in systems where the tem-perature has been measured, or estimated by some other means, andwhere the pressure is known.

3 For a closed system at constant T and p, the minimum energystate is:

Geq = minn

G(T ,p,n) ,

A∆n = 0 ⇔ ∆n = Nξ .



17 Simultaneous . . .Minimum Gibbs energy (2)

4 The only difference between these two formulations is in the massbalance term, but the difference is big enough to mean that differentsolution algorithms are required. 5 There was a breakthrough in this field in the 1960s, as a direct

result of the US space programme.6 The historical development of the algorithms is discussed in sub-chapters 17.4 and 17.5, but before we embark on that discussion wemust first derive the equilibrium conditions in a form that is conducive tosolving the equations numerically.



17 Simultaneous . . .§ 133 § 132 § 134

Minimize G(n) in order to show that µ = ATλ andNTµ = 0 are two equivalent criteria for multicomponentchemical equilibria based on Lagrange multipliers andthe reaction stoichiometry of the system respectively.



17 Simultaneous . . .About Lagrange’s method

1 2 The Lagrange[a] method is a topic that belongs under multivariateanalysis, and so we will limit ourselves to a quick glance at the underly-ing principles.

[a] Joseph Louis Lagrange, alias Giuseppe Luigi Lagrangia, 1736–1813. French math-ematician. In brief, the method involves finding the stationary points of the

Lagrange function

L(λ,n) = G (n) − λTA∆n ,

where λ is a vector of the Lagrange multipliers.

3 Based on classical optimisation theory it can be demonstrated thatall Gibbs energy minima must satisfy the following conditions:

∂L∂n

= ∂G∂n− ∂(λTA∆n)

∂n= 0 ,

∂L∂λ = ∂G

∂λ −∂(λTA∆n)

∂λ = 0 ,

where ∂/∂n = ( ∂/∂N1 ∂/∂N2 . . . ∂/∂Nn )T and 0 =

( 0 0 · · · 0 )T. Our next task is to rewrite the partial deriva-tives of L in a form that is relevant to thermodynamics.



17 Simultaneous . . .§ 133 Lagrange multipliers

1 We immediately recognise ∂G/∂n as being the chemical potentialvector µ, and if we give it a little bit of thought it is clear that ∂G/∂λ = 0,since G(T ,p,n) is not dependent on λ.

2 Now what we need to do is to differentiate λTA∆n with respect toλ and n.

3 Let us start by writing the inner product in summation notation

f = λTAn =∑i

∑j

Aijλi∆Nj

before differentiating.



17 Simultaneous . . .§ 133 Lagrange multipliers (2)

4 Finally the results can be brought together in the following matrix–vector expressions

∂f∂n

=

∂f∂N1∂f∂N2...∂f∂Nn

=

∑i

Ai1λi∑i

Ai2λi

...∑i

Ainλi

= ATλ , ∂f∂λ =

∂f∂λ1∂f∂λ2...∂f∂λm

=

∑j

A1j∆Nj

∑j

A2j∆Nj

...∑j

Amj∆Nj

= A∆



17 Simultaneous . . .§ 133 Lagrange multipliers (3)

5 The conditions for a stationary point can be summarised as follows:

∂L∂n

= 0 ⇒ ATλ = µ , (17.8)∂L∂λ = 0 ⇒ A∆n = 0 , (17.9)

with the two equations representing the criterion for equilibrium and themixture’s atom balance respectively. 6 Together, they define the conditions for reaction equilibrium in a

closed system at a given T and p.



17 Simultaneous . . .§ 133 Extent of reaction ENGLISH text is missing.Multipliers1 In thermodynamics, Lagrange’s method of undetermined multipli-

ers is not as obscure as you might think, and in the example above λcan be seen as a vector of the chemical potentials associated with theatoms that comprise matrix A; see Theme 136 on page 909 for a com-prehensive discussion of this.

973 / 1776

1 Alternatively, the criterion for equilibrium can be derived by lookingat the independent extents of reaction as free variables.

2 For a system with n compounds and m atom balances, there willbe r = n −m independent reactions.

3 For each independent reaction there is precisely one extent of re-action, as shown in Theme 131 on page 887.



17 Simultaneous . . .§ 133 Extent of reaction (2)

4 The extents of reaction are linearly independent, and the minimumGibbs energy can therefore be calculated by finding the stationary pointof the implicit function G(n(ξ)). The chain rule for partial differentiationgives us

(∂G∂ξ

)T ,p

=

∑i

∂G∂Ni

∂Ni

∂ξ1∑i

∂G∂Ni

∂Ni

∂ξ2...

∑i

∂G∂Ni

∂Ni

∂ξr

= 0 .

5 Next we introduce the stoichiometric matrix NT = ∂Ni/∂ξj and thechemical potential vector µ = (∂G/∂Ni)T ,p.



17 Simultaneous . . .§ 133 Extent of reaction (3)

6 This gives us the following criterion for equilibrium:

NTµ = 0 . (17.10)

7 It is worth noting that the solution obtained using Lagrange multi-pliers 17.8 also satisfies the condition for equilibrium 17.10.

9 8 We can show this by combining the two equations, which gives usNTATλ = 0. Since N is designed to ensure that AN = 0, this will be true for all

possible (equilibrium) values of λ.



17 Simultaneous . . .Newton–Raphson–Lagrange iteration

1 In order to obtain a numerical solution to the equilibrium Eq. 17.8,µ and λmust be iterated simultaneously subject to the constraints givenby the atom balance 17.9.

4 2 There are a number of algorithms for solving this type of problem,but it is natural to use the Newton–Raphson method of iteration if themixture has a moderate number of chemical components.3 The weakness of the Newton–Raphson method is that it requires

you to invert a matrix with n2 elements, but for systems with 10–30compounds this isn’t a problem in practice. Linearisation of Eq. 17.8 gives us:

ATλk + AT(λk+1 − λk ) = µk + G(nk+1 − nk ) .

5 The matrix G = (∂µ/∂n)T ,p is often referred to as the Jacobian ofµ, but since µ = (∂G/∂n)T ,p we can alternatively express it as G =

(∂2G/∂n∂n)T ,p.

6 The matrix, and other (symmetrical) matrices containing secondderivatives of scalar functions of state, are called Hessian matrices.



17 Simultaneous . . .Newton–Raphson–Lagrange iteration (2)

10 7 The symmetry of G results in n(n − 1)/2 Maxwell relations, but inthis case they are of no practical importance.8 However, the fact that Hessian matrices are symmetrical is impor-tant in relation to convergence analysis and optimisation theory.9 The discussion of the Newton–Raphson method of iteration in Ap-pendix 27 is an example of this. Combining the linear version of the condition for equilibrium with

Eq. 17.9, which is naturally linear, gives us the following iterative modelfor the equilibrium problem:

NkRT Gk AT

A 0

1Nk

[nk+1 − nk ]

−1RT [λk+1 − λk ]

=

−1RT [µk − ATλk ]

0

. (17.11)

11 The expression has been written in dimensionless form, in orderto emphasise the fact that the iteration sequence is independent of thesize of the system Expanded footnote [a]:

[a] It is easy to see that N-1k [nk+1 − nk ] is a dimensionless measure of the changein the composition of the system, and that this vector actually expresses the changein the mole fractions in the mixture if the number of moles remains constant duringthe reaction: Nk+1 = Nk . For the matrix Nk (RT)-1Gk the situation is more complex,and here it is important to understand how the Gibbs energy G(n) relates to Euler’shomoegeneous function theorem. From Chapter 5 we know that G is a homogeneousfunction of degree one, whereas µ is a function of degree zero. By differentiatingyet again we can obtain G, which is a homogeneous function of degree minus one.Multiplying by Nk thus gives us a matrix that is independent of the size of the system.The effect of this scaling on the thermodynamic properties in the set of equations isinvaluable to our basic understanding of thermodynamic equilibrium calculations. .

12 In order to start the iteration we need an initial value n > 0 consis-tent with the system’s atom balance. 13 In practice this will be the composition of the reactor feed stream,

or the mixture of reactants if the reaction is taking place in a closedspace.14 As such, the initial estimate is not a big problem, but in certaincontexts we only know the system’s atom balance, or the number ofmoles of some of the components may be zero. There are two ways to approach

the matter:



17 Simultaneous . . .Newton–Raphson–Lagrange iteration (3)

1 components of which there are zero moles are given an arbitrary smallvalue ǫ and the inaccuracy in the atom balance Ank − b is moved to theright-hand side of the set of equations in the hope that the discrepancywill become smaller in the next iteration.

2 use of linear programming to calculate a feasible n > 0 that isconsistent with the atom balance.

vmode (bug): hmode expected 15 The first option is perhaps more intuitive, but the discrepancy in theatom balance can in some cases lead to the iterations diverging.16 The second option requires greater mathematical expertise, buthas the benefit of having better convergence properties.17 Incidentally, Appendix 29 gives a brief introduciton to linear pro-gramming.



17 Simultaneous . . .§ 136 § 135 § 137 § 1341 Eq. 17.11 represents the classical non-stoichiometric formulation

of an equilibrium problem in a homogeneous phase, i.e. in a gas, liquidor solid solution. Rewrite the equation in such a way that it explicitlyapplies to an ideal gas. Use the structure of the model to simplify theset of equations as much as possible.

980 / 1776

§ 134 RAND (method)1 The equilibrium is not affected by the state space model used, butif some of the mole fractions become very small (which they typicallydo in combustion equilibria), the inverse matrix coefficient can becomenumerically unstable.

2 This is because the diagonal elements in Gk are inversely propor-tional to yi when yi → 0.

3 We are therefore going to rearrange the set of equations in orderto make it more stable.

4 When an ideal gas is introduced into Eq. 17.11 the chemical po-tential vector and Hessian matrix can be written

1RT µk = 1

RT µ⋆ + ln yk ,

NkRT Gk = y-D

k − eeT ,

where yk = N-1k nk and µ⋆ = µ + RT ln(p/p).

5 The reference potential µ⋆ is in this case independent of the itera-tion sequence, as the equilibrium requires T and p to be constant.

6 Also note that y-Dyk ≡ e and that eTyk = 1, which means thatGk yk = 0.

7 This relationship is a consequence of the homogeneity of G, cf.Theme 34 on page 188, and is true of all Gibbs energy functions, al-though it is easiest to verify for an ideal gas.

8 In the above example, the homogeneity means that Gk [yk+1−yk ] =

Gk yk+1.

9 This enables us to solve the equations for the new state nk+1 with-out having to use the update nk+1 − nk .

10 Furthermore, the update λk+1 −λk can be replaced by the solutionλk+1.

11 This is made possible by subtracting ATλk from both sides of theequation, and is a general property of all minimisation problems withlinear constraints.

12 All of this means that the equation can be rewritten as:

y-Dk − eeT AT

A 0

1Nk nk+1

−1RT λk+1

=

−1RT µ⋆ − ln yk

Ayk

. (17.12)

Unlike in the case of Eq. 17.11, the solution to Eq. 17.12 obtained byiteration is only valid for an ideal gas.

13 In spite of its limited validity, this equation is a good starting pointfor learning more about the homogeneity of the Gibbs energy function.

981 / 1776

ENGLISH text is missing.RAND (ideal gas)1 Working out what to do next with Eq. 17.12 requires a certain

amount of cunning and a bit of imagination.

2 What is obvious is that we cannot perform a normal Gaussian elim-ination on the coefficient matrix, as the upper left-hand block y-D

k − eeT

is singular. This can easily be shown by multiplying by yTk .

3 However, it is still possible to multiply the first row of the set ofequations by yD

k , and then substract the result from the last row.

4 This results in the modified equation

I − yk eT yDk AT

Ayk eT −AyDk AT

1Nk nk+1

−1RT λk+1

=

−yDk [

1RT µ⋆ + ln yk ]

Ayk + AyDk [

1RT µ⋆ + ln yk ]

,

from where the set of equations can be split into two sub-problems. Thefirst row of the above equation gives us the expression

1Nk nk+1 = yk 1

Nk eTnk+1 + yDk [

1RT ATλk+1 − ( 1

RT µ⋆ + ln yk )] . (17.13)

5 Our job is then to determine the inner product eTnk+1 = Nk+1

independently of nk+1.

6 On the face of it this is impossible, but it turns out that multiplyingthe first row of the equation by eT allows us to split it, because theequation can now be rewritten as:

0 yTk AT

Ayk AyDk AT

1Nk eTnk+1

1RT λk+1

=

yTk [

1RT µ⋆ + ln yk ]

Ayk + AyDk [

1RT µ⋆ + ln yk ]

.

(17.14)

7 The elements in the vector on the right-hand side can alternativelybe written as gıg

k /RT and bk + AyDk µ

ıgk /RT . The result will be the same.

8 A trivial change of rows and columns moves the zero element fromthe top left down to the bottom right corner, which ensures that Gaus-sian elimination is stable for the coefficient matrix without it first need-ing to be pivoted.

9 The presence of small mole fractions does not result in any majorproblems during inversion, at least if a row transformation is performedon A to give [I B], with the minor components being placed in matrix B.

10 The equilibrium problem itself has incidentally been split into twosub-problems.

11 First we need to solve Eq. 17.14 with respect to the Lagrange mul-tipliers λk+1 and the size of the system eTnk+1.

12 Then these results are substituted into Eq. 17.13, which gives thenew composition vector nk+1.

13 What is interesting is that this calculation always converges[a] pro-vided that we ensure that none of the numbers of moles in Eq. 17.13are negative.

14 In practice we can do that by specifying that τ ≤ 1, and hence thatthe mole number vector nk + τ(nk+1 − nk ) > 0.

982 / 1776

[a] Minimising the Gibbs energy of an ideal gas is an example of a convex problem.

§ 1351 Computerise the function [n] = MinGibbs(n,T ,p,µ,A) using

Eq. 17.14 as your starting point.983 / 1776

§ 135 RAND (implementation)1 See Function 2.2 in Appendix 31. This is a one-to-one correspon-

dence between this function and Eq. 17.14.

2 To enable us to handle components of which there are zero moles,we give them an arbitrary, small value before starting to iterate. Mean-while, the inaccuracy in the mass balance is moved to the right-handside of the set of equations.

3 This simple strategy works well in many cases, but not for stoichio-metric mixtures in which the equilibrium is far to the right or left.

4 In those cases the precision of the computer may affect the out-come of the calculation.

5 Problems associated with limitations in computer precision canlargely be eliminated by the transformation A = ( I B ), with the mi-nor components being placed in matrix B.

984 / 1776

A dimensional analysis of µ = ATλ in Eq. 17.8 revealsthat λ can be interpreted as a chemical potential vectorfor the atoms Expanded footnote [a]:

[a] Or, in a more general context, the system’s independent extensive variables. in the system. Assume that µ is knownfrom a previous equilibrium calculation. How can you

then determine λ, and what does this mean in practice?



17 Simultaneous . . .§ 136 Chemical potentials of atoms

1 From the equation µ = ATλ it follows that µ is located in the columnspace of AT.

3 2 This is an overspecified problem because dim(µ) > dim(λ), butsince the equilibrium state does in fact exist, any invertible sub-systemcan be used to calculate λ from µ. In order to avoid using random basic variables, it is also possible to

use the least square solution

λ =(AAT

)-1Aµ .

4 After minimising the Gibbs energy you can check whether any new(and more important) minor components have been formed that werenot originally expected.

5 Let the stoichiometric matrix for the new compounds be B.



17 Simultaneous . . .§ 136 Chemical potentials of atoms (2)

6 The equilibrium potential is by definition π = BTλ and the minormole fractions become

yi =pp exp

(πi−πiRT

),

assuming that the mixture is an ideal gas. It is now easy to decidewhether any of the new compounds should be included in the minimi-sation.



17 Simultaneous . . .Newton–Raphson null space iteration

1 Any numerical solution to the equilibrium Eq. 17.10 on page 905must be found in the null space of the stoichiometric matrix A. 2 The Newton–Raphson–Lagrange method used in Eq. 17.11 also

has this property, but only as a result of a loose assumption that theo-retically does not always hold.3 This time we are going to solve NTµ = 0 in such a way that themass balance is correct after each iteration. Linearis-

ing the equilibrium equation gives us

NTµk + NTGk (nk+1 − nk ) = 0 .

Here nk+1 − nk has to occupy the null space of A, which means thatnk+1 − nk = N∆ξ.

4 This leads us to the equation

∆ξ = −(NTGk N

)-1NTµk .



17 Simultaneous . . .Newton–Raphson null space iteration (2)

7 5 The composition is updated by nk+1 = nk + τN∆ξ, where τ ≤ 1 isused to ensure that nk+1 > 0.6 Alternatively we can write the equation in dimensionless form, toemphasise the fact that the iteration sequence is independent of thesize of the system. Meanwhile, the solution ∆ξ is substituted into the update of nk+1:

1Nk

nk+1 = yk − N(NT Nk

RT Gk N)-1

NT 1RT µk . (17.15)

8 Eq. 17.15 shows the classic iteration sequence for the stoichiomet-ric formulation of an equilibrium problem, but unlike in the case of theequivalent non-stoichiometric formulation in Eq. 17.11, assuming idealgas properties will not significantly simplify the result.9 In practice Eq. 17.15 will run into problems if any of the mole frac-tions are very small.10 The reason for this is the same as for the RAND algorithm inTheme 134 on page 909.11 In order to preempt potential problems, it is helpful to perform acolumn transformation on N, to give us [I B]T, where the minor compo-nents are placed in I.



17 Simultaneous . . .CombustionENGLISH text is missing.Null space vs Lagrange1 In terms of calculation efficiency, the non-stoichiometric formulation

generally gives you fewer equations to solve if n >> rang A, whereas thestoichiometric formulation is more efficient if n is a big number and thenumber of reaction equations is strongly limited by kinetic conditions,i.e. when rang N << n.

2 This factor must be weighed up against the fact that Eq. 17.11 hasone matrix product and one inverse, whereas Eq. 17.15 has four matrixproducts and one inverse.

3 The stoichiometric formulation therefore involves 2–3 times moreoperations than the non-stoichiometric formulation if A and N haveroughly the same rank.

4 However, the iteration sequence for Eq. 17.15 is the same asfor 17.11, provided that the latter starts with the constraints providedby the atom balance being met.

5 This may appear to be a surprising result, but the two algorithmshave the same starting point and use the same linearisation.

6 There is therefore no reason why their iteration sequences shouldbe different.

990 / 1776

§ 1371 Computerise the function [n] = MinGibbsNull(n,T ,p,µ,A) fromEq. 17.15.

991 / 1776

§ 137 Null space (implementation)1 See Function 2.3 in Appendix 31.

2 There is a one-to-one correspondence between this function andEq. 17.15, but for the sake of efficiency the basis for N, i.e. [I B]T, isonly updated if there are significant changes in the composition.

3 During each iteration the mole fractions are sorted by increasingvalue, and are compared with the previous sorted vector.

4 If one or more elements in the new vector differ from those in theprevious vector by a factor of at least 106, the new list of components isused to calculate a new basis for N.

5 Components of which there are zero moles are handled with theaid of linear programming, but note that the initialisation requires morecode than the equilibrium algorithm itself.

6 It is often the case that trivial matters such as initialisation are morevulnerable to numerical problems and require more time and effort thanthe main calculation.

992 / 1776

1 2 For high-temperature combustion it is realistic to assume ideal gasbehaviour, such that µi = µi + RT ln(pyi/p). Below we will assume that combustion takes place at a tempera-

ture of 1500 K. 3 We will use the JANAF tables[a] to determine the standard poten-tial µi (1500 K) for all of the compounds mentioned in Theme 130 onpage 883.

[a] JANAF thermochemical tables. 3rd edition. Part I,II. J. Phys. Chem. Ref. Data,Suppl., 14(1), 1985. Once again we have two options: Either we can use a

set standard temperature of 298.15 K

i) µi (1500 K) = ∆fhi (298.15 K) +

1500 K∫

298.15 K

cp,i (T) dT

︸︷︷︸hi (1500 K)−hi (298.15 K)

︸︷︷︸hi (1500 K)

− 1500

si (298.15 K) +

1500 K∫

298.15 K

cp,i(T)

T dT

︸︷︷︸si (1500 K)

, (17.16)



17 Simultaneous . . .Combustion (2)

or we can use a “variable” reference temperature T :

ii) µi (1500 K) = ∆fµi (1500 K) . (17.17)

4 The JANAF tables contain the data that is summarised in Ta-ble 17.2 on the following page.

5 The left-hand column is equivalent to option ii) and the right-handcolumn to option i).



Figure 17.2: Thermodynamic data from JANAF at T = 1500 K.

∆fµi (T) ∆fhi (T)T∫

T

cp,i dT si (T)µi (T) =

hi (T) − Tsi (T)

CH4 74918 −74873 78153 279.763 −416364.5N2 0 0 38405 241.880 −324415.0O2 0 0 40599 258.068 −346503.0Ar 0 0 24982 188.427 −257658.5CO −243740 −110527 38850 248.426 −444316.0CO2 −396288 −393522 61705 292.199 −770115.5H2O −164376 −241826 48151 250.620 −569605.0H2 0 0 36290 178.846 −231979.0OH 16163 38987 36839 232.602 −273077.0H 136522 217999 24982 148.299 20532.5

NO 71425 90291 38729 262.703 −265034.5



17 Simultaneous . . .§ 138 § 137 § 139

The two µi functions 17.16 and 17.17 that are shownin Table 17.2 are very different, but they nevertheless

result in the same equilibrium distribution. Explainwhy this is. Use practical calculations to show that

the equilibrium distribution is indeed the same.



17 Simultaneous . . .§ 138 Reference state

1 2 The difference between the two options lies in the nature of theformation reactions. Option ii) uses a changing reference state in which the stable state

of each element at temperature T and 1 bar has zero enthalpy, whereasoption i) uses a fixed reference temperature of 298.15 K.

3 However, the energies of the chemical elements involved in each ofthe formation reactions cancel each other out when the standard Gibbsenergy is calculated for the equilibrium reactions, and therefore have noimpact on the outcome of the calculation.

6 4 In other words, the standard state affects the values of ∆fµi ,whereas the values of ∆rxGr for the equilibrium reactions remain un-changed.5 This is related to Hess[a]’s law, which states that the total heat offormation of a system is equal to the sum of the heats of formation ofall of the intermediate reactions.

[a] Germain Henri Hess, 1802–1850. Russian physician and chemist. The most direct way to interpret this is to add to the enthalpy of allof the compounds through a virtual change ∆λ in the zero point energyof the individual atoms:

µ⋆ = µ+ AT∆λ . (17.18)



17 Simultaneous . . .§ 138 Reference state (2)

7 Then let us substitute µ⋆ into the equilibrium Eq. 17.10.

8 Since NTµ = 0 at equilibrium, and as we know that N is located inthe null space of A, we can conclude that

NTµ⋆ = NTµ+ NTAT∆λ = 0 . (17.19)

9 In other words, a change in the zero point energy of the atoms (orelements) does not affect the system’s criterion for equilibrium.



17 Simultaneous . . .§ 139 § 138 § 140

Assume that the pressure in the burner is 1 bar and thatthe methane–air ratio varies ±0.1 about the stoichiometric

mixture ratio. Calculate the composition of the smokeas a function of the methane–air ratio, and representthe result graphically in a suitable diagram. Explainwhat happens to the concentrations of the various

components as you approach the stoichiometric point.



17 Simultaneous . . .§ 139 Combustion

1 2 We will use data from Table 17.2 on page 916. The Gibbs energy has been minimised using the Function 2.2; seeProgram 1.27 in Appendix 31 for details.

3 The results for selected compounds are shown on the left-handside of Figure 17.3 on the following page.

4 Note the transition from reducing to oxidising conditions at the sto-ichiometric point.

5 The similarity with an acid–base titration curve is striking. 6 That is not really surprising, when you think about it, as themethane gas is being titrated with oxygen— in a solution of nitrogen.



0.9 1 1.110

−10

10−5

100

0.9 1 1.110

−8

10−6

10−4

CO, O2, H, OH NO (1 and 20 bar)

Mol

efr

actio

n

Air–methane stoichiometryAir–methane stoichiometry

Figure 17.3: Concentrations of CO, O2, H, OH (left sub-figure) and NO (rightsub-figure) as functions of the excess combustion air. The NO content hasbeen calculated at both 1 and 20 bar to show the pressure dependence of theequilibrium.



17 Simultaneous . . .Stationary–adiabatic § 1401 Repeat the calculation that you did in Theme 139, but this timeuse p = 20 bar. Compare the NO concentration with your previouscalculation and comment on the results.

1002 / 1776

§ 140 Le Chateliér[1]’s principle1 The pressure doesn’t affect the NO concentration much if there is

excess air in the system.

2 In that case the main reaction is N2 +O2 = 2 NO, which maintainsthe number of moles of the reactants.

3 If there is a shortage of air, the dominant reaction is N2 +2 H2O =

2 NO+2 H2, and the quantity of NO declines with increasing pressure,as shown on the right-hand side of Figure 17.3 on the previous page.

1003 / 1776

[a] Henri Louis Le Chatelier, 1850–1936. French chemist.

1 An energy balance for stationary, adiabatic conditions shows thatthe energy entering the system per time unit, which is a combination ofmatter and pV work, must also leave the burner at the same rate:

~Hın = ~Hout ,

~Mın = ~Mout .

~H is the enthalpy rate, i.e. the amount of enthalpy carried by the feedand product streams per time unit.

5 2 In spite of the unit of ~H being [J s-1], we cannot use the term powerto describe it.3 Power is energy per time unit supplied in the shape of heat or work,

in other words energy transfers that do not involve mass, whereas ~Hinvolves the physical movement of mass.4 As it happens, the zero point energy is unknown, and hence the

value of ~H is also unknown. The only thing that we do know is that the enthalpy rates into andout of the burner are the same, but we know nothing about the absolutequantity of energy (enthalpy) that flows with the mass over the sameperiod of time.



17 Simultaneous . . .§ 141 § 140 § 142

Write the energy balance for an adiabatic burner insuch a way that the flame temperature is an explicit

variable. Assume stationary conditions.



17 Simultaneous . . .§ 141 Adiabatic temperature I

1 2 Enthalpy and mass are conserved quantities, but how can we com-bine the two balance equations?3 A molar analysis won’t work because the number of moles of com-ponents leaving the burner are greater than the number entering it.4 One approach is to use specific values, but that is not really basedon chemistry.5 The solution is to take the middle road. By selecting the time frame in such a way that the mass entering

the burner exactly matches the mass of the components on the left-hand side of the total reaction equation

a CH4 +b O2 +c N2 + · · · ⇒ d CO2 +e CO+f H2O+ · · ·

the energy balance can be expressed on an ordinary molar basis asAr∑

CH4

Nın,i hi(Tın) =NO∑CH4

Nout,i hi(Tout) , (17.20)

where Ni is the number of moles of each compound in accordance withthe total reaction.



17 Simultaneous . . .§ 141 Adiabatic temperature I (2)

6 Note that for an ideal gas, the partial molar enthalpy is independentof the composition:

hıgi (Tout) = ∆fh

i (298.15 K) +

Tout∫

298.15 K

cp,i (T) dT . (17.21)

7 The combustion is adiabatic, and the outlet temperature Tout is bydefinition equal to the adiabatic flame temperature.

8 This quantity can thus be calculated from the energy balance, pro-vided that the inlet temperature and feed and product streams areknown.

9 If the product stream is unknown, the energy balance must besolved along with the equilibrium problem in an adiabatic equilibriumcalculation.10 We will come back to that later, but for the moment it will suffice

to say that equations 17.20 and 17.21 together represent one equationwith one unknown, assuming that Tın and all Nın,i and Nout,i are knownquantities.



17 Simultaneous . . .§ 142 § 141 § 143

In Themes 139 and 140, the reaction temperature hasbeen set to 1500 K, but by solving the mass and energy

balances simultaneously, the actual temperature canbe calculated. First we will simply estimate an upper

and lower bound for the (adiabatic) flame temperature.Assume a stoichiometric mixture of methane and air,and set the inlet temperature to 25 C. Simplify thecomposition of the air to 80% N2 and 20% O2, and

assume complete combustion to CO2 and H2O. Hint:The heat capacity of the gas rises monotonically with

the temperature.



17 Simultaneous . . .Simplified energy balance

1 Given the abovementioned assumptions, the energy balance forthe burner can be written as ~Hın = ~Hout, and from equations 17.20and 17.21 on page 926 we get the expression

∑CH4,N2,O2

Nın,i∆fhi (Tın) =

∑CO2,H2O,N2

Nout,i

[∆fh

i (Tın) + cp,i(Tout − Tın)

],

where cp,i = (Tout − Tın)-1

∫ Tout

Tıncp,i dT is the mean heat capacity for the

temperature range T ∈ [Tın,Tout].

2 The adiabatic combustion temperature Tout is indpendent of theflow rate and we can, for instance, use one mole of methane as ourbasis, i.e. Nın,CH4 = 1 mol.



17 Simultaneous . . .Simplified energy balance (2)

3 Meanwhile, let us simplify the composition of the air to 20% O2 and80% N2, in order to make the stoichiometry as simple as possible:

CH4 +2 O2 +8 N2︸︷︷︸reactants

= CO2 +2 H2O+8 N2︸︷︷︸products



17 Simultaneous . . .§ 142 Adiabatic temperature II

1 Introducing the mean heat capacity of the reaction products en-ables us to solve the energy balance explicitly with respect to the outlettemperature

Tout = Tın − ∆rxH(Tın)

CP

,

where the following quantities have been used:

∆rxH (Tın) = ∆fhCO2

+ 2∆fhH2O −∆fh

CH4− 2∆fh

O2︸︷︷︸

calculated at Tın

,

CP = cp,CO2+ 2cp,H2O

+ 8cp,N2︸︷︷︸calculated at Tın<T<Tout

.



17 Simultaneous . . .§ 142 Adiabatic temperature II (2)

3 2 In order to find a lower and upper limit for the flame temperature,we need an upper and lower limit respectively for the individual cp,i inthe product stream. Since the heat capacity in this case is a monotonic function of tem-

perature, the upper temperature limit can be estimated by using cp,i ≈cp,i(298.15 K), giving an estimate of 2677 K.

4 The lower temperature limit follows from the conservative assump-tion that cp,i ≈ cp,i(2700 K) > cp,i(2677 K) > cp,i(298.15 K), which givesan estimate of 2021 K.

6 5 These calculations require the enthalpy values shown in Table 17.2and heat capacities taken from JANAF[b].

[b] JANAF thermochemical tables. 3rd edition. Part I,II. J. Phys. Chem. Ref. Data,Suppl., 14(1), 1985. When substituted into the temperature equations, the following nu-

merical results are obtained:Tout

[K]< 298.15 − −393522 − 2 · 241826 + 74873

37.129 + 2 · 33.590 + 8 · 29.124= 2677 ,

Tout

[K]> 298.15 − −393522 − 2 · 241826 + 74873

61.802 + 2 · 54.723 + 8 · 36.801= 2021 .



17 Simultaneous . . .§ 142 Adiabatic temperature II (3)

11 7 This simplified procedure enables us to estimate a conservativetemperature range, but no more than that.8 An adiabatic temperature calculation is and will always be a non-linear problem, and using the mean heat capacity doesn’t make theproblem any simplier.9 It is possible to narrow the temperature range by making suc-cessive improvements to the mean heat capacity, but in practice itis easier to calculate the outlet enthalpy per 100th of a degree Cel-sius and compare the result with the reaction enthalpy ∆rxH(Tın) =

−393522 − 2 · 241826 + 74873 = −802301 J.10 When the sum of the integral of the heat capacity and the reactionenthalpy is zero, we have reached our goal. We can use the JANAF tables to produce the following table:

T [K]∫ T

298.15CP,prod dτ T [K]

∫ T

298.15CP ,prod dτ

2000 686115 2400 868381

2100 731226 2500 914501

2200 776756 2600 960817

2300 822454 2700 999304

12 Linear interpolation of∫ T

298.15CP,prod dτ = 802301 J gives us Tout =

2256 K.13 Note that this temperature is safely within the estimate range Tout ∈[2021,2677]K.14 For the sake of comparison, the actual flame temperature[b] hasbeen measured to be approximately 2150 K.

[b] Lange’s Handbook of Chemistry. McGraw-Hill, 5th edition, 1961. p. 825.



17 Simultaneous . . .Flame temperature

1 In order to calculate the exact flame temperature, we need to simul-taneously solve the energy balance 17.20 and equilibrium criteria 17.8–17.9, and if necessary equilibrium Eq. 17.10.

2 It should be remembered that heat capacities are temperature de-pendent, and that the dissociation equilibria become increasingly im-portant at higher temperatures.

6 3 This means that T and n must be updated simultaneously, or thatT must be calculated in an external loop based on the properties of asystem that is always at chemical equilibrium (convergeng in an internalloop).4 The first of these methods is preferable for simulations where ro-

bustness and flexibility are very important, but there are pedagogicaladvantages to the second method: It highlights the difference betweenthe heat capacity CP at constant composition and the change in en-thalpy (dH /dT)p,b along the equilibrium manifold (see below).5 In certain circumstances there can be a significant difference be-

tween their values, and in such cases the temperature calculation willgive highly divergent results whether or not the product composition isassumed to be constant. In the example above, the difference between the rigorous solution

(2246 K) and the interpolated value at constant product composition(2256 K) is insignificant, and as such it is a poor example, as the dis-sociation equilibria are negligible at temperatures below approximately2500 K.



1000 1500 2000 2500−8

−6

−4

−2

0

1

2

3

4

5−10

1000 5000 90000.5

1

1.5

2

2.5

3

T k [K] T [K]

(dH/

dT) p,b

Iteration sequence Heat capacity

Hk[M

J]

Hın

Figure 17.4: The left sub-figure shows the iteration sequence of an adia-batic calculation where an upper limit of 400 K has been applied to the steplength. The enthalpy values along the ordinate axis represent the combustionof 10 mol methane in air. The right sub-figure shows (dH /dT)p,b along theupper curve and CP = (∂H/∂T)p,n along the lower curve (normalized values).



17 Simultaneous . . .Looking back§ 142 Adiabatic temperature III1 An iterative calculation of the flame temperature can be derived

from equilibrium Eq. 17.10, in other words NTµ = 0.

2 Any change in the equilibrium state must satisfy NT dµ = 0, and ifwe substitute in the total differential of the chemical potential we get

NT dµ = NT[−s dT + v dp + G dn]= 0 ,

where G = ∂2G/∂Ni∂Nj.

3 The equation is generally valid, both for open and closed systems,but in the latter the component balance puts significant constraints onthe behaviour of the system.

4 This is because in a closed system (dn)b = N dξ, as shown inEq. 17.7. Substituting this into the above expression gives us the rela-tion

(dξ)p,b

=(NTGN

)-1NTs dT ,

which tells us how the extent of reaction changes with temperature,assuming complete chemical equilibrium.

5 In order to calculate the change in enthalpy we need the differential

dH = CP dT +(∂H∂p

)T ,n

dp + hT dn ,

and provided that (dn)b = N dξ we can then say that (dH)p,b = CP dT +

hTN dξ or(dH)p,b

dT = CP + hTN(NTGN)-1NTs .

6 Linearisation gives us the following Newton–Raphson iteration forthe flame temperature:

Tk+1 = Tk − Hk − Hın[CP + hTN(NTGN)-1NTs

]k .

7 You may have noticed that (dH /dT)p,b , (∂H/∂T)p,n = CP , butwhat is the physical explanation for this difference?

8 The answer lies in the internal degrees of freedom of the system.

9 It therefore makes a big difference whether it is n or b that is con-stant.

10 If it is the former, then all of the numbers of moles will be constant,but if it is the latter, the composition may vary as a result of chemicalreactions in the gas (the number of moles of each atom, however, willremain constant).

11 The effective heat capacity of the system is therefore affected bythe reaction as well as by CP .

1015 / 1776

ENGLISH text is missing.Iteration sequence1 The graph on the left-hand side of Figure 17.4 on the previous pageshows the iteration sequence for an adiabatic temperature calculationin which Gibbs energy is minimised at each step.

2 However, it is not entirely typical in that there is an upper limit onthe step length of 400 K, in order to ensure a suitable distribution of thepoints[b].

3 In reality the calculation converges much quicker, because theheat capacity of the combustion gas is virtually constant in the rele-vant range.

4 According to the graph on the right-hand side, this is true up totemperatures of 2–3000 K, at which point the dissociation equilibria be-come important[b] and the reaction starts to make the dominant contri-bution to (dH /dT)p,b.

5 Note that constant heat capacities have been used in these calcu-lations, and that the non-linearity is solely due to the varying composi-tion of the combustion gas.

6 The reaction equilibria are, incidentally, the same as in Theme 139,and once again the calculations have been performed using Pro-gram 31:1.27.

1017 / 1776

[b] A pragmatic solution to the problem is clearly to calculate the enthalpy for a numberof temperatures (e.g. for every one hundred degrees) until the outlet enthalpy exceedsthe inlet enthalpy, and then repeat the process with increasingly small temperatureincrements.

[b] In the range 3000–6000 K there is a transition from low to high enthalpy, and theinternal degrees of freedom of the system result in the characteristic hump shape of(dH / dT)p,b as shown in 17.4. The proposed Newton–Raphson algorithm requires(d2H / dT2)p,b to have a constant sign in the range where we are iterating. This condi-tion ceases to be met at approximately 4000 K, resulting in divergence.

1 Atom (or formula) matrix: A = [ atoms x compounds ]

2 Atom (mass) balance: A∆n = 0

3 Null space solution: AN = 0

4 Independent reactions:(

AT N)

has full rank

5 Extent of reaction: ∆n = Nξ

6 Reaction equilibrium: minAn=b

G(n)T ,p ⇔ minξ

G(n(ξ))T ,p

7 Linearization: µk+1 ≈ µk + Gk (nk+1 − nk )

8 Lagrange functional: L(n,λ) = G(n) − λT(An − b)

9 N–R iteration:

NkRT Gk AT

A 0

1Nk

∆n

λk+1

= −

1RT µk

0



17 Simultaneous . . .Looking back (2)

10 Reaction manifold: NTµ = 0

11 N–R iteration: 1Nk

∆n = −N(NT Nk

RT Gk N)-1

NT 1RT µk

12 Adiabatic (flame) temperature: h(T ,p,n)out = h(T ,p,n)ın


News Thermodynamic cycles Power plants The steam engine Olds

Part 18

Heat engines




18 Heat . . .New concepts

1 Thermodynamic cycles

2 Work and heat diagrams

3 The basic 6: Otto, Brayton, Carnot, Stirling, Ericsson & Rankine

4 Thermodynamic efficiency

5 Thermal power plants

6 The classic steam engine

7 Modern steam turbines



18 Heat . . .Heat power ENGLISH text is missing.Energy1 Our modern industrial society requires huge amounts of mechani-

cal and electrical energy.

2 Ready access to raw materials and cheap energy has fuelled eco-nomic growth and ever-improving standards of living.

3 Given the limits on available natural resources, it is clear that in-dustrial growth must sooner or later come to a halt, but the technologi-cal insight that we have developed since the emergence of modern sci-ence in the 17th century will always inform our understanding of the lim-its on human activity and development.

1022 / 1776

ENGLISH text is missing.Energy consumption1 Our standard of living is a direct function of our energy consump-

tion.

2 In practice, consumption of 100 W per capita is the lower limit forhuman survival. This is equivalent to the energy we require in the formof food.

3 Consumption of 104 W per capita is in many ways an upper limit onwhat is possible, albeit unsustainable at a global level. This is equivalentto Euro-American lifestyles today.

4 Idealistically, if we want a sustainable society, 1000 W per capitawould be a reasonable compromise. Assuming 10 billion people onEarth, this would mean 10 TW of demand, which is in turn equivalent to4.5?? times the total amount of hydroelectricity generated in the world,or 3.3 times the energy dissipated by tides, or 1.1 times the energyproduced by photosynthesis on Earth.

1023 / 1776

ENGLISH text is missing.Fossil fuels1 Since the farming and slaving societies of the past, we have re-

alised that we can (more than) replace raw muscle power by convertingchemical and thermal energy into mechanical energy.

2 Ultimately, chemically bound energy is most important to us, andabove all we exploit it through the combustion of fossil fuels, i.e. oil, coaland gas.

3 The development of an alternative technology based on directlyconverting chemical energy in fuel cells is still at the experimental stage,and in practice the transformation of the chemical energy in the fuel stillinvolves two steps.

4 The first step of this tried and tested technology is combustion ina boiler or burner, while the second stage involves transforming thereleased heat in a heat engine (a mechanical device that converts heatinto work).

1024 / 1776

1 The discovery of the “mechanical equivalent of heat”, and henceof the principle that work could be converted into heat, and vice versa,was an important prerequisite for the industrial revolution of the 19thcentury.

4 2 This discovery also gave a big impetus to the subject of thermody-namics, and particularly to our understanding of thermodynamic cycles.3 Here we will look at a small selection of closed cycles, where an

ideal gas is used as the working medium. The 1st and 2nd laws ofthermodynamics are important underlying principles, as is the conceptof reversibility. Based on Eq. 3.8 in Chapter 3:3.5, the 1st law Expanded footnote [a]:

[a] Since (dU)n = T dS − p d V , then ∮ p dV = ∮ T dS. The projected area of thecycle is thus the same in T , s and p, v coordinates, provided that the physical unitsform part of a coherent system of units, as is the case in Figure 18.2 on page 948 andFigure 18.3 on page 950. can be written

∮(dU)n =

∮(δQ − δW) = 0 ,

6 5 while the 2nd law tells us that the entropy must also be conservedin a reversible change of state. Put another way, δQ rev = T dS and δW rev = p dV .

8 7 In a closed system, the entropy is a function of only two state vari-ables i.e. s(T ,p), s(T , v) or s(p, v). Under isentropic conditions ds = 0, which can be used to calculate

the temperature functions T(s, v) and T(s,p), and the pressure func-tions p(T , s) and p(v , s).



18 Heat . . .Heat power (2)

9 Together, there are four primary variables s, v ,T and p giving riseto 4 · 3/2 = 6 different coordinate systems, as shown in Appendix 25.

10 An ideal heat engine is defined by the fact that it has a rectangularcycle in one of these systems of coordinates.

12 11 We are not going to say much about the practical implications ofthese coordinate systems but in principle the four branches of the cyclecan be separated either in time or in space. Which means the cycle will show either a time-dependent state de-

scribing a closed loop over the period τ or it will show four time inde-pendent states existing simultaneously at different locations in the ma-chinery.

15 13 The two alternatives lead to a dynamic or a stationary process de-scription, respectively, but such details will not be of prime concern forour understanding of the matter.14 It is clearly possible to design countless alternative, non-trivial cy-cles, but that lies outside the scope of this chapter. The six fundamental cycles have been given names that link them

to more or less famous historical figures:



18 Heat . . .Heat power (3)

External combustion: Stirling[1] (T ,V ) and Ericsson[2] (T ,p)Internal combustion: Otto[3] (S ,V ) and Brayton[4] (S ,p)

Limiting case: Carnot[5] (T ,S)Vapour–liquid: Rankine[6] (p,V )

Expanded footnote :

[a] Robert Stirling, 1790–1878. Scottish reverend.

Expanded footnote :

[b] John Ericsson, 1803–1889. Swedish engineer.

Expanded footnote :

[c] Nikolaus August Otto, 1832–1891. German engineer.

Expanded footnote :

[d] George Brayton, 1830–1892. American engineer.

Expanded footnote :

[e] Nicolas Leonard Sadi Carnot, 1796–1832. French engineer.

Expanded footnote :

[f] William John Macquorn Rankine, 1820–1872. Scottish physicist.



18 Heat . . .Efficiency

1 The thermodynamic efficiency of these cycles is defined as beingη = W/Qh.

4 2 For a real, physical machine, W and Qh are measured under con-trolled conditions in a laboratory.3 By looking at an ideal cycle, we avoid having to take measure-

ments, but it is important to realise that the additional definitions setout below only apply to reversible changes of state and frictionless ma-chines:

W = ∮ p dV , (18.1)

Qh −Qc = ∫

dS>0

T dS + ∫

dS<0

T dS = ∮ T dS . (18.2) Note that Qc is positive if heat is removed from the cycle.

5 This goes against the normal sign convention, but it has the practi-cal advantage that all heat flows are considered positive, whether theyare entering or leaving the cycle.



18 Heat . . .Maximum efficiencyENGLISH text is missing.Inexact differentials1 Also note that T(dS)n and p(dV)n in isolation are inexact differen-

tials.

2 This is in spite of the fact that both S and V are state functions, sodS and dV can actually be written as exact (total) differentials.

3 The explanation for this lies in the system’s degrees of freedom.

4 At constant composition, the cycles have two degrees of free-dom — let us call them x and y — which means that each of the in-tegrands T(x , y) and p(x, y) has one degree of freedom that is not be-ing used if only the differential change in (dS)n or (dV)n is known.

5 Generally we can say that there is no state function f(x , y) with atotal differential of df = g(x , y) dx or df = h(x , y) dy.

6 Since the system has two degrees of freedom, we must insteadexpect the total differential to be of the form df = g(x, y) dx +h(x , y) dy.

7 It is, in other words, impossible to express all possible variations inf(x , y) as functions of only dx or dy.

8 Thus, the closed path integrals in equations 18.1 and 18.2 do notrepresent state functions, and their values are therefore non-zero.

9 Conversely, e.g. T(dS)V ,n = CV and p(dV)S,n are exact differen-tials, because the degrees of freedom are locked in such a way that thechanges take place in the T ,S and p,V planes respectively.

10 T and p are themselves state functions, which means that∮T(dS)V ,n = 0 and

∮p(dV)S ,n = 0, as each of the closed path in-

tegrals can be simplified to two line integrals, one in each direction.1029 / 1776

1 Thermodynamic efficiency is defined by: η = W/Qh = (Qh − Qc)/

Qh.

3 2 Naturally this value, and particularly its maximum, is of great prac-tical and theoretical interest. From the above definitions we can see that Qh > Qc > 0, and that

η→ 1 if Qh →∞ and Qc → 0.

4 This limit is of no practical significance, but if the machine is of fi-nite size (meaning the entropy undergoes a finite change during the cy-cle), and there is also a maximum and minimum temperature for heatexchange with the surroundings, the question becomes more interest-ing.

6 5 What happens then is graphically illustrated in Figure 18.1. Consequently the optimal design of the cycle has a rectangularpath when represented on a T , s diagram, with η = (Th − Tc)/Th; this iscommonly known as the Carnot cycle. 7 This cycle is independent of the working medium, since its design

means that the heat integrals are only dependent on the temperatureson the cold and hot sides (the two isentropes do not contribute to theintegral ∫ T dS).8 Provided that the temperatures do not alter the composition, theCarnot cycle will achieve a given efficiency regardless of the mediumused.9 No other cycles have this property, but then again the Carnot cyclecannot be achieved in practice, since isothermal heat transfer requiresthe system to have heat exchange properties that are physically impos-sible (infinite heat exchange surface or infinite heat conductivity).



A0

A1

A2

A∞

Tc

Th

S1 S2

Figure 18.1: Optimal design of a closed cycle in T , s coordinates. The heatintegrals satisfy the relation Ai+1 > Ai . Combined with the physical constraints,this implies that Qh,i+1 > Qh,i and Qc,i+1 < Qc,i . Consequently, ηi+1 > ηi whereη∞ = ηCarnot.



18 Heat . . .Thermodynamic cyclesENGLISH text is missing.Mechanical energy1 To explain the Carnot cycle we assume the existence of a thermal

reservoir.

2 For reasons of symmetry, it should also be possible to define amechanical reservoir, but does one actually exist?

3 The answer is yes, if we assume the existence of an externalsource of mechanical work that is not affected by its interaction withthe system.

4 A frictionless drive or a resistance-free source of voltage are twoexamples of this. Exploiting the potential energy of a lossless hy-dropower reservoir is an example with a thermodynamic dimension.

5 The hydrostatic pressure in the system can be used to drive anideal turbine generator that converts p, v work into electrical energy.

6 This is a flow process, which has little in common with closed cy-cles, but we can still imagine using hydrostatic pressure to drive a hy-pothetical cyclic process with a rectangular path in p, v coordinates.

7 The efficiency of this cycle is η = W/Wh = (Wh −Wl)/Wh = (ph −pl)/ph, assuming that

Wh =∫

dV>0

p dV = ph(vh − vl) and

Wl =∫

dV<0

p dV = pl(vh − vl) .

1032 / 1776

1 To perform a mathematical analysis of a thermodynamic cycle, wefirst project the cycle onto T , s and p,V diagrams.

4 2 This projection, or perhaps it is better to call it a mapping, trans-forms the cycle’s original system of coordinates to curvilinear[g] coor-dinate systems, one of which describes the heat leaving and enteringthe system, and the other of which describes the work delivered and re-ceived.3 In each of these coordinate systems, the cycle can be represented

by a closed graph in which the temperature and pressure vary in a man-ner that is characteristic of the cycle and working medium.

[g] The curvilinear coordinates are derived from the original (orthogonal) coordinatesusing a (locally) invertible transformation (one–to–one mapping) at each point alongthe graph. This means that each point in the orthogonal coordinate system can betransformed to just one point in the curvilinear coordinate system, and vice versa. Asthe word curvilinear implies, the original coordinate axes are curved by the mappingprocess. Interestingly, the slopes of the graphs T(s) and p(v), as seen in

the T , s and p, v diagrams, always satisfy the criteria(∂T∂s

)v = T

cv>

(∂T∂s

)p =

Tcp

> 0 , (18.3)(∂p∂v

)s =

−cp

cv vβ <(∂p∂v

)T = −1

vβ < 0 (18.4)

6 5 regardless of the working medium. In other words, an isochor is always steeper than an isobar, and anisentrope is always steeper than an isotherm in the two diagrams.

7 Here it does not make any difference whether the medium is anideal gas or some other fluid (or even a solid).



18 Heat . . .Thermodynamic cycles (2)

8 This broad conclusion is related to the fact that cv > 0, cp > 0 andβ > 0 are three fundamental properties of any stable thermodynamicsystem.

9 Equations 18.3–18.4 will significantly assist our discussion of theheat and work diagrams.



18 Heat . . .§ 143 § 142 § 144

Draw realistic T , s and p, v diagrams for the ideal(frictionless) cycles postulated by Otto, Brayton, Carnot,Stirling, Ericsson and Rankine. Let the working medium

be one mole of an ideal, diatomic gas. Two of thecycles are incompatible (in practice) with the working

fluid being an ideal gas. Which two?



170 180 190 2000

3

6

9

12

10 20 30 40 500

3

6

9

170 180 190 2002

4

6

8

10 20 30 40 501

1.4

1.8

2.2

170 180 190 2002

3

4

5

6

7

1 5 10 50100 5000.04

0.4

4

40

Qh = 169Qc = 95.6

W = 73.7

Qh = 135Qc = 112

W = 22.9

Qh = 174Qc = 80.1 W = 93.9

CarnotCarnot

OttoOtto

BraytonBrayton

T[K/1

00]T[K/1

00]T[K/1

00]

s [J/K mol]

p[b

ar]

p[b

ar]

p[b

ar]

v [dm3/mol]

Figure 18.2: T , s diagrams (left) and p, v diagrams (right) for one mole ofideal gas (γ = 1.40). The cycles move through their respective states in theclockwise direction. The characteristics of the cycle are: Compression of coldgas (green), external heating (red), expansion of hot gas (yellow) and externalcooling (blue). The displaced volume of gas, and its entropy change, is thesame in all four cases except for the p, v diagram for the Carnot cycle, whichfor practical reasons is drawn on a logarithmic scale.



18 Heat . . .§ 143 Otto . . . Rankine

1 2 The six cycles have been drawn in Figures 18.2 and 18.3 using theProgram 1.30 in Appendix 31. Note that the changes in volume and entropy Expanded footnote [g]:

[g] The change in volume is, incidentally, a direct measure of the size of the machine,whereas the change in entropy is an indirect measure of the machine’s heat exchangearea. are the same in all

of the diagrams, with the exception of the p, v diagram for the Carnotcycle, and the T , s diagram for the Rankine cycle.

3 Any attempt to specify both v and s makes these diagrams over-specified from a thermodynamic point of view.

4 These diagrams also show that the Carnot and Rankine cycles areincompatible with using an ideal gas as the working medium.

5 The integration ranges are 3.5 decades of pressure and 1 decadeof temperature respectively, so it is difficult to avoid considerable frictionor heat losses. 6 It is therefore worth knowing that, with a simple modification, the

Stirling and Ericsson cycles can achieve the same efficiency as theCarnot cycle (see Theme 144), while the Rankine cycle is equivalentto the classical steam engine (see Subchapter 18.4).



170 180 190 2002

3

4

5

6

7

10 20 30 40 500

1

2

3

4

5

170 180 190 2002

3

4

5

6

7

10 20 30 40 501

1.4

1.8

2.2

150 170 190 2101

5

9

13

10 20 30 40 501

1.4

1.8

2.2

Qh = 144

Qc = 104W = 39.8

Qh = 131

Qc = 113W = 18.2

Qh = 293

Qc = 257W = 36.2

EricssonEricsson

RankineRankine

StirlingStirling

T[K/1

00] T[K/1

00]T[K/1

00]

s [J/K mol]

p[b

ar]

p[b

ar]

p[b

ar]

v [dm3/mol]

Figure 18.3: T , s diagrams (left) and p, v diagrams (right) for one mole ofideal gas (γ = 1.40). The cycles move through their respective states in theclockwise direction. The characteristics of the cycle are: Compression of coldgas (green), external heating (red), expansion of hot gas (yellow) and externalcooling (blue). The displaced volume of gas, and its entropy change, is thesame in all four cases except for the T , s diagram for the Rankine cycle, whichfor practical reasons is drawn on a different scale.



18 Heat . . .§ 144 § 143 § 145

Show that the Stirling and Ericsson cycles can achievethe same efficiency as the Carnot cycle if the workingmedium is an ideal gas, and assuming perfect heatrecovery in the non-isothermal part of the process.



Th

Th

Th

Tc

Tc

Tc

v1

v1

v2

v2

p

s

s

W

Qh

Qc

a)

b)c)

d)

Figure 18.4: Figure a illustrates the Stirling cycle in the natural variables Tand v. Figures b, c and d show the projected work and heat diagrams whereW = Qh−Qc. The cycles move through their respective states in the clockwisedirection (which means W =

∮p dv > 0). The heat integral is divided into

four sections as shown in Figure c, the coloured parts of which cancel for anideal gas with perfect heat recovery (because the heat capacity is then onlya temperature function). The net (external) heat requirement is therefore asshown in Figure d.



18 Heat . . .§ 144 Stirling & Ericsson

1 The Stirling cycle describes a rectangle in the T ,V plane, asshown in Figure 18.4a.

3 2 The projections onto the T ,S and p,V planes are shown in Fig-ures 18.4b and 18.4c,d. The work integral is consequently defined by

W ıg =∮

pıg dV

=V2∫

V1

NRThV dV +

V1∫

V2

NRTcV dV = NR(Th − Tc) ln

(V2V1

),

but in order to integrate T dS we need to know the differential of S as afunction of the free variables T and V :

dS =(∂S∂T

)V dT +

(∂S∂V

)T dV =

CV

T dT +(∂p∂T

)V dV .



18 Heat . . .§ 144 Stirling & Ericsson (2)

4 Along the various isopleths, it is also true that

T(dS)ıgV = C ıg

V dT ,

T(dS)ıgT = T

(∂p∂T

)ıg

V dV = NRTV dV .

5 The heat integral for the hot side can now be calculated as follows:

Q ıgh =

∫

dS>0

T dS ıg

=Th∫

Tc

C ıgV dT +

V2∫

V1

NRThV dV = C ıg

V (Th − Tc) + NRTh ln(V2V1

),

Finally we can calculate the integral for the cold side:

Q ıgc =

∫

dS<0

(−T) dS ıg



18 Heat . . .§ 144 Stirling & Ericsson (3)

=Th∫

Tc

C ıgV dT +

V2∫

V1

NRTcV dV = C ıg

V (Th − Tc) + NRTc ln(V2V1

).

7 6 For your own benefit you should check that W ıg = Q ıgh −Q ıgc . Also note that the term C ıgV (Th − Tc) appears in the same context

and with the same sign in both Q ıgh and Q ıg

c .

8 If this heat is temporarily stored in a recuperator, the consumptionof Q ıg

h is reduced by an equivalent amount, and assuming perfect heatrecovery the efficiency rises to

ηıgStirling →

NR(Th−Tc) ln(V2V1

)

NRTh ln(V2V1

) = Th−TcTh

= ηCarnot (18.5)



18 Heat . . .Heat recovery

1 2 Here Qh contradicts the simple definition contained in Eq. 18.2 onpage 942, and a new graphical representation (that only applies in theevent of perfect heat recovery) follows from Figure 18.4d. An equivalent argument also applies to the Ericsson cycle, where

Q ıgh = C ıg

P (Th −Tc) +NRTh ln(p2p1

)and Q ıg

c = C ıgP (Th −Tc) +NRTc ln

(p2p1

).

3 Here the heat that needs to be recovered is C ıgP (Th − Tc).

4 We can draw the conclusion that the Stirling and Ericsson cyclesreach the same efficiency as the Carnot cycle, provided that there isperfect heat recovery in the heating and cooling steps, and that theworking medium is an ideal gas:

ηıgStirling and ηıg

Erıcsson → ηCarnot(Th) . (18.6)

5 The Carnot cycle has the same theoretical efficiency whatever theworking medium, but unlike the Stirling and Ericsson cycles, it would notbe possible to implement it using an ideal gas as the working medium.



18 Heat . . .§ 145 § 144 § 146

Find an expression for the thermodynamic efficiency ofthe Otto cycle (an idealised petrol engine) as a function

of the temperature Th at the end of the compressionstroke, i.e. immediately before ignition. Set out anequivalent expression for the Brayton cycle (a gas

turbine). In both cases assume that the working mediumis an ideal gas. Compare your results with the Carnot

cycle.



18 Heat . . .§ 145 Otto & Brayton

1 2 The ideal Otto cycle describes a rectangle in the S ,V plane, but allwe need to know is the projection of this rectangle in the T ,S plane. This is because, assuming constant heat capacity, the work integral

can be solved as follows:

W ıg =∮

pıg dV =∮

T dS ıg

=Ta∫

Th

C ıgV dT +

Tc∫

Te

C ıgV dT

= C ıgV(Ta − Th) − C ıg

V(Te − Tc) .



18 Heat . . .§ 145 Otto & Brayton (2)

3 The associated heat integral is

Q ıgh =

∫

dS>0

T dS ıg =Ta∫

Th

C ıgV dT = C ıg

V (Ta − Th) ,

which gives us the efficiency

ηıgOtto = W ıg

Q ıgh

= 1 − Te−TcTa−Th

,

where Ta, Th, Te and Tc are the respective temperatures after combus-tion, compression, expansion and cooling. 4 The Otto cycle uses isentropic compression, where sc = sh, and

isentropic expansion, where sa = se, and as the volume ratio is thesame for these subprocesses, we get the relation

(ThTc

)ıg

S,n = (TaTe

)ıg = (V1V2

)1−γ ,

in line with Theme 61 on page 382. From this we can concludethat T ıg

a = TeTh/Tc.

5 The final expression for the efficiency is thus

ηıgOtto = 1 − Te−Tc

ThTeTc−Th

= Th−TcTh

= 1 − (V2V1

)1−γ.



18 Heat . . .§ 145 Otto & Brayton (3)

6 The efficiency is the same for a hypothetical Carnot cycle in whichheat is added at (an assumed) constant temperature Th immediatelyafter the completion of the compression stroke.

8 7 An equivalent conclusion can be drawn for the Brayton cycle, byreplacing cv with cp in our calculation:

ηıgBrayton = 1 − Te−TcThTeTc −Th = Th−TcTh = 1 − (p1p2

)(γ−1)/γ . We can draw the conclusion that

ηıgOtto = ηıg

Brayton = ηCarnot(Th) (18.7)



Figure 18.5: The thermodynamic efficiency of the six ideal (rectangular) cy-cles described in the main text can be expressed as η = 1 − ∆cx

∆hx . ∆hx and∆cx denote the change in the state variable x over the hot and the cold sidesof the cycle. The definition of x for each of the cycles is shown in the table be-low.

x x x

Stirling (∆u)v − T(∆s)T Otto (∆u)v Carnot T

Ericsson (∆h)p − T(∆s)T Brayton (∆h)p Rankine (∆u)v − (∆h)p



Figure 18.6: Thermodynamic work W = ∆hx −∆cx for the six ideal (rectan-gular) cycles described in the main text. ∆hx and ∆cx denote the change inthe state variable x over the hot and the cold sides of the cycle. The definitionof x for each of the cases is shown in the table below.

x x x

Stirling (∆a)v Otto (∆u)v Carnot T(∆s)T

Ericsson (∆g)p Brayton (∆h)p Rankine p(∆v)p



18 Heat . . .Thermal power plants ENGLISH text is missing.Corner coordinates1 The two preceeding sections show that equations 18.1 and 18.2

can be integrated analytically if the working medium is an ideal gas.

2 However, it is likely that the approach we have used becomes un-necessarily complex if the ideal gas is replaced with a real fluid.

3 If we think about it, all of the cycles discussed in this chapter arerectangular in their respective characteristic coordinate systems.

4 It should therefore be possible to express the efficiency of eachcycle as a function of its corner coordinates alone.

5 All we need are analytical expressions for the partial differentials

(dU)V ,n = (dH)p,n = T dS ,

which can then be integrated in such a way that∮

T dS is given as asimple difference in the function values of the corners of the cycle.

6 This approach gives us general expressions for the efficiency ofthe cycles, without taking into consideration the working medium itself;see Table 18.5.

1049 / 1776

ENGLISH text is missing.Corner co-ordinates II1 In addition to knowing the efficiency of each of the cycles, it is also

interesting to know how much work they can perform.

2 For the s, v and s, p cycles, W =∮

T dS is equal to the changes(∆u)v and (∆h)p on the hot and cold sides, cf. the integration of Eq. 5.

3 The adiabatic (isentropic) changes of state do not contribute to theintegral of T dS.

4 For the T , v and T ,p cycles the calculation is more complicated,but the problem can be solved by integration by parts.

5 From∫

T ds = Ts −∫

s dT we can say that∮

T ds = −∮

s dT .Note that the Ts term is a state function that makes a zero contributionto the closed path integral.

6 Similarly, we know that∮

p dv = −∮

v dp from the work diagram,but that is of no importance here.

7 All we now need are the partial differentials

(dA)V ,n = (dG)p,n = −S dT ,

so that we can integrate them to obtain∮(−S)dT expressed as a simple

difference in the function values of the corners of the cycle.

8 This approach gives us general expressions for the work performedby the cycles; see Table 18.6.

9 The table also states values for the Carnot and Rankine cycles,which come directly from the T , s and p, v diagrams respectively.

1050 / 1776

ENGLISH text is missing.Thermomechanical en-

gines1 We can draw the conclusion that the four cycles — Otto, Brayton,

Stirling and Ericsson— are abstract representations of thermomechani-cal engines that convert changes in internal energy, enthalpy, Helmholtzenergy and Gibbs energy into mechanical work.

2 The Carnot and Rankine cycles are not related to any of the canon-ical thermodynamic potentials; rather they can be considered purelythermal or mechanical engines.

3 Here it should be added that the Carnot cycle is independent ofthe working medium, and that it could alternatively be conceived as athermoelectric or thermomagnetic engine.

4 The Rankine cycle, on the other hand, can be seen as a purelymechanical device that merely exploits pressure–volume forces withoutthem necessarily being linked to a thermal change of state.

5 Examples ??1052 / 1776

1 At a thermal power plant, a proportion of the energy in the fuel willbe lost with the hot flue gases produced by the plant.

2 If no electricity is being generated, and assuming adiabatic condi-tions, the temperature of the flue gases will be Ta, but when the plant isdoing work, the temperature falls.

3 Consequently, the efficiency is lower than the Carnot efficiency im-plied by the adiabatic flame temperature Ta.



18 Heat . . .§ 146 § 145 § 147

A thermal power plant produces flue gases of temperatureTa and has one Carnot engine available. Heat is added

to the engine at the boiler temperature Th and is removedat the cooling water temperature Tc. Calculate theoverall efficiency of the power plant as a function ofTh. Determine the optimal boiler temperature. The

flue gases have constant heat capacity.



18 Heat . . .§ 146 Single stage power plant

1 The work performed by the Carnot engine can be expressed as

W = Th−TcTh

Qh ,

Qh = CP(Ta − Th) ,

3 2 where Ta is the adiabatic flame temperature. The total heat content of the flue gases is Qtot = ∆rxH = CP(Ta −Tc) where CP is the heat capacity of the mixture.

5 4 Note that Qtot ≥ Qh and that the difference between the two heatflows can be attributed to the emission of hot flue gases from the plant. The overall efficiency of the power plant is therefore lower than the

maximum theoretical Carnot efficiency:

ηtot =W

∆rxH = WQtot

= Th−TcTh· Ta−Th

Ta−Tc= ηCarnot · ηthermal .



18 Heat . . .§ 146 Single stage power plant (2)

7 6 The efficiency of a single-engine gas-fired power plant with a cool-ing water temperature of 300 K is shown in Figure 18.7 on page 969, cf.Program 1.28 in Appendix 31. Still assuming constant heat capacity, we can obtain an approxi-

mation for (Th)opt, and this closely matches the optimal operating tem-perature of the boiler:

∂ηtot

∂Th=

Th−(Th−Tc)

T2h

· Ta−ThTa−Tc

− Th−TcTh· 1

Ta−Tc= 0 .

8 We can simplify the equation to TcTh

(Ta − Th) = Th − Tc, or putslightly differently:

Topt = (Th)opt =√

TaTc . (18.8)

9 If we substitute this into the expression for the total efficiency ηtot,we get

ηopt =√

TaTc−Tc√TaTc

· Ta−√

TaTcTa−Tc

.



18 Heat . . .§ 146 Single stage power plant (3)

10 Let us multiply the numerator and the denominator by√

Tc, andthen factorise the right-hand side:

ηopt =√

Ta−√

Tc√Ta

·√

Ta(√

Ta−√

Tc)(√

Ta−√

Tc)(√

Ta+√

Tc)=√

Ta−√

Tc√Ta+

√Tc

=Topt−Tc

Topt+Tc



18 Heat . . .Curzon–Ahlborn optimum

1 According to the definition in the above equation, ηopt gives theoverall efficiency of the plant.

2 How does this compare with the Carnot efficiency of the plant?

3 The effective Carnot efficiency of the plant is given by (Th−Tc)/Th.

4 If we substitute in the optimisation condition from Eq. 18.8, this canbe expressed as ηCA = 1 −

√Tc/Ta, following Curzon and Ahlborn Expanded footnote [g]:

[g] Adrian Bejan. Advanced Engineering Thermodynamics. Wiley, 1988. .

7 5 At the limit case when x = (Ta − Tc)/Ta approaches a small value,then

limTa→Tc ηopt = 1−√1−x1+√1−x = x4−x ≃ 14 · ηCarnot(Ta) .

6 Here we have used the expansion √1 + x = 1+ 12 x− 18x2+ 348 x3−· · · The optimal efficiency can also be expressed as a function of Topt =√

TaTc from Eq. 18.8. At the limit case when ∆T = Topt − Tc is small,we get

limTopt→Tc

ηopt =12 ·

Topt−Tc

Topt= 1

2 · ηCarnot(Topt) . (18.9)



0 400 800 1200 1600 2000 24000

0.2

0.4

0.6

11.4

23

6

20

T [K]

ηηCarnot

ηopt =Topt−Tc

Topt+Tc

Figure 18.7: Overall efficiency of a single-engine natural gas-fired powerplant (with temperature-dependent ideal gas heat capacity) with a variableair : methane ratio. 1.0 is the stoichiometric ratio.



18 Heat . . .§ 147 § 146 § 148ENGLISH text is missing.Exhaust heat1 This means that a thermal power plant that extracts its heat from

flue gases, or more generally from a hot process stream, will have anefficiency that is lower than if it had extracted the heat from a heatreservoir of the same temperature as the flue gas.

2 According to the above equations, the efficiency will be 25% or 50%of the equivalent Carnot efficiency, depending on whether you baseyour calculation on the flame temperature or the optimal boiler temper-ature.

3 We should hasten to add that these statements are only valid forlow process temperatures.

4 At high temperatures, the differences between the various efficien-cies decreases, until finally they disappear when T →∞.

5 This, however, is a purely mathematical limit, which is of no practi-cal relevance.

1060 / 1776

Show that if the power plant in Theme 146 has n > 2engines available, then optimal operation will occur

when Ti =√

Ti+1Ti−1, where i − 1, i and i + 1 representthree engines connected in series on the hot side. What

will the maximum available work be when n→∞?



18 Heat . . .§ 147 Exergy

1 Let us assume that the heat capacity of the flue gases is constant(in practice calculated as the effective average value).

2 As previously mentioned, the Carnot engines are connected in se-ries on the hot side, and have a common cold reservoir.

3 The total amount of work that they can perform is the sum of theirindividual contributions

Wn = T1−TcT1

CP (Ta − T1) +T2−Tc

T2CP (T1 − T2)

+ · · ·+ Ti−TcTi

CP (Ti−1 − Ti) +Ti+1−Tc

Ti+1CP (Ti − Ti+1) + · · · (18.10)

if Ti−1 is the inlet temperature for engine i and Ti is the outlet tempera-ture of the same engine (which in turn is equal to the inlet temperaturefor the next engine).



18 Heat . . .§ 147 Exergy (2)

4 Tc is the temperature of the common reservoir on the cold side ofthe engine.

5 The optimal operating conditions are given by

1CP

∂Wn∂Ti

=Ti−(Ti−Tc)

T2i

(Ti−1 − Ti) − Ti−TcTi

+Ti+1−Tc

Ti+1= 0 .

6 Multiplying by T2i Ti+1 gives us

Ti+1Tc(Ti−1 − Ti) − TiTi+1(Ti − Tc) + T2i (Ti+1 − Tc) = 0 .

7 This can be expressed as the tridiagonal Expanded footnote [g]:

[g] Refers to the structure when written in matrix form. system of equations Ti =√Ti−1Ti+1; show that this has the general solution Tn+1

i = T icTn+1−i

a .

8 When n → ∞, the sum of Eq. 18.10 can be expressed as an inte-gral.



18 Heat . . .§ 147 Exergy (3)

9 The fraction (Ti − Tc)/Ti then takes the variable value (T − Tc)/T ,while the factor Ti−1 − Ti can be seen as a differential temperaturechange −dT :

W∞ = −CP

Tc∫

Ta

T−TcT dT = −CP

Tc∫

Ta

dT + TcCP

Tc∫

Ta

dTT

= CP(Ta − Tc) − TcCP ln(TaTc

)

= ∆H − Tc∆S . (18.11)

10 Constant heat capacity has been assumed in the above derivation,but in Chapter 19 we showed that W∞ = ∆H −Tc∆S (called the exergyof the flue gases) is also valid as a general result, based on a thermo-dynamic analysis of flow processes.



300 400 500 6000

0.2

0.4

0.6

η

Boiler temperature [K]

Figure 18.8: Thermodynamic efficiencies of Newcomen (blue) and Watt(green) steam engines, and of a steam turbine with superheated steam (red).The dashed curve shows the corresponding efficiency of the Carnot cycle,where the boiler and the condenser (40 C) act as thermal reservoirs. Themeasurements are from the years 1740–1960.Tore Haug-Warberg (Chem Eng, NTNU) Termodynamic methods TKP4175 02 September 2013 1072 / 1776


0 40 80 120 1603

5

7

9

0.018 26.020.01

0.5

10

500

0 40 80 120 1603

5

7

9

0.018 26.020.01

0.5

10

500

0 40 80 120 1603

5

7

9

0.018 26.020.01

0.5

10

500

Thomas NewcomenThomas Newcomen

James WattJames Watt

Ideal steam turbineIdeal steam turbine

T[K/1

00]

T[K/1

00]

T[K/1

00]

s [J/K mol]

p[b

ar]

p[b

ar]

p[b

ar]

v [dm3/mol]

a) b)

c) d)

e) f)

Figure 18.9: T , s diagrams (left) and p, v diagrams (right) for a single mole ofwater and steam. The cycles that describe the Newcomen and Watt enginesmake use of saturated steam whereas the turbine works with superheatedsteam. The volumes of water and steam at 1 bar and 40 C are marked on theabscissas in the p,V diagrams. The enlarged portion of the diagrams focuson the details on the liquid side. Otherwise the notation is the same as inFigures 18.2 and 18.3.



18 Heat . . .Vapour–liquid cycles ENGLISH text is missing.Limit case1 The efficiency of the plant can be expressed as η∞ = W∞/∆H =

1−Tc ·∆S/∆H, where ∆H is the heat needed to raise the temperaturefrom Tc to Ta.

2 If we assume constant heat capacity, η∞ can be expanded[g] in xwhere x = (Ta − Tc)/Tc:

η∞ = 1 − ln(1+x)x = x

2+x − 23x

( x2+x

)3 − 25x

( x2+x

)5 − · · ·

At the limit case where x is small, we get

limTa→Tc

η∞ = x2+x = Ta−Tc

Ta+Tc ≃12 · ηCarnot(Ta) .

3 At low process temperatures the efficiency is twice that of a singleCarnot engine, cf. the result on page 968.

1066 / 1776

[g] Using ∆S = CP ln(Ta/Tc) = CP ln(1 + x) and ∆H = CP(Ta − Tc) = TcCPx, and

substituting in the series 12 ln(1 + x) = x2+x + 13

( x2+x

)3+ 15

( x2+x

)5+ · · · .

ENGLISH text is missing.Heron’s æolipile1 The trusty old steam engine has long-since become passé for prac-

tical applications, but it retains its exalted position, high above all of theother machines, in the engineering hall of fame.

2 Not only did it pave the way for Western industrialisation, but it alsoinspired the emergence and development of thermodynamics[g] in theearly 19th century.

3 However, from a thermodynamic point of view, the development ofthe steam engine was something of a coincidence.

4 The story starts with Heron[g]’s æolipile, and ends with the closelyrelated reaction turbine 2,000 years later, but there is no direct linkbetween these two machines.

5 Heron’s invention was far ahead of his time, and remained a merecuriosity.

6 There are many possible explanations for this, but the most impor-tant one is probably that in Antiquity people kept slaves, and didn’t needmachines to perform mechanical work.

7 It is unclear whether the æolipile had any practical application otherthan as a “miracle of nature” in one of the temples in Alexandria.

1067 / 1776

[g] As so often in history, the results came first, and the theory was developed subse-quently.

[g] Heron fra Alexandria, ca. 10–70. Greek engineer and mathematician.

ENGLISH text is missing.Horror vacui1 The first true steam engine was not developed until the 18th cen-tury, first as a low-pressure steam engine, which means that atmo-spheric pressure— rather than the steam pressure— performs the me-chanical work.

2 Heron’s æolipile, on the other hand, had been a high-pressure en-gine, and high pressure had later been used in another context to pro-duce effective guns, but in the early 17th century people started to un-derstand that vacuums were also associated with great forces, and thatthese forces could be harnessed to perform mechanical work.

3 In order to properly appreciate this rebellious thought, we mustrefer back to Aristotle[g] and his thesis that nature harbours a “fear ofthe empty space” (lat. horror vacui).

4 He considered the vacuum (or aether) to be a fifth, divine elementalongside earth, air, fire and water.

5 Aristotle’s works disappeared in the West, and were only rediscov-ered in the 12th century, but his ideas were a potential threat to Catholicdoctrine.

6 It was the Church’s great philosopher and theologian ThomasAquinas who reconciled Christian doctrine with the Aristotelian world-view.

7 This reinforced Aristotle’s position in the West, and the dogma ofthe horror vacui remained unchallenged right until 1643, when Torri-celli [g] created the first barometer, demonstrating that the hydrostaticpressure of the atmosphere could balance the weight of a column ofmercury.

8 Only then did people develop a scientific understanding of howancient inventions such as the water pump and the siphon worked.

1068 / 1776

[g] Aristotle, 384–322 f.Kr. Greek philosopher.

[g] Evangelista Torricelli, 1608–1647. Italian physicist and mathematician.

ENGLISH text is missing.Otto von Guericke1 In 1654, Guericke[g] demonstrated his “Magdeburg hemispheres”,thereby showing that it was possible to use the atmosphere to controlforces that were far greater than human muscle power.

2 This demonstration came at a time when there was a great deal ofinterest in scientific innovation in Europe.

3 Meanwhile, the mining industry in England needed greater pump-ing capacity.

4 The mine water was causing major technical and financial prob-lems, and in view of the scientific progress that had recently been made,it was natural that an attempt was made to remove the water using neg-ative gauge pressure, either directly or in combination with some devicethat could be used to pump the water in the traditional way.

1069 / 1776

[g] Otto von Guericke, alias Gericke, 1602–1686. German engineer and poltician.

ENGLISH text is missing.Dennis Papin1 All that remained was a small practical difficulty — how could you

create the negative pressure?

2 A solution to this intricate problem was first demonstrated by Pa-pin[g] in 1690.

3 By alternately boiling and condensing water (steam) in a metalcylinder tightly sealed with a piston, he created a periodic negativegauge pressure that was able to raise a mechanical load using a spool,but the idea did not catch on in the contemporary scientific community.

4 Savery[g], on the other hand, was granted a patent in 1698 for hissteam-driven water pump designed to “raise water by fire”.

5 This was a pistonless device, in which the steam came into directcontact with the water that it was lifting, which was first raised 4–5 me-tres using negative pressure, and then pumped 10–12 metres usingpositive pressure.

1070 / 1776

[g] Denis Papin, 1647–1714. French physician.

[g] Thomas Savery, ca. 1650–1715. English inventor.

ENGLISH text is missing.Savery, Newcomen &

Watt1 Savery had understood the need to separate the boiler from the

pump, but The Miner’s Friend was still dreadfully inefficient.

2 Papin proposed (1706) that the steam should be separated fromthe water with a hollow metal piston.

3 This was a significant improvement, and also meant that the pumpwould only use positive pressure, but once again he was out of luck.

4 That was because the previous year, Newcomen[g] had reachedan agreement with Savery, and had patented the first working atmo-spheric steam engine based on a combination of Papin’s original steamengine, and Savery’s boiler.

5 Subsequent improvements included directly injecting water to con-dense the steam, although the temperature in the cylinder still varied.

6 It was only once Watt[g] patented the condenser in 1769 that thesteam engine achieved its final design, with the boiler, engine and con-denser becoming three separate units.

7 Watt worked systematically to further improve the engine, and in1782 he introduced partial (isentropic) expansion of the steam in thefinal part of the piston stroke.

8 This doubled the theoretical efficiency in the relevant temperaturerange 373–383 K, as shown in Figure 18.8.

9 The coloured spots reflect technological progress[g] [g] in this area.1071 / 1776

[g] Thomas Newcomen, ca. 1664–1729. English blacksmith.

[g] James Watt, 1736–1819. Scottish engineer.

[g] R. L. Hills. Power from Steam, pages 59,103–112,131. Cambridge Univ. Press,1989.

[g] Adrian Bejan. Advanced Engineering Thermodynamics. Wiley, 1988.

ENGLISH text is missing.High pressure engines1 However, Watt stuck to the principle of using atmospheric pressure,

without realising that this left huge room for improvement.

2 The problem was that an improvement of this kind required a highworking pressure, which in turn would require multiple expansion stagesand a far stronger construction of the boiler.

3 The development of multiple expansion engines therefore took along time, culminating around the same time that Parsons[g] launchedthe first reaction turbine in 1884.

4 Shortly afterwards (1887), de Laval[g] created the first impulse tur-bine, and in 1890 he equipped it with a supersonic nozzle.

1073 / 1776

[g] Charles Algernon Parsons, 1754–1931. Irish engineer.

[g] Carl Gustaf Patrik de Laval, 1845–1913. Swedish engineer.

1 For a long time the steam engine remained the gold standard forrailways and mechanical industry, while the steam turbine was univer-sally adopted in ships and for electricity generation.

2 This rapid transition was due to the fact that steam turbines aresmaller, have a higher operating temperature and can make effectiveuse of the isentropic expansion of superheated steam.

3 This latter quality is particularly important and, as shown in Fig-ure 18.8, it raises the efficiency significantly above that of a steam en-gine using saturated steam.

4 In fact, using superheated steam results in an engine that is theo-retically perfect; see Theme 150 on page 979.



18 Heat . . .Vapour–liquid cycles (2)

5 The theoretical difference between a basic steam engine — withand without steam expansion— and a one-stage steam tubine can beseen from the T , s- and p, s diagrams in Figure 18.9.

6 In the literature the general term Rankine cycle is used regardlessof the engine, but note that only the Newcomen cycle has a rectangularpath in p,V coordinates.

7 The Watt cycle is more like the Carnot cycle, while the steam tur-bine cycle (as shown here) is really a Brayton cycle, with a rectangularpath in p,S coordinates.



18 Heat . . .State diagramsENGLISH text is missing.Practical calculations1 All of the diagrams (including Figure 18.8) have been drawn using

the Program 1.31 in Appendix 31.

2 This is based on rigorous calculations involving an advanced equa-tion of state[g] with 232 parameters.

3 One practical disadvantage of this equation of state is that it has anunknown number of volumes that satisfy the expression f(T ,p, v) = 0at a given pressure and temperature.

4 By comparison, the Van der Waals equation has either one or threereal solutions.

5 You therefore need good initial values for the calculations— partic-ularly along the phase equilibrium curve and close to the critical point.

6 In this particular case there are correlations between the saturationproperties of water and steam[g] that get us out of our predicament, butin general the problem of obtaining adequate initial values is difficult tocircumvent.

1077 / 1776

[g] A. Saul and W. Wagner. J. Phys. Chem. Ref. Data, 18(4):1537–1564, 1989.

[g] A. Saul and W. Wagner. J. Phys. Chem. Ref. Data, 16(4):893–901, 1987.

1 2 At first glance the difference between the p,V diagrams 18.9b,d,f

appears to be relatively insignificant, and we will therefore use the restof this chapter to discuss some important details. Firstly, it is important to note that the saturation states are in dif-

ferent locations in the diagrams — the steam engines convert liquid tosaturated steam at a high temperature, whereas the steam turbine con-verts saturated steam to liquid at a low temperature.

3 In the case of the two steam engines, both heating and cooling takeplace within the two-phase region, whereas the steam turbine exploitsisentropic compression and expansion in the one-phase region.

4 Consequently, the steam turbine can make use of some of theenergy contained in the fuel at a higher temperature than the boilertemperature.

6 5 This can be seen clearly from the T , s diagram in Figure 18.9e. The efficiency of the turbine therefore exceeds the Carnot effi-ciency of the system based on the boiler temperature. 7 The steam turbine is really a Brayton cycle with alternate isobaric

and isentropic changes of state, but as can be seen from Figures 18.9e

and 18.9f, the isentropic compression of the water produces so littlechange in volume as to make it, to all intents and purposes, isochoric.8 The steam engine, meanwhile, is limited by the Carnot efficiency,but there is still a big difference between Newcomen’s and Watt’s de-signs.9 Figure 18.8 shows that Watt’s steam engine is almost as good as atheoretical Carnot engine, which is pretty good considering that it wasdeveloped half a century before Carnot set out his ideas and a wholecentury before the emergence of modern thermodynamics.



18 Heat . . .§ 150 § 149 § 151 § 1481 Derive a simplified expression for the efficiency of Newcomen’s at-mospheric steam engine. The cylinder is completely filled with satu-rated steam, and both evaporation and (residual) condensation occurat constant pressure (and temperature), while heating and cooling oc-cur at constant volume.

1079 / 1776

§ 148 Newcomen’s engine1 The heat added is qh = (∆u)v + (∆h)p ; also see Figure 18.9a and

Table 18.5 on page 961.

2 Under normal operating conditions, the isochoric heat is much lessthan the isobaric heat (of vaporization), and qh ≈ ∆h is therefore a goodapproximation of the total heat requirement.

3 The work performed is w = (∆p)v(∆v)p ; see Figure 18.9b, butsince the pressure in the condenser is much lower than atmosphericpressure, and the steam is virtually an ideal gas, we can say that w ≈psat

h vvaph ≈ RTh.

4 This means that the efficiency is ηNewcomen ≈ RTh/∆h.1080 / 1776

§ 1491 Repeat Exercise 148, but this time for Watt’s steam engine, includ-

ing partial (isentropic) expansion of the saturated steam.1081 / 1776

§ 149 Watt’s engine1 The heat requirement is the same as in Theme 148, but the cycleperforms the expansion work

∫p(dv)s in addition to the compression

work (∆p)v(∆v)p .

2 The expansion takes place when the saturation curve for the steamand dv are dominated by the change in p.

3 This means that p(dv)s ≈ p(∂v∂p

)T dp ≈ v ıg dp.

4 The Clausius–Clapeyron relation in Chapter 14 gives us dp =

∆h/(T∆v) dT ≈ ∆h/(Tvvap) dT , which means that∫

p(dv)s ≈−∆h

∫dln T = ∆h ln(Th/Tc) and ηWatt ≈ ηNewcomen + ln(Th/Tc).

5 This trebles the efficiency of the engine at Tc ≈ 313 K.1082 / 1776

Show that the maximum work that can be performed byan ideal steam turbine is equal to the exergy content

w = ∆h − Tc∆s of the superheated steam.



18 Heat . . .§ 150 The ideal steam turbine

1 Figure 18.9e and Table 18.5 imply that w = (∆hh)p − (∆ch)p.

2 All of the heat is removed at constant (low) temperature i.e.(∆ch)p = Tc(∆cs)p, but from Figure 18.9e it follows that ∆cs = ∆hs, sothe work performed can therefore be expressed as w = ∆h − Tc∆s, inline with Eq. 18.11 (the ∆ changes refer to the limits of the cycle).



18 Heat . . .Looking back

1 Closed cycle:∮(dU)n =

∮(δQ − δW) = 0

2 Work = heat:∮

p dV =∮

T dS

3 Heating: Qh =∫

dS>0T‖dS ‖

4 Cooling: Qc =∫

dS<0T‖dS ‖

5 Thermodynamic efficiency: η = W/Qh = (Qh −Qc)/Qh

6 Carnot efficiency: ηCarnot = (Th − Tc)/Th

7 Stirling (and Ericsson) with recuperator:

ηıgStirling →

NR(Th−Tc) ln(V2V1

)

NRTh ln(V2V1

) = Th−TcTh

= ηCarnot



18 Heat . . .Looking back (2)

8 Otto (and Brayton) after compression stroke:

ηıgOtto = ηıg

Brayton = ηCarnot(Th)

9 Single stage power plant: ηtot = W∆rxH = Th−Tc

Th· Ta−Th

Ta−Tc= ηCarnot ·

ηthermal

10 Optimal design: ηopt =√

Ta−√

Tc√Ta+

√Tc

=Topt−Tc

Topt+Tc

11 Infinite number of stages: W∞ = −CP

Tc∫

Ta

T−TcT dT = ∆H − Tc∆S


News The Clausius inequality The Gouy–Stodola theorem Available energy Olds

Part 19

Entropy production and available work




19 Entropy . . .New concepts

1 Reversible heat

2 Clausius’s inequality

3 Entropy production

4 Lost work

5 Gouy–Stodola’s theorem

6 Available energy

7 Exergy



19 Entropy . . .Chemical energy

1 Chemical energy can be converted directly into work at the molec-ular level, as a coupled electron–ion transport process in a fuel cell, orindirectly by first transforming it to thermal energy in a thermal powerstation.

4 2 Fuel cell technology is undoubtedly the most elegant option, butit has to be acknowledged that power stations are much more widelyused.3 The two methods of conversion are not only different at a practicallevel— the very concepts behind them are also different. In a fuel cell, the ionic transport through the electrolyte is coupled

to the chemical reactions on the surface of a membrane and to thetransfer of an equivalent number of electrons in an external circuit .

6 5 In principle, this involves a lossless conversion process. In a thermal power station, the chemical energy is first transformedinto thermal energy, which is then converted into mechanical energyusing suitable machinery.

9 7 The first stage of this transformation is in theory lossless, whereasthe second stage is subject to the 2nd law of thermodynamics, whichlimits the potential efficiency of the process.8 The fact that combustion reactions are irreversible means thatlosses are inevitable during the conversion process. This chapter is about thermal energy in an abstract sense — the

exact details of the processes involved are beyond its scope.



19 Entropy . . .Chemical energy (2)

10 We will build on a number of concepts that we have previously stud-ied: Mass and energy balances, state functions, conditions for equilib-rium, reference states and Carnot efficiency.

11 In addition, we will introduce some new concepts: Entropy produc-tion and energy availability.

13 12 The reward for our efforts will be a more general energy equa-tion that allows high-level analysis of chemical and mechanical energytransformation. It should be pointed out that by itself our analysis only tells us what

processes are theoretically possible — there is no explanation of thenature of the process Expanded footnote [a]:

[a] In general we cannot say how long the process takes, whether it requires a largeamount of space, or if there are any other physical limitations. However, we can ensurethat the 2nd law of thermodynamics is fulfilled, and that the mass and energy balanceequations can be solved.

14 or, if you prefer, its physical design.15 The difference between analysis and design is not self-evident, andone of the subsidiary aims of this chapter is precisely to clarify the dis-tinction.



19 Entropy . . .The Clausius inequality

1 The Clausius Expanded footnote [a]:

[a] Rudolf Julius Emanuel Clausius, 1822–1888. German physicist. inequality applies to changes in thermal energy in aclosed control volume, and put simply it states that the maximum valueof

∫T -1δQ occurs in a reversible process.

4 2 This is an independent thermodynamic hypothesis, which doesn’taffect our understanding of the thermodynamic state itself.3 What we are interested in here is the interaction between the sys-

tem and its surroundings.

1

2

ırrrev

Figure 19.1: Two alternative paths illustrating the Clausius inequality. This can be expressed in terms of equations as

∮T -1δQ ≤ 0 ,

∮T -1δQrev =

∮dS = 0 .

5 Note that 1/T plays the role of an integrating factor for δQrev in thelast expression.

7 6 A more thorough explanation of this concept is given on page 109under the discussion of Pfaff?? differentials. The thermodynamic cycle in Figure 19.1 has one reversible and

one irreversible path, which leads to the inequality Expanded footnote [a]:

[a] From calculus we have ∫ 21 T -1 δQrev = limn→∞∑ni T -1i ∆Q+i , where ∆Q+i = ~Q+i dt+i .For a process to be reversible, the integral must have the same absolute value, butthe opposite sign, when the process is reversed to establish the initial state 1 from thefinal state 2. In other words, for each heat term ∆Q+i = ~Q+i dt+i in the forward pro-cess, there is an equivalent and opposite heat term ∆Q−i = ~Q−i dt−i in the reverseprocess, and hence all temperatures Ti assuming i ∈ [1,n] stay the same in thetwo processes. The net effect of the reversible heat added, and subtracted, fromthe system is ∑ni T -1i ∆Q+i +∑in T -1i ∆Q−i , and since ∆Q− = −∆Q+ this can be written∑ni (T -1i − T -1i )∆Q+i = 0. On integral form we can say that ∫ 21 T -1 δQrev + ∫ 12 T -1 δQrev =

0. However, the return to the initial state along the irreversible route does not havethis property. For an irreversible process, it is generally imposssible to force it in abackward direction. Here we must assume that the physical events occur in reverseorder and have the opposite signs for the reversed process. Which means time it-self is being reversed. This is not amenable to real physics but for our hypotheti-cally reversed process we may consider to write ∫ 12 T -1 δQırr = ∫ 21 T -1 δQırr, because∆Q−i = − ~Q+i (− dt+i ) = ∆Q+i when time is reversed. The point being that the integralalong the irreversible route has a natural direction of integration and that the rule ofsign reversal, valid for ordinary integrals, does not apply to this process. :



19 Entropy . . .The Clausius inequality (2)

∮T -1δQ =

2∫

1

T -1δQrev +

1∫

2

T -1δQırr ≤ 0 .

8 The integral of the reversible heat transfer (divided by the temper-ature) defines entropy as a thermodynamic state variable, and we canstate that

1∫

2

T -1δQırr ≤ −

2∫

1

T -1δQrev =

1∫

2

T -1δQrev = S1 − S2 = ∆2,1S .

9 Up to this point everything seems normal but we have to be extracareful if we are ever planning to change the direction of integration inthe inequality above.10 The change of sign on both sides of the inequality sign gives

−1∫

2

T -1 δQırr ≥ ∆1,2S ,

but we cannot on this basis conclude that ∫ 2

1 T -1 δQırr ≥ ∆1,2S. Thiswould then imply ∮ T -1 δQ ≥ 0 which is contradictory to the Clausiusinequality.11 The reason of the contradiction stems from the fact that the irre-versible integral has only a single direction of integration which is phys-ically meaningful, see footnote on the previous page.12 To avoid confusion it is therefore better to write

∫ T -1 δQırr ≤ ∆S

and at the same time make clear that the inequality is valid for two pro-cesses being physically meaningful, and sharing the same end states,while having entirely different process routes.



19 Entropy . . .Entropy is time’s arrow

1 Based on the integral inequality we can, for differential changes inentropy, conclude that T dS ≥ δQ.

3 2 Agreeably, it is tacitly assumed that T > 0 but this does hardlycause any frustration in the current context. For an isolated system, or a closed system that only experiences

adiabatic changes of state, there is the special condition δQ = 0 whichleads to

(T dS)isolated ≥ 0 .

4 Since T > 0 and dt > 0, then it is also true that (dS /dt)isolated ≥ 0.

5 The entropy of an isolated system therefore increases monotoni-cally with time. 6 This has given rise to the claim that entropy is time’s arrow [a],

and to speculation about the link between time and irreversible phys-ical events[a].

[a] Arthur Stanley Eddington. The Nature of the Physical World. MacMillan, 1928.

[a] What if time is inversely proportional to the frequency of physical events and the “BigBang” is a singularity in the grand scheme of things? If so, creation took place infinitelylong ago. Alternatively, the universe may have a finite age, but in that case what wasits state prior to creation? One critical discussion of the 2nd law of thermodynamics

and how it relates to our understanding of time is presented by Jos Uffink. Stud. Hist.Phil. Mod. Phys., 32(3):305–394, 2001.



19 Entropy . . .The Gouy–Stodola theorem

1 In all real processes there are irreversible losses.

4 2 Everyone has a subjective notion of what loss involves[a], but whatshould really be considered a loss in thermodynamic terms?3 Whatever it is, it isn’t the loss of energy, mass, impulse or charge,

as they are all subject to fundamental conservation laws.

[a] E.g. financial loss, memory loss, loss of youth etc. A loss must relate to something that is transient, and in thermody-namics we speak about lost work.

5 This refers to a reduction in the system’s ability to perform mechan-ical work.

6 According to the Gouy Expanded footnote [a]:

[a] Louis Georges Gouy, 1854–1926. French physicist. –Stodola Expanded footnote [a]:

[a] Aurel Boleslav Stodola, 1859–1942. Slovenian physicist and engineer. theorem Expanded footnote [a]:

[a] G. Gouy. Journal de Physique, 8:501–518, 1889.


[a] A. Stodola. Z. VDI, 32(38):1086–1091, 1898. , Wlost is proportionalto the irreversible entropy production Sırr in the control volume.

7 The factor of proportionality is the temperature T of the surround-ings, or more specifically: The temperature of the thermal reservoir thatexchanges heat with the system.



19 Entropy . . .The Gouy–Stodola theorem (2)

8 The Gouy–Stodola theorem can be expressed as an equation:Wlost = TSırr. However, it is not yet clear how the entropy produc-tion term can (or should) be interpreted.

10 9 Entropy is inherently a difficult concept, and facile claims that en-tropy is a measure of disorder or chaos, or that entropy is an indicatorof time, do not help us to understand it any better[a], and if anythingthey further confuse the matter.

[a] We are not forgetting about the Boltzmann equation S = k lnΩ. There is a finiterelationship between entropy on the one hand, and the number of microstates of thesystem on the other, but this is deeply rooted in statistical–mechanical theory, and itisn’t designed to shed light on the concept of entropy at a macroscopic level. There is every reason to fear that a derived quantity such as en-

tropy production will only make things worse, but in some ways it is eas-ier to conceive that entropy can be produced in a physical process thanit is to ascribe any deep philosphical meaning to the entropy as an ab-stract subject.



19 Entropy . . .§ 151 § 150 § 152 ENGLISH text is missing.Simplest case1 In order to illustrate the Gouy–Stodola theorem we will study a sim-ple control volume with two thermodynamic states, and we will limit our-selves to verifying that the entropy production in this system is actuallya measure of the lost work.

2 At the same time we will demonstrate that the concept of entropyproduction is underpinned by an intricate model that soon has us solv-ing complex partial differential equations.

1096 / 1776

Let us consider an isolated system consisting of twosolid blocks of temperatures Th and Tc. Each blockis a diabatic heat conductor, with a constant heat

capacity and volume. Heat is transferred from the hotblock to the cold block in the presence of a thermal

resistance with negligible (thermal) capacity. At t = 0the temperatures of the two blocks are Th(0) and Tc(0)

respectively. Your task is to find an expression for dSdt and

to explain how it relates to the lost work of the system.



19 Entropy . . .Towards equilibrium

1 The blocks have uniform temperatures, and individually they pro-duce no entropy.

Th Tc

~Q

Figure 19.2: Production of entropy caused by heat conduction in an isolatedcontrol volume.

2 The thermal resistance is to blame!



19 Entropy . . .Towards equilibrium (2)

3 It drains valuable thermal energy from the system without produc-ing any work, and without the thermal energy being used to produceanything other than less valuable heat.

4 As the thermal resistance has negligible (thermal) capacity, we canleave it out of the balance equations, and hence the change in entropyof the system can be expressed as follows:

dSdt = dSh

dt + dScdt .

5 For each block it is true that (dU)V ,n = T dS, where (dS)V ,n =CV

TdT , which means that (dU)V ,n = CV dT assuming the volume of theblock does not change with temperature (it said to be constant in thiscase).



19 Entropy . . .Towards equilibrium (3)

6 The energy balances for the whole system and for each of theindividual blocks can be stated as follows:

System : dUtotdt = dUh

dt + dUcdt = 0 ,

Hot : dUhdt =−~Q ,

Cold : dUcdt = ~Q .

When they are substituted into the entropy equation we get

dSdt =

~QTc− ~Q

Th.



19 Entropy . . .Differential equation

1 2 What remains to be done is to relate ~Q to the temperature gradientbetween the two blocks. The thermal resistance is directly proportional to the temperature

difference, which means that the heat flux fulfils the empirical equation~Q = κ(Th − Tc).

3 Let us first determine Tc by substituting the differentials dUh = Ch

dTh and dUc = Cc dTc into the energy balance of the isolated system.That yields Ch dTh + Cc dTc = 0 which can be integrated to give Tc =

Tc(0) + ChC -1c (Th(0) − Th).

4 Substituting the expression for Tc into the heat flux equation ~Q =

κ(Th − Tc), and then substituting ~Q into the energy balance of the hotblock, produces an equation that describes the change in the state ofTh:

dThdt = − κ(Th−Tc)

Ch= − κ

Ch

[(1 + Ch

Cc)Th −

[Tc(0) +

ChCc

Th(0)]].



19 Entropy . . .Differential equation (2)



19 Entropy . . .Time–temperature curve

1 This is a separable differential equation that can be expressed inthe form dTh /(aTh − b) = −dt .

2 Integration between the two states Th(0) and Th gives us the inter-mediate result

ln( aTh−baTh(0)−b

)= −at .

4 3 If we write the above equation in exponential form, we get an ex-plicit expression for Th = Th(t), and a bit of routine manipulation is allwe need to determine a and b, but as the intermediate steps are not ofany great value we will leave them out[a].

[a] In spite of the fact that in general it is a stated aim of this book to avoid shortcuts ofthe kind “from which it trivially follows that. . . ” The final result is

Th =CcTc(0)+ChTh(0)

Cc+Ch+

Cc(Th(0)−Tc(0))Cc+Ch

exp(−κ(C -1

h + C -1c )t

).

From symmetry reasons we expect a similar result for Tc but the exactformula is not so important for the current discussion.



19 Entropy . . .§ 151 Production of entropy

1 Now what remains to be done is to insert realistic values for aphysical system into the final equation from the previous section, andit probably makes most sense to use values that tell us about typicalsystem behaviour, rather than getting hung up in complexities relatingto geometry and mass distribution.

3 2 Our motto is maximum insight with the minimum of effort. We will therefore choose the symmetric state description Cc =

Ch = 2κτ, where τ is the system’s time constant.

4 That gives us the manageable temperature equation Th =12(Tc(0) + Th(0)) + 1

2(Tc(0) − Th(0)) exp(−tτ ) which can be further sim-

plified by introducing T = 12(Th(0) + Tc(0)) and ∆T = 1

2(Th(0)− Tc(0)).



19 Entropy . . .§ 151 Production of entropy (2)

5 Bearing these two definitions in mind, we can state that

Th = T +∆T exp(−tτ ) ,

Tc = T −∆T exp(−tτ ) .

6 If we combine these with the heat flux ~Q = κ(Th − Tc) = 2κ∆Texp(−t

τ ) and substitute them into the entropy equation dS /dt = Q/Tc −Q/Th, we get

dSdt = 2κ∆T exp(−t

τ )

[1

T−∆T exp(−tτ )− 1

T+∆T exp(−tτ )

].



19 Entropy . . .§ 151 Production of entropy (3)

7 The final expression Expanded footnote [a]:

[a] One little extra exercise for anyone who is interested: S is a state function and theincrease in entropy ∆S = ∫ ∞0 ( dSdt )dt from the initial state to the final state can also be

expressed as ∆S = Ch ln( TT+∆T ) + Cc ln( TT−∆T ) = 2κτ ln(1 − (∆T/T)2)-1. Verify thatthe two alternative expressions for ∆S are identical. for dS

dt is obtained by finding a common de-nominator for the fractions and then dividing both the numerator anddenominator by ∆T2 exp(−2t

τ ):

dSdt = 4κ(

T∆T exp(−t

τ )

)2

−1

∈ 〈0,∞〉 .

8 It is important to note that 0 < ∆T ≤ T and 0 < exp(−tτ ) ≤ 1.Applied to the above equation this demonstrates that dSdt ∈ 〈0,∞〉.9 Remember that the control volume is entirely isolated from its sur-roundings.10 Hence the change in the system’s entropy must be attributable tothe irreversible production of entropy in the control volume Sırr = dSdt .



19 Entropy . . .Lost work

1 Finally we will show that the control volume fulfils the Gouy–Stodolas theorem Wlost = TSırr = T dS

dt .

2 Let us first replace the conductor with a perfect insulator.

3 Next we will allow heat transfer from the hot block to a thermalreservoir of temperature Tc < T < Th.

4 On its way, the heat ~Qh > 0 flows past a Carnot engine.

5 For Th > T the power of this engine will be ~Wh = (1 − TT -1h )~Qh.

6 A similar argument can be made with respect to the cold block.

7 The heat ~Qc < 0 flowing between the reservoir and the block is inthis case ~Wc = (1 − TT -1

c )~Qc.



19 Entropy . . .Lost work (2)

8 If the two engines are synchronised such that ~Qh + ~Qc = 0, thesystem as a whole will behave in the same way as the isolated controlvolume Expanded footnote [a]:

[a] But unlike in the case of the isolated system, in this case the surroundings will beaffected. , except that it will produce the following amount of net work

per unit time:

~WCarnot = ~Wh − ~Wc = Th−TTh

~Qh − Tc−TTc

~Qc

= T( ~Q

Tc− ~Q

Th

)

= T dSdt .




Th

T

~Qh

~Q

~Whη = Th−TcTh

Figure 19.3: Extraction of valuable heat in a Carnot engine.




9 In other words, T dSdt represents the mechanical work that can be

done using two external Carnot engines and one thermal reservoir.

10 If this work is not performed, it will be permanently lost as a resultof the irreversible heat transfer within the control volume.



19 Entropy . . .Process analysis

1 To analyse the available energy in a dynamic system, or the exergyof a flowing fluid, we need to look at the energy, entropy and massbalances together.

4 2 At first sight this may appear to be a case of mixing apples andoranges, but in fact it is possible to combine the equations if they arefirst rewritten in a common form (in this case energy per unit of time).3 This section deals with a rather esoteric topic, so it makes senseto take on the challenges that it poses gradually, even if that makes thebook a bit longer. Initially we will therefore limit ourselves to looking at a single chem-

ical component.

5 Moreover, the potential and kinetic energy contributions can beignored, which means that the energy balance can be written dU

dt =~Uın − ~Uout + ~Q − ~W .

6 The work performed can be split into pressure–volume work andmechanical work, i.e. ~W = ~Ws + p~V + p dV

dt .

7 The final term allows for the fact that the size (volume) of the controlvolume can change.



19 Entropy . . .Process analysis (2)

8 The pressure in the surrounding atmosphere has been given thesymbol p.



19 Entropy . . .Energy balance

1 Introducing the enthalpy function H = U + pV makes it possible toexpress the energy balance in the form

d(U+pV)dt = ~Hın − ~Hout + ~Q − ~Ws .

2 Note that the energy balance does not require information aboutthe chemical components involved in the flows of material.

3 It is therefore also valid for multicomponent systems.



19 Entropy . . .Entropy balance

1 In the case of reversible processes, entropy will be a conservedquantity, but in the general case there is no such conservation law.

2 In order to concoct an entropy balance, we therefore need to relyon the source term Sırr, which quantifies the entropy production of theirreversible subprocesses in the control volume.

4 3 Here the difference between design and analysis becomes impor-tant. For high efficiency, Sırr must be as low as possible (optimal design),

while all potential processes that result in Sırr < 0 can immediately berejected (based on the 2nd law).

5 In addition to entropy production, we need to take into account anycontribution from convection (~Sın and ~Sout) and any change in entropydue to heat transfer.



19 Entropy . . .Entropy balance (2)

6 This gives us the following balance equation: dSdt = ~Sın − ~Sout +∫

ΩT -1nT~q dΩ+ Sırr.

8 7 The integral in this equation sums the contributions from thechanges in heat δ~Q = nT~q dΩ, weighted by T -1, over the surface areaof heat transfer Ω. In this case n ∈ R3 is the unit normal vector to the surface element

dΩ [m-2], and ~q [W/m2] is the associated heat flux.

9 Note that the entropy equation can be multiplied by the ambienttemperature T , 0 without it changing the nature of the equation inany way:

T dSdt = T~Sın − T~Sout + T

∫

Ω

T -1δ~Q + TSırr .

12 10 What this operation means (and the motivation behind performingit) will become clearer in due course.11 Here let it suffice to say that the equation has not fundamentallychanged as a result of being scaled up. Like the energy balance, the entropy balance is also valid for mul-

ticomponent systems.



19 Entropy . . .Mass balance

1 If the inflows and outflows consist of one, and only one, chemicalcomponent, the mass balance can alternatively be expressed by themole balance dN

dt = ~Nın − ~Nout.

2 As we did with the entropy equation, we will now scale this equa-tion, but this time by the factor µ , 0:

µ dNdt = µ~Nın − µ~Nout .

3 In due course it will become clear that µ can be viewed as a ref-erence for chemical potential, but right now that is of secondary impor-tance.



19 Entropy . . .Available entropy

1 For the moment we will stick to our goal of combining the energy,entropy and mass balances.

2 Most of the groundwork has already been done Expanded footnote [a]:

[a] Also check that the three equations have the same physical units. The real reasonwhy we multiplied the entropy and mass balances by T and µ respectively was toachieve this! , but it will help

us to split the heat flux integral into two parts using the identity ~Q =∫δ~Q ≡

∫(1 − TT -1) δ~Q +

∫TT -1

δ~Q.

3 If we substitute this into the entropy equation we get

T dSdt = T~Sın − T~Sout + ~Q −

∫

Ω

(1 − TT -1) δ~Q + TSırr .

4 By subtracting this result from the energy balance, we can elimi-nate the heat flux ~Q.



19 Entropy . . .Available entropy (2)

5 We will also choose to subtract the mass balance, in spite of thefact that none of the terms cancel, as it gives us a more balanced resultfrom a thermodynamic point of view:

d(U−TS+pV−µN)dt = (~H − T~S − µ~N)ın − (~H − T~S − µ~N)out

+∫

Ω

(1 − TT -1) δ~Q − ~Ws − TSırr .



19 Entropy . . .Euler homogeneity

1 In Chapter 5, specifically on page 176, we used homogeneousfunctions to which Euler’s theorem applies to derive the integral expres-sions U = TS − pV + µN and H = TS + µN.

2 Applied to the above equation, this allows us to define a new statefunction F(x) = F(x) − gT

x = (g − g)Tx.

3 This assumes that x is the homogeneous set of variables of theoriginal function F , and that the gradient vector g defines the tangentplane to F at the reference point x.

4 This shows us that F has the same properties as an ordinary ther-modynamic energy function, except that the function’s gradient haschanged from g to g − g.



19 Entropy . . .Euler homogeneity (2)

5 Whereas the ordinary energy functions U,H,G,A etc. have an ar-bitrary zero point, in the case of U, H, G, A etc. it is defined exactly.

7 6 When performing a thermodynamic analysis of an open control vol-ume in contact with a thermal, mechanical and chemical reservoir, thisproperty is important, because it defines the absolute amount of avail-able energy in the system. With this in mind, we can choose to express the combined energy,

entropy and mass balance as

dUdt = Hın − Hout +

∫

Ω

(1 − TT -1) δ~Q − ~Ws − TSırr . (19.1)

8 Here U is the available energy of the system, and H is the exergyof the process flows (admittedly we have lost the rate symbol of ~H, buteven LATEX has a limited selection of function attributes).



19 Entropy . . .§ 152 § 151 § 153

Find the maximum achievable work per unit time(power) that can be extracted from a stationary control

volume under adiabatic conditions.



19 Entropy . . .§ 152 Process work

1 Stationary and adiabatic imply that dU /dt = 0 and ~q = 0 respec-tively.

2 When the heat flux is zero, the heat flux integral∫Ω(1 − TT -1) δ~Q

will also be zero, which in turn means that ~Ws = Hın − Hout − TSırr.

3 Since T > 0 and Sırr ≥ 0, the maximum power can be given asmax( ~Ws) = Hın − Hout.

4 Let us next assume tha that the outlet temperature Tout is equal tothe temperature T of the process’s surroundings. If we also select Expanded footnote [a]:

[a] Yes, we can indeed choose our own zero point. For each independent variable weare free to select one zero point for either the enthalpy, internal energy, Gibbs energyor any other energy function. Hence, in the case of a one-component system we canalways say that Hout = 0. However, it is only if Tout = T that we also get the simpleresult µout = µ.

µ = µout, then Hout = 0.



19 Entropy . . .§ 152 Process work (2)

5 It follows that the exergy Hın is a direct measure of the maximumachievable power:

max( ~Ws) = Hın .

6 This result is generally valid if the outflow can be viewed as a wastestream with no exergy.

8 7 Logically, the inflow would then have positive exergy, and it is im-possible to do better than to use 100% of it. The only thing we need to do is to (re)define T and µ in such a

way that their values reflect the outflow state.



19 Entropy . . .§ 153 § 152 § 154ENGLISH text is missing.A little bit of philosophy1 First we assume that the outlet temperature is the same as the

ambient temperature and then we conclude that the result is generallyvalid if the outflow has no exergy.

2 But what does this really mean? Are we really entirely free to definethe zero point of an energy exchange process as whatever we want?

3 Yes, we can indeed, provided that we stick to independent, possiblestates. We, and not the principles of thermodynamics, decide wherezero lies.

4 This is also at the heart of the difference between analysis anddesign.

5 It is only possible to perform a thermodynamic analysis on the pro-cess once all of the premises for the design are in place.

6 This is particularly true of practical issues such as the choice oftemperature and pressure levels, the exchange of heat and work in theprocess, and the characterisation of the product and waste streams.

1124 / 1776

Determine the entropy production of a stationary,adiabatic control volume that does not perform any

mechanical work.



19 Entropy . . .§ 153 Isenthalpic flow)

1 One formal answer is that TSırr = max( ~Ws) as set out inTheme 152.

2 But the process parameters also mean that the control volume isisenthalpic, which means that ~Hın = ~Hout.

3 Moreover, we also know that ~Nın = ~Nout.

4 We can therefore conclude that the entropy production TSırr =

(Hın − Hout)~H,~N = T(~Sın − ~Sout) is equal to the difference between theexergy flows into and out of the control volume.

5 It is important to note that this difference is independent of thereference state of H (also note how H is transformed into ~S via ~H).



19 Entropy . . .§ 154 § 153 § 155

Determine the maximum possible work (per cycle)for a closed process if heat is added at a constanttemperature Th > T and removed at the ambient

temperature Tc = T.



19 Entropy . . .§ 154 Carnot work

1 For a closed process we know that Hın = Hout = 0.

2 Since the process is cyclical, it is also true that∫ τ

0(dU /dt) dt = 0,

where τ is the period of the cycle.

3 That leaves us with the expression ~Ws =∫Ω(1−TT -1) δ~Q −TSırr.

4 The amount of work performed per cycle can be determined byintegrating ~Ws over the period τ.

5 We have been told that all of the heat is added to the process at atemperature of Th > T and is removed at a temperature of Tc = T.

6 Let us therefore divide the cycle into one period of time t ∈ [0, τ1]

during which heat is added to the process, and another period of timet ∈ 〈τ1, τ] during which heat is removed.



19 Entropy . . .§ 154 Carnot work (2)

8 7 If substituted into the heat flux integral, the latter will not contributeany useful work. Moving heat from within the system to the surroundings does noth-

ing, since there is no temperature difference. Hence we can concen-trate on what happens during the addition of heat.

9 We know that T = Th > T for this part of the cycle, and sinceTSırr ≥ 0 the maximum work that can be performed by the cycle is:

max(Ws) =τ1∫

0

∫

Ω

(1 − TT -1h ) δ~Q dt =

τ1∫

0

(1 − TT -1h )~Q dt = Th−T

ThQh .

10 It shouldn’t surprise anyone that this result is identical to the workdone by the Carnot cycle, at least if you give it a little bit of thought, but itis worth noting that we have found a formula that covers both stationaryflow processes (previous section) and cyclical closed processes (thissection).11 These two extremes are relatively simple to describe, but it is also

possible to analyse dynamic processes, as we will do below.



Figure 19.4: Excerpts from the steam tables of water. The reference state ofzero enthalpy and zero entropy is liquid water at its triple point.

T[C]

p[bar]

ρ[kg/m3]

s[kJ/K kg]

h[kJ/kg]

0.01 0.00611 0.0048 999.80 9.1575 0 2501.6 0

7 0.01001 0.0078 999.90 8.9762 0.1063 2514.4 29.41

25 0.03166 0.0230 997.11 8.5592 0.3670 2547.3 104.77

100 1.01325 0.5977 958.13 7.3554 1.3069 2676.0 419.06

225 25.501 12.763 833.89 6.2461 2.5641 2801.2 966.89

285 69.186 741.45 36.062 5.8220 3.1146 2774.5 1263.2

325 120.56 654.07 70.472 5.4969 3.5008 2688.0 1194.0

374.15 221.20 315.46 4.4429 2107.4



19 Entropy . . .Looking back § 1551 Steam is used for heating and as an energy buffer in many chem-

ical process plants. But what is the energy density of a boiler, and whatis its thermodynamic efficiency if it is heated with electricity? Start withthe maximum amount of work that can be performed under the adia-batic depressuring and emptying of a pressurised system. Let the vol-ume of the boiler be V = 1 m3, and assume that initially it is filled withwater at saturation point at a temperature of T = 225 C. Use data fromTable 19.4.

1130 / 1776

ENGLISH text is missing.Process dynamics1 If no mass is added, and the process takes place under adiabatic

conditions, then Hın = 0 and∫Ω(1 − TT -1) δ~Q = 0.

2 It follows that ~Ws = − dUdt − Hout − TSırr.

3 As in the previous two sections TSırr ≥ 0, and ideally the processshould be reversible.

4 But what next? How can we maximise ~Ws = − dUdt − Hout?

5 This is an open question, which can be solved in a variety of ways,one of which is to find the lower limit value for Hout.

6 If there is such a limit, max( ~Ws) can be calculated by integration.1132 / 1776

ENGLISH text is missing.Energy minimum

1 From the definition H(~S, p, ~N) = ~H − T~S − µ~N we can, by differ-entiating, derive (∂H/∂~S)p,~N = (T − T) and (∂H/∂~N)p,~S = (µ − µ).

2 In other words, at fixed ambient pressure p, H has a stationarypoint at T = T and µ = µ, with the associated function H = 0.

3 It can be shown that this stationary point is a local minimum, seeChapter 21 on page 1312, and since (∂H/∂p)~S ,~N = (∂~H/∂p)~S ,~N = ~V > 0

it follows that H = 0 is actually a global minimum for the depressuringprocess[a].

4 Our best bet is to ensure that the outlet conditions remain at alltimes T = T, p = p and µ = µ(T,p).

5 This is equivalent to requiring reversible mass and energy ex-change with the surroundings[a]. By way of contrast, imagine that T ,T. The temperature difference could in that case be used by an exter-nal heat engine, which would mean that our process was not optimal.

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[a] Depressuring to atmospheric pressure requires the condition p > p to be met.

[a] It is the outlet from the energy transformation process that needs to meet thisrequirement— not the outlet from the boiler.

§ 155 Boiler1 If all of the outlet parameters are adhered to, the maximum power

is max( ~Ws) = − dUdt .

2 Integrated over the whole duration of the process, this gives us thefollowing simple result for available work[a]:

max(Ws) =∞∫

0

~Ws dt = −0∫

U

dU = U

= (T − T)S − (p − p)V + (µ − µ)N .

3 We are now in a position to determine how much work the boilercan perform.

4 For waste streams that go straight to landfill, or that are releasedinto the air or water, it is natural to choose T ≈ 280 K and p ≈ 1 atm.

5 The waste stream from our boiler fits into that category, and theoptimal solution will be to depressurise and cool the water (reversibly)until it finally can be released into the recipient at T and p.

6 Note that we do not have any degrees of freedom with respect tothe choice of µ.

7 The chemical potential is uniquely determined if both the tempera-ture and pressure are known.

8 Now what remains to be done is to determine the values of p,S, µand ρ at 225 C and µ at 280 K and 1 atm.

9 The steam table 19.4 on the previous page contains some informa-tion that will help us to do that.

10 The values for enthalpy and entropy in the table relate to liquidwater at the triple point.

11 In the above equation, on the other hand, the entropy has 0 K asits reference, which means that we need to rewrite the expression.

12 To do so, we will apply the fundamental relation µ = h − Ts, in thisinstance using specific values.

13 We will also take care to state the final expression in terms of vol-ume, since we have been asked to find the available energy in a cubicmetre of water:

U = (T − T)S − (p − p)V + (µ − µ)M= TsM − TsM − (p − p)V + (h − Ts)M − (h − Ts)M ,

max(Ws)V = ρ(h − h) − ρT(s − s) − (p − p) .

14 Based on the data from Table 19.4, the maximum available energyis thus V -1 max(Ws) = 833.90 · [966890 − 29410 − 280.15 · (2564.1 −106.3)] − (25.501 − 1.01325) · 105 = 205.13 MJ m-3.

15 This can be converted to 57 kW h m-3, which is equivalent to thedaily electricity consumption of a normal Norwegian detached house.

16 Also note that a (microscopic) error has been introduced into thesecalculations by using the saturation properties of water at 0.010012 bar,rather than the actual properties at 1 atm.

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[a] Note that U(t →∞) = 0 in the final state. This is a direct result of the self-imposedlimits T = T, p = p and µ = µ at t → ∞. Hence it is not necessary to empty thetank completely— it is sufficient for there to be no available energy in the contents ofthe tank.

ENGLISH text is missing.Efficiency1 The thermodynamic efficiency of the boiler can be defined as the

work performed (delivered) divided by the energy supplied (electricity),but this immediately raises the question: Should the energy supplied becalculated assuming stationary or dynamic conditions?

2 In this case we will assume the former, since boilers normally op-erate under stationary conditions, but it should be emphasised that thisis not by definition true.

3 For an electric boiler, the energy supplied is equal to the electricpower supplied, and from the energy balance we can deduce that Wel ≈Vρ(h − h).

4 The approximately equal to symbol has been used to emphasisethe fact that the equation is holds if: i) there is no heat loss duringheating and compression, ii) conditions are stationary; and iii) that Hın =

0.

5 Based on these assumptions, the efficiency is η = max(Ws)/Wel =

max(Ws)/Vρ(h − h) = 205.13/781.76 = 0.262.

6 Given this poor theoretical efficiency, and bearing in mind the lowenergy density, there is no reason to use the boiler as anything otherthan a thermal energy store.

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1 Clausius’s inequality:∮

T -1δQ ≤ 0

2 Reversible heat:∮

T -1δQrev =

∮dS

3 Entropy production: (T dS)isolated ≥ 0

4 Gouy–Stodola’s theorem: Wlost = TSırr

5 Available energy: U = U − TS + pV − µN

6 Exergy: H = ~H − T~S − µ~N

7 Available energy balance equation:

dUdt = Hın − Hout +

∫

Ω

(1 − TT -1) δ~Q − ~Ws − TSırr


Titlepage List of symbols Thermodynamic concepts...

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