Physical Transformations of Pure Substancescourses.chem.psu.edu/chem451/Lecture14_large.pdf · ·...

Prof. Mueller Chemistry 451 - Fall 2003 Lecture 14 - 1

Physical Transformations of Pure Substances

Chapter 6 of Atkins

Sections 6.1- 6.7

Phase Diagrams

Stabilities of PhasesPhase BoundariesThree Typical Phase Diagrams

Phase Stability and Phase Transitions

Thermodynamic Criterion of EquilibriumDependence of Stability on the ConditionsLocation of Phase BoundariesEhrenfest Classification of Phase Transitions


Phase Transformations

We now move from some of the relatively abstract concepts of the firstfive chapters to some direct applications of thermodynamics: Apply thelaws of thermodynamics to physical transformations of pure substances

How do we know the phase of a substance at a given pressure andtemperature ?

The Gibbs energy can give us the answer to this question sincespontaneous processes have negative changes in Gibbs free energy

For example, if you want to know the most stable phase of water at agiven pressure and temperature, you can calculate the Gibbs energies foreach of the phases, compare them, and pick to lowest one

Chemical potentials are equivalent to Gibbs energies for puresubstances, so the phase with the lowest chemical potential will be themost stable phase !


Stabilities of Phases

dG = Vdp − SdT

∂G∂T

⎛ ⎝ ⎜

⎞ ⎠ ⎟

p

= −S


Stabilities of Phases

Phase : form of matter that is uniform throughout in both chemical composition and physical state

Phase Transition : spontaneous conversion of one phase into another, occurring at a characteristic temperature for a given pressure

Consider water at p = 1 atm:

Ice is stable phase when T < 0 oCµ ice < µ liquid water when T < 0 oC

Liquid water is stable phase when T > 0 oCµ ice > µ liquid water when T > 0 oC

Transition Temperature , T trs :Temperature when chemical potentials areequal; for example, µ ice = µ liquid water at T = 0 oC


Rates of Phase Transitions

Distinguish between:

Spontaneity of transition Rate of transitionThermodynamics may predict the Kinetics predicts the rate at which aspontaneity of the physical transition occurs, but not thetransformation, but not the rate at spontaneitywhich it occurs

Consider the phase transition from diamond to graphite:

Normal pressure and temperature, µ diamond > µ graphite

For change to take place, C atoms must exchange positions, which is anincredibly slow process for solids (except at high temperatures)

In gases and liquids, these changes can take place rapidly, but in solids,thermodynamic instability may be inherent in the system

Metastable phases : Thermodynamically unstable phases that persistdue to kinetic hindrance


Phase Boundaries

Phase Diagram : Shows regions of pressure and temperature where phases are thermodynamically stable

Vapour PressureSublimination Vapour Pressure of a gasPressure in equilibrium withPressure of gas in the liquid phase equilibrium with solid phase

Phase BoundariesSeparate regions andshow p and T wheretwo phases exist inequilibrium

Triple Point Critical Temperature

Vapor pressures increase with higher T, as the Boltzmann distribution populates the higher energystates with increasing temperature


Critical Points and Boiling Points

In an open vessel , the temperature where the vapour pressure is equalto the external pressure, free vapourization occurs throughout the liquid,and vapour can expand freely into the surroundings. Free vapourization is known as boiling

The normal boiling point , T b , is the temperature of free vapourization ata pressure of 1.0 atm

The standard boiling point is the temperature of free vapourization at apressure of 1.0 bar (0.987 atm) (e. g., water: 99.6 o C for T b = 100 o C)

In a closed vessel , boiling does not occur - ratherthe vapour pressure and density of the vapourincrease with increasing temperature(a) equilibrium, (b) increasing density & (c) twodensities equal, surface between phases disappears

Critical temperature, Tc and critical pressure , p c :where interphase surface disappearsSupercritical fluid : fills container and interface nolonger exists (figure c)


Melting Points and Triple Points

The temperature at a specified pressure where the liquid and solidphases coexist in equilbrium is called the melting temperature (orfreezing temperature or fusion temperature )

The normal freezing point , Tf , is the temperature of freezing ata pressure of 1.0 atm

The standard freezing point is the temperature of freezing ata pressure of 1.0 bar (0.987 atm)

The point where all three phases exist in equilibrium is called the triplepoint , T3 (three phase boundaries meet)

•T3 occurs at a specific temperature and pressure which are characteristic of a substance

•T3 marks lowest pressure in which a liquid phase of a substance can exist

Example: Water, T3 = 273.16 K and 611 Pa (6.11 mbar, 4.58 Torr)


Phase Diagram of Water

•The liquid- vapour boundary Negative slope: m.p.shows variation of vapour decreases as p increasespressure with temperature (decrease in V on melting)

•The solid- liquid boundary Scale changeshows variation of melting indicates enormouspoint with temperature pressures are

required to bring•Glacier motion may result about significantfrom decrease in m. p. with changesincreasing p

• At extremely high pressures,different phases of ice comeinto existence, as bondsbetween molecules aremodified by huge stresses

•Multiple triple points exist withinthe diagram other than thelowest where vapour, liquid andice I coexist


Phase Diagram of Carbon Dioxide

Positive slope of solid- liquid boundary ischaracteristic of most substances (m. p.increases as pressure is increased)

Triple point lies above 1 atm, implying thatliquid CO2 cannot exist at normalpressures (solid sublimes at normalpressures, hence the name, “ dry ice ”)

To obtain liquid CO2 , p = 5.11 atm is theminimum. For a cylinder of CO2 (l) at25 o C, the pressure of the gas must begreater than 67 atm - gas comes out of aJoule- Thomson throttle, emerging into p=1 atm, condenses into a snow-like solid

Supercritical CO2 (highly compressed) isused in supercritical fluid chromatographywhich separates lipids and phospholipids,fuel oils into components and todecaffeinate coffee


Phase Diagram of Helium (4He)

• At low temperatures, helium behavesunusually - solid and gas are never inequilibrium: He atoms are so light that theyvibrate with high amplitude motions which“shake the solid apart”

• Solid helium can only be obtained at lowtemperature and very high pressure - hcpand bcc denote hexagonal close packingand body-centered cubic packing

• The λ-line marks a special phase transitionwhere heat capacity becomes infinite,marking the fluid-superfluid transition ofliquid Helium

• The He-II liquid phase is a superfluid, as itflows without any viscosity

• 4He and 3He have different phase diagrams;in fact, the entropy of 3He(l) is less than thatof 3He(s), so melting is exothermic


Phase Stability and Phase Transitions

The thermodynamic criterion of equilibrium is:At equilibrium, the chemical potential of a substance is the samethroughout a sample, regardless of how many phases are present

For example, when solid and liquid phases are inequilibrium, neither has a higher chemical potential (samethroughout the solid and the liquid)

If amount dn is transferred from 1 to 2, Gibbs energychanges by -µ1 dn in location 1, and by +µ2 dn in location 2

If µ 1 > µ2, ∆G is negative, process is spontaneousIf µ1 = µ2, no change in ∆G, system at equilibriumIf ∆Ssys + ∆Ssur= 0, system at equilibrium

µ1 µ2

dG = µ2 − µ1( )dn


Temperature Dependence of Phase Stability

Temperature dependence of Gibbs energy is expressed in terms ofentropy, (∂G/∂T)p = -S. Since the chemical potential of a pure substanceis the same as the molar energy, we can write

As T raised, µ lowered, Sm > 0, so, slope of µ vs. T is negative

Plot of µ vs. T is steeper for gases than liquids, since the Sm(g) > Sm(l).For solids and liquids, usually Sm(s) < Sm(l)

Solids and Liquids: Liquids and Gases:

Steep negative slope of µ(l) falls µ(g) plunges downwards asbelow µ(s) when the temperature is temperature is raised (Sm(g) high !),high enough (liquid becomes stable so the gas is the stable phase andphase and solid melts) the liquid vaporizes

∂µ∂T

⎛ ⎝ ⎜

⎞ ⎠ ⎟

p

= −Sm


Melting Response to Applied Pressure

Most substances melt at a higher temperature when subjected topressure (the pressure prevents the formation of a less dense liquidphase) - the exception is water, with ρ(l) > ρ(s)

The variation of chemical potential with pressure is

so, the slope of chemical potential vs. pressure is equal to molar volume

Increase in pressure raises thechemical potential of most puresubstances, because Vm must begreater than zero

In (a), the freezing temperatureincreases, and in (b) the freezingtemperature is lowered (these areplots of negative entropy)

∂µ∂p

⎛

⎝ ⎜

⎞

⎠ ⎟

T

= Vm


Pressure and Chemical Potential

Example: Calculate the effect on chemical potentials of ice and water ofincreasing the pressure from 1.00 bar to 2.00 bar at 0oC. The density ofice is ρ = 0.917 g cm-3, and of water is ρ = 0.999 g cm-3

∆µ =Vm ∆ p for change in chemical potential of an incompressiblesubstance when pressure is changed by ∆p; thus,

∆ µ(ice) = (1.802 x 10-2 kg mol-1) x (1.00 x 105 Pa) = +1.97 J mol-1917 kg m-3

∆ µ(water)=(1.802 x 10-2 kg mol-1) x (1.00 x 105 Pa) = +1.80 J mol-1999 kg m-3

∆µ(ice) rises more sharply than ∆µ(water), so if they are at equilibrium at1.00 bar, there is a tendency for ice to melt at 2.00 bar

∆µ =M∆p

ρ


Applied Pressure and Vapor PressureWhen pressure is applied to a condensed phase, the vapor pressurerises (molecules are squeezed out and escape as a gas)

Pressure can be exerted on the condensedphase: (a) mechanically or (b) subjecting it toan inert pressurizing gas

In case (b), vapor pressure is the partial pressure of the vapor in equilibrium with the condensed phase -this is the partial vapor pressure of the substance (it is possible that some of the inert gas can dissolve in thecondensed phase if it is a liquid, or perhaps solvate some of the liquid molecules but we will not treat these complications)

The relationship between vapor pressure, p, of a liquid, and the original vapor pressure (p*), after an appliedpressure change of ∆ P is

Vapor pressure increases when pressure acting on condensed phase increases

p = p∗eVm∆P RT


Justification

At equilibrium, µ(l) = µ(g); so, any change that preserves equilibrium, theresulting change in µ(l) must be equal to that in µ(g) (dµ(l)=dµ(g))

When pressure P on liquid is increased by dP, dµ(l) = Vm(l) dP, anddµ(g) = Vm(g) dp (dp = change in vapor pressure). If vapor is a perfectgas, Vm(g) = RT/p, and

If changes in chemical potentials of vapor and liquid are equated:

The pressure P experienced by the liquid is equal to the normal vaporpressure, p*; so when P = p*, p = p* as well. When there is additionalvapor pressure ∆P on the liquid, P = p + ∆P, the vapor pressure is p(which we are trying to find) – replace p in p + ∆P by p* itself

rearrange, q.e.d.

dµ g( )=RTdp

p

RTdppp∗

p

∫ = Vm l( )dP or p∗

p∗ +∆P

∫ RT ln pp∗

⎛

⎝ ⎜

⎞

⎠ ⎟ = Vm∆P

RTdpp

= Vm l( )dP


Estimating Effect of Pressure

Example: Derive an expression from the equation below that is valid forsmall changes in vapor pressure and calculate the fractional increase ofvapor pressure of water for an increase in pressure of 10 bar @ 25oC

If Vm ∆p/RT « 1, then exponential function on RHS is approx. 1 + Vm∆p/RT

rearranging:

For water, ρ = 0.997 g cm-3 at 25o C and Vm = 18.1 cm3 mol-1

Since Vm ∆p/RT « 1, (p-p*)/p*=0.0073, an increase of only 0.73%.

p = p∗eVm∆P RT ex =1+ x +

x 2

2+

x 3

6+L , so if x <<1, ex ≈1+ x

p − p∗

p∗ =Vm∆P

RT

p = p∗ 1+ Vm∆P RT( )

Vm∆PRT

=1.81×10−5 m3 mol-1( )1.0 ×106 Pa( )

8.3145 J K-1 mol-1( )298 K( )= 7.3 ×10−3


Location of Phase Boundaries

Locations of phase boundaries are found by recognizing that when twophases are in equilibrium, chemical potentials are equal (phase α and β):

Solve the equation for p in terms of T to get the boundary equation

When pressure is applied to a system in whichthe two phase are in equilibrium (at a) theequilibrium is disturbed.

It can be restored by increasing thetemperature (moving the system to state b)

So, there is a relationship between dp and dT,ensuring that the system stays in equilibriumas each variable is changed

µα p,T( )= µβ p,T( )


Slopes of Phase Boundaries

p and T are changed infinitesimally such that two phase stay in equilibrium with one another - the chemical potentials are initially equal, and remain equal after these changes (i.e., dµα = dµβ). For each phase:

thus,

hence,

which rearranges to give the Clapeyron equation:

Where ∆trsS = Sβ,m –Sα,m and ∆trsV = V β,m –Vα,,m are entropy and volume of the transition (exact expression for the slope of a phase boundary -applies to phase equilibrium of any pure substance)

dµ = −SmdT + Vmdp

−Sα ,mdT + Vα,mdp = −Sβ ,mdT + Vβ ,mdp

Vβ ,m −Vα,m( )dp = Sβ ,m − Sα ,m( )dT

dpdT

=∆ trsS∆ trsV


Solid - Liquid Boundary

Melting (fusion) has an enthalpy change ∆fusH occurs at T. Molar entropyof melting at T is ∆fusH/T, and we write the Clapeyron equation as

∆fusV is change in molar volume on melting

Enthalpy of melting is positive (except 3He) andvolume change is usually positive and small, sodp/dT is steep and positive.

Formula for boundary is obtained by integrationassuming that ∆fusH and ∆fusV are constant(very small changes):

dpdT

=∆ fusHT∆ fusV

dpp∗

p

∫ =∆ fusH∆ fusV

dTTT ∗

T

∫


Solid - Liquid Boundary, 2

After integration, an approximate equation of the solid-liquid phaseboundary is written as

When T is close to T* (the melting temperature when pressure is p*)

because ln (1+x) ≈ x when x « 1. We can rewrite the above equation:

Plotting this equation as a function of T yields a steep straight line

p = p∗ +∆ fusH∆ fusV

ln TT∗

⎛ ⎝ ⎜

⎞ ⎠ ⎟

ln TT∗

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = ln 1+

T − T∗

T∗

⎛

⎝ ⎜

⎞

⎠ ⎟ ≈

T − T∗

T∗

p = p∗ +T − T∗( )∆ fusH

T∗∆ fusV


Liquid - Vapor Boundary

Entropy of vaporization at temperature T is equal to ∆vapH/TThe Clapeyron equation for the liquid- vapor boundary is

∆vapH is positive, ∆vapV is large and positive, so dp/dT is positive, but notas steep a slope as for the solid-liquid boundary:

dT/dp is large, boiling temperature is more responsive topressure than the freezing temperature

To estimate the size of the effect of increasing p on the b.p. of liquids:∆vapV = Vm(g) -Vm(l) ≈ Vm(g), (sinceVm(g) »Vm(l))

This is 0.034 atm K-1, so dT/dP = 30 K atm-1 (+0.1 atm pressure changealters the boiling point by approximately +3 K)

dpdT

=∆ vapHT∆ vapV

dpdT

=85 J K−1 mol−1

25 x 10-3 m3 mol−1 = 3.4 x 103 Pa K−1


Liquid - Vapor Boundary, 2

Since Vm(g) » Vm(l), ∆vapV ≈ Vm(g). If gas behaves perfectly, Vm(g) = RT/p

since dx/x = d ln x. If ∆vapH is independent of temperature, thenintegration of this equation yields

where p* is the vapor pressure whentemperature is T*, and p for T (as shown inthe plot) - this plot will not extend beyond Tc,since there is no liquid phase at that point

dpdT

=∆ vapH

T RT p( ) , d ln p

dT=

∆ vapHRT 2

p = p∗e−χ , where χ =∆ vapH

R1T

−1T∗

⎛ ⎝ ⎜

⎞ ⎠ ⎟


Solid - Vapour Boundary

This is treated exactly in the manner as the liquid- vapor boundary,replacing ∆vapH with ∆subH, the enthalpy of sublimation

Since ∆vapH < ∆subH, the equation predicts a steeper slope for thesublimation curve than the vaporization curve at similar temperatures

Note the very similar slopes where thesolid and liquid boundaries meet

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