Metallurgical Thermodynamics
Click here to load reader
description
Transcript of Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Metallurgical Thermodynamics
MT – 2102 Credit:04
Instructors:
Dr. C. K. Behera and
Mr. J. K. Singh
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Marks Distribution
Total Marks 100
Sessional Test - I 15
Sessional Test - II 15
Assignments / Attdn. 10
End Semester 60
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Grading System
S 90 -100
A 80-89
B 70-79
C 60-69
D
E
F
50-59
40-49
< 40
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Course Content
Basic Principles
Extensive and intensive properties, thermodynamic systems and processes. First Law of Thermodynamics, enthalpy, Hess’ Law, heat capacity, Kirchhoff’s law. Second Law of Thermodynamics, entropy, entropy change in gases, significance of sign change of entropy. Trouton’s and Richard’s rules. Driving force of a chemical reaction, combined statement of first and second laws of thermodynamics, Helmholtz and Gibbs free energies. Ellingham diagram, Equilibrium constants, van’t Hoff’s isotherm, Le Chatelier principle. Clausius-Clapeyron equation. Maxwell’s equations, Third Law of Thermodynamics.
Solution Thermodynamics
Solution, mixture and compound. Raoult’s law: activity, ideal solution, standard state. Partial molar quantities, Gibbs-Duhem equation, chemical potential, fugacity, activity and equilibrium constant. Free energy of mixing, excess and integral quantities. Regular solutions, -function. Dilute solutions: Henry’s and Sievert’s laws. Alternative standard states. Gibbs-Duhem integration
Statistical concept of entropy. Elements of Gibbs Phase Rule and its applications.
Experimental Techniques
Determination of thermodynamic quantities by different techniques, viz. calorimetry, chemical equilibria, vapour pressure and electrochemical: aqueous, fused and solid electrolytes; formation, concentration and displacement cells.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Suggested Reading
1. D.R. Gaskell: Introduction to Metallurgical Thermodynamics, McGraw-Hill.
2. L.S. Darken and R.W. Gurry: Physical Chemistry of Metals, McGraw-Hill.
3. G.S. Upadhyaya and R.K. Dube: Problems in Metallurgical Thermodynamics and
Kinetics, Pergamon.
4. J. Mekowiak: Physical Chemistry for Metallurgists, George Allen & Unwin.
5. J.J. Moore: Chemical Metallurgy, Butterworths.
6. R.H. Parker: An Introduction to Chemical Metallurgy, Pergamon.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Scope, Basic Concepts andDefinitions
Thermodynamics is that branch of science which deals with the study of the
transfer and conversion of energy from one form into other and its
conversion to work.
It deals with only conventional forms of energies like electrical, mechanical,
chemical. etc.
The non-conventional energy like nuclear energy related to atomic and sub-
atomic particle forms has to be dealt separately because in that case all
matter would have to be considered as per the famous Einstein’s equation :
E = mC2 .
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Here the subject matter of discussion is chemical and/or metallurgical
thermodynamics alone.
The systems under discussions consisting of large no. of particles i. e
macro systems.
Classification
Thermodynamics may be broadly classified into three :
Classical: it treats a substance as continuum, ignoring behavior of
atoms and molecules. It consists of first, second and third laws of
thermodynamics
Scope, Basic Concepts andDefinitions
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Statistical thermodynamics: The application of probability theory,
quantum theory and statistical mechanics allowed it to arrive at
macroscopic thermodynamic relations from atomistic point of view.
Irreversible Thermodynamics: Irreversible thermodynamics deals
with the application of thermodynamics to irreversible processes such
as chemical reactions.
The thermodynamics generally means classical thermodynamics
Scope, Basic Concepts andDefinitions
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Chemical thermodynamics is based on the three laws of thermodynamics
systematically applied to various physico-chemical processes in physical
chemistry.
Broadly speaking, the application primarily chemical thermodynamics to
metals and materials lead to the development and growth of Metallurgical
thermodynamics or its later generalization as Thermodynamics of
materials.
Processing of ceramics and metals is carried out primarily at high
temperatures which led to the development of metallurgical
thermodynamics.
Scope, Basic Concepts andDefinitions
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
It is the ability to do work. This is too mechanical an answer.
The broader definition is : it is the capacity to bring about changes in the
existing materials as per the requirements.
Forms of energies
Mechanical: Kinetic, potential and configurational.
Thermal: Heat exchanged.
Electrical: Electrical energy = current x time x potential difference.
Chemical: Chemical energy = no. of chemical bonds x bond strength
Energy
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
System and Surrounding
Any portion of the universe selected for consideration is known as the system
or thermodynamic system.
A thermodynamic system must, of necessity be stable with respect to its
chemistry during its study. If the system is undergoing continuously some
chemical change cannot be considered as system
For example a live animate body like tree and human being are not system. All
inanimate aggregates are called systems as they have their fixed chemistry
The rest of the universe excluding system is called surrounding.
Scope, Basic Concepts andDefinitions
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Classification of thermodynamic systems
Thermodynamic Systems
In terms of interaction with surrounding
Based on Material Distribution
Based on composition
Isolated
Neither matter nor energy is exchanged with surrounding
Open
Both matter and energy exchange occurs
Closed
Can exchange energy not the matter
Hom
ogen
eous
Het
erog
eneo
us
Single component
Multi component
Hom
ogen
eous
Het
erog
eneo
us
Hom
ogen
eous
Het
erog
eneo
us
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Homogeneous and Heterogeneous system
Homogeneous system consists of single phase only.
Heterogeneous consists of more than one phase.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
State of a system
As the position of a point in the space is described by its coordinates w.r.t
some prefixed axes, similarly the state of a system is described by some
experimentally determinable parameter which can lead to the complete
reproduction of the system.
These parameters are temperature, pressure and volume.
The minimum number of variable required to describe the state of the
system are called independent state variables.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Incase of a multicomponent system the independent state variables are
i) composition ii) two of the three variables T, P, V.
All other variables whose values get fixed with the specification of
independent state variables are referred to as dependent state
variables.
State variables are also known as state properties or state functions.
State of a system (cont.)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Extensive and Intensive State properties
If a state variable, whether dependent or independent, is a function of the
mass of the system it is known as extensive state property. For example:
Mass, Volume, weight, length, energy etc.
If a state variable is independent of the mass or size of the system it is
called intensive state property. For example: temperature, pressure,
conductivity, density, colour, odor, malleabilty, hardness, specific heat,
molar volume etc.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Product of an intensive and an extensive state variable is also
an extensive state variable.
Ratio of two extensive properties yields an intensive
properties. For example: Density = m / V
The general convention in chemical thermodynamics is to go
for molar properties, which are intensive and become
independent of quantity of matter and hence of more general
applicability.
Extensive and Intensive State properties
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Equation of State
The state of the system can be described in the form of some
mathematical equation involving some state variables. The
analytical form as applicable to the system under consideration
is known as equation of state e.g. for an ideal gas PV = nRT.
The above relationship is an expression which correlates the P,
T and V. In fact this is found to be true in case of solids as well
as liquids though exact form of this relationship may not be
known.
The same can be described in generalized form:
F(P,V,T) = 0
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Thermodynamic Processes
A material system may under go a change, under externally or
internally imposed constraints, in terms of their state variables
from the existing one to some different values. Such a change in
the state of the system is known as thermodynamic processes.
For example: Expansion of a gas from V1 to V2 may be called as
a thermodynamic process. Many a times such processes are
carried out under additionally imposed conditions and are
named accordingly. Such processes are:
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Isothermal processes: These are the processes which proceed
without any change in temperature of the system e. g. melting
of ice or metal. dT = 0
Isobaric process: These are the processes which proceed
without any change in pressure of the system e.g. processes
carried out in open atmosphere. dP = 0
Isochoric Processes: These are the processes which proceed
without any change in volume of the system e.g. the processes
carried out in vessel of known volume . dV = 0
Thermodynamic Processes
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Adiabatic processes: These are the processes which
proceed without any exchange of heat by the system with
its surrounding e. g. the system is completely insulated
from the surrounding . dq = 0. for ideal gas PVγ = const. γ
= Cp/Cv
Polytropic Processes: Those processes which obey equation
PVn = const where n is any positive number between 1 and
γ.
Thermodynamic Processes
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Path and State Functions
• The property whose change depends on only the initial and
final states of the system not on the path adopted to bring about
the change is called state function. Mathematically therefore if
the property is a state function (X) then in a cyclic process,
when system under goes a change and returns to original state
then
Ф dX = 0
• If Y is not a state function
Ф δY ≠ 0
So Y is called a path function
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of State Function
• If a system has two independent variables say x and y and any other function or property can be expressed in its total differential form
dz = Mdx + Ndy where M and N may be function of x and y then the function z is
a state variable if and only if • For an ideal gas T = PV/R wherein P and V are independent
variables and T as dependent variable. It can be expressed in total differential as
YXx
N
y
M
dPP
TdV
V
TdT
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Relationship among state variablesLet there be three state functions X, Y and Z and two of these as
independent state variables. Once Z and once Y as dependent
variables. Thus
Z = Z (X, Y)
Y = Y (X, Z)
Total differential can be written as
And
dYY
ZdX
X
ZdZ
XY
dZZ
YdX
X
YdY
XZ
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Putting the value of dY in first expression leads to
equating the coefficients of dX and dZ on both
Sides, we get
dZZ
Y
Y
ZdX
X
Y
Y
Z
X
ZdZ
XXZXY
ZXYZXY
X
XXX
X
Y
Y
Z
X
Z0
X
Y
Y
Z
X
Zand
Z
Y
1
Y
Z1
Z
Y
Y
Z
Relationship among state variables
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Similarly,
and
Finally after substitution we can write
The above two expressions correlate the partial differentials of three state
variables w.r.t one another and are popularly called reciprocity theorem.
Z
Z
Y
X
1
X
Y
Y
Y
Z
X
1
X
Z
1X
Z
Z
Y
Y
X
similarly
1Z
X
X
Y
Y
Z
YXZ
YZX
Relationship among state variables
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Thermodynamic Equilibrium
Mechanical Equilibrium: if there is no pressure gradient in the
system.
Thermal Equilibrium: if there is no temperature gradient in the
system.
Chemical Equilibrium: if the rate of forward reaction is equal
to rate of backward reaction.
Complete thermodynamic equilibrium is thus that situation
where the system is in equilibrium with respect to all such
potentials like mechanical, thermal and chemical.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Reversible and Irreversible Processes
A process that can be reversed in its direction by an
infinetesimal change in one or more of the state variables is
said to be a reversible process.
A classical example of this is the gas cylinder and piston. If
the pressure of the gas is say P atm and (p+dp) is exerted
from outside on the piston, the gas inside the piston shall be
compressed. However, if the external pressure is (p-dp) then
the gas shall expand.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
On the contrary a matchstick when it burns, the process
can not be reversed by changing one or other parameters.
Once burnt can not be reproduced by reversing the
process. This is typically a Irreversible process.
Other examples are mixing of two gases, mixing of two
liquids to form a solution or flow of electric current
through resistor.
All natural processes are irreversible.
Reversible and Irreversible Processes
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Experimental Evidence Leading to First Law
• For number of thermodynamic cycle each consisting of number of processes
• If , however work and heat measured in same unit then for a thermodynamic cycle
W = q
184.4
i
i
q
W
It is impossible to produce energy of any kind or form without the disappearance of an equivalent amount of energy
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
System goes from state I to state II by
various path and returns to the initial
state by path r only. Then we can write
Wa +Wr = qa + qr
or Wa – qa = (qr - Wr)
Simillarly along other paths
Wb – qb = (qr - Wr)
Wc – qc = (qr - Wr)
Wd – qd = (qr - Wr)
Or qa – Wa = qb – Wb=
qc – Wc= qd – Wd
The difference between q and W shall remain constant as long as initial and final states are not changed.
Experimental Evidence Leading to First Law
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
From the previous discussion, the following conclusions can be drawn:
i) The amount of heat exchanged and work done for taking the system from state I to II is different for different paths thus these functions are path functions. These are denoted by symbol δq and δw for infinetesimal change.
ii) The difference between the heat input q and work done W can be equated to change in another variable, say U i.e
q - W = U
Since q-W is independent of path, U is a state function. It has
further been proved that for a thermodynamic cycle
ΔU = 0
It is denoted by dU for infinetesimal change.
Experimental Evidence Leading to First Law
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Let us consider that the system goes from state I to II by absorption of
heat from the surrounding and doing work on the surrounding. As we
know that q and W cannot be equal we can consider two distinct cases:
i) q < W: system partly imparts energy for the work done.
ii) q > W: The heat is partly retained by the system itself and partly
returned to the surrounding in the form of work done.
In both the cases system acts as a reservoir of energy. This stored energy
in the system which can change during a thermodynamic process is called
internal energy and denoted by symbol U.
Experimental Evidence Leading to First Law
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Internal Energy
It consists of
macroscopic kinetic energy due to motion of the system as a whole.
potential energy of the system due to its position in the force field.
kinetic energy of atoms and molecules in the form of translation, rotation and vibration.
energy of interaction amongst atoms and molecules
columbic energy of interaction amongst electrons and nucleii in atoms
energy contents of the electrons and nucleii of atoms
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
In conventional chemical thermodynamics which we shall be concerned
with only kinetic energies of atoms and molecules and interaction
amongst atoms and molecules i. e. items (3) and (4) are considered to be
important since changes occurring in them principally contribute to ΔU.
Absolute value of the energy is not known. All we can determine is
change in internal energy.
Internal energy will depend on temperature for a material of fixed mass,
composition and structure.
U is function of Temperature only.
Internal Energy
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Statement of First law
For an infinetesimal process, the statement is:
dU = δq – δW
‘Sum of all forms of energy exchanged by a system
with its surrounding is equal to the change in internal
energy of the system which is a function of state’.
‘Energy can neither be created nor can be destroyed’
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Significance of the First Law
It is based on the law of conservation of energy.
It brought in the concept of the internal energy.
It separates heat and work interactions between the system
and surroundings as two different terms.
It treats internal energy as a state property is an exact
differential.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Internal energy in terms of Partial Derivatives
For given homogeneous system consisting of given amount
of substance of fixed composition;
U = F(P,V,T)
If any two variables are independent third will be
automatically fixed. This also therfore can be stated as:
U = F(P,V); U = F(V,T) ; U = F(P,T)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
From the theorem of partial derivatives
Internal energy in terms of Partial Derivatives
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Enthalpy
If pressure is maintained constant during change of system
from state I to state II the work done
From the first law:
UII – UI = δq – P(VII – VI )
On rearranging these terms
(UII + PVII ) - (UI + PVI ) = δq
Or HII - Hi = ΔH = δq
Where H = U + PV is called entalpy. The heat content at
constant pressure is called enthalpy.
III
II
I
II
IVVPdVPPdVw
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Enthalpy is a state property but heat is not.
As we know
dH = dU + PdV + VdP
As U is a function of state so one can write
Hence
= M dP + N dV (Say)
VPUH
dVV
UdP
P
UdU
PV
dVV
UPdP
P
UVdH
PV
Enthalpy
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Differential of coefficient of dP w.r.t V at constant P is given by the
relation
Differential of coefficient of dV w.r.t P at constant V is given by the
relation
Hence the above two partial differentials are equal proving thereby that
the equation for dH is an exact differential equation and thus leading to
the conclusion that H is a function of state.
PV
U
V
M 2
1
VP
U
P
N 2
1
Enthalpy
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
As Internal energy of an ideal gas is a function of temperature only.
At constant temperature dT = 0 and therefore dU = 0 so
As we know:
For an ideal gas at Constant T, PV = constant, d (PV) = 0
0
TP
U
VPUH
TTT
P
VP
P
U
P
H
Enthalpy
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
In other words enthalpy of an ideal gas is independent of
pressure at constant temperature. Similarly, enthalpy is
independent of volume. Hence H is a function of T only for
the fixed mass of the substance.
0P
H
TTP
U
Enthalpy
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Internal Energy Vs Enthalpy
Internal energy is all that energy stored in the system. But, What does
enthalpy mean? This can best illustrated by the example of calcination of
calcium carbonate.
CaCO3 = CaO + CO2
The enthalpy change of the process will be equal to invested
bond energy to break the bond between CaO and CO2 or internal energy
provided both CaO and CO2 are solids.
However if CO2 is allowed to form gas then breaking of one mole of CaCo3
nearly 22.4 ltrs. of CO2 will be formed. The expansion in volume will take
place.
In this process of expansion the system will do work equivalent to ∫ PdV as
mechanical work on the surrounding.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Hence in addition to the requirement of energy for breaking the bond
of Cao-CO2 additional energy equivalent to ∫ PdV will have to be
supplied to the system making a total of U + ∫ PdV and this is the
enthalpy change of the system on calcination.
For chemical processes where there is no significant change in volume
as in
2CaO3. SiO2 = 2CaO + SiO2
The ∫ PdV is practically absent and U and H are almost the same.
In chemical and metallurgical world, even if term ∫ PdV is absent, it
worthwhile to refer to H which is more appropriate.
Internal Energy Vs Enthalpy
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Heat Capacity
In chemical and metallurgical processes, materials get heated or
cooled and therefore it is necessary to know the amount of heat
required to heat or amount of heat liberated on cooling a material
over a certain temperature range.
Different materials require different amounts of heat to get
heated through the same temperature rise. This is because the
materials have different heat capacities.
This is so because of the variation in the crystal structure of the
materials and their related parameters.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The heat capacity of a substance is the amount of heat
required to raise its temperature by one degree.
In thermodynamics the molar heat capacity (C) i.e. heat
capacity of 1 g-mole of a substance is most widely
employed. Thus
The molar heat capacity at constant volume is given by
Since at constant volume δq = dU.
T
qC
VV
V T
U
T
qC
Heat Capacity
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Similarly molar heat capacity at constant pressure is given by
Since at constant pressure δq = dH. Further we can write
dH = CP dT
Or
CP > CV since CP includes heat required to do work against
pressure also besides raising temperature
PP
P T
H
T
qC
2
112
T
T PTT dTCHH
Heat Capacity
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Interrelationship of CP and CV
And we know
differentiating w.r.t T at constant P we get
PVP
VVVP
VP
T
VP
T
U
T
U
T
U
T
PVU
T
U
T
HCC
dTT
UdV
V
UdU
VT
VPTPT
U
T
V
V
U
T
U
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Putting the value of we get
For an ideal gas rate of change of U with V is zero at const T.
For ideal gas
So CP – CV = R
PT
U
PV
U
T
V
T
VP
T
V
V
UCC
TPPPT
VP
PT
VCC
P
VP
P
TRV
Interrelationship of CP and CV
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Application of First law to Thermodynamic Processes
With the help of the first law, one is able to calculate the changes
taking place in internal energy and enthalpy of a system during a
thermodynamic process.
Processes which are frequently studied and to which this law will
be applied include: i) Isothermal process ii) Adiabatic process iii)
polytropic processes.
In all these cases working substance of the system shall be
considered to be an ideal gas
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
i) Isothermal Process: The process is carried out when dT = 0
thus dU = 0 . Hence
δq = δW = PdV
It is also true that in this case
If Vf > Vi i.e. for expansion of gas q is positive i.e.
the system will absorb heat from its surroundings
and produce an equivalent amount of work.
f
i
f
i
f
i
f
i
V
Vi
f
V
V
V
V
V
V
V
VTRVdTR
V
dVTRdV
V
TRPdVwq
lnln
Application of First law to Thermodynamic Processes
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Application of First law to Thermodynamic Processes
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
If Vf < Vi i.e. for compression of gas q is negative i.e. the system will impart
heat to the surroundings at the same time absorbing mechanical energy from
them in the form of mechanical work done on it.
The change in enthalpy of the system will be:
Tf = Ti in the isothermal process.
Thus in an isothermal process with an ideal gas internal energy and enthalpy
remains unchanged and work done is equal to the heat exchanged.
0)(
)(
if
iiffif
TTR
VPVPUHHH
Application of First law to Thermodynamic Processes
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
ii) Adiabatic Process: There is no exchange of heat between the system and
surrounding i. e. δq = 0. Hence first law takes the form
dU = - δW = - P dV
Or dU = CV dT = - P dV
This indicate that system will perform work at the cost
of its internal energy and therefore the lowering of
temperature of the system will result.
Adiabatic work done (w)
Change in Internal energy = CV (Tf – Ti)
ifV
T
TV TTCdTCf
i
Application of First law to Thermodynamic Processes
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Relationship between P and V and T and V in
adiabatic process.
Or
Integration of this equation under the limiting
condition
V = Vi at T = Ti
And
V = Vf at T = Tf
V
dVTRdTCV
V
dV
T
dT
R
CV
Adiabatic Process
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
And substituting the expression for R in above
Expression
Or where
Or Or
Also since by definition H = U + P V
Or dH = dU + P dV + V dP
As dU = - P dV
So dH = V dP
i
f
i
f
VP
V
V
V
T
T
CC
Clnln
11 iiff VTVT
V
P
C
C
ConstVT 1ConstVP
Adiabatic Process
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Enthalpy change in a adiabatic process is
As for such process.
This on integration lead to the expression
f
i
f
i
V
V
P
PdVPdPVH
ConstVP
11
1
f
iii
V
VVPH
Adiabatic Process
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Work done in reversible adiabatic expansion can
be deduced as follows
Or
Where
1
11
12
2
1
2
1
VVm
V
dVmPdVw
V
V
V
V
1
1122 VPVPw
.constamVP
Adiabatic Process
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
iii) Polytropic process: The general expression for this process is
The work done in this process is similar to adiabatic process but
.constVP n
11
11n
f
iii
ffii
V
V
n
VP
n
VPVPw
n
Polytropic Process
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Change in Internal energy = CV (Tf – Ti)
Internal energy can also be expressed in terms of P and V as follows:
From the first law heat exchanged q is obtained as
For n = γ; q = 0 obtained from the above expression which is true for adiabatic
process.
1
1
1n
f
iiiV
ii
ffiiViiffV
ifV
T
TV
V
V
R
VPC
VP
VP
R
VPC
R
VP
R
VPC
TTCdTCUf
i
11
11n
f
iiiV
V
V
nR
VPRnCq
Polytrophic Process
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Change in enthalpy of the system is given by
Substituting the expression for ΔU in above expression
Or
Above equation show that for n = 1 i.e. isothermal process with ideal gas ,
both
change in internal energy and enthalpy are equal to zero.
iiff VPVPUH
1
1n
f
iiiV
V
V
R
VPRCH
1
1n
f
iiiP
V
V
R
VPCH
Polytropic Process
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Summary of ThermodynamicProcesses
Process characteristics P-V-T relationship
Work done Heat exchange
Isothermal dT = 0 PV = const RT ln (Vf / Vi) RT ln (Vf / Vi)
Isochoric dV = 0 P/T = const 0 Cv (Tf – Ti)
Isobaric dP = 0 V/T = const Pi(Vf – Vi) Cp (Tf – Ti) + Pi(Vf – Vi)
Adiabatic q = 0 PVγ = const (PfVf - Pi Vi) / (1- γ) 0
polytropic - PVn = Const (PfVf - Pi Vi) / (1- n) Cv (Tf – Ti).(γ - n)/(1-n)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Thermo chemistry
It may be defined as the branch of science which deals with the
study of heat exchanges associated either with chemical reaction
or physical changes in the state of matter such as melting,
sublimation or evaporation etc.
If heat is produced by a chemical reaction it is denoted by – ve
sign (Exothermic reaction)
If heat is absorbed during a chemical reaction it is denoted by
+ve sign (endothermic reaction).
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Hess’s Law
It states that the total heat exchanged in a given chemical reaction, which may take place
under constant pressure, volume or temperature is the same irrespective of the fact
whether it is made to take place over a path involving formation of number of
intermediate products or over the one involving the formation of final product from the
reactant directly in one stage.
ΔH =ΔH1+ΔH2+ΔHx+ΔHy
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Hess’s Law
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Variation of heat Capacity with Temperature
Experimental data consists of CP as function of temperature.
For the liquid and solid, The P V term is very small. Hence H is
taken as equal to U and CP as equal to CV.
In other words no distinction is made between CP and CV so far as
applications are concerned.
It has been found that experimental CP Vs. T data for elements
and compounds fit best with an equation of type:
CP = a + b T + c T-2
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Where a, b, c are empirically fitted constant and differ
from substance to substance.
The last term is the smallest and therefore often ignored.
In some cases, such as liquid metals, both bT and cT-2 are
usually ignored.
The above expression is also valid for diatomic and
polyatomic gases as well.
Variation of heat Capacity with Temperature
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Variation of Enthalpy with Temperature
Change in enthalpy during the course of process (ΔH) is
equal to the heat supplied to the system (q) at constant
pressure.
The constant pressure restriction is mostly not important
for example H is not function of P for ideal gases.
Energies of solid and liquids are hardly affected by some
changes in pressure due to their very small molar volumes.
In other words, the VdP term is negligible.
In most metallurgical and materials processing, gases are
ideal and pressure is maximum a few atmosphere.
Therefore ΔH = q approximation is quite all right.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Classification of Enthalpy change
Absolute value of enthalpy change of a substance is not
known. Only we can measure are changes of enthalpy.
Enthalpy change occur due to various causes.
Sensible heat: enthalpy change due to change of
temperature of a substance is known as sensible heat. It is
divided into:
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
i) change in enthalpy without any change in aggregation of the
substance:
As a universal convention, thermo chemical data books take
sensible heat at 298 K (250C) as zero for any substance. Hence
sensible heat at temperature T, per mole of a substance is given as:
298 K is known as reference temperature
dTCHHT
PT 298298
Classification of Enthalpy change
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Putting the expression for heat capacity, we get
Where A, B, C and D are lumped parameters and functions
of empirical constants a, b, c
DTCTBTA
cTTa
dTTcTbaHH
Tb
T
T
12
2981122
2
298
2298
298298
Classification of Enthalpy change
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
ii) Enthalpy change due to changes in state of aggregation of
substance : These are isothermal processes. By convention
enthalpy change s for all isothermal processes are designated
by ΔH. For example,
ΔHm = enthalpy change of one mole of solid due to melting (i. e.
latent heat of fusion per mole )
ΔHv = enthalpy change of one mole of liquid due to vaporization
(i. e. latent heat of vaporization per mole )
Classification of Enthalpy change
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Consider a pure substance A, which is undergoing
following changes during heating from 298 to T K
A (Solid) → A (Liquid) → A (Gas) → A (Gas) at 298 K at Tm at Tb at T
Tm and Tb are the melting and boiling points of A.Then
Cp (s), Cp (l) and Cp (g), are for solid liquid and
gaseous A. it is only applicable for pure substance.
dTgCH
dTlCHdTsCHH
m
b
b
m
m
T
T PV
T
T Pm
T
PT
)(
)()(298298
Classification of Enthalpy change
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Heat of reaction (ΔH): This is the change of enthalpy that occurs
when a reaction takes place. By convention reaction is considered to
be isothermal. Consider the following reaction occurs at
temperature T:
A (pure) + BC (pure) = AB (pure) + C (pure)
Then
ΔH (at T) = ΔHAB (at T) + ΔHC (at T) - ΔHA (at T) - ΔHBC (at T)
Where ΔHAB , ΔHBC, ΔHA and ΔHc are molar enthalpies
of pure Ab, BC, A and C, respectively.
Classification of Enthalpy change
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Heat of mixing (ΔHmix): This is the change of enthalpy that
occurs when a substance is dissolved in solvent.
This process is generalized as
A (pure) = A (in solution)
This process is also accompanied by a change of enthalpy
(ΔHmix), where
(ΔHmix) = HA (in solution) – HA (pure)
Again by convention, the process is assumed to be isothermal for
thermodynamic calculations
Classification of Enthalpy change
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Some comments:
• For calculation of enthalpy change, reactions, dissolutions and
phase transformations has been assumed to occur
isothermally. Since enthalpy is a state property, it depends
only on initial and final states and not the path.
• For a reversible isothermal process, the temperature remains
constant all through. If the process is not reversible, then
temperature at the beginning and end of a process would be
same. In between, temperature can vary significantly.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Sign convention for ΔH:
The process accompanied by liberation of heat are called
exothermic. This happens if the enthalpy in the final state
(state 2) is lower than the initial state (State 1) i. e. H2 < H1 so
for the process
state 1 → state 2
we have ΔH = H2 – H1 < 0
Therefore ΔH is negative. The opposite is an
endothermic process which is characterized by
absorption of heat and positive value of ΔH.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Standard state of enthalpy
• The stable state of a substance changes with the temperature. The
stable state of H2O is ice which is below 0 oC, liquid water 0-100 oC and a stable gas at 1 atm pressure above 100 oC.
• Considering all these points a standard state has been defined as a
pure element or compound at its stable state at the temperature
under consideration and at 1 atm pressure. Thus the standard state
of H2O at 50 oC is pure water at 1 atm pressure.
• By convention enthalpy changes at standard state are denoted by
subscript ‘0’ e. g. 00TT HandH
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
As already mentioned, 298 K is the universal reference
temperature for compilation of sensible heats. By this
convention, sensible heat at standard state of a substance is
arbitrarily taken as zero at 298 K. This is solely for calculation
of sensible heats not ΔH0 for a process occurring at 298 K
Standard state of enthalpy
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Kirchoff’s Law
Utility:
It allows us to calculate the enthalpy changes at various
temperatures, provided the enthalpy is known at some other
temperature, and the heat capacity of the reactants and
products are known in the range of temperatures under
consideration.
Derivation:
Consider a chemical reaction at temperature T1 whose enthalpy
change is ΔH1. Calculate the enthalpy change ΔH2 at
temperature T2.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The product of this reaction at T2 can be obtained in many
different ways. Let us, however, consider the reaction to be
carried out first by reacting the reactants (x+y) at T1 and then
raising the temperature of the products from T1 to T2 along ABC.
Kirchoff’s Law
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The heat absorbed by this process will be
The second way of obtaining the same result is to raise the
temperature of the reactants from T1 T2 and then react them
together at T2 i. e. along ADC.The heat absorbed by this process is
2
11
T
T P dTCHz
2
1
yx
T
TPP2 dTCCH
Kirchoff’s Law
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
According to Hess’s law the heats absorbed during the two ways of
producing z must be the same since the initial conditions of the
reactants and final conditions of the products are in each case the
same.
=
Or
Or
Or
2
1
yx
T
TPP2 dTCCH 2
11
T
T P dTCHz
2
112
T
T PPP dTCCCHHyxz
2
112
T
T P dTCHH
2
112
T
T P dTCHH
Kirchoff’s Law
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Since all the terms are known in the right hand of the equation, ΔH2
can be calculated. We know
Or
This is the statement of the Kirchoff’s law. It means the rate of
change of enthalpy of a process or a reaction with temperature is
given by the difference of the heat capacity at constant pressure of
products and reactants taking part in the reaction
2
112
T
T P dTCHH
P
P
CT
H
Kirchoff’s Law
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem-1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem-2
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Enthalpy change:
H (a → c) = U (a→c) + (Pc Vc – Pa Va)
= - 9.13 + n R (Tc – Ta)
= - 9.13 + 4.09 x 8.3144 x (119 - 298)
= - 9.13 - 6.0870 kJ = - 15.2170 kJ
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem-3
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem - 5
Calculate the heat of the following reaction at 1000K
Fe2O3(s) + 3C (s) = 2Fe(s) + 3CO(g) from the following data:
Cp(, Fe) = 4.18 + 5.92 x 10-3 T caldeg-1mol-1
Cp(CO) = 6.79 + 0.98 x 10-3T – 0.11 x 105 T-2 caldeg-1mol-1
Cp(Fe2O3) = 4.10 + 1.02 x 10-3 T – 2.10 x 105T-2 caldeg-1mol-1
The heats of formation of Fe2O3 and CO at 298 K are
-197000 and – 26400 cal/mol. respectively.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
Cp = Cp (product) – Cp (reactant) = 2.Cp(-Fe)+ 3 Cp(CO) – Cp(Fe2O3) – 3Cp(C)
2 x Cp(-Fe) = 8.36 + 11.84 x 10-3T
3 x Cp(CO) =20.37 + 2.94 x 10-3 T -0.33 x 105 T-2
__________________________________________
Cp (product) = 28.73 + 14.78 x 10-3T – 0.33 x 105 T-2
Cp = 23.49 + 18.60 x 10-3T – 3.55 x 105 T-2
3Cp(C) = 12.30 + 3.06 x 10-3T – 6.30 x 105 T-2
__________________________________________
Cp (reactant) = 35.79 + 21.66 x 10-3T -4.85 x 105 T-2
Cp = -7.06 - 6.88 x 10-3T + 9.52 x 105 T-2
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
= 117800 – 7.06(1000-298) -6.88. (10002 – 2982) x 10-3 – 9.52 x 105 (1000-1 – 298-1)
= 117800 – 5850 = 111950 cals.
dTCHH p .1000
298
2981000
( )∫1000
298
253 dT)T10x52.9+T10x88.606.7+117800=
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem-6
The standard heat of formation, of ammonia gas is -11.03
kcal/mol. at 250C utilizing the data given below derive a
general expression for heat of formation applicable in the
temperature range 273 – 1500K.
112633 deg10728.010787.7189.6, molcalTxTxNHC p
112632, deg100808.010414.1450.6 molcalTxTxNC p
112632, deg104808.0102.0947.6 molcalTxTxNC p
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
The reaction is
= -7.457 + 7.38 x 10-3T – 1.409 x 10-6T2
molcalHgNHgHgN /110302
3
2
1 0298322
223
, 2
3
2
1HCCCC pNpNHpp
∫298
273p273298 dT.CΔ+HΔ=HΔ
dTCΔ-HΔ=HΔ ∫298
273p298273
( )∫298
273
263 dTT10x409.1T10x38.7+457.7--11030
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
=-11030 + 7.457 (298 – 273) - x 7.38 x 10-3 (2982-2732) + x 10-6 (2983-2733)
= -11030 + 186.4 – 52.7 +2.9 = -10893 cal.
= -10893 – 7.457 T + 3.69 x 10-3T2 – 0.47 x 10-6T3 – (-7.457 x 273 + 3.69 x
10-3 x 2732 – 0.47 x 10-6 2733)
= -10893 +2036 – 275 + 10 – 7.457 T + 3.69 x 10-3T2 – 0.47 x 10-6T3
= - 9122 – 7.457T + 3.69 x 10-3T2 – 0.47 x 10-6 T3
= -9122 – 7.457T + 3.69 x 10-3T2 – 0.47 x 10-6 T3 cal/mol
For H1000K solution T = 1000 in the above equation
H1000K = -9122 -7457 + 3690 – 470 = -13359
T
T dTTxTxHH273
263273 10409.11038.7457.7
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
= - 11030 – 7.457 (1000-298) + 3.69 x 10-3 (10002-2982) – 0.47 x 10-6 (10003-2983)
= -11030 – 5234.8 + 3362.3 – 457.6 = -13360.1 cal/mol
∫1000
298p2981000 dTCΔ+HΔ=HΔ
( )dTT10x409.1T10x38.7+7457.7+11030-= ∫1000
298
363
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem- 4
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
CT-ISolution
Solution for Q-2: Fe + ½ O2 = FeO
Cal63200-=HΔ=HΔ FeO298
∫∫1123
10332ptr
1033
2981p2981123 dTCΔ+HΔ-dTCΔ+HΔ=HΔ
2532OpmagFepFeop1p T10×470-T10×424-93=C
2
1-C-C=CΔ ...,,,,
2532OpnonmagFepFeop2p T10×470-T10×51+920-=C
2
1-C-C=CΔ ...,,,,
kcal20863-=HΔ 1123 .
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
CT-ISolution
Solution for Q-3:
(a)
As U, T, ans S are dependent state functions
We can express G in the following manner
dG = dU + PdV + VdP – TdS – SdT
( ) ( ) dV+dP=dU PV∂U∂
VP∂U∂
( ) ( ) dV+dP=dS PV∂S∂
VP∂S∂
( ) ( ) dV+dP=dT PV∂T∂
VP∂T∂
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
CT-ISolution
( ) ( ) ( ) ( ) ( ) ( )NdV+MdP=
dPS-T-V++dVS-T-P+=dG VP∂T∂
VP∂S∂
VP∂u∂
pV∂T∂
pV∂S∂
pV∂u∂ ][][
( ) ( ) ( ) ( ) ( ) )( ∂P∂VT∂
pV∂T∂
vP∂S∂
∂P∂VS∂
pV∂S∂
vP∂T∂
∂P∂VU∂
vP∂M∂ 222
S--T--1+=
( ) ( ) ( ) ( ) ( ) )( ∂P∂VT∂
pV∂T∂
vP∂S∂
∂P∂VS∂
pV∂S∂
vP∂T∂
∂P∂VU∂
PV∂N∂ 222
S--T--1+=
So G is a function of state
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
CT-ISolution
(b)
dH = TdS + VdP
( )[ ] ( )NdV+MdP=
dVT+dPV+T=dH P∂VS∂
v∂PS∂
( ) ( ) ( ) 1+T+= ∂P∂VS∂
VP∂S∂
PV∂T∂
PV∂M∂ 2
)(
( ) ( ) ( ) )( ∂P∂VS∂
PV∂S∂
VP∂T∂
VP∂N∂ 2
T+=
So H is not a function of state
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Second law of Thermodynamics
System which are away from equilibrium, upon initiation, shall
move towards equilibrium and such processes are referred to as
natural or spontaneous or irreversible processes.
Examples: heat flow from hotter to colder body or
depressurisation of inflated tube in a low pressure surroundings
or free fall under gravity and so on.
Introduction
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The spontaneous change from an existing state to equilibrium
state is possible because in the existing state the system happens
to be at higher potential which is the driving force for the
change to occur.
Higher temperature is therefore the higher thermal potential
which makes heat flow from higher to a lower temperature.
Similarly higher pressure is a higher mechanical potential and
so on.
Second law of Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
If the system is in equilibrium and if it is to be moved in the
reverse direction in the above examples it would be termed as
unnatural or non-spontaneous processes.
Therefore water can not be raised to an overhead tank unless
energy in the form of motor pump set is provided to the system
from the surroundings.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Similarly heat can not flow from a colder to a hotter body unless
aided by the surroundings in the form of compressor energy as in
the refrigerator.
But in chemical processes it is not readily possible to assess as to
what is a natural or unnatural process. This can only be evaluated
from the equilibrium state of the system.
The knowledge of equilibrium CO / CO2 ratio in contact with Fe
or FeO can only guide us as to how to prevent oxidation of iron or
effect reduction of iron oxide by providing a suitable CO / CO2
gas mixture as surroundings.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Heat Engines
•Heat engines operate in a cycle, converting heat to work then returning to original state at end of cycle.
•A gun (for example) converts heat to work but isn’t a heat engine because it doesn’t operate in a cycle.
•In each cycle the engine takes in heat Q1 from a “hot reservoir”, converts some of it into work W, then dumps the remaining heat (Q2) into a “cold reservoir”
HOT
COLD
Engine
Q1
Q2
W
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
This whole process for conversion of heat to work necessarily
produces a permanent change in the cold reservoir by way of
release of heat into it. This is called compensation.
It means the entire heat that is absorbed initially is not
converted to useful work but a part of it is rejected to a cold
reservoir as of necessity.
In other words ‘in a cyclic process it is just not possible to
convert all heat into mechanical work. In a non-cyclic process it
may be possible.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
HOT
COLD
Engine
Q1
Q2
W
Efficiency of a heat engine
Definition:
cycleper input heat cycleper donework
efficiency
1QW
•Because engine returns to original state at the end of each cycle, U(cycle) = 0, so W = Q1 - Q2
•Thus:12
11
21QQ
QQQ
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Efficiency of a heat engine
•According to the first law of thermodynamics (energy conservation) you can (in principle) make a 100% efficient heat engine.
BUT………….
•The second law of thermodynamics says you can’t:
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Kelvin Statement of Second Law:
“No process is possible whose SOLE RESULT is the complete conversion of heat into work”
William Thomson, Lord Kelvin (1824-1907)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
HOTCOLD
HOTTERCOLDER
WARMWARM
Q
HOTCOLDQ
Both processes opposite are perfectly OK according to First Law (energy conservation)
But we know only one of them would really happen – Second Law
Heat flow
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Clausius Statement of Second Law:
“No process is possible whose SOLE RESULT is the net transfer of heat from an object at temperature T1 to another object at temperature T2, if T2 > T1”
Rudolf Clausius (1822-1888)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
How to design a “perfect” heat engine
1)Don’t waste any workSo make sure engine operates reversibly (always equilibrium conditions, and no friction).
2)Don’t waste any heatSo make sure no heat is used up changing the temperature of the engine or working substance, ie ensure heat input/output takes place isothermally
Sadi Carnot (1796-1832)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
HotSource
T1 Q1
Piston
Gas
a
b
Working substance (gas) expands isothermally at temperature T1, absorbing heat Q1 from hot source.
The Carnot Cycle (I): isothermal expansion
T1
P
V
T1T1
a
b
Q1Q1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The Carnot Cycle (II): adiabatic expansion
Gas isolated from hot source, expands adiabatically and temperature falls from T1 to T2.
Piston
Gas
b
c
Gas isolated from hot source, expands adiabatically, and temperature falls from T1 to T2
P
V
T1T1
a
b
T2T2c
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The Carnot Cycle (III): isothermal compression
Piston
Gas
d
c
Q2
ColdSink
T2
Gas is compressed isothermally at temperature T2 expelling heat Q2 to cold sink.
T2
V
P
T1
a
b
T2cd
P
T1T1
a
b
T2T2cd
Q2Q2
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The Carnot Cycle (IV):adiabatic compression
Gas is compressed adiabatically, temperature rises from T2
to T1 and the piston is returned to its original position. The work done per cycle is the shaded area. Piston
Gas
a
d
Gas is compressed adiabatically, temperature rises from T2 to T1 and the piston is returned to its original position. Work done is the shaded area.
V
P
T1
a
b
T2cd
W
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Efficiency of ideal gas Carnot engine
12
11
21QQ
QQQ
•We can calculate the efficiency using our knowledge of the properties of ideal gases
V
P
T1
a
b
T2cd
W
Q2
Q1
V
P
T1
a
b
T2cd
W
V
P
T1T1
a
b
T2T2cd
WW
Q2Q2
Q1Q1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
a
bab V
VnRTWQ ln11
Isothermal expansion (ideal gas)
P
V
T1T1
a
b
Q1Q1
Va Vb
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Isothermal compression (ideal gas)
d
ccd V
VnRTWQ ln22
V
P
T1
a
b
T2cd
Q2
V
P
T1
a
b
T2cd
Q2
V
P
T1
a
b
T2cd
P
T1T1
a
b
T2T2cd
Q2Q2
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
a
b
d
c
VV
nRT
VV
nRT
ln
ln11
1
2
1
2
(2)
(1) 1
21
1
12
11
da
cb
VTVT
VTVT
Adiabatic processes
)2()1( d
c
a
b
VV
VV
2
2
1
1
1
21
TQ
TQ
TT
Efficiency of ideal gas Carnot engine
V
P
T1
a
b
T2cd
W
Q2
Q1
V
P
T1
a
b
T2cd
W
V
P
T1T1
a
b
T2T2cd
WW
Q2Q2
Q1Q1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
So……..
h
c
TT
1
0h
h
c
c
h
h
c
c
TQ
TQ
TQ
TQ Conservatio
n of “Q/T”
For all Carnot Cycles, the following results hold:
What about more general cases?????
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
0h
h
c
c
h
h
c
c
TQ
TQ
TQ
TQ
Was derived from expressions for efficiency, where only the magnitude of the heat input/output matters. If we now adopt the convention that heat input is positive, and heat output is negative we have:
0h
h
c
c
TQ
TQ
The expression
In other words0
cycle
TdQ
For any reversible cycle:
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The Carnot CycleThe Importance of the Carnot Engine
1.All Carnot cycles that operate between the same two temperatures have the same efficiency.
2. The Carnot engine is the most efficient engine possible that operates between any two given temperatures.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Entropy
To emphasise the fact that the relationship we have just derived is true for reversible processes only, we write:
0cycle
rev
TdQ
We now introduce a new quantity, called ENTROPY (S)
TdQ
dS rev
Entropy is conserved for a reversible cycle
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Is entropy a function of state?
P
V
A
B
path1
path2
For whole cycle:
0 cycle
rev
TdQ
S
2 path1 path
2 path1 path
0
B
A
revB
A
rev
A
B
revB
A
rev
TdQ
TdQ
TdQ
TdQ
ABBABA SSSS 2) (path1) (path
Entropy change is path independent → entropy is a thermodynamic function of state
Reversible cycle
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Irreversible processes
Carnot Engine
h
c
revh
cc T
TQ
Q
11
Irreversible Engine
h
c
irrevh
c
TT
Q
Q
11irrev
For irreversible case:
h
c
irrevh
c
h
c
irrevh
c
TT
Q
Q
TT
Q
Q
0
irrevc
c
irrevh
h
TQ
TQ
h
h
irrevc
c
TQ
T
Q
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Irreversible processes
Following similar argument to that for arbitrary cycle:
0cycle
irrev
TdQ
P
V
A
BPath 1(irreversible)
Path 2(reversible)
For irreversible cycle
0)(
)(
)(
)(
rev
rev
irrev
irrev
A
B
rev
B
A
irrev
TdQ
TdQ
)(
)(
)(
)(
rev
rev
irrev
irrev
B
A
rev
B
A
irrev
TdQ
TdQ
Irreversible cycle
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Irreversible processes
)(
)(
)(
)(
rev
rev
irrev
irrev
B
A
rev
B
A
irrev
TdQ
TdQ
)(
)(
)(
)(
rev
rev
irrev
irrev
B
A
B
A
irrev dST
dQ
TdQ
dS irrev
T
dQdS
Equality holds for reversible change, inequality holds for irreversible change
General Case
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
“Entropy statement” of Second Law We have shown that:
T
dQdS
For a thermally isolated (or completely isolated) system, dQ = 0
0dS
“The entropy of an isolated system cannot decrease”
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
What is an “isolated system”
The Universe itself is the ultimate “isolated system”, so you sometimes see the second law written:
“The entropy of the Universe cannot decrease”(but it can, in principle, stay the same (for a reversible process))
It’s usually a sufficiently good enough approximation to assume that a given system, together with its immediate surroundings constitute our “isolated system” (or universe)………
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Entropy changes: a summary
For a reversible cycle:S (system) = S (surroundings) = 0S (universe) = S (system) + S (surroundings) = 0
For a reversible change of state (A→B):S (system) = -S (surroundings) = not necessarily 0S (universe) = S (system) + S (surroundings) = 0
For an irreversible cycle S (system) = 0; S (surroundings) > 0S (universe) = S (system) + S (surroundings) > 0
For a irreversible change of state (A→B):S (system) ≠ - S (surroundings) S (universe) = S (system) + S (surroundings) > 0
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Other Types of EnginesSchematic diagram of a refrigerator:
Refrigerator removes heat from cold reservoir, puts it into
surroundings, keeping food in reservoir cold.
Heat transfer takes place from cold to hot body
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Dependence of entropy on temperature
Dependence of entropy on temperature can becalculated at constant
volume as follow:
Dependence of entropy on temperature can becalculated at constant
pressure as follow:
TlndCSSS
TlndCdTT
C
T
PdV
T
dU
T
qdS
V
T
T12V
VVrev
2
1
TlndCSSS
TlndCT
dTCdH
T
1PVUd
T
1PVddU
T
1
T
qdS
p
T
T12P
PPrev
2
1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Volume dependency of entropy
We know;
and
So
As ‘S’ is function of state therefore the above
expression forms an exact differentials. Hence
PVddUT
1
T
qdS rev
dVV
UdT
T
UdU
TV
dVPV
U
T
1dT
T
U
T
1dS
TV
VTTVT
P
V
U
T
1
TT
U
T
1
V
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
PV
U
T
PT
TV
On simplification the above expression yields:
So we can write
And under isothermal condition dT = 0 lead to the following
expression:
dVT
PdT
T
U
T
1dS
VV
dVT
PdS
V
T
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
In case of gas one can directly integrate the above expression to calculate
the change in entropy during a process.
However, in case of solid and liquids, the equation of state is not known
and can also not be determined experimentally. Hence in such cases
this partial differential is required to be expressed in terms of some
experimentally determinable parameters. For this purpose the reciprocity
theorem is used to write the following:
VT
P
TPVV
P.
T
V
T
P
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Isobaric coefficient of volumetric thermal expansion of a material ()
is given as
Isothermal compressibility of a material () is given as
So
PT
V
V
1
TP
V
V
1
dVdS
Which on integration will yield the change in entropy for a change in volume
from V1 to V2 under isothermal condition
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Pressure dependency of entropy
We know:
From the definition of the enthalpy, one can get
Therefore,
The total differential dH can also be expressed as
dVPdUT
1dV
T
PdU
T
1
T
qdS rev
VdPdVPdUdH
dPT
VdH
T
1dS
dPP
HdT
T
HdH
TP
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
After substitution, we get
The equation is an exact differential because entropy being function of
state. Therefore,
On simplification this yields
dPT
V
P
H
T
1dT
T
H
T
1dS
TP
PTTPT
V
P
H
T
1
TT
H
T
1
P
PPTT
VV
P
H
T
1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
• So after substitution we can write
• For an isothermal process the above expression yields,
• On integration, one can calculate the change in entropy for a
change in pressure from P1 to P2 under isothermal conditions.
dPT
VdT
T
H
T
1dS
PP
dPVdPT
VdS PT
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Relationship between CP&CV
• We have derived the following relationship:
PTT
VTV
P
H
P
T
PT
V
U
VT
TPVV
P.
T
V
T
P
PT
V
V
1
T
P
V
V
1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
• We know the following expression
• Elimination of from the above equation leads to the relation
• Or
TV
U
PV
VP T
V
T
PTCC
TV
CC2
VP
PV
U
T
V
T
VP
T
V
V
UCC
TPPPT
VP
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Entropy Change in Gases
Ideal gas
dq/T = Cv .dT/T + R dV/V = dS
S = S2 –S1 =
On integration S = Cv ln + R ln
Entropy change can be calculated when temperature and volume change take place on heating an ideal gas
S1 p1
v1 T1
Change
S2 p2
v2 T2
∫ ∫2
1
2
1
T
T
v
vv v
dvR+
T
dTC
1
2
T
T
1
2
v
v
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
S = Cv ln (T2/T1) + R ln (T2p1/T1p2)
= Cv ln (T2/T1) + R ln (T2/T1 )+ R ln (p1/p2 )
Or S = Cp ln (T2/T1 )– R ln p2/p1
In an isothermal process i.e. at constant temp T = T1 = T2.
ST = - R ln p2/p1 = R ln p1/p2 = R ln (v2/v1)
For an isobaric process p1 = p2 = p
Sp = Cp ln (T2/T1)
For an isochoric process v1 = v2 = v
Sv = Cv ln (T2/T1)
Entropy Change in Gases
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
As the system absorbs heat, its entropy will increase e.g. with
melting and boiling.
S= S2 – S1 =
S is the area beneath the
curve between temperature,
T1 and T2(graphical method).
2
1
lnT
T
TCpd
Entropy Change in Gases
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Significance of sign of entropy change of a process in closed system
• Two identical copper vessel touching each other, full of water and each at different temperature.
• Vessels are completely insulated so that heat can neither enter nor leave this system.
• T1 – T2 is the measure of irreversibility of this process. This also denote the how much heat will flow
I II
T1 T2
q
Insulation
T1 > T2
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Entropy change of vessel I = - q1 / T1
Entropy change of vessel I = q / T2
Total change of entropy will be the sum of entropy changes of the two
vessels. Thus
When q is +ve and T1 – T2 is +ve then the entropy
change for a real irreversible process in a closed
system must also be positive.
21
21ocessPr
1221ocessPr
TT
TTqS
T
q
T
qSSS
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
T1 > T2 ; Sirr > 0 (+ve) : The entropy change of a real process is greater
than zero(+ve).
T1 = T2 ; Srev = 0 (zero): Dynamic equilibrium exists between two vessels
and there is no heat transfer.
T1 < T2 ; Sirr < 0 (-ve). : Entropy change is negative and the process
proceeds in the reverse direction.
Sign of entropy change shows the direction of flow of heat energy.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Calculation of entropy change
• Entropy is a state property. Hence the basis and procedure for
calculation of entropy changes are similar to those for the
enthalpy changes.
• Hess’s law and Kirchoff’s law are applicable here too.
• A pure substance at its stablest state also constitutes standard state
for entropy at that temperature which is designated as .0TS
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Following significant difference are to be noted
between entropy and enthalpy.
– Entropy changes are to be calculated only through the
reversible path. This restriction is not there for any other state
property, including enthalpy.
– Absolute value of entropy can be determined. Thermodynamic
data sources provide the values of entropy of a substance
for pure substance. This is in contrast with the energy where
only changes are available in the data sources.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Entropy changes associated with phase transformation
For a pure substance, reversible phase changes
(melting, boiling etc.) at a constant pressure occurs at
a constant temperature. Therefore,
for melting
for boiling
in general for phase trans.
m
0m0
m T
HS
V
0V0
V T
HS
tr
0tr0
tr T
HS
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
A (Solid) → A (Liquid) → A (Gas) → A (Gas)
at 298 K at Tm at Tb at T
Tm and Tb are the melting and boiling points of A. Then
dT
T
gC
T
H
dTT
lC
T
HdT
T
sCSS
T
T
0P
V
0V
T
T
0P
m
0m
T
298
0P0
2980T
b
b
m
m
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Entropy changes in chemical processes
By nature, reactions and mixing are irreversible. Like enthalpy
changes, entropy changes are considered only for the isothermal
process. However, they cannot be calculated from enthalpy as done for
the phase transformation in view of irreversibility. Consider the
reaction A + BC = AB + C
The entropy change of the reaction at temperature T
)Tat(S)Tat(S)Tat(S)Tat(SS 0BC
0A
0C
0AB
0T
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The entropy change of the reaction with temperature
can be calculated as
dTT
CS
dTT
CdT
T
CS
SSSSSS
2
11
2
1
2
11
121212
T
T
P0T
T
T
ttanreacPT
T
productP0T
ttanacRe
0T
0T
product
0T
0T
0T
0T
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Various interpretations of entropy
for an infinitesimal, isothermal reversible process. Entropy is times arrow i. e. a fundamental indicator of
time. Entropy has the relationship with heat not available for
work. Entropy is a measure of disorder of a system.
T
qdS rev
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Combined statement of first and second law
If a system is capable of doing only mechanical work the first law equation can be put
as:
dU = δq – P dV. It is not exact differential. Holds true for the reversible processes.
The second law for the reversible process gives
dS = δq rev / T or δq = T dS
The two laws therefore can lead to
dU = TdS – P dV or TdS = dU + PdV
This is the combined statement of first and second law.
It includes only those terms which are state functions only and
hence is exact differential equation.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
For irreversible process second law gives
dS > 0
Or therefore, TdS > dU + PdV
For a unnatural or non spontaneous process
dS < 0
Or therefore TdS < dU + P dV
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Thermodynamic Potentials
A system by itself, in isolated state or in contact with
surrounding shall stay in equilibrium unless acted upon by some
constraints.
If a system tends to move as a result of being not in equilibrium
or as a result of external constraint, there must be a driving
force making the system move from within itself or under the
applied constraints.
This driving force is referred to as thermodynamic potential
driving the system to change to a new state.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Is this driving force or potential the same under all conditions of the system or
it varies from condition to condition of the system ?
Heat flow s from higher to lower temperature. The flow is possible due to
higher thermal potential.
Similarly the higher pressure is the driving mechanical potential.
Also higher electrical voltage is driving the electrons under the influence of
electrical potential as voltage.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
However the question remains as to what is that potential, forcing a
thermodynamic process to take place?
It is easy to imagine that a system with higher associated energy will be relatively
unstable as compared to the one having lower energy.
In other words, when a natural changes take occurs the system moves from higher
to lower energy levels or moves from higher to lower thermodynamic potentials.
It also means that the energy is a potential driving the system for change to occur.
It also means that for equilibrium to exist the potential of all the systems or sub-
systems within it must be at the same potential.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
So far the internal energy, enthalpy and entropy have been evolved as
energy parameters but it is not yet known as to whether these act as
potential in driving a particular process?
It must be noted that for natural process the system moves from lower
to a higher entropy level hence entropy can not be considered as a
driving force.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Can internal energy and enthalpy qualify as potentials capable of driving the
process towards equilibrium if already at higher levels ?
If internal energy and enthalpy are potential terms then changes in them will have
to be zero at equilibrium and these will have to decrease for natural changes and
increase for unnatural changes.
Let us therefore see if criteria can be evolved mathematically to evaluate what
constitute as thermodynamic potential and if so under what conditions?
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Potentials under constant volume and constant entropy conditions
Mathematically these conditions are expressed as
dV = 0 and dS = 0
i) TdS > dU + P dV for natural process
So under above condition we can write
dU < 0
ii) TdS = dU + P dV for the equilibrium process
So under above conditions we can write
dU = 0
iii) TdS < dU + P dV for unnatural process
So under above conditions we can write
dU > 0
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
As internal energy decreases for natural process, remains same at
equilibrium and increases for unnatural process thus fully qualifies to be
called as thermodynamic potential under the conditions of constant
volume and entropy.
As a corollary, the internal energy must be the function of entropy and
volume can be expressed:
U = F (S, V)
Total differential of internal energy is given by
dVV
UdS
S
UdU
SV
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
We know: dU = TdS – P dV
On comparison this leads to
and
By specifying U as function of S and V it is possible to evaluate T and
P thereby describing the system fully.
By describing U as a function of any other two variables, it is not
possible to describe the system fully.
TS
U
V
PV
U
S
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Potentials under constant pressure and constant entropy conditions
Mathematically these conditions are expressed as
dP = 0 and dS = 0
i) TdS > dU + P dV for natural process
So under above condition we can write
dU + d (PV ) < 0
Or d (U + P V) < 0
Or dH < 0
ii) TdS = dU + P dV for the equilibrium process
So under above conditions we can write
dU + d (PV ) = 0
Or dH = 0
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
iii) TdS < dU + P dV for unnatural process
So under above conditions we can write
dU + d (PV ) > 0
Or dH > 0
So on the whole dH > = < 0 for unnatural, equilibrium
and natural processes as the case may be.
Therefore, the term enthalpy qualifies as being called
a potential term under constant pressure and entropy
conditions.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Enthalpy can therefore be described mathematically as: H = F (P, S)
Total differential of enthalpy can be written as
Also by definition: dH = d (U + P V)
= dU + P dV + V dP
= T dS + V dP
On comparison we get
dPP
HdS
S
HdH
SP
VP
HandT
S
H
SP
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Mathematically these conditions are
expressed as
dV = 0 and dT = 0
i) TdS > dU + P dV for natural process
So under above condition we can write
dU - TdS < 0
Or d ( U – T S) < 0
Let us define new mathematical term
A = U – T S
then dA < 0
Potentials under constant volume and constant temperature conditions
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
ii) TdS = dU + P dV for the equilibrium process
So under above conditions we can write
d ( U – T S) = 0
Or dA = 0
iii) TdS < dU + P dV for the unnatural process
So under above conditions we can write
d ( U – T S) > 0
Or dA > 0
The newly defined function qualifies for being referred as a
potential term under constant volume and temperature
conditions
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
‘A’ being the function of all state variables ‘A’ also
must be state function. Expressing as before
A = F (V, T)
Total differential of ‘A’
Now A = U – T S
dA = dU – TdS – S dT
And since, TdS = dU + P dV
Therefore dA = - S dT – P dV
On comparison we get
dVV
AdT
T
AdA
TV
PV
AandS
T
A
TV
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
This function ‘A’ was first defined by Helmholtz and
therefore is known as Helmholtz free energy.
We know dA = dU – T dS
And for reversible process qrev = T dS
An therefore qrev = dU – dA
From the First law dU = qrev – W
And hence qrev = dU + W
Comparing the two equations for qrev we get
- dA = W
Or dA = - W
That the change in Helmholtz free energy in a process is equal
to the amount of work done by the system on the surrounding
or is equal to the amount of work the system is capable of doing.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
These are by far the most commonly adopted
conditions in chemical and metallurgical engineering
practices. It means
dP = 0 and dT = 0
i) TdS > dU + P dV for natural process
So under above condition we can write
dU + d (P V) - TdS < 0
Or d ( U + P V– T S) < 0
Let us define new mathematical term
G = U + P V– T S
then dG < 0
Potentials under constant pressure and constant temperature conditions
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
ii) TdS = dU + P dV for the equilibrium process
So under above conditions we can write
d ( U + P V– T S) = 0
Or dG = 0
iii) TdS < dU + P dV for the unnatural process
So under above conditions we can write
d ( U + P V– T S) > 0
Or dG > 0
The function defined above thus qualifies for being
called as a potential term under constant pressure
and temperature conditions
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
‘G’ is a function of all state variable. so it is a state function.
The function ‘G’ is referred to as Gibbs free energy.
Now G = U + P V – T S
And H = U + P V
Then G = H – T S
Or G = H – T S for a finite change
Or dG = dH – T dS – SdT in differential form
At constant pressure dP = 0 and dH = q = TdS
And hence dG = - SdT
Or
STd
Gd
P
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Again dG = dH – T dS – SdT
On putting dH = dU + P dV + V dP
dG = dU + P dV + VdP – T dS – SdT
Putting in this dU + P dV = T dS
dG = T dS + V dP – T dS – S dT
= V dP – S dT
At constant temperature when dT = 0 ; dG = V dP
Or
Since G is a state function any change in G can be represented as G = U – T S + (P V)
= - Wrev + (P V)The change of Gibbs free energy during a process is equal and opposite in sign to the net reversible work obtainable from the process occurring under isothermal condition when corrected for change in volume under isobaric condition.
VPd
Gd
T
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Since G is a state function any change in G can be
represented as
G = U – T S + (P V)
= - Wrev + (P V)
The change of Gibbs free energy during a process is
equal and opposite in sign to the net reversible work
obtainable from the process occurring under
isothermal condition when corrected for change in
volume under isobaric condition.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Important thermodynamic relations
For closed and isolated systems have fixed mass and
composition and the reversible work is done against
pressure.
dU = T dS - PdV
dH = T dS + V dP
dA = - S dT – P dV
dG = - S dT + V dP
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Criteria for thermodynamic equilibria
Differential form Finite difference form
(dU)S, V = 0
(dH)S, P = 0
(dA)T, V = 0
(dG)T, P = 0
(U)S, V = 0
(H)S, P = 0
(A)T, V = 0
(G)T, P = 0
Since it is easy to maintain temperature and pressure
constant the Gibbs free energy criteria is employed in
chemical and metallurgical processes. However, in other
areas, other criteria are also employed.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Maxwell’s equations
YXx
N
y
M
From the properties of the exact differential
equation: dZ = M dX + N dY,
PT
VT
PS
VS
T
V
P
S
T
P
V
S
S
V
P
T
S
P
V
T
Maxwell’s relations are used
frequently in thermodynamics
for the calculation of changes in
thermodynamic variables for
different processes.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The Driving Force of a Chemical Reaction
Why does a chemical reaction take place?
It was believed that the energy change accompanying a
reaction can be measured directly by the enthalpy change
at constant pressure, or change in intrinsic energy (U) at
constant volume.
The reasoning behind this would be apparent – if a system
losses energy as a result of a chemical reaction, that
reaction will take place spontaneously – and the greater
the quantity of heat lost, the greater the driving force
behind the reaction.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
C (s)+ O2 (g) = CO2(g) with evolution of heat energy, i.e. cal/mol which happens to be a driving force in this
case. 2 Fe (s) + O2 (g) = 2FeO(s) and
3FeO(s) + 2 Al (l) = 2Fe(l) + Al2O3(s) Both have negative heats of reaction at 16000C and both take place
spontaneously On the other hand, if we consider the reaction; ZnO(s) + C(s) = Zn(g) + CO (g) and This reaction will not take place at 250C but if the system is heated
to 11000C, carbon will reduce zinc oxide to produce zinc metal.
940500298 H
8320001373 H 568000
298 H
The Driving Force of a Chemical Reaction (cont.)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
we cannot use heats of formation as a criterion of their
tendency to take place. We must therefore search for a
more consistent rule for the driving force of a reaction.
Consider the reaction: ZnO(s) + C(s) = Zn(g) + CO (g)
At 11000C, i.e above the boiling point of Zn (9070C), the
reaction will take place between phases indicated above.
At 250C, zinc is a solid (melting point 419.60C) so that the
reaction is
ZnO(s) + C(s) = Zn (s) + CO (g)
The Driving Force of a Chemical Reaction (cont.)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The most obvious difference in these reactions apart from
their temperature, is the difference in the physical state,
means a difference in the state of order of a system and
consequently in the entropy of the system.
S0 at 250C should be much less than that 11000C, because
at 11000C two molecules of gas are being produced from
two solid molecules whereas at 250C two solid moles only
produce one gaseous mole.
The Driving Force of a Chemical Reaction (cont.)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
A solid to gas transformation means a comparatively large entropy increase and the following figures for the reduction of ZnO by C:
Why not use entropy change of a reaction as a measure of the driving force behind a chemical reaction?
The idea is immediately contradicted if we consider the formation of the oxide of any metal:
We know that this reaction is spontaneous – an oxide film forms readily on iron at room temperature. Thus a positive entropy change is not the criterion of a chemical reaction.
degmol/cal68=SΔanddegmol/cal46=SΔ 1373298
molcal17-=SΔsFeO=gO2
1+sFe 0
2982 deg/,)()()(
The Driving Force of a Chemical Reaction (cont.)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The Second Law of thermodynamics states that a spontaneous process is always accompanied by an increase in entropy of the system and its surroundings.
The surroundings receive e a quantity of heat - H at constant temperature and pressure
entropy increase of the surroundings = If S is the entropy change of the system, total entropy
change of the system and surroundings = S –
For the 2nd Law to be obeyed (S - ) must be +ve
T
H
T
H
T
H
ve+=T
HΔ-SΔT
The Driving Force of a Chemical Reaction (cont.)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
we can say that (H – T S) must always be –ve for a reaction to proceed spontaneously in order that the total entropy change of the system and surroundings can be +ve, when the reaction proceeds.
We should examine the factor (H – T S) for the above three reactions in question. For the reduction of ZnO by C at 250C
The reaction cannot proceed because this factor is +ve, and the total entropy change of the system and its surroundings is –ve
This is negative so that the reaction will take place spontaneously at 11000C. This explains why zinc oxide smelting must be carried out at temperature of the order of 11000C in order that reduction of the oxide by carbon can proceed.
( ) molcal43100+=46298-56800=SΔT-HΔ 298298 deg/
( ) cal10200-=681373-83300=SΔT-HΔC1100at 137313730
The Driving Force of a Chemical Reaction (cont.)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
For the oxidation of iron at 250C
Again it is negative, and the reaction takes place spontaneously at 250C.
Consistent Rule The driving force of a reaction can be calculated as (H -
TS); the more negative this factor, the greater the driving force and if the factor is +ve, the reaction will not proceed spontaneously.
G = H - TS called Gibbs Free energy
.571401729863200298298 calSTH
The Driving Force of a Chemical Reaction (cont.)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
It is the maximum work available from a system at
constant pressure other than that due to a volume change.
Most metallurgical processes work at constant pressure.
We had already seen the fundamental importance of the
factor (H-TS), so that G is a measure of the ‘driving
force’ behind a chemical reaction. For a spontaneous
change in the system, G must be negative, the more
negative, the greater the driving force.
The Driving Force of a Chemical Reaction (cont.)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Thermodynamic Mnemonic Square
In order to memorize important thermodynamic relationship
involving the different thermodynamic potentials Max Born
suggested a diagram known as Thermodynamic Mnemonic
Square.
Four sides of the square are leveled
in alphabetical order starting from the
top and moving in clock-wise direction
with four thermodynamic potentials
namely A, G, H, U.
The other four primary functions namely V, T, P, S are placed at the four corners in such a way
that each thermodynamic potential is surrounded by the condition under which it acts
U
V T
S PH
G
A
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
We can write the total differentials of each of
potentials with help of this diagram as follows:
Differential of any potential is equal to the sum of the differentials of its
adjoining variables with their coefficients equal to diagonally opposite
variables.
This coefficients are taken to be positive if arrow points away from the
variable and negative if it points towards the variable.
dG = - S dT + V dP
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Maxwell’s relations can also be read from the
diagram as follows:
It is required to consider the corners only
The square is rotated in anti clock-wise direction in such a way that the
function whose partial differential is to be arrived it appears on the top
left hand corner of the square.
The partial derivative of this variable w.r. t variable on the bottom left
hand corner keeping the lower right hand corner variable as constant is
equal to the partial derivative of the top right hand corner variable w.r.t
bottom right hand corner variable with bottom left hand corner variable
as constant.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Negative sign on the right hand side of the equation is put if the arrows
are placed unsymmetrically with respect to vertical axis drawn from
centre of the square on rotation.
VSS
P
V
T
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Standard state of free energy
Standard state is pure element or compound at 1 atm pressure and at its stablest state at the temperature under consideration.
Free energy change of the reaction are generallycalculated when the reactants and products arepresent in their standard states is called the standard free energy change.
The standard free energy is designated by G0T.
Like enthalpy, we cannot measure the absolute value of free energy but change in free energy is quite possible to measure.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Therefore, there must be some reference point with respect to
which the actual values of various substance can be calculated.
The free energies of stable form of the elements at 298 K and 1
atm pressure are arbitrarily assigned as zero value.
The free energy of formation of compound are calculated on
the basis of the above assumption and the value is described as
standard free energy of formation.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
This quantity is generally reported at 298 K and for compound say
MO, it would be written as G0298, MO.
The Hess’s law is applicable for calculation of free energy change as it
is a state property.
The standard free energy of formation of compound and the standard
free energy of compound are same.
M + ½ O2 = MO
G0298 = G0
298,MO - G0298,M - ½ G0
298, O2
G0298,MO = G0
298,MO
00
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
knownbemustproductsandttanreacofenergiesfreedardtansthe
reactionaofchangeenergyfreedardtansthecalculatetoinorder
GbGaGdGc
GGG
bygivenisK298atGchangeenergyfreedardtansthe
DdCcBbAa
reactiontheFor
0B,298
0A,298
0D,298
0C,298
0ttanacRe,298
0oductPr,298
0298
0
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Some thermodynamic relationships
By definition G = H – TS = U + PV –TS
differentiating dG =dU + PdV + VdP – TdS – S.dT (1)
Assuming a reversible process involving work due only to
expansion at constant pressure
According to First Law: dU = dq – PdV and from Second
Law: dq = TdS
dU = TdS – PdV (2)
From (1) and (2) we get dG = (TdS – PdV) + PdV + VdP –TdS –
SdT
or dG = VdP – SdT (3)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Some thermodynamic relationships
At constant pressure, dP = 0, so that = - S
and at constant temperature, dT = 0 and dG = VdP
from PV = RT for ideal gas, dG =
Integrating between the limits PA and PB at constant
Temperature
if GA is the free energy of the system in its initial state and GB
the free energy in its final state when the system undergoes a
change at constant pressure
PT
G
dPP
RT
)4(ln BP
AP A
BAB P
PRT
P
dPRTGGG
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
but G = GB – GA and S = SB – SA and so that d (G) = -S.dT
on substitution
Some thermodynamic relationships
dTSdGanddTSdG BBAA .( ) ( )dTS-S=G-Gd ABAB
ST
G
P
TPT
GHSTHG
,
known on Gibbs – Helmholtz equation
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Calculation of G0 at high temperature
It is possible to calculate G0 of a reaction at high
temperature from the H0 and S0 values at 298 K
(available in the literature ) in this way
It must be remembered that if any transformation
takes place between 298 and T K in reactants and
products must be introduced in the above equation.
T
298
P0298
T
298
P0298
0T
0T
0T
0T
dTT
CSTdTCHGor
STHG
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Fugacity
The variation of Molar Gibbs free energy of a
closed system of fixed composition, with pressure
at constant temperature is given by
dG = V dP
For one mole of ideal gas
dG = (RT / P) dP = RT d ln P
For isothermal change of pressure from P1 to P2
at T
G (P2, T) – G (P1, T) = RT ln (P2 / P1)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Fugacity contd……
AS Gibbs free energy do not have absolute values,
it is convenient to choose an arbitrary reference
state from which changes in Gibbs free energy can
be measured. This reference state is called
standard state and chosen as being the state of
one mole of pure gas at one atm pressure and the
temperature of interest.
The Gibbs free energy of the ideal gas in the
standard state G (P = 1, T) is designated as G0 (T).
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Fugacity contd..
Thus the Gibbs free energy of 1 mole of gas at any
other pressure P is given as
G (P, T) = G0 (T) + RT ln P
Simply
G = G0 + RT ln P
So the molar Gibbs free energy of the ideal gas is
a linear function of the logarithm of the pressure of
the gas
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Fugacity contd…
If the gas is not ideal then the relation of molar free energy and
logarithm of pressure is not linear.
A function is invented which when used in place of pressure gives a
linear relationship.
This function is called ‘Fugacity ( f )’ is partially defined as
dG = RT dln f
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
In addition, as the real gas and the ideal gas
behave the same at very low pressure, it is
obvious that (f / P) → 1 as P → 0.
after integration from standard state to any arbitrary state we can write
G - G0 = RT ln (f / f0)
Where G0 is the Gibbs free energy of the real gas
in the standard state. f is the fugacity at a
specified state and f0 is standard state fugacity.
Fugacity is the measure of escaping tendency of a gas.
Fugacity is the effective pressure corrected for non-ideality
Fugacity contd…
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Activity
Activity (a) is defined as
So we can write
0i
ii0 f
faor
f
fa
i0ii
0 alnRTGGOralnRTGG
Partial molar free energy of component i in the state of interest. free energy of component I
in the standard state.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The standard state of a substance are generally chosen as the pure solid
or liquid form of the substance at 1 atm pressure and temperature under
consideration or as the gases at 1 atm pressure at the temperature under
consideration.
The activity of a substance in its standard state is seen to be unity.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Equilibrium Constant
Let us consider a general chemical reaction at
constant temperature and pressure
Capital letters for chemical element and small
letters for the number of gram mole. The general
free energy change of the reaction may be written as
RrQqMmLl
MLRQ GmGlGrGqG
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The standard free energy change can be written as
Subtracting these two equations,
Or
Or
0M
0L
0R
0Q
0 GmGlGrGqG
M0ML
0L
R0RQ
0Q
0
GGmGGl
GGrGGqGG
ML
RQ0
alnRTmalnRTl
alnRTralnRTqGG
JlnRTaa
aalnRTGG
mM
lL
rR
qQ0
The parameter J is called activity quotient
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Or
Which is known as van’t Hoff isotherm.
Let us consider the equilibrium in which all the reactant and
products are in equilibrium. In this case the activity product
defined as K, the thermodynamic equilibrium constant.
At equilibrium G = 0,
Or
eq
mM
lL
rR
aa
aaK
KlnRTJlnRTG .eqat0
K
JlnRTG
JlnRTGG 0
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Since the standard state of a substance is at 1atm
pressure the G0 of a reaction is function of
temperature only. Therefore, the equilibrium
constant K also function of temperature only.
(J / K) < 1 reaction is in forward direction.(J / K) > 1 reaction is in backward direction.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
For this process, mass, composition and pressure are
assumed to be constant only variation of temperature
is considered.
Derivation of Gibbs – Helmholtz Equation
P22
2
P
P
T
G
T
1
T
H
T
G
obtainweTbybothsidesDeviding
T
GTHSTHGr0
ST
G
Gibbs – Helmholtz Equation
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
This is one of the form of Gibbs – Helmholtz Equation.
The alternative form can be derived as follows.
2
P
22P
T
H
T
TGor
T
H
T
G
T
G
T
1or
2P
2
P
2
P
PP
P
T
G
T
G
T
1TG
T
G
T
1T
T
1
TGor
GT1
T
T
G
T
1G
T1
G
T
1
T
1
TG
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
This is the alternative form of Gibbs – Helmholtz Equation .
We can write Gibbs – Helmholtz Equation for a process as
H
T
HT
T
1
TG2
2
P
H
T
1
TG
T
H
T
TG
P
2P
Utility Evaluation of enthalpy of reaction from
free energy change. Evaluation of free energy change from
calorimetric data.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
.etemperaturoneatH
andGorestemperaturtwoatGeitherofknowledgethefrom
edminerdetareThese.ttanconsegrationintareIandHWhere
TIT
1
2
cT
2
bTlnTaH
TdT
c
2
b
T
a
T
HT
TdTdTcTbaT
1
T
HT
TdTdCT
1
T
HTGor
TdT
H
T
G
0
00
0
20
320
222
0
P2200
T
2
0T
0T
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Trouton’s and Rechard’s RulesRechard’s rule: it states that the ratio of latent heat of melting
to the temperature of the normal melting point of the F.C.C
metal is approximately 9.61 J / K and that of BCC metal is
approximately 8.25 J / K. i.e
Trouton’s Rule: it states that the ratio of latent heat of boiling to the temperature of
the normal boiling point of the F.C.C metal is approximately 87 J / K.
metalC.C.BforK/J25.8T
H
metalC.C.FforK/J61.9T
H
m
m
m
m
K/J87T
H
V
V
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
The initial state of one mole of an ideal gas is P = 10 atm and
T = 300 K. Calculate the entropy change in the gas for
a) Reversible isothermal decrease of pressure to 1 atm.
b) Reversible Adiabatic decrease of pressure to 1 atm.
c) Reversible constant volume decrease of pressure to 1 atm.
Solution
.ltrs62.241
462.210
P
VPVVPVP
processisothermalperasK300TTandatm1P)a
.ltrs462.210
30008207.0
P
TRVTRVP
K300Tandatm10P
2
1122211
122
1
11111
11
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
K/J14.19462.2
62.24ln314.8
V
VlnR
dVV
RdV
T
P
T
dVPUd
T
qS
1
2
V
V
V
V
2
1
2
1
rev2
1
2
1
processadiabaticfor0qas0T
qdS)b rev
rev
K/J71.28300
30ln314.8
2
3
T
TlnC
dTT
C
T
Ud
T
dVPUd
T
qS
K3008207.0
462.21
R
VPT,R
2
3C.,atm1P)c
1
2V
T
T
V2
1
2
1
2
1
rev
222V2
2
1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
Calculate the entropy change of the universe in isothermal
freezing of 1 gm-mole of super cooled liquid gold at 1250 K
from the following data for gold.
Tm = 1336 K,H0m = 12.36 kJ/mol,
CP(s) = 23.68 + 5.19 x 10-3 T J / mol / K, CP(l) = 29.29 J / mol / K
gSurroundinSystemUniverse SSS
Solution
.propertystateaisentropy
cesinchoosewepathreversiblewhichtoasdifferencenomakesIt
.pathreversibleaalongonlycalculatedbetoisS,But
.etemperaturfreezingmequilibriutheatoccuringnotisitcesin
,leirreversibbutisothermalisprocessfreezingcasethisIn
S)i
System
System
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
)2State()1State(
K1250atK1336atK1336atK1250at
)s(Au)s(Au)l(Au)l(Au
ispathreversiblesimplestTheCoolingFreezingHeating
1
1250
1336
31336
1250
1250
1336
P
m
0m
1336
1250
P12System
KJ327.9
TdT
T1019.568.23
1336
12360Td
T
69.29
TdT
)s(C
T
HTd
T
)l(CSSS
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Hence.K1250atreversiblyfreezingduringreleasedheatabsorbsngSurrroundi
.capacityheatiniteinfhasitandK1250atisgsurroundintheifonlypossibleisThis
.K1250atbetoassumedbeenhavesystemtheofstatefinalandinitialtheBoth
S)ii gSurroundin
T
H
T
qS System
gSurroundin
J12460
TdT1019.568.2312360Td69.29
Td)s(CHTd)l(CHHH
1250
1336
31336
1250
1250
1336
P0m
1336
1250
P12System
1SystemgSurroundin KJ967.9
1250
12460
T
HS
1Universe KJ64.0967.9327.9S
Suniverse is positive since the process is irreversible.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
one gram of liquid ThO2 at 2900 0C is mixed with 5 gm of ThO2
at 3400 0C adiabatically. a) what is the final temperature
b) What is the entropy change of the system and surrounding
Combined c) is the process spontaneous? Assume CP to be
Independent of temperature.
f
H
f
C
T
T
PH
T
T
PC
f
fPfP
coldPHotP
T
TdC5)S(bodyhotofchangeentropy
T
TdC)S(bodycoldofchangeentropy
K3589T
T3673C13173TC5
TdCmTdCm
bodycolderbyabsorbedheatbodyhotterbyreleasedHeat)a
Solution
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
P
C
fP
H
fP
T
T
P
T
T
P
CHSystem
C00752.0
T
TlnC
T
TlnC5
T
TdC
T
TdC5
SS)S(systemtheofchangeentropyTotal)b
f
C
f
H
eoustansponisprocesstheSo
0SSS
.llyadiabaticaoutcarriedisprocessas
zeroisgsurroundintheofchangeEntropy)c
gSurroundinSystemUniverse
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
Calculate the standard entropy change of the following reaction
at 1000 K.
Pb (l) + 0.5 O2(g) = PbO (s)
and also calculate the entropy change of the Universe.
Given: Tm, Pb = 600K, H0m, Pb = 4812 J mol-1, H0
PbO,298 = -219 kJ
S298, PbO = 67.78 J K-1, S298, Pb = 64.85 J K-1, S298, O2 = 205.09 J K-1
CP, PbO(s) = 44.35 + 16.74x10-3 T J K-1mol-1
CP,Pb(s) = 23.55 + 9.75X10-3 T J K-1mol-1
CP,Pb(l) = 32.43 – 3.09X10-3 T J K-1mol-1
CP,O2 = 29.96 + 4.184X10-3 T - 1.67x 105 T-2 J K-1mol-1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
1
0O,298
0Pb,298
0PbO,298
0298
2
P0298
2
01000
KJ615.9909.2052
185.6478.67
S2
1SSS
)2.......(..........)s(PbO)g(O2
1)s(Pb
reactiontheforvaluesCandSfrom
)1......(..........)s(PbO)g(O2
1)l(Pb
reactiontheforS
evaluationofconsistsphysicallyoblemPrThe
Solution
2
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
11253P
11253P
molKJT10836.0T1089.481.5)2(C
molKJT10836.0T1072.1705.3)1(C
1
22
53
22
5
30298
1000
600
P
m
0m
600
298
P0298
01000
KJ079.99
07.0956.196.202.835.035.513.2615.99
600
1
1000
1
2
10836.060010001089.4
600
1000ln81.5
600
4812
298
1
600
1
2
10836.0
2986001072.17298
600ln05.3S
TdT
)1(C
T
HTd
T
)2(CSS
writecanwe,K600atmeltsPbthatNoting
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
kJ244.218
73.5580.15642324481220.14179.240210.21910219
600
1
1000
110836.06001000
2
1089.4
600100081.54812298
1
600
110836.0
2986002
1072.1729860005.3H
HHTd)2(CHTd)1(CHH
isK1000atreactionofheatorreactionthebyreleasedHeat
3
5223
5
223
0298
0PbO,298
0298
1000
600
P0m
600
298
P0298
01000
1
gSurroundinSystemUniverse
101000
gsurroundin
KJ119218079.99
SSS
KJ2181000
244.218
T
HS
.capacityheatearglhavingK1000atisoundingsurrthethatgminAssu
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
From the data given for different phases involved, calculate the
change in entropy of one mole of manganese when it is heated
from 298 K to 1873 K under one atmospheric pressure.
(CP)-Mn = 5.7 + 3.4 x 10-3 T – 0.4 x 105 T – 2 cal / deg/ mol
(CP)-Mn = 8.3 + 0.73 x 10-3 T cal / deg/ mol
(CP )γ-Mn = 10.70 cal / deg/ mol
(CP )δ-Mn = 11.3 cal / deg/ mol
(CP )Mn, l = 11.0 cal / deg/ molTransformation Temperature / K Latent heat in Cal / mol.
→→γ
γ →δ
δ→l
990
1360
1410
1517
535
525
430
3500
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The total change in entropy will be equal to the sum of the
changes in entropy due to the rise of temperature and those
due to the phase transformation occurring on heating. Thus
dT
T
C
1517
3500
dTT
C
1410
430dT
T
C
1360
525dT
T
C
990
535dT
T
CS
1873
1517
MnlP
1517
1410
MnP1410
1360
MnP
1360
990
MnP990
298
MnP
MnlMnMnMnMn 151714101360990
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
dTT
113071.2dT
T
3.113049.0
dTT
7.103860.0dT
T
T107.03.8
5404.0dTT
T104.0T104.37.5S
1873
1517
1517
1410
1410
1360
1360
990
3
990
298
253
.mol/K/cal3655.193188.23071.28265.0
3049.03863.03860.08945.25404.04010.9S
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
Using the following values, calculate the standard free energy change per
mole of the metal at 1000 K for the reduction of molybdic oxide and chromic
oxide by hydrogen:
.mol/kcal5.45G
.mol/kcal0.120G
.mol/kcal5.205G
0)g(OH,1000
0)s(MoO,1000
0)s(OCr,1000
2
3
32
mol/kcal5.45G;OHO2
1H
mol/kcal0.120G;MoOO2
3Mo
.mol/kcal5.205G;OCrO2
3Cr2
Solution
01000222
0100032
01000322
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
.veischangeenergyfreetheasetemperatur
thisatreducedbecanmolybdenum,But.etemperatur
thisathydrogenbyreducedbenotcanit,veisoxide
chromiumofreductiontheforchangeenergyfreethecesin
.mol/kcal5.161205.453G
OH3MoH3MoO
.mol/kcal5.342
5.2055.453G
OH3Cr2H3OCr
aswrittenbecanreactionreductionThe
01000
223
01000
2232
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
The standard heat of formation of solid HgO at 298 K is – 21.56
kcal / mol. The standard entropies of solid HgO, liquid Hg and
O2 at 298 K are 17.5, 18.5 and 49.0 cal /deg/mol, respectively.
Assuming that H0 and S0 are independent of temperature,
calculate the temperature at which solid HgO will dissociate
into liquid Hg and O2.
/moldeg//cal49S
/moldeg//cal5.18S
/moldeg//cal5.17S
mol/kcal56.21H
Solution
0)g(O,298
0)l(Hg,298
0)s(HgO,298
0)s(HgO,298
2
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
C572K8455.25
21560TT5.25215600
mol/calT5.2521560STHG
0GwhendissociatetostartwillHgOsolidThe
STHGwritecanWe
ToftindependenareSandHcesin
KJ5.25492
15.185.17
S2
1SSS
.mol/kcal56.21HH
)s(HgO)g(O2
1)l(Hg
aswrittenbecanreactionThe
0
0298
0298
0T
0
0298
0298
0T
00
1
o)g(O,298
o)l(Hg,298
o)s(HgO,298
0298
0)s(HgO,298
0298
2
2
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
For the reaction WO3 + 3H2(g) =W(s) + 3H2O(g)
a)Calculate G0 and K at 400, 700 and 1000K
b)What is the maximum moisture content of H2 needed for the
reaction given:
WO3(s) = W(s) +
3H2 +
Solution:
WO3(s) = W(s) +
H2+
WO3(s) + 3H2(g)W(s) + 3H2O(g),
calsTTTGgO 7.91log2.10201500),(2
3 02
.1.1358900,)()(2
1 022 calsTGgOHgO
calsTTTG 4.52log2.10248000
calsTTTGgO 7.91log2.10201500),(2
3 02
..,)()( calsT339++176700-=GΔgOH3=gO2
3 022
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem-2
a) i)
ii)
iii)
K400= 1.26x10-8
K700= 2.5x10-2
K1000= 0.22
b)
KatG 00 400 .144604004.52400log400.10248000400 calsxxG
KatG 00 700 .cals5122=GΔ 0700
KatG 00 1000 .cals3000=GΔ 01000
RTGeKorRT
GKeiKRTG
00
0 ln..ln
3
2
2
H
OH
p
pK
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
solution
+ = 1
(0.00247+1) = 1
% moisture = 0.24 %
At 700K, K700 = 2.317x10-3
( ) 3-318-31400
H
OH10x4722=10x2531=K=
p
p
2
2 ..
OHp2 2Hp
2Hp
...
atm99750=0024721
1=p
2H
atm0024660=p∴and OH2.
( ) 2960=10x52=p
p 312-
H
OH
2
2 ..
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
solution
(0.296 +1) = 1
at 1000K K1000 = 0.22
pH2O= 0.3776
% H2O at 1000K = 37.76 %
2Hp
22830=pand77160=1321
1=p OHH 22
...
%.% 8322=K700atOH∴ 2
( ) 60670=220=p
p31
H
OH
2
2 ..
62230=60671
1p =H2
..
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem-3
What is the maximum partial pressure of moisture which can
be tolerated in H2 – H2O mixture at 1atmospheric total
pressure without oxidation of nickel at 7500C?
SolutionEquation under question Ni + H2O = NiO(s) +H2
= 450 + 10.45 T cals.
..)()()( calsT5523+58450-=GΔsNiO=gO2
1+sNi 0
12
..)()()( calsT113+58900-=GΔgOH=gO2
1+gH 0
2222
02
01
0r GΔ-GΔ=GΔ
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
solution
G 01023 = 450 + 10.45 x 1023
= 11140
= - RT lnK
ln k = -
=-5.48006
K = + 0.0041669 =
from 1 & 2 :
1023987.1
11140
OH
H
p
p
2
2
(1) 0041669.022 OHH pp (2) 1
22 OHH pp
1=p+p 00416690 OHOH 22. 99585.0
0041669.1
1
2OHpor
00415.02 Hp Hence maximum tolerable partial pressure of
moisture is 0.99585 atm
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
problem
The standard free energy change for reaction at 1125K NiO(s) +
CO(g) = Ni(s) + CO2(g) is -5147 cals. Would an atmosphere of 15% CO2,
5% CO and 80% N2 oxidise nickel at 1125K?
Solution:
G0 = -RTlnK = -4.575 x 1125 log K = -5147
K = 10
K for the above reaction is
In the given atmosphere i.e J/K is <1 so no oxidation of nickel
0001=1125x5754
5147=K .
.log
NiO
Ni
CO
CO
a
a
p
p=K 2 . NiONi aa 1
10=p
p=
CO
CO2
J=3=p
p
CO
CO2
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
For the reaction NiO(s) + H2(g) = Ni(s) + H2O(g)
calculate the equilibrium constant at 7500C from the following
data:
Could pure nickel be annealed at 7500C in an atmosphere
containing 95% H2O and 5% H2 by volume without oxidation?
..),()()( Tcals5523+58450-=GΔsNiO=gO2
1+sNi 0
2
..),()()( Tcals113+58900-=GΔgOH=gO2
1+gH 0
222
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
solution
G0 for the required reaction would be G0 =
G0 = -RT lnK = -4.575T log K
(aNi = 1 = aNiO)
J/K <1 rxn in forward direction
So no oxidation will take place
01
02 GG
2
2
2
2
H
OH
NiO
Ni
H
OH0
p
p=
a
a
p
p=
T5754
GΔ=K .
.log
382=1023x5754
11140= .
.
222
2 240..240 HOHH
OH ppeip
pK
J=19=5
95=
p
p given
2
2
H
OH:
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
Calculate the standard enthalpy and entropy changes at 298 K
for the reaction
2 Cu (s) + ½ O2 = Cu2O (s)
G0 = - 40500 -3.92 T log T + 29.5 T cal / mol
T
702.1
T
40500
T
TG
,atingDifferenti
5.29Tln702.1T
40500
T
G
T5.29TlnT702.140500
T5.29TlnT4343.092.340500G
Solution
2
0
0
0
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
K/cal10.18
298ln702.1798.27S,K298at
Tln702.1798.27Sor
Tln702.1798.27
5.29Tln702.1702.1ST
G
.mol/kcal99.39
29870.140500HK298at
mol/calT70.140500Hor
T
702.1
T
40500
T
H
writecanwe,equationHGBy
0298
0
00
0298
0
22
0
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
Enthalpy of the solid platinum at any temperature is given by
the expression:
Derive an expression for the change in free energy when one
mole of platinum is heated from 298 K to any temperature T K.
.mol/cal1785T1064.0T8.5HH 23298T
CTdT
H
T
G
.Tetemperatur
atsystemanyofequationHGtoAccording
Solution
2T
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
mole/cal69.9903C298T1064.0TlnT8.5TCGor
C2982981064.0298ln29885.51785HG
GGG
bygivenbethereforewillK298from
TetemperaturtoheatingonGchangeenergyfreeThe
TCT1064.0TlnT8.51785HGor
CT1064.0Tln8.5T
1785H
T
Gor
CdTT
17851064.0
T
8.5
T
H
T
G
Thus
23
23298298
298T
23298T
3298T
23
2298T
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem-1
Calculate (1) the elevation of the boiling point of zinc when the external pressure is 2 atm. and (ii) the depression of the freezong point when the external pressure is 50 atm. The latent heat of vaporization of zinc is 27.3 kcal/g atom and the normal temperature of boiling is 9070C. The corresponding heat of fusion is 1.74 kcal/gatom and the normal temperature of fusion is 419.50C. The density of solid zinc is 7.0 g/cc and that of liquid zinc is 6.48 g/ce at 1 atm. pressure.
Solution
Using the Clausius-Clapeyron equation for this problem we have
To solve this problem we have also to know the values of the volume of 1
gatom of zinc in the vapour and liquid states. The volume occupied by 1 gatom of a perfect gas at STD in 22400 cc.
V
S
dT
dP e
zincliquid
ofatomg1 ofvolumetheis vandvapourzincofgatom1ofvolumetheisvwhere) g l
lge
e
vvT
L
dT
dP
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The volume occupied by 1 g.atom of zinc vapours at 9070C
(1180K) is
Volume occupied by 1 gatom of liquid zinc =
Substituting these values in the equation dP = 1 atm.
1 cc atm = 0.024212 cals.
( ) .cc1180273
22400=vg
.09.1048.6
38.65cc
( )e
lge
L
v-vTdP=dT
.
glglg vvvvv
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
solution
cal 273×27300
Katmcc 1180×22400×1180×1=
..
K3101= cal 273×27300
Kcal 0242120×22400×1180×1180×1= .
.
7
38.65
,48.6
38.65 (ii)
s
lsl
f
f
v
vvvT
L
dT
dP
slf
f
vvL
TdPdT
.
.024212.0 1.1740
0.7
38.65
48.6
38.655.69249
calatmcccal
Katmcc
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem-2
( ) 0242120×349-089101740
5692×49= ...
.
0242120×750×1740
5692×49= ..
.
= 0.354 K Ans
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem-2
The latent heat of vaporization of zinc is 27.3 k cal/mole at the boiling point of
9070C. Find the vapour pressure over pure zinc at 8500C.
Solution:
for l g transformation we have
P
RTvvv
vT
L
vvT
L
dT
dPlg
ge
e
lge
e
; .
KTpKTatmpRT
LP
dT
dP1123,?,1180.,1 22112
21
12
1
22
ln1
TT
TT
R
L
p
por
TR
L
P
dP
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
ln p2 – 0 = = -0.5909869
p2= 0.55378 atm
11231180
11801123
987.1
27300
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Lechatelier’s Principle
Statement
It states that if an equilibrium in a system is upset,
the system will tend to react in a direction that will
reestablish the equilibrium.
Factors
i) Concentration of reactant and product
ii) Pressure
iii) Temperature
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Effect of concentration
Consider the reaction
A + B = C + D
At equilibrium for which
If the concentration of any one of the reactant or product is altered the
concentration of others must alter to keep K constant.
If for example [D] is increased by adding more D to the system then [C] will
decrease (by reaction with D to form A and B) preserving the value of K.
In general, If the product is added, the system will shift towards left to reestablish
the equilibrium and converse is true for the reactant.
BA
DCK
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
If the reactant is removed, the equilibrium will shift towards reactant and
same is true for the product.
This shift of equilibrium may be used to advantage in some commercial
processes since continuous removal of product will drive the reaction to
completion. This is more easily achieved if one of the product is a gas e.g.
CaCO3 = CaO + CO2
In the lime kiln removal of CO2 in an air current drives the reaction
towards right.
Other example is the production of magnesium:
2 MgO(s) + Si(s) = 2Mg (g) + SiO2(s)
At the temperature employed the magnesium vapour can be evacuated.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Effect of Pressure
Changes in pressure have no effect on the position of equilibrium when
only solids or liquids are involved or in gaseous reactions involving no
volume change e.g.
FeO(s) + CO(g) = Fe(s) + CO2((g)
Since equal volumes of gas appear in both sides.
However, the equilibrium position for a reaction in which changes in
gaseous volume occur may be displaced by pressure
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
We can consider following examples
N2(g) + 3 H2(g) = 2 NH3(g)
1 vol + 3 vol = 2 vol
Here in this case there is an overall volume decrease( 4 → 2).
If the system is subjected to a pressure increase the system will move
in such a direction as to lessen this increase in pressure .
By moving to the right the volume diminution results in reduction in
pressure i. e increasing pressure drives the reaction to the right.
Ammonia is commercially produced at 350 atm.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Effect of Temperature
‘K’ is unaffected by concentration and pressure change but dependent
on temperature.
Exothermic reaction:
N2 +3 H2 = 2 NH3 H = - 92.37 kJ
If the temperature is increased the system reacts in such a way as to
oppose this constraint by removing heat. Which can be done by
shifting to the left.
Exothermic reactions are favoured by low temperature
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Endothermic reaction:
ZnO + C = Zn + CO H = +349 kJ Increase in temperature will again shift the equilibrium in the
direction which absorbs heat
i. e. to the right in this case. Endothermic reactions are therefore, favoured by high temperatures.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Clausius-Clapeyron Equation
It is extremely important for calculating the effect of change of pressure on the equilibrium transformation temperature.Derivation Let us consider a single solid substance in equilibrium with its liquid at its
temperature of melting and under one atm pressure.
There is a natural tendency for the molecule to pass from solid into liquid or vice versa.
The number of atoms passing at any time from one state to other will depend on the temperature and pressure.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Let us assume that the molar free
Energy of the solid at constant T, P
Is GA and of the liquid is GB; then
If GA > GB the solid metal can
decrease its free energy by
dissolving, i. e. if
GA > GB Solid → Liquid; solid melts since G = -ve
GA = GB Solid = Liquid; Equlibrium since G = 0
GA < GB Solid ← Liquid; liquid solidifies since G = +ve
Me (s)
GA
Me (l)
GB
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The condition for a dynamic equilibrium between the solid and the liquid metal is
GA = GB
and dGA = dGB
The change in the free energy of either phase may caused by the change in temperature and pressure of the phases. dG = f ( T, P)
dP.P
GdT.
T
GdG
T
A
P
AA
dP.P
GdT.
T
GdG
T
B
P
BB
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
At equilibrium dGA = dGB
Or
Or
Or
Or
dP.P
GdT.
T
G
T
A
P
A dP.P
GdT.
T
G
T
B
P
B
dPvdTSdPvdTS BBAA
dTSSdPvv ABAB
dT.SdP.v
tr
tr
T
HSas
V
S
Td
Pd
ABtr
tr
vvT
H
Td
Pd
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Or
If v is +ve as is normally in the case of melting and Htr will be +ve quantity i. e as
the pressure increases the transition temperature should also increase. In other
words the equilibrium melting point shall increase with pressure and vice versa.
However, for ice – water system v is – ve and hence (dP / dT) is –ve and which is
fully exploited in the game of skating on ice wherein the skate pressure increase
shall decrease the melting point of ice and hence help skating by providing fluid for
lubrication.
vT
H
Td
Pd
tr
tr
Clausius – Clapeyron Equation
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Applications of Clausius – Clapeyron Equation
.lyrespectiveliquidand
vaporofvolumesmolarthearevandvand
ionvapourizatofheatlatenttheisHwhere
vvT
H
Td
Pd
EquilibiaVapourLiquid
liqvap
V
liqvap
V
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Applications of Clausius – Clapeyron Equation
Contd….
2V
2V
vap
vap
V
liqvap
TR
H
Td
Plnd
TR
HP
Td
Pd
getweonsubstitutiafterP
TRv
gasideal
anasbehavesvaporthethatgminAssu
vT
H
Td
Pd
hencevv
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
ttanconsegrationintisCwhere
CTR
HPln
.eqaboveofegrationint
thenttanconsisHgminAssu
V
V
concernedervalinttemptheovervalue
meanthebewillmethodthisbycalculatedmetal
liquidofionvapourizatofheatThe.Ciserceptint
andR
His
T
1.vsPlnplottheofslopeThe V
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
.etemperaturoftindependen
isHthatgminAssu.lyrespectiveTandT
tempsthetoingcorrespondPandPitslimthe
withindonebecanequationCCofnIntegratio
V21
21
.knownareionvapourizatofheatand
etemperaturanotheratpressurevapourtheifetemperaturany
atpressurevapourthecalculatetousedbecanequationThis
T
1
T
1
R
H
P
Plnor
T
Td
R
HPlnd
12
V
1
2
T
T2
V
P
P
2
1
2
1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solid – Vapour (sublimation) Equilibria
On the basis of assumptions similar to those made
in liquid – vapour equilibria, one can obtain the
similar expression for solid – vapour equilibria.
Where Hs is the heat of sublimation
2S
TR
H
Td
Plnd
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solid – liquid (Fusion) equilibria
Applying C – C equation to so solid liquid equilibria
Hf is the molar heat of fusion, vliq and vsolid are the
molar volumes of liquid and solid respectively.
This equation may be applied to calculate the change
in melting point of a metal with change of pressure.
f
solidliqf
solidliqf
f
H
vvT
Pd
Td
vvT
H
Td
Pd
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solid – Solid equilibria
The rate of change of transition temperature at
which two crystalline forms of a solid are in
equilibrium with pressure is given by following
expressions:
Where Htr is the molar heat of transition, v and
v are the molar volume of the indicated forms of
solid measured at Ttr.
trT
tr
tr
tr
tr
H
vvT
Pd
Td
vvT
H
Td
Pd
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Ellingham Diagram
Ellingham diagrams are basically graphical representation of G0 vs. T
relations for the chemical reactions of chemical and metallurgical
engineering interest.
H. J. T. Ellingham in 1944, was first to plot the standard free energy of
formation of oxides against temperature and these later became known as
Ellingham diagram.
Later on the same plotting was applied for sulphides, chlorides, fluorides
etc.
Oxide diagrams are mostly used in metallurgy.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Features of Ellingham diagram
1) Formation reaction for oxides may be generalized as
322
2
yx
yx2
OAl3
2)g(OAl
3
4
NiO2)g(OiN2
areexamplesSpecific
.lyrespective,oxidesmetaland
metalforsymbolsgeneralareOMandM.compound
specifictheondependwillyandxofvaluestheWhere
OMy
2gOM
y
x2
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
activity
unitat.g.estatespureorstatedardtans
theirinareproductand)metal(ttanreac
themeansenergyfreedardtanS.oxygenof
moleperareGofvaluetheThus.reaction
formationoxidetheinbasisthesconstitute
Oofmole1diagramEllinghamtheIn
.involvedmolesof.notheondepend
wouldreactionaofGvaluehence
propertyextensiveanisenergyFree
0
2
0
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
2) Since G0 = H0 - T S0
The variation of H0 not being large, it can be treated as constant over a wide
range of temperatures. S0 however, changes with temperature particularly
when gaseous phases are involved with the condensed phases.
The plot of G0 against temperature is a straight line. So the slope of the line is
- S0 . Let us consider the reaction
M(s) + O2 (g) = MO(s)
S0 = S0 (oxide) - S0 (metal) - S0 (oxygen gas)
since S0 (oxide) and S0 (metal) are practically the same, the
entropy change (S0) arises predominantly due to disappearance of one mole of
oxygen gas and hence it is -ve. So the slope of the line is positive.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The entropy changes of various such metal oxidation process are expected to
be substantially of the same value. Therefore, the most of the oxide lines
slope upwards and parallel to each other.
3) The plot of G0 against temperature is a straight line as long as there is no
phase change (melting, boiling, phase transformation) in either the reactant
and products. The reason is that when such phase change takes place, there
is a change in entropy and since the change entropy is the slope of the line,
therefore when such changes takes place, the straight line will change its
slope. Let us consider the reaction
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
let us consider the melting of reactant. The reaction is
M(l) + O2(g) = MO(s)
S0 = S0 (oxide) - S0 (metal (l)) - S0 (oxygen gas)
Since S0 (metal (l)) > S0 (metal (s))
S0 = more negative than when both were solid
Hence the slope of the line will be further upwards from
the temperature of melting.
if the reactant boils. We can write
S0 (metal (g)) > S0 (metal (l))
S0 = more negative than when metal oxide was solid
and metal was liquid
Hence the slope of the line will be further upwards from
the boiling point..
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
let us consider the melting of product. The reaction is
M(s) + O2(g) = MO(l)
S0 = S0 (oxide(l)) - S0 (metal (s)) - S0 (oxygen gas)
Since S0 (oxide (l)) > S0 (oxide (s))
S0 = les negative than when both were solid
Hence the line will be bend downward with +ve slope from
the temperature of melting.
if the product boils. We can write
S0 (oxide (g)) > S0 (oxide (l))
S0 = less negative than when metal oxide was liquid
and metal was solid
Hence the line will be bend further downward with +ve slope from the temperature
of melting.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
4) The intercept of the straight line with the ordinate at absolute zero gives
approximately the value of H0, since from the equation G0 = H0 - T S0
when T = 0, G0 = H0
5) since G0 must have a negative value for the reaction to take place, it can be seen
from this plot that all metals shown below the negative areas of G0 are oxidized
spontaneously by oxygen while those above are not e. g. gold
6) When a line touches the positive regions of G0 this means beyond that
temperature no further oxidation can take place or the oxide formed in that
region is unstable or oxides will start decomposing when G0 = 0. i. e Ag2O will
decompose at 15000C.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
7) The stability of an oxide is directly related to its - G0 values; less stable
oxides have a small - G0 and more stable oxides have a high - G0 value.
8) An oxide can be reduced by only those metals below it in Ellingham
diagram; the reverse cannot take place. For example at 8000C, Cr2O3 can be
reduced by aluminium but Al2O3 cannot reduced by chromium:
4/3 Al + O2 = 2/3 Al2O3 G0 = - 800 kJ
4/3 Cr + O2 = 2/3 Cr2O3 G0 = - 500 kJ
2/3 Cr2O3 + 4/3 Al = 2/3 Al2O3 + 4/3 Cr; G0 = - 300 kJ
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
9) G0 of an oxide may be greater than another at low temperature but
becomes less than the other at higher temperature. Thus while the MnO is
reduced by Na below 2200C, the reverse is true above that temperature.
10) The line of the reaction C + O2 = CO2 runs nearly horizontally on the chart
i.e its slope is zero or practically there is no entropy change for this
reaction. This can be seen from the fact that the initial and final volumes
are practically the same which is the one volume of oxygen gas and one
volume of CO2, respectively; entropy of the solid being negligible.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
11) The line for the reaction 2 C + O2 = 2CO runs downwards i.e. it has a negative
slope, this is due to the large increase in entropy: two volume of gaseous CO are
formed from one volume of oxygen gas. In this case G0 become more negative
as the temperature increases.
12) That the line for CO formation runs downward is of great importance in
pyrometallurgy. It enables almost all the metal – metal oxide lines meet C – CO
lines at high temperature. This makes most the oxides unstable beyond point of
intersection. This is called reversion of stability. This means that reduction of
most metal oxides by carbon is possible at high temperature.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
13) Carbon monoxide can reduce all oxides above the CO2 lines. For example at 7000C NiO can be reduced by CO
2 CO + O2 = 2 CO2 G0 = - 95 kcal.
2 Ni + O2 = 2 NiO G0 = - 75 kcal.
2CO + 2NiO = 2Ni + 2CO2 G0 = - 20 kcal.
14) All oxides above H2O lines can be reduced by H2 e.g at 7000C
2 H2 + O2 = 2 H2O G0 = - 92 kcal.
2 Co + O2 = 2 CoO G0 = - 75 kcal.
2CoO + 2 H2 = 2Co+ 2 H2O G0 = - 17 kcal.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Nomographic scale on Ellingham diagram
The diagram already described can be made more useful
by superimposing grids or nomographic scales around
them.
1. PO2 grid : Consider the reaction involving the oxidation of pure solid metal to
pure solid oxide
2 M (s) + O2 (g) = 2 MO (s)
G0 = - R T ln (1 / PO2) = R T ln PO2
Hence knowing G0 and T the corresponding equilibrium PO2 can be calculated
for any such reaction.
When G0 = 0, PO2 = 1 atm and the equilibrium PO2 values radiate from the point
‘O’ on the G0 axis. For the fixed values of PO2 the G0 is plotted as a function
of T
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
2) CO / CO2 grid: For the reaction
2 CO + O2 = 2 CO2
G0 = - 135000 + 41.57 T cal / mole
G0 = - R T ln K = - R T ln (P2CO2 / P2
CO . PO2)
RT ln PO2 = G0 - 2 R T ln (PCO / PCO2)
= - 135000 + 41.57 T - 2 R T ln (PCO / PCO2)
RT ln PO2 can be plotted as a function of temperature for fixed values of CO /
CO2. When T = 0 , RT ln PO2 = - 135000 cal / mole. The equilibrium (PCO /
PCO2) values radiate from the point ‘C’ on the G0 axis.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
3. H2 / H2O grid: For the reaction
2 H2 + O2 = 2 H2O
G0 = - 118000 + 26.57 T cal / mole
RT ln PO2 = G0 - 2 R T ln (PH2 / PH2O)
= - - 118000 + 26.57 - 2 R T ln (PH2 / PH2O)
RT ln PO2 can be plotted as a function of temperature for fixed values of H2 / H2O.
When T = 0 , RT ln PO2 = - 118000 cal / mole. The equilibrium (PH2 / PH2O) values
radiate from the point ‘H’ on the G0 axis.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Use of Nomographic Scale
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Disadvantages of Ellingham diagram
1) The diagram is applicable only for the substances present in their standard states. But practically activity may not be unity.
2) Compound whose formation lines are represented in the diagram are assumed to be stoichiometric only which often is not true.
3) The information regarding rate of the reaction can not be obtained.
4) The diagram do not show the condition under which the reaction tend to occur.
5) Where oxide formation lines in the diagram are close together accurate measurement and subsequent calculation is difficult.
6) The possibility of formation of intermetallic compound between reactants and products is ignored
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Non-standard state condition
If any one of the constituents of the reaction is not present in its standard state is called non-standard state condition.
The activity of that substance will be less than one. It can be best illustrated by the reduction MgO by silicon.
2 MgO(s) + Si(s) = 2 Mg(g) + SiO2(s)
G01473 = + 272 kJ
So there appears to be very little chance of using silicon as a reducing agent to produce magnesium from magnesia.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
From the van’t Hoff isotherm, the actual free energy change of the this
reaction is given by
If pMg and aSiO2 can be lowered sufficiently, G can be made negative even
though G0 is positive.
PMg is lowered by working at a pressure of 10 – 4 atm.
aSiO2 is lowered by adding sufficient CaO to form orthosilicate (2 CaO. SiO2). A
basic slag would give aSiO2 < 0.001.
Si2MgO
SiO2Mg0
a.a
a.plnTRGG 2
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Third law of Thermodynamics
Enunciation of third lawThe change in entropy with constant pressure and constant
volume is given by following equations
Integration of these two leads to the following
dTT
CdSand
dTT
CdS
V
P
V,0
T
0
V
P,0
T
0
P
SdTT
CS
SdTT
CS
Where S0, P, and S0,V are the integration constant.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
These equations also state that for calculation of absolute value of entropy of
any substance, these constants should be known.
In order to evaluate the constant, Nernst analyzed the low temperature data
on free energy and enthalpy of chemical reactions.
The analysis lead to the formulation of first, the Nernst heat theorem and later
the third law of thermodynamics.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Nernst heat theorem
• Nernst collected enthalpy of
formation of various reactions
by calorimetric method and
determined free energies as a
function of temperature by
emf method.
• These free energies and
enthalpy when plotted against
temperature yielded curves as
shown.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
It is seen from the figure that at low temperature that both
enthalpy and free energy of a reaction approaches each
other asymptotically with the asymptote laying parallel to
temperature axis. This lead Nernst to arrive at the following
conclusions:
As
Therefore, one can write
0
T
HLimii
0T
GLimi
0T
0T
S
T
G
P
0SLim0T
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Statement of Nernst heat theorem
It states that both entropy change and temperature
coefficients of enthalpy and free energy changes for the
reaction tends to zero as temperature approaches 0 0K.
Entropy of a pure substance
A pure substance is a pure element or pure compound.
Consider the formation of compound AB from elements A
and B i.e
A(s) + B(s) = AB(s)
For which
At absolute zero temperature S0 = 0 it is possible provided
either i)
Or ii)
0B
0A
0AB
0 S-S-S=SΔ
)( 0B
0A
0AB S+S=S
zerolyindividualallareSandSS 0B
0A
0AB ,
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
From a probability point of view, the first alternative is very unlikely as a general feature. It may be true, by chance, in a few cases.
Hence alternative (ii) is accepted as of general validity and it may be concluded that “entropies of pure solids at T = 0 are zero”
Therefore, the absolute value of entropy of a pure substance at any temperature can be determined by taking T = 0, S = 0 as the lower limit.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Various statements of third law
“All reactions involving substances in the condensed state
S is zero at absolute zero”
“The entropy of any homogeneous crystalline substance
which is in complete internal equilibrium may be taken as
zero at 0 0K”
“The entropy of any phase whose quantum states and
atomic arrangement corresponds to a unique lowest energy
state at the absolute zero is zero”
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Experimental verification of the third law
Third law is not so obvious as first and second law; hence it is required to
describe some of the evidence by which it was established.
The so called direct verification of the third law involves the application of the
second law, namely the principle that the entropy of the system is a function of
state and the net entropy change experienced by a system undergoing a cycle is
therefore zero.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
• Let us consider the cycle shown in the figure.
• The starting point is a system composed of reactants at the absolute zero.
• The first step of the cycle consists in heating the reactants to temperature T.
• The reaction is allowed to completion at T in the second step.
• The products are cooled to absolute zero in the third step.
• The fourth step which imaginary, the reverse reaction is allowed to proceed
00
% reacted
I
II
III
IV
T
100
0K
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
We have from the second law
SI + SII + SIII + SIV = 0
Or SIV = - (SI + SII + SIII )
Since the entropy change in the first three steps measured experimentally, the
entropy change at absolute zero can be determined.
Let us consider the following transformation reaction
)white(Sn)gray(Sn C190
EU04.11Tlnd)white(CS
EU85.1292
541
T
QS
EU11.9Tlnd)gray(CS
0
292
PIII
r
rII
292
0
PI
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The entropy change at absolute zero ‘S0’ is
S0 = SIV = - (SI + SII + SIII )
= - (9.11 + 1.85 – 11.04) = 0.08 EU
This is probably smaller than the experimental error and thus it has been
shown that in this case S0 is zero within the experimental error.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Consequences of the third law
Third law leads to following important consequences
1. CP, CV and will be zero at absolute zero.
0C0Tat,finiteisT
Sas
T
STCwritecanwe
CT
HandS
T
GknowweAs
ST
ST
T
H
T
G
STHG
P
P
P
P
P
PP
PPP
Prove that CV is zero at absolute zero
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
00T
V
0P
Sso0S,0Tas
P
S
T
V
equations'wellmaxtoaccording
T
V
V
1
P
0T
TP
P
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
2. Unattainability of absolute zero temperature
let us consider a carnot cycle in which heat is absorbed at temperature T (T > 00K)
delivered to cold reservoir maintained at 00K. Entropy change during such a cycle
will be equal to zero i. e.
SI + SIII = 0
where SI is the entropy change during isothermal expansion and SIII is the entropy
change during isothermal compression.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
According to third law the entropy change during isothermal
volume change (SIII ) at 00K is zero. Hence SI will be zero.
But this not true as both heat exchange and temperature for
this is finite non-zero and positive.
Hence one cannot design a cycle in which the temperature of
the working substance descend to the absolute zero.
Thus it can be concluded that the temperature of 00K cannot
be attained.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
3. Calculation of entropy and free energy from calorimetric data.Entropy of a substance at constant volume is given by
From the knowledge of heat capacity which can be
determined by calorimetric techniques, one can
calculate the absolute value of entropy of a component
at any temperature T.
T
0
VV
V,0
T
0
V,0VV
TlndCS,therefore
0SlawthirdperAs
STlndCS
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Third law also helps in the calculation of free energy during the
chemical reaction from the calorimetric data obtained in the
form of enthalpy of a reaction as a function of temperature. We
know G = H – T S
Where H0 is the enthalpy of reaction at 00K.
Or
All the quantities present in the above expression can be
determined by calorimetric techniques. Thus, one can calculate
the change in free energy of any reaction at desired temperature
T
0
P0
T
0
P TdCHHandTdT
CS
T
0
PT
0
P0 TdT
CTTdCHG
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
ProblemCalculate the standard free energy change G0 and free energy change G for the following reaction at 1000 K. Ca (a = 0.9 ) = Ca (a = 0.5 )
SolutionG0 is calculated by measuring the free energy change of both reactants and products in their standard states. Since calcium is solid at 1000 K the standard state for both reactant and products is pure solid Calcium as indicated and thus G0 = 0
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
we know G = G0 + RT ln K
= G0 + RT ln (aproduct / aproduct)
= G0 + RT ln (aCa0.5 / aCa0.9) = 0 + 8.314 x 1273 ln (0.5 / 0.9) G = - 4.88 kJ / molProblemConsider the equilibrium reaction of pure solid CaSi at 1000 K with Ca and Si dissolved in a solvent. Suppose that activity of calcium is 0.5 in the solution and CaSi is pure. Find
i) aSi in equilibrium with Ca (a = 0.5) and pure CaSi
ii) PCa in equilibrium with dissolved Ca.
iii G1000K of the reaction: Ca (a = 0.5) + Si(a= 0.4) = CaSi (a = 0.8)
Given : Ca(s) + Si(s) = CaSi (s); G01000k = - 172 kJ / mol
P0Ca = 14.6 N / m2
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
i)For the reaction: Ca(s) + Si(s) = CaSi (s)
G0 = - RT ln K
K = exp (- G0 / RT)
= exp( 172000 / 8.314 x 1273 ) = 8.92 x 108
Or K = (aCaSi / aCa. aSi) = 8.92 x 108
since CaSi exists in pure states, aCaSi = 1 and
putting aCa = 0.5 we get aSi = 2.24 x 10-9
ii) It is given that aCa = 0.5, since aCa = PCa / P0Ca .
We can obtain PCa if the value of P0Ca is known.
Since our standard states for Ca is pure solid at
1000K, P0Ca is simply the vapour pressure of pure Ca at
1000 K. It is given that P0Ca = 14.6 N / m2, So PCa =
7.3 N / m2
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
iii) For the reaction
Ca (a = 0.5) + Si(a= 0.4) = CaSi (a = 0.8)
G = G0 + R T ln K
= - 172000 + 8.314 x 1273 ln (0.8 / 0.5 x 0.4)
= - 160 kJ / mol
Problem
Solid TiO2 is converted into gaseous TiCl4 by treatment with
chlorine in presence of carbon. Calculate the thermodynamic
utilization of chlorine gas at 10000C. Assume the oxidation of
carbon into CO and total pressure as 1 atm.
Given: TiO2(s) + 2 C (s) + 2 Cl2 (g) = TiCl4(g) + 2 CO (g)
G0 = - 318 kJ / mol at 10000C.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
G0 = - R T ln K
Or – 318 000 = - 8.316 x 1273 ln K
Or K = 1.11 x 1013 = (P2CO . PTiCl4) / (aTiO2 . a2
C . P2Cl2)
Assuming TiO2 and carbon as pure solid aTiO2 = 1= aC
From the stoichiometry of the reaction production of one mole
of TiCl4 leads to the production of 2 moles of CO i.e.
PCO = 2 PTiCl4
Making all the substitution
P3TiCl4 / P2
Cl2 = 0.275 x 1013 i.e PTiCl4 >> PCl2
So as a first approximation let us take PTiCl4= PT = 1atm. Then
we get PCl2 = 6 x 10 -7 atm. This is truly negligible. Therefore,
the thermodynamic utilization of Cl2 is 100%.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
A gas consisting of 60.2% H2 , and 39.8 % H2O at 1
atmospheric pressure is in equilibrium with pure γ – iron at
9100C. At the same temperature the gas composition in
equilibrium with an iron – nickel alloy (0.71 atom fraction of
iron) is 51.9% H2 and 48.1% H2O. Determine the activity of
iron in the alloy.
SolutionIn equilibrium with pure iron the reaction is:
Fe (s) + H2O(g) = FeO(g) + H2(g)
For which: K = (aFeO . PH2) / (aFe . PH2O)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Fe and FeO are in their standard state (aFe = 1 = aFeO)
K = (PH2 / PH2O)Fe, Pure
The corresponding reaction with the alloy is:
Fe (Fe - Ni) + H2O(g) = FeO(g) + H2(g)
The activity of pure FeO is again unity. But the activity of iron is
less than unity in the alloy. Thus :
K = (PH2 / aFe . PH2O)Fe – Ni
However the value of the equilibrium constant is unchanged on
the solution of the iron in the alloy, then:
(PH2 / PH2O)Fe, Pure = (PH2 / aFe . PH2O)Fe – Ni
Or aFe = (PH2 / aFe . PH2O)Fe – Ni / (PH2 / PH2O)Fe, Pure
Substituting the gas composition
aFe = (0.519 / 0.481) / (0.602 / 0.398) = 0.713
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
ProblemChromium plates are bright annealed at 7270C in wet
hydrogen atmosphere. The pressure of wet hydrogen is 1 atm.
i) Calculate the permissible water content in the hydrogen if there is to be no oxidation
at 7270C.
ii) Will annealed chromium plates be oxidized when cooled to 2270C in the furnace
atmosphere as calculated in (i).
Given: 2 Cr (s) + 3 H2O(g) = Cr2O3(s) + 3 H2
G0 = - 91050 + 22.80 T cals.
Solution
i)For the reaction: 2 Cr (s) + 3 H2O(g) = Cr2O3(s) + 3 H2
G01000K = - 68250 cal = - R T ln K
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
K = exp( - G01000K / 1.987 x 1000) = 8.26 x 10 14
K = (PH2 / PH2O)3 = 8.26 x 10 14
(PH2 / PH2O) = 9.384 x 10 4
As PH2 + PH2O = 1
After solving above equations we get:
PH2O = 1 / (1 + 9.384 x 10 4 ) = 1.065 x 10 – 5 atm.
Since total pressure is 1 atm, the volume percent of H2O
is1.065 x 10 – 3. This the maximum water content of H2 to
avoid any oxidation of chromium.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
ii) G0500K = - 79650 cal = - R T ln K
K = 6.57 x 10 34
The activity quotient is :
J = (PH2 / PH2O)3 = (1 / 1.065 x 10 – 5 )3 = 8.25 x 1014
J / K = 8.25 x 1014 / 6.57 x 10 34 < 1 so forward reaction
will oxidise the Cr
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
ProblemThe lowest temperature at which copper oxide (Cu2O)can
dissociate in a vacuum of 10 – 5 mm Hg.
Given : Cu2O(s)= 2 Cu(s) + ½ O2(g)
G0 = 40500 + 3.92 T log T – 29.5 T cal
Solution:
For the reaction: Cu2O(s)= 2 Cu(s) + ½ O2(g)
G = G0 + (RT /2) ln PO2 = G0 -2.288 T log PO2
PO2 = (0.2 x 10 -5 ) / 760 = 2.635 x 10 – 9 atm.
At equilibrium G = 0
G0 = -2.288 T log (2.635 x 10 – 9 ) = 19.6 T
Or 40500 + 3.92 TD log TD – 29.5 TD =19.6 TD
Or 3.92 TD log TD = 49.1 TD – 40500
This is best solved graphically and TD = 1085 K.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problemi. At what temps will carbon reduce a) SnO2(s) b) Cr2O3 (s) and c) SiO2(s)?
ii. Steam blown through hot coke gives rise to the fuel gas mixture called water gas (CO+H2): C + H2O = CO + H2. calculate the temperature the coke must be maintained for the reaction to be feasible?
iii. At what temperature the Ag2O just begin to decompose at one atmospheric pressure?
iv. In what temperature range can hydrogen be used to reduce SnO2 to Sn?
v. Deduce the standard free energy change for the reduction of Al2O3 by Mg at 1000 0C.
vi. Explain the reasons for the change in slope of the following lines:
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
2Mg + O2 = 2 MgO at 1100 0C.
2Pb + O2 = 2 PbO at 1470 0C
4 Li + O2 = 2 Li2O at 1300 0C
VII. Estimate the standard free energy change for the following reaction at 1200 K. a) reduction of copper (I) oxide with hydrogen and b) reduction copper oxide (I) with carbon.
VIII. At what temperature is the reaction: 4/3 Cr + O2 = 2/3 Cr2O3
at equilibrium when PO2 is 10-14 atm.
IX. Calculate the equilibrium oxygen pressure between Al2O3 and Al at 1000 K. Could a vacuum of 10-10 mm Hg prevent the oxidation of aluminum?
X. Suppose CaO is placed in a vacuume in which the partial pressure of oxygen is 10-5 mm Hg. Will CaO be reduced.
XI. Which is the suitable material for the steam pipe Ni or Al?
XII. What is the equilibrium CO / CO2 ratio for the following reaction: MnO + CO = Mn + CO2
XIII. At what temperature is the reaction PbO + H2 = Pb + H2O at equilibrium when the H2 / H2O ratio is 1/104?
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
ProblemCalculate the vapour pressure of liquid silver at its melting
point, 960 0C, making use of Trouton’s rule. Boiling point of
silver is 2210 0C.
Solution:
atm10459.246.10)273960(314.8
216021plnor
46.10TR
HplnSo
.46.10betooutcomesC
K2483Tat1PconditionthebyevaluatedbecanC
CTR
Hpln
equilibriavapourliquidtheforknowWe
.mol/Jk021.216248387HHence
K/J87T
H
rules'TroutontoAccording
5
V
V
V
b
V
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
ProblemThe equilibrium vapour pressure in atm. of a solid is given by the
expression: ln p = - 4.085 + 1.57x10 -4 ln T – 2.83x10- 3 T
Calculate the heat of sublimation of solid at 300 0C.
Solution:
getweionaboveequattheinthisputting
1083.2T
1057.1
Td
plndSo
T1083.2
Tln1057.1085.4pln
solidtheforthatgivenisit
Td
plndTRH
TR
H
Td
plnd
equilibriavapoursolidtheforknowWe
34
3
4
2S2
S
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
ProblemCalculate the melting point of ice under a pressure of 2 atm.
From the following data.
Density of water at 00C = 1.0 gm /cc
Density of ice at 00C = 0.9174 gm /cc
Latent heat of fusion of ice = 80 cal / gm
Assume that the above data to be independent of temperature
and pressure and also that the ice melts at 0 0C under 1 atm.
Pressure.
K573at.cal085.1846isationlimsubofHeat
.cal085.1846
1083.2107399.2573987.1
1083.2)273300(
1057.1RTH
372
34
2S
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
CPH
VVTln
equilibrialiquidsolidtheforknowWe
c.c62.199174.0
18
density
eightmolecularwC0iceofvolumeMolar
c.c181
18
density
eightmolecularwC0atwaterofvolumeMolar
M
sl
0
0
6095.5CorC1107.2273lnSo
atm1P,K273TconditiongiventheFrom
CP107.2
CP293.411880
62.1918Tln
atmc.c293.41cal1
atmc.ctocalfromconversionthefor
5
5
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
ProblemCalculate the rate of change of melting point of iron with
pressure with the help of the following:
M. P of iron = 1535 0C.
Density of solid iron at melting point = 7.86 gm/cc
Density of liquid iron at melting point = 7.55 gm/cc
Latent heat of fusion of iron = 3.3 kcal / mol
Atomic weight of iron = 56
K98.272Tor
6094.56095.5)2107.2(Tln
getweandequationabovethein.atm2Pand
6095.5CofvaluetheputcanWe..atm2
aticeofintpomeltingofncalculatiotheFor
5
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
ProblemRhombic sulphure transform to monoclinic sulphure at 368.5
K. with an enthalpy changes of 96 cal / mol. test the validity of
the third law of thermodynamics for this transition from the
following data:
i) CP of the rhombic sulphure at 15 K = 0.3 cal/deg/mol
ii) Area under the curve drawn between (Cp / T ) and T for rhombic sulphure between T = 15 K to 368.5 K is 8.71 cal/deg/mol.
.atm/reedeg1088.31808
293.413300
1246.74172.7T
H
VV
Pd
Td
equilibrialiquidsolistheforknowwe
atmc.c293.41cal1
mol/cc4172.755.7
56
density
.wtmolecularironliquidofvolumeMolar
mol/cc124.786.7
56
density
.wtmolecularironsolidofvolumeMolar
3m
m
sl
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
iii) For monoclinic sulphure: S368.5 – S0 = 9.07 cal /deg / mol
Assume that the entropy contribution of rhombic sulphure
below 15 K follows Debye’s equation.
Solution:
0
0
% reacted
I
II
III
IV
T
100
0K
SR → SM
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
5.368
15
P15
0P
5.368
15
P15
03
5.368
15
P15
0
2
5.368
15
P15
0
3
I
3P
5.368
15
P15
0
P
dTT
CC
3
1
dTT
CTa
3
1
dTT
CdTaT
dTT
CdT
T
TaS
TaCK15belowobyedislaws'Debye
dTT
CdT
T
C
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
5.368
15
P15
0
3
I
3P
5.368
15
P15
0
P
5.368
0
P05.368I
dTT
CdT
T
TaS
TaCK15belowobyedislaws'Debye
dTT
CdT
T
C
dTT
CSSS
stepfirstduringchangeEntropy
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
2605.05.368
96
T
HS
mole/deg/cal81.871.83.03
1
dTT
CC
3
1
dTT
CC
3
1
tr
trII
5.368
15
P15P
5.368
15
P15
0P
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
.verifiedislawthirdtheThus.zerois
zeroabsoluteatreactiontheofchangeentropythethatmeansit
moldeg//cal0005.0SSSS
lawondsectoAccording
07.9)SS(SSS
IIIIIIIV
05.3685.03680III
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Statistical Interpretation of Entropy
Statistical Thermodynamics: It deals with the interpretation and derivation of thermodynamic
properties based on the properties of the particles constituting
the system and their distribution in the system.
Elementary Statistical ConceptsMicrostates: In a gaseous system, each particle shall have
three position coordinates namely x, y, and z and three velocity
coordinates vx, vy, and vz. If at any instant one is able to specify
value of all these six coordinates for each of the particle
constituting the system, such a complete specification is said to
define the microstate of the system.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
• A set which defines all the possible microstates of a system is called an ENSEMBLE.
• A set of microstates having same characteristics is called a MACROSTATE.
let us consider an example a system consisting of three
distinguishable balls A, B and C contained in a box consisting
of two compartments I and II. Let us construct an ensemble.
Position of
the Particles
Microstates No.
1 2 3 4 5 6 7 8
Compartment I A, B,C A,B A,C B,C A B C -
Compartment II - C B A B,C A,C A,B A,B,C
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
These eight microstates can be grouped into the following four
macrostates:
I. All the three particles lie in compartment I – only one No.1 microstates
corresponds to this macrostates.
II. Two particles present in compartment I and only one in compartment II:
three microstates namely 2, 3, 4 correspond to this macrostates.
III. one particle present in compartment I and two in compartment II: three
microstates namely 5, 6, 7 correspond to this macrostates.
IV. All the three particles present in compartment II corresponds to Microstate
no.8
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
In actual practice a system consists of a large no. of particles and hence for every
macrostates there are large no. of microstates and no. of such macrostates are also
large. For the calculation microstates for a macrostates, one uses the technique of
permutation and combination.
Next, how to calculate the no. of microstates or maximum no. of possible
arrangement of particles in a system.
This can be illustrated by an example: let us consider first where we have, say,
eight object numbered 1 through 8 and four boxes.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Let us attempt to answer the question: in how many different ways can these
objects are placed in these boxes assuming that each box is large enough to contain
the all of the objects.
We may put object labeled 1 in any of the four boxes. Similarly, since each event is
independent we may do likewise for the object 2 etc.
The total no of ways of putting object 1 is four . The total no of ways of putting
object 2 is four, etc up through object 8.
The total no of ways of arranging all the balls is therefore 48.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
In the situation we will be dealing with (atoms in crystals) we can differentiate
atoms. We can in principle distinguish a gold atom from a copper atom. However
we can not distinguish within the given type one from other.
The situation discussed above therefore is not completely applicable to our case: we
shall not be interested in which object is in given box but, how many of the same
type of object are in each box.
To illustrate, let us calculate the number of ways for which there are two balls in
each box. Such a total is called the total no. of microstates in the given macrostate.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Thus we need to calculate how many ways we have of arranging two objects in box 1 two of the remaining six in box 2 and two of the remaining four in box 3 etc.
The first step is to calculate the no of ways of picking two objects out of eight for box 1. This is done by combination of N objects taken n at a time i.e 8C2 = 8! / (2! 6!).
We have six objects left. So no. of ways of putting two object out of remaining six in box 2 is 6C2 = 6! / (2! 4!).
Similarly for box 3 we have four objects left and 4C2 = 4! / (2! 4!) and so on..
Now we are ready to answer: the total no of ways putting two balls in each box is (W)
W = 8C2 x 6C2 x 4C2 x 2C2 = N! = 8! = 2520
n1!n2!n3!n4! (2!)4
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
What is the probability of achieving the macrostate in which we have two objects in each box. The answer is:
2520 / 48.
The probability of a system to be present in a particular macrostate is directly proportional to the no. of microstates which it has.
In other wards we can say a system will spend highest time fraction in the macrostate which has highest no of microstates (W).
It can be further simplified as it is this macrostate of highest probability which determines all the macroscopic properties of the system.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Entropy and Most Probable Macrostate
Properties of a system depends on the most probable macrostate i. e the
one having highest no. of microstates (W).
It has been proved that for the isolated system of constant volume have
maximum entropy at equilibrium.
So one can conclude that both W and entropy must be related to each
other.
Let such a relation be expressed as:
S = f(W)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
In order to arrive at the analytical form of this function, let us
consider a system divided into two sub-system A and B. The
no of microstates W in any macrostate will be equal to the
product of the microstates WA and WB of the sub-system.
Hence
W = Wa. WB
Total entropy of the system is
S = SA+ SB
As we can write
SA = f (WA)
SB = f (WB)
So f(WA. WB) = f(WA) + f(WB)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
B
B
BA
BAA
B
B
B
A
B
BA
BA
BA
B
W
Wf
WW
WWfW
yieldstionsimplificaonwhich
W
Wf
W
Wf
W
WW
WW
WWf
givesWtorespectwithantionDifferenti
)(
).(
).(.
)()().(
).(
).(
0
.
)(
).(
).(
).(
).(.
.
)(
).(
).().(
).(
).(.
...
2
2
2
2
2
2
A
B
BBA
B
BA
BAB
BA
BAA
BA
B
BA
BA
A
BA
BA
BAA
A
W
Wf
WWW
Wf
WW
WWfW
WW
WWfW
or
WW
Wf
WW
WWf
W
WW
WW
WWfW
givesWtrwationDifferenti
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
for a perfect crystal S = 0 for W = 1, we get C2 = 0
and C1 can be determined from the throttling of one mole of ideal gas. The gas expands by throttling process to double its volume. The throttling process is isenthalpic process.
21
21
2
2
2
2
ln
ln)(
..sec
0)(
)(
)(
)(.
0).(
).(
).(
).(.
CWCSThus
CWCWf
iseqdifforderondthetosolutiongeneralA
W
Wf
W
WfW
or
WW
WWf
WW
WWfWW
BA
BA
BA
BABA
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Let us calculate the W in the initial and final state
2ln
2/
ln
RSSTherefore
PPthrottlingofprocessthisIn
P
PRdP
P
RSSS
isprocessthisduringchangeentropyThe
if
fi
i
fP
P
if
f
i
WKSgetweCofvaluetheputting
KN
RC
NNN
NNNNNN
Cor
WWCSS
NNNNeiionapproximatsterlingtoAccordingN
NW
NN
NW
ifif
i
f
ln
22ln
2
22ln
2ln
2ln
;lnln
ln!ln.
1!
!
!)2/(!)2/(
!
1
11
1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Entropy of Mixing for a Binary Solution
Component A(N0 –n) atoms
Component Bn atoms
SolutionN0 atoms
(N0-n) A + n B = solution [(N0- n) A, n B]
Sm
Sm = SA,B – SA – SB
Or Sm = K (ln WA,B – ln WA – ln WB)
In 1 mole of solution there are N0 lattice sites
(Avogadro's no.)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
WA,B is the no.of ways of arranging (N0 – n) A atoms and n B atoms on N0 sites so
Since, in pure metal the atoms are indistinguishable so
WA = WB= 1. Thus
entropy of mixing
of ideal solution
)!(!
!
0
0, nNn
NW BA
BBAAm
BA
m
m
xxxxRSSo
N
nxand
N
nNxas
nN
N
N
nN
N
n
N
nKNS
ionapproximatsterlingApplying
nNn
NKS
lnln
lnln
)!(!
!ln
00
0
0
0
0
0
000
0
0
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
THERMODYNAMICS OF SOLUTIONS
A solution may be defined as a homogeneous phase composed of
different chemical substances, whose concentration may be varied
without the precipitation of a new phase.
It differs from a mixture by its homogeneity and from a compound
by being able to possess variable composition.
Solution may be gaseous, liquid or solid. It may be classified as
binary or ternary solution depending on whether it contains two or
three components.
A binary solution has two chemical substances (elements or
compounds), e.g. molten cadmium and zinc miscible in all
proportions.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Composition of solution
Composition of solution may be expressed in a number of ways. In metallurgy usually by weight, mol or atom percent, for example, if w1 and w2 are weights of the solvent and solute in the solution:
where WA and WB are weights of components A and B having atomic/molecular weights MA and MB, respectively.
100%21
2
ww
wsolutewt
100%/%
B
B
A
A
A
A
Mw
Mw
Mw
Amolatom
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Composition of a solution
If nA number of moles of A and nB number of moles of B form a solution, A-B. atom fractions of A and B are given as
and xA + xB = 1
BA
AA nn
nx
BA
BB nn
nx
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Raoult’s law
The law states that the relative lowering of the vapour pressure a solvent due to the addition of a solute is equal to the mole fraction of the solute in the solution.
Imagine A and B forming a solution, each one exerts its own vapour pressure, pA and pB in the solution, respectively.
Suppose p is the total pressure of the solution and xA and xB are
atom/mole fractions. If p0A and p0
B are the partial pressure of
pure A and pure B, respectively at the same temperature at which solution exists.
0
0
BA
AA xp
pp
0
0
AB
BB xp
pp
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Ideal Solution
B
A
A xp
p
01
BB
BAB
A
A xp
pandxx
p
p
001
)(. and
)(...0
B
0
BBBB
AAAAA
xporxpp
xporxppei
A solution which obeys Raoult’s law is called an ideal solution.
BBAA BBAA 2
1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Ideal Solutions
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Non-ideal or real solutions
Deviations from Raoult’s law occur when the attractive forces between the molecules of components A and B of the solution are stronger or weaker than those existing between A and A or B and B in their pure states.
For example, if there were a stronger attractive force between components A and B in solution than the mutual attraction between molecules of A-A and of B-B, there would be less tendency for these components to leave the solution is observed.
In this case the vapour pressure whould be less than that predicted by Raoult’s law . This is called negative deviation from Raoult’s law.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Non-ideal or real solutions
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Non-ideal or real solutions
Using the same argument if the attractive force between A and
B was less than that between A and A or B and B in their pure
states there would be a greater tendency for these components
to leave the solution as a gas, thereby increasing the vapour
pressure above the liquid. This is represented in Fig and is
called positive deviation from Raoult’s law.
Immiscible liquids exhibit positive deviation since the
attractive force between the components in the liquid is low.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Activity
We have already defined activity as
For ideal solutions we have:
For an ideal solution it will be seen that aA = xA, xA the mole fraction of A in the solution.
if the solution deviates from Raoult’s law we write
where γA is a fraction greater or less than unity for a positive or negative deviation, respectively is called Raoultian activity coefficient.
For a pure substance: xA = 1 and γA = 1, so that we have unit activity of the substance A which is said to be in its ‘standard state’
0A
AA p
pa
AAA xpp .0
AAA xa .
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Activity
If we return to our original argument that the vapour pressure
of a substance is a measure of its attraction to the solution in
which it exists and hence a measure of its availability for
reaction, perhaps with another phase, we can state that this
fundamental definition of activity, as that fraction of molar
concentration “available” for reaction, is universally
applicable.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Activity
Let us now consider the physical significance of both Raoult’s
ideal and non-ideal behaviour of binary systems.
1. Raoult’s ideal behaviour: If in a solution composed of A and B (atoms or
molecules) the attractive forces between A and B are the same as between
A and A or between B and B, then the activities of A and B in the
solution at all concentrations will be equal to their mole fractions and the
solution is said to be ideal. The system Bi-Sn can serve as an example of
such a solution at a particular temperature. In this case the net attractive
force between Bi and Sn in the solution can be represented by the
equation: BBAABBAABA
2
1)(
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Activity
2. Positive deviation: When the net attractive force between the substances
A and B is less than between A-A and B-B, then the solution of A and B
exhibits positive deviation from Raoult’s law. In this case the Raoultian
activity coefficient is always greater than unity except when approaching
the concentration of xA 1 Pb and Zn liquid solutions show such
behaviour at temperature above 1071K. The heat of solution in systems
showing positive deviation is endothermic. In general a solution, A-B
exhibiting positive deviation has
))}({(2
1BBAABA
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Activity
3. Negative deviations occur when the attractive force between the two
components A and B is higher than between A-A and B-B. For example
Mg-Bi system shows such a behaviour. Negative deviations generally
indicate a tendency for compound formation (Mg3Bi2). The heat of
solution for systems exhibiting negative deviatias is usually exothermic.
Occasionally both negative and positive deviations from Raoult’s law
occur in the same binary system. An outstanding example of this
behaviour is found in the Zn-Sb, Cd-Bi and Cd-Sb systems. In general in
case of negative deviation in the system, A-B we have
BBAABA 2
1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Activity
Fig.: Relationship between activity and mole fraction of substance
A in three solutions-one ideal, the second showing a positive
deviation and the third a negative deviation from ideal behavior.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
PARTIAL MOLAR QUANTITIES
In dealing with solutions, one of the first questions which naturally arises is
how to express molar quantities of a substance in solution when two liquids
are mixed, the total volume of the solution is, in general, not equal to the sum
of the individual volumes before mixing, this reflects the difference of the
interatomic forces in the pure substance and in the solution.
The problem presented is not solved directly but is avoided by the
introduction of partial molar quantities. Since the same general treatment is
applicable to any extensive thermodynamic quantity such as volume, energy,
entropy and free energy, we shall use the symbol Q to represent any one of
these.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
PARTIAL MOLAR QUANTITIES
Prime is used to indicate any arbitrary amount of solution rather than one mole, molar quantities are represented as unprimed.
Thus Q is the total quantity of solution, Q the molar. If n1, n2 , n3, ………are number of moles of components 1,2
3,……. respectively in the solution, we have
)1(..........321
nnn
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
PARTIAL MOLAR QUANTITIES
Let us imagine that to an arbitrary quantity of a solution an infinitesimal number of moles, dn1, of component 1 is added at constant temperature and pressure without changing the amount of other constituents. The corresponding increment in the property Q is dQ, the ratio
Is known as the as the partial molar quantity of components 1 and designated as
......3,2,,1 nnTPn
Q
)2(,.......3,2,,1
1 nnTPn
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
PARTIAL MOLAR QUANTITIES
may be represented equally well as the increment of Q when 1 mol of
the first component is added to a very large quantity of the solution for
example, if the volume increase accompanying the addition of 1 gm
mol/atom of Cu to a large quantity of a liquid alloy is observed to be 8.5
cc, the partial molar volume of copper in the alloy at the particular
composition, temperature and pressure is 8.5 cc. This is written as
We know Q’ =Q’(T,P, n1, n2, ……)
From the fundamentals of partial differentiation we have at constant
pressure and temperature
1Q
ccVCu 5.8
)3(..........2
3,121
3,21
dn
n
Qdn
n
QQd
nnnn
)4...(..............................' 332211 dnQdnQdnQdQ
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
PARTIAL MOLAR QUANTITIES
If we add to a large quantity of solution n1 moles of component 1, n2
moles of component 2 etc the increment in Q after mixing is given as
If we now mechanically remove a portion containing n1 + n2 + n3 ….moles,
the extensive quantity Q for the main body of solution is now decreased
by ( n1 + n2 + …) Q.
Since at the end of these processes the main body of the solution is the
same in composition and amount as it was initially, Q has the same value
finally as initially and the increment in Q accompanying the individual
addition is equal to the decrement accompanying their mass withdrawal
...........2211 QnQn
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
PARTIAL MOLAR QUANTITIES
Dividing by (n1 + n2+………) and noting that
on multiplying by (n1+ n2 + n3 ……..) we can write as
In a binary solution we can write equations 5 and 6 as
.......................)( 221121 QnQnQnn
ii x
nnn
n
321
)5(..........2211 QxQxQ
)6..(....................2211 QnQnQ
)6(
)5(
2211
2211
QnQnQ
QxQxQ
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
PARTIAL MOLAR QUANTITIES
On differentiating equation 6, we get
Subtracting (4) from (7) we get
Equation (9) is known as one of the forms of Gibbs – Duhem equation.
)7(22112211 dnQdnQQdnQdnQd
)8(02211 QdnQdn
)9(02211 QdxQdx
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Method of obtaining partial molar quantities from molar quantities
Differentiate equation (5)
Combining equation (9) and (10)
Multiplying (11) by (x1/dx2)and putting dx1 = -dx2,(x1 + x2 = 1)
Adding this to (5) we get
)10(22112211 dxQdxQQdxQdxdQ
)11(2211 dxQdxQdQ
2111212
111
21 . QxQxQx
dx
dxQx
dx
dQx
222212
1 QQxQxdx
dQxQ
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Method of obtaining partial molar quantities from molar quantities
Similarly,
)12(12
22
12 dx
dQxQ
dx
dQxQQ
)12(11
11 adx
dQxQQ
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Tangent intercept method
1Q
2Q
BD F
G
H
0
EA C
↑Q
1 x2 2
CHCFCHCF
x
CFCHxCF
dx
dQxQQ
xx
yyslope
x
CFCH
dx
dQ
22
222
12
12
22
1).1(
)1(
)(1
AGADAGAD
x
ADAGxAD
dx
dQxQQ
xx
yyslope
x
ADAG
dx
dQ
11
111
12
12
11
1).1(
)1(
)(1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Method of obtaining one partial molar quantities from another partialmolar quantities
Integration of this equation from x1 =1 to x1 = x1 will result
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Method of obtaining one partial molar quantities from another partialmolar quantities
If the input in the calculation is the functional relationship
between and ⎯ x2, then
If the input in the calculation is the functional relationship
between and ⎯ x1, then
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Obtaining Molar propertyfrom Partial molar properties
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
The enthalpies of mixing of Cd-Sn alloys at 5000C are given below:
xCd 0 0.1 0.3 0.5 0.7 0.9 1.0
HMCal/mole 0 298.2 652.4 800.0 620.5 251.5 0
Calculate the values of the partial molar enthalpies of mixing of cadmium and tin in a Cd-Sn alloy containing 60 at % cadmium.
Solution:
To find the value of partial molar enthalpies of mixing of cadmium,
Draw a tangent at xCd = 0.6. The values of and are obtained by intersection of the tangent with the axes at xCd = 1 and xSn = 1, respectively.
CdMM
SnMCd xvsHplotHeitinofthatandH ,..
MCdHM
SnH
./1360/320 molcalHandmolcalH SnCd
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
In the formation of liquid brass: (1-x) Cu(l) + x Zn(l) = Cu-Zn (l), the molar heat of formation is given by H = - 7100 x (1-x) cals, where x is the atom fraction of zinc. Derive expressions for partial molar heat of mixing of Cu and Zn in the liquid brass as a function of composition.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
H = - 7100 x (1-x)
=
=
Zn
ZnZn x
HxHH
.1
xxdx
d
x
H
x
H
Zn
17100
xxxdx
d2171007100 2
xxxxH Zn 217100117100
xxx 21710071001
2171001171002117100 xxxxxx
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
solution
= -7100 x (1-x) – x [ -7100 (1-2x)]
= -7100 x [1-x – (1-2x)]
= -7100 x [1-x -1 + 2x] = -7100 x (x) = -7100 x2 Ans.
x
HxH
x
HxH
x
HxH
x
HxHH
ZnZn
CuZn
CuCuCu
.1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
A 30% mole by methanol –water solution is to be prepared.
How many m3 of pure methanol (molar volume
=40.7x10-3m3/mol) and pure water (molar volume = 18.068x10-
6m3/mol) are to be mixed to prepare 2m3 of desired solution.
The partial molar volume of methanol and water in 30%
solution are 38.36x10-6 m3/mol and 17.765x10-6 m3/mol
respectively.
Problem
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Methanol =0.3 mole fraction
Water=0.7 mole fraction
V=0.3 x38.36x10-6+0.7x17.765x10-6
=24.025x10-6 m3/mol
For 2 m3 solution
mol10246.8310025.24
2 36
Solution
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Number of moles of methanol in 2m3solution=83.246x103x0.3= 24.97x103mol
Number of moles of water in 2m3solution
=83.246x103x07= 58.272x103mol
Volume of pure methanol to be taken
= 24.97x103 x 40.7x10-3 =1.0717 m3
Volume of pure water to be taken
= 58.272x103 x 18.068x10-6 =1.0529 m3
Solution
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Find weather the equation given below is thermodynamically consistent
)xx10(xxx150x100G 212121
111 x
G)x1(GG
112 x
GxGG
101x16x35x18G 12
13
11
150x8x18G 21
312
Problem
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
16x70x54dx
Gd1
21
1
1
12
11
2 x16x54dx
Gd
0)1(1
21
1
11
dx
Gdx
dx
Gdx
0)x16x54)(x1()16x70x54(x 12
1112
11
It satisfies the GD equation, the above equation is consistent.
G D equation
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The Free Energy Change due to the Formation of a solution
The pure component i, occurring in a condensed state at the
temperature T exerts an equilibrium partial pressure, pi0 . When
occurring in a condensed solution at temperature T, it exerts a
low equilibrium pressure pi. Consider the following steps:
1. evaporation of 1 mole of pure condensed i to pure i at the
pressure pi0 at T.
2. change in the pressure of 1 mole of vapour i from pi0 to pi at the
temperature T.
3. Condensation of 1 mole of vapour i from the pressure pi into the
condensed solution at T.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The Free Energy Change due to the Formation of a solution
The difference in molar free energy between pure i and i in
the solution = Ga + Gb +Gc.
However as steps a and c are equilibrium processes, Ga and
Gc are both zero.
The overall free energy change for the three step process thus
equals Gb which can be written as
Gb = Gi(in sol)-Gi (pure) = RT ln pi – RTln pi0
= RT ln (pi /pi0) = RT ln ai
but Gi (in solution) is simply the partial molar free energy of i
in the solution, (pure) is the molar free energy of
pure i, Gi0 .
iG and iG
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The Free Energy Change due to the Formation of a solution
The difference between the two is the free energy change accompanying the dissolution of 1 mole of i in the solution.
The quantity is designated as the partial molar free energy
of the solution of i. Hence , If at constant T and P, nA number of mole of A and nB of moles
of B are mixed to form a binary solution, free energy before mixing = nAGA
0 + nBGB0
free energy after mixing = The free energy change due to mixing, GM, referred to the
integral free energy of mixing, is the difference between the two quantities, i.e.
MiGΔ
iiiM
i aRTGGG ln0
BBAA GnGn
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
The Free Energy Change due to the Formation of a solution
= RT (nA ln aA + nB ln aB)
In terms of one mole of solution,
i. e. integral molar free energy
for an ideal solution ai = xi, GM,id = RT (xAln xA + xB ln xB)
0000BBBAAABBAABBAA
M GGnGGnGnGnGnGnG
MBB
MAA GΔn+GΔn=
BBAAM axaxRTG lnln
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of RaoultianIdeal solution
Change in volume accompanying the formation of ideal solution.
GM,id = RT (xAln xA + xB ln xB)
We can write,
A
idMA xRTG ln
,
B
idMB xRTG ln
,
i
CompT
iV
P
G
,
0
,
0
i
CompT
i VP
G
0
,
0)(ii
CompT
iiVV
P
GG
M
i
CompT
Mi
VΔ=P∂
GΔ∂
,
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of RaoultianIdeal solution
As xi is not a function of pressure, then
Integral volume of the solution (mixing)
i
idMi xRTGsolutionidealanfor ln,
,
0,
idM
iV
0,,, idM
BB
idMAA
idM VxVxV 0=VΔor idM ,
00BBAABBAA
M VnVnVnVnV
( ) )( 0BBB
0AAA V-Vn+V-Vn=
0=VΔn+VΔn= MBB
MAA
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of RaoultianIdeal solution
The heat of formation of an ideal solution
For a component in the solution, G - H equation
For the pure component
2, -
/
T
H
T
TG i
compPi
2
0
,
0
T
HH
T
T
GG
ii
CompP
ii
( )2
oi
compP
0i
T
H-=
T∂
TG∂,
/
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of RaoultianIdeal solution
is the partial Molar heat of solution (mixing) of i In an ideal solution,
Integral molar heat of mixing,
2
,
T
H
T
TG
Mi
CompP
Mi
MiH
i
idMi xRTG ln
,
2
,
,
ln
T
H
T
xRidM
i
CompP
i
0,,, idM
BB
idMAA
idM HxHxH
0,
idMiHor
0, idMH
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of RaoultianIdeal solution
The entropy of formation of an ideal solution
For the formation of a solution
For an ideal solution
GM,id = RT (xAln xA + xB ln xB)
ST
G
ompP
,
M
ompP
M
ST
G
,
BBAA
ompP
idMidM xxxxR
T
GS lnln
,
,,
iidM
i xRS ln,
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of Real Solution
Real solutions do not obey Raoult’s law (ai = γi. xi)
Integral molar free energy of mixing :
Partial molar free energy of mixing:
BBAAM axaxRTG lnln
idMXS
BBAABBAAM
GG
xxxxRTxxRTG,
lnlnlnln
idM
iXS
i
ii
iM
i
GG
xRTRT
aRTG
,
lnln
ln
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of Real SolutionChange in volume accompanying the formation of Real solution.
Partial molar volume of mixing
Integral Molar Volume of mixing
P
aRT
P
GV
i
CompT
M
iM
i
ln
,
P
ax
P
axRTV B
BA
AM lnln
MBB
MAA
M VxVxV
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of Real Solution
Change in Entropy accompanying the formation of real solution.
Partial molar entropy of mixing
Integral molar entropy of mixing
i
i
CompP
M
iM
i
aRP
aRT
T
GS
lnln
,
BBAAB
BA
AM axaxR
T
ax
T
axRTS lnln
lnln
MBB
MAA
M SxSxS
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of Real Solution
Change in enthalpy accompanying the formation of real solution.
Partial molar enthalpy of mixing
2
,
T
H
T
TG
Mi
CompP
Mi
CompP
iMi
CompP
i
Mi
T
aRTH
T
aR
T
H
,
2
,
2
ln
ln
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of Real Solution
Integral molar enthalpy of mixing
MBB
MAA
M HxHxH
compP
BB
AA
M
T
ax
T
axRTH ,
2 lnln
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Regular Solution
In the case of non ideal solutions it is still possible to assume random
mixing in certain cases but the enthalpy of mixing will no longer be zero
because there will be heat changes due to changes in binding energy.
This assumption of random mixing can only be made where there is a
small deviation from ideal behaviour, so that the enthalpy of mixing is
quite small. Solutions of this type are called regular solution. For regular
solutions the entropy of mixing is the same as for ideal solution, so that
where and are the actively coefficient of A and B, respectively.
MMM STGH BBAA axaxRT lnln BBAA xxxxRT lnln
BBAA xxRT lnln
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Regular Solution
In 1895 Margules suggested that activity coefficients, A and B of a binary solution at any given temperature can be represented by power series equations as follows:
By the application of the Gibbs – Duhem equation xA d lnγA = - xB d ln γB, it can be shown, for the system to hold over the entire range of composition, .
By similar comparison of co-efficients of the power series, Margules further demonstrated that the variation of the γ’s can be represented by the quadratic terms only, when
..........3
1
2
1ln
..........3
1
2
1ln
33
221
33
221
AAAB
BBBA
xxx
xxx
011
22
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Regular Solution
For regular solutions Hildebrand in 1929 established that
α- function is defined as:
From comparison of the above expressions we get
αA = αB = α and
α is independent of compositionwhich indicates that Tl – Sn system follows regular solution model.
2BA
2AB xα′=γRTandxα′=γRT lnln
22 lnln ABBBAA xandx
RT
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of Regular Solution
Thermodynamic properties of solutions may be divided into two parts: ideal and excess.
Properties of regular solutions can be discussed by the excess parts.
QReal = Qid + QXS The change in property during mixing:
Hence we can write for integral molar free energy of mixing as
XSidMRM Q+QΔ=QΔ ,,
XSidMRM G+GΔ=GΔ ,,
idMRMXS GΔ-GΔ=GΔ ,,
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of Regular Solution
it may be shown as
ΔGxs = ΔHxs - TΔSxs = (ΔHM,R – ΔHM,id ) – T (ΔSM,R – ΔSM,id )
= ΔHM,R (as ΔHM,id =0 and ΔSM,R = ΔSM,id )
ΔGxs = ΔHM,R
XS
p
XS
ST
G
idMRMXS GΔ-GΔ=GΔ ,,
BBAA axaxRT lnln BBAA xxxxRT lnln
RMBBAA HxxRT ,lnln
XSBB
XSAA GxGx
MBB
XSB
MAA
XSA HRTGandHRTG lnln
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of Regular Solution
Hence for regular solution
This means GXS for a regular solution is independent of
temperature
BBAAXS xxRTG lnln
22ABBA xxxxRT
BABA xxxxRT
BAxxαRT=
BAMXS xxRTHG
RTwherexx BA
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of Regular Solution
This can also be shown as
As SXS for a regular solution is zero, then GXS and ΔHM are independent of temperature. Similarly, is also independent of temperature.
This equation is of considerable importance and of practical use in connecting activity data at one temperature to activity data at another temperature. Hence for a regular solution we have.
XS
p
XS
ST
G
XSG
22211 lnln BAA
XSA xTRTTRTG
2
1
1
2
ln
ln
T
T
eTtemperaturat
eTtemperaturat
A
A
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of Regular Solution
2211
2
1
1
2
2
1
12
22
2
2
1
1
2
)(
)(
Tat )1(
ln
Tat )1(
ln
)1(;ln
ln
TTorT
T
T
T
T
T
x
x
getwexbydividingT
T
Tetemperaturat
Tetemperaturat
A
A
A
A
AA
A
For strict adherence to this model αT should be independent of temperature. Thus Tl – Sn is not strictly regular in behaviour
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of Regular Solution
Change in volume accompanying the formation of regular solution.
Partial molar volume of mixing
Integral molar volume of mixing
ompT
i
CompT
M
iM
i
P
aRT
P
GV
,
,
ln
MBB
MAA
M VxVxV
ompT
BB
AA
M
P
ax
P
axRTV
,
lnln
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of Regular Solution
Change in Entropy accompanying the formation of real solution.
Partial molar entropy of mixing
Integral molar entropy of mixing
i
M
i xRS ln
BBAAM xxxxRS lnln
MBB
MAA
M SxSxS
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Properties of Regular Solution
Change in enthalpy accompanying the formation of real solution.
Partial molar enthalpy of mixing
Integral molar enthalpy of mixing
XSii
Mi GRTH ln
MBB
MAA
M HxHxH
BBAAM xxRTH lnln
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Regular solution behaviour:
1. linear variation of ln γA vs xB2 (slope = ) at a given
temperature indicates regular solution behaviour of the system: A-B.
2. However for strict adherence to the model, T should be independent of T but not so in many cases. In general we find ln varying linearly with x but iT decreases with temperature.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Thermodynamic properties of Solutions
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Thermodynamic properties of Solutions
Kk QxQ .
idMMXS QQQ ,
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
GIBBS – DUHEM INTEGRATION
Thermodynamic equations for calculation of excess free energy and
integral molar free energy of a solution need activity coefficient and
activity of all the components of the solution. However,
experimental techniques viz. chemical equilibria, vapour pressure
and electrochemical can measure activity of only one component.
In order to get activity of the second component in a binary solution
we must couple activity and atom/mole fractions of both the
components with the aid of Gibbs-Duhem equation as follows:
Q is any extensive property. ,0iQdXi
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
GIBBS – DUHEM INTEGRATION –METHOD -I
Since activity of a component is related to the partial molar free energy, we can write Gibbs-Duhem equation as under:
)1(0 MB
MAA GxdGdx
)2(0lnln BBAA adxadx
)(lnln 3adx
x=ador B
A
BA
AA
A
AA
xx
x
BA
BxxA ad
x
xa
1
ln|ln
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
GIBBS – DUHEM INTEGRATION METHOD -I
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
GIBBS – DUHEM INTEGRATION – METHOD -II
(4)
Eq.(4) Eq.(2)
(5)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
GIBBS – DUHEM INTEGRATION
A
BBAB x
xbutaxas 0ln,1,1
0ln,0,0 A
BBBBB x
xforfinitealsoisandfiniteisaxas
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
GIBBS – DUHEM INTEGRATION – METHOD -III
As a further aid to the integration of the Gibbs Duhem equation, the is introduced as
-function is always finite because For the components of a binary solution
B is known as a function of composition:
On differentiation we get
21
ln
i
ii
x
11 ii xas
,ln
2B
AA x
2
ln
A
BB x
and
2ln ABB x
function
,..2ln 2BAAABB dxdxxd
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
GIBBS – DUHEM INTEGRATION – METHOD -III
substituting this into
On integration
BA
BA d
x
xd lnln
BABABBBAA
BAAB
A
BA dxxdxxdx
x
xdxx
x
xd .2..2ln 2
xAx
x
xxat
xat
ABABABBA
A
A
AAB
AB
dxxdxx1 1
)(2ln
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
GIBBS – DUHEM INTEGRATION – METHOD -III
(A)
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
GIBBS – DUHEM INTEGRATION – METHOD -III
Thus ln at xA=xA is obtained as – xBxAB minus the area under the plot of
B vs xA from xA=xA to xA = 1. Since B is everywhere finite, this
integration does not involve a tail to infinity.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
The activity coefficient of zinc in liquid Cd-Zn alloys at 4350C
have been expressed as ln Zn = 0.87 x2 Cd – 0.30 x3 Cd
(a) Calculate the activity of cadmium in a 30 at % Cd at 4350C.
(b) Develop a corresponding equation for the activity coefficient of cadmium in the alloy system at this temperature.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
From Gibbs-Duhem equation we can write as
xZn d ln Zn +xCd d ln Cd = 0
on integration we get
Zn
xx
x Cd
ZnCd d
x
xCdCd
Cd
lnln1
CdCdCd
xx
x Cd
Cd dxxxx
xCdCd
Cd
2
1
90.0-74.1-1
CdCdCd
xx
x
Cd dxxxxCdCd
Cd
2
1
90.0-74.1)-1(
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
= -1.74(0.3-0.1) + 1.32 (0.09-1)-0.3 (0.027-1)
Cd
x
x
CdCd dxxxCd
Cd
3.0
1
290.064.2-74.1
301
3Cd
2CdCd x30+x321-x741-= .]...[
[ ] 30
13Cd
2CdCd x30-x321+x741-=
....
3090=2920+2011-2181=γCd ....ln
362.1 Cd..... Ans40860=30×3621=xγ=a CdCdCd
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Zn
xx
x Cd
ZnCd d
x
xb
CdCd
Cd
lnln)(
Solution
CdCdCd
xx
x Cd
Cd dxxxx
xCdCd
Cd
2
1
90.0-74.1-1
CdCdCd
xx
x
Cd dxxxxCdCd
Cd
2
1
90.0-74.1)-1(
CdCdCdCdCdCd dxxxdxxx )9.064.2-74.1()9.0-74.1(-1 2
CdCd
xx
x
Cd dxxxCdCd
Cd
)90.0-64.274.1-( 2
1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
Cdx=Cd
1=Cd
x
x3Cd
2CdCD x30-x321+x741- ]...[
formdesiredthegettoxxput
xxx
xxx
ZnCd
CdCdCd
CdCdCd
1
3.032.174.172.0
)3.032.174.1(3.032.174.132
32
3Zn
2ZnZn x-130-x-1321+x-1741-720= )(.)(.)(..
)331(3.032.164.232.174.174.172.0 322ZnZnZnZnZnZn xxxxxx
720+741-321+30-741+642-90x+321+90-x+x30= Zn2Zn
3Zn ....)...()..(.
... Ansx420+x30= 2Zn
3Zn
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
Alternatively, we can also solve as
)-)(9.0-74.1()9.0-74.1(ln 2 CdCdZnCdCdCdCd
ZnCd dxxxdxxx
x
x
ZnZn
Zn
CdCd
Cd
xx
x
ZnZnZnZnZn
xx
x
ZnCdZn dxxxdxxdxxx01
)-1(9.0-74.1).9.0-74.1(
ZnZnZnZnZnZn dxxdxxdxx 29.09.074.1
320
3
022 3.042.0]
3.90.0[]42.0[9.084.0 ZnZn
xZnxZnZnZnZnZn xx
xxdxxdxx ZnZn
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
At 746K the activity coefficient of lead in liquid Pb-Bi alloy is expressed as Making use of the Gibbs-Duhem equation develop the corresponding equation for the activity coefficient of bismuth in the alloy at 746K. (a) Calculate the activity of lead at 746K and 1000K in the Pb-Bi alloy containing 50at % lead (b) Calculate the integral molar heat of mixing /excess free energy of the alloy containing 40at% Pb at 746 K (c) What is the integral molar free energy of mixing of the above alloy in (b) at 1000K (d) Calculate the difference in change in free energy when 1 gatom of lead dissolves in a very large amount of the above alloy at 746 and 1000K
.)(.ln 2PbPb x-1740-=γ
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
(a) For Pb-Bi system the Gibb-Duhem equation may be written as:
on integration we get
0=γdx+γdx PbPbBiBi lnln
Pb
xxat
xat Bi
PbBi d
x
xBiBiPb
BiPb
ln-lnln
1ln
PbPbPb
PbPb
dxxd
x
)1(48.1ln
)1(74.0ln 2
PbPbPb
Pb dxxx
x-148.1
-1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
= - 0.74(1-xBi)2
So the solution is a regular solution
)1(
0
2
1 248.1-48.1
BiPb
Pb
Bi
Bi
xx
x
Pb
x
x
PbPb
xdxx
.)746(415.083.05.0,83.0)746( AnsKaK PbPb
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
for regular solution
22 Pb
Bi
Bi
PbBiPb x
γ=
x
γ=α=αb
lnln)(
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
at 746 K
= RT xAxB
= - 1.987 746 0.74 0.4 0.6
= -263.3 cal/mole. Ans.
2Bi
2BiPb x740-=xα=γ .ln
2Pb
2PbBi x740-=xα=γ .ln
BibIPbPbMXS xxRTHG lnln
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
(C) for regular solution 1T1 = 2T2
1 = -0.74 (746 = T1)
At 1000K
550-=1000
746×740-=
T
Tα=K1000α
2
112 .
.)(
4350=50x87150=a
87150=γ==>x550 -=xα=γ
Pb
Pb2Bi
2BiPb
...
..ln
4350=50x87150=a
87150=γ==>x550-=xα=γ
Bi
Bi2Pb
2PbBi
...
..ln
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
=1.987 1000 (0.5 ln 0.435 + 0.5 ln 0.435)
= 1.987 1000 ln 0.435 = 1987 (-0.8324)
= -1654 cal/m. Ans.
(d) difference in free energy =
= RT ln aPb (1000K) –RT ln aPb(746K)
= 1.987 (1000 ln 0.435 – 746 ln 0.415)
= 1.987 [1000(-0.8325) – 746 (-0.8795)]
= 1.987 (-176.4) = -350.5 cal/mole. Ans.
BiBiPbPbM axaxRTG lnln
)()( K746GΔ-K1000GΔMPb
MPb
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
Al-Zn alloys exhibit the following relationship at 4770C:
RT ln γZn = 1750 (1-xZn)2 where R and T an expressed in cal/deg.gatom
and K, respectively.
i) Develop the corresponding expression ln Al
(ii) Calculate the heat of mixing of the alloy containing 40 at % Zn at 4770C. What would be excess molar free energy of the alloy at this temperature?
(iii) Calculate the integral molar free energy of the above alloy at 5070C.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
appropriate form of the Gibbs-Duhem equation.
dx )x-(1RT
3500- ln,)1(
1750ln ZnZnZn
2 dxRT ZnZn
0lnln ZnZnAlAl dxdx
ZnAl
ZnAl d
x
x ln.ln
ZnZn
xx
x Zn
Zn dxxRTx
xAlAl
Al
)1(3500
11
AlAl
Al
AlxZnx
Znx
xx
x
ZnZnAl
x
RTdxx
RT 1
2 )1(
02.
35003500
2)1(1750
AlxRT
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
(ii) The above form of equation show that is independent of composition. Thus Al-Zn system follows regular solution model, hence
xZn = 0.4, xAl = 0.6
= 1750 0.24 = 420 cal/gatom Ans.
AlZn
Al
Zn
ZnZn RTx
1750
1
ln2
ZnAlXSM xxRTGH
4.06.01750
. RT
RT
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
(iii) ).ln( ZnZnAlAlM axaxRTG
.780273507,750273477
solutions,regular for
21
2211
KTKT
TT
174.1987.13
7
750
17501750
)1(
ln2750
RRTxZn
Zn
129.1987.178
175
780
750
750
1750
780
750750780
xx
R
27801
ln
Zn
Zn
x
xZn=0.4
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
γAl= 1.19798
aAl = 1.19798 0.6 = 0.71879
40644.036.0129.11ln 2780)780( ZnZn x
50146.1)780( Zn
6005.04.050146.1. ZnZnZn xa
2780 )1(
ln
Al
Al
x
18664.016.0129.1)6.01(ln 2780 Al
)6008.0ln4.0718979.0ln6.0( RTG M
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
=1.987 780 (0.6 (-0.3301856) + 0.4 (0.5075)
= 1.987 780 [-0.1981 – 0.2038] = 1.987 780 (-0.4019)
= -622.8 cal.mole. Ans.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
In liquid Fe-Ni solution at 1873 K the activity of nickel as a function of
composition is listed below:
xNi 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
aNi 0.067 0.137 0.208 0.287 0.376 0.492 0.620 0.776 0.89
(i) Calculate the activity of iron in an alloy containing 60 at% iron by Gibbs - Duhem integration.
(ii) Calculate in the above alloy at 1873 K.XS
iMi
MFe
MNi GGGG ,,,
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
Solution:
By third method of integration calculate Ni
xNi 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
aNi 0.067 0.137 0.208 0.287 0.376 0.492 0.620 0.776 0.89
γNi 0.67 0.685 0.69 0.72 0.75 0.82 0.986 0.96 0.99
(1-xNi)2 0.81 0.64 0.49 0.36 0.25 0.16 0.09 0.04 0.01
Ni -0.49 -0.59 -0.75 -0.92 -1.14 -1.24 -1.35 -1.07 -1.01
Making use of function we have
= - (-0.92 0.6 0.4 – 0.08) = 0.301 i.e. γNi = 1.35, aFe = 0.6 1.35 = 0.81
FeFe
Fe
xx
x
FeNiNiFeNiFe dxxx1
..ln
][ FeNiNiFeNi dxxx
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
./19438)287.0ln(1873314.8ln moleJaRTG Ni
MNi
./3281)81.0ln(1873314.8ln moleJaRTG Fei
MFei
./9744)3281(6.0)19438(4.0)( moleJGxGxG FeFeNiNiM
)35.1ln6.072.0ln4.0(1873314.8)lnln(, FeFeNiNiMid
MXS xxRTGGG
= 758 J/mole
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
At 1200K the activity coefficient of zinc in liquid brass is
expressed as
ln γZn = - 1.929 (1-xZn)2
(a) Calculate the integral molar heat of mixing and the excess free energy of brass containing 40 at % copper at 12000 K.
(b) What is the integral molar free energy of mixing of the above alloy at 1300 K. Check your answer.
(c) Calculate the difference in change of free energy when one g. atom of liquid Zinc dissolves in a large amount of liquid brass at 1200 and 1300 K.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
The variation of activity coefficient with composition shows that Zn =-1.929, is independent of composition. This indicates that brass behaves regularly at 1200K.
Zn=Cu
we know that for regular solution we have
Integral molar heat of mixing (or excess molar free energy) of brass,
T = 1200 K , xCu = 0.4, xZn = 0.6
929.1)1(
ln
)1(
ln22
CuCu
Cu
Zn
ZnZn xx
)lnln( BBAAXSM xxRTGH
)lnln( CuCuZnXSM xxZnRTGH
30864.04.0929.1929.1ln 22 CuZn x
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
= - 2384.4 (0.6 0.30864 + 0.4 0.6944)
= -2384.4 0.4629 = -1103.8 cal/mol.
or GXS= (RT) xCu xZn = (-4600) 0.24 = -1104 Ans.
4406.06.07344.07344.0 ZnZn aand
6944.06.0929.1929.1ln 22 xxZnCu
1997.04.04993.04993.0 CuCu aand
)ln4.0ln6.0(1200987.1 CuZnXSM GH
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
(b) for brass is needed at 1300 K, hence aCuand aZn should be first calculated at 1300K by
making use of properties of regular solution.
MBBAA
M GceaxaxRTG sin),lnln(
2211)1200(929.1 TTandKatCuZn
1300.1200. 13001200
781.113
12929.1
13
12.12001300
2849.0)4.0(781.1.ln 221300 CuZn x
45125.06.07520.07520.0 ZnZn aand
6412.06.0781.1.ln 221300 ZnCu x
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
= 1.987 1300 (-1.10034) = -2842.3 cal/mol.
21071.04.05267.0,5267.0 CuCu a
)21071.0ln4.045125.0ln6.0(1300987.1 MG
5.110324.0)781.1(1300987.1.)1300( ZnCuXS xxRTG
)(Re./1104)1300()1200( solutiongularmolcalGG XSXS
./5.1738
)4.0ln4.06.0ln6.0(1300987.1)lnln(,
molcal
xxxxRTG CuCuZnZnidM
./8.1103)5.1738(3.2842 molcalGGHG Mid
MMXS
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
Difference in change in free energy =
=RT ln aZn(1300) – RT ln aZn (1200)
=1.987 (1300 ln 0.4513 – 1200 ln 0.4406)
=1.987 (-1034.31 + 984.54) = -98.9 cal/mol. Ans.
)()( 1200G-1300GMZn
MZn
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Dilute Solution –Henry’s Law
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Henry’s Law
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Henry’s Law
The constant is equal to the slope of the curve at zero concentration of A, designated by coefficient of the solute A at infinite dilution).
Like Raoult’s law, Henry’s law is valid within a concentration range where the extent varies from one system to another, but it is valid only at low concentration.
AAAAAAA
AA
AA
AA
xaorxconstaeixp
k
p
phavewepbydividing
kxpei
xp
...
)1(..
000
activity(0A
AAA xa .0
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Henry’s Law
aA
xA
γ0A → Henry’s law constant
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Henry’s Law
In concentratrated solution the standard has been defined as unit atmospheric pressure and unit activity i.e. pure substance at any temperature.
In dilute solutions relative standard states other than pure substance being used. Henry’s law offers two such standard states, called alternative standard states.
(1) Infinitely dilute, atom/mole fraction standard state.
(2) Infinitely dilute, wt% (w/o or %) standard state.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solubility of Gases
It is important to note that the validity of Henry’s law depend upon the proper choice of solute species. For example, consider
(a) solution of nitrogen in water
(b) solution of nitrogen in liquid iron
In the first case nitrogen dissolves molecularly as N2
As solubility of N2 in water is low according to Henry’s law we have
)waterin dissolved(22 NgN 2
2
N
N
p
aK
22 NN xka 2
2
N
N
p
kxKor
222)lub( NNN pkp
k
Kilitysox
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solubility of Gases
Thus the solubility of nitrogen in water is proportional to the partial pressure of nitrogen gas in equilibrium with water.
Solubility can be expressed as mole fraction, cc per 100 g of water or any other unit.
(2) In the second case under consideration nitrogen dissolves atomically in solid or liquid metals:
)(2)(2 FeinNgN
22
2)(
2 )(
N
N
N
inFeN
p
kx
p
aK
22')lub()( NNinFeN pkp
k
Kilitysoxor
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solubility of Gases –Sievert’s law
Since all the common diatomic gases N2, O2, H2 etc. dissolve atomically in metals, the general expression for solubility is given as:
This is known as Sievert’s law and can be stated as – solubility of diatomic gases in metals is directly proportional to the square root of partial pressure of the gas in equilibrium with the metal.
2NpkS
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
At 8000C, 100g of silver dissolves 3.3 cm3 (STP) of oxygen at one
atmosphere pressure. How much oxygen does silver dissolve from air at
8000C?
Solution:
According to Sieverts law:
In air pO2= 0.21 atm,
2OpkS 3.313.3.. kkei
351.121.03.32
cmpkS O
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
The following solubility of oxygen in 100 g of silver at 10750C have been
measured.
a) Show whether these observations agree with Sievert’s law for the
solubility of oxygen in metals.
b) How much oxygen does 100g of silver absorb at 10750C from air?
c) What pressure of air corresponds to one atm of O2 with respect to the solubility of oxygen in silver at 10750C.
1203760488128)(2
mmHgpO
Oxygen dissolved cm3/100gAg 81.5 156.9 193.6 247.8
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
2
solubilitytan'
O
sp
ktconssSievert
2037.7128
5.81Sk
1025.7488
9.156sk
0226.7760
6.193sk
1444.71203
8.247Sk
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
Almost constant values of ks demonstrate that solubility of oxygen in
silver is proportional to the square root of partial pressure of oxygen in
equilibrium. Hence the observations agree with Sievert’s law. Average
value of ks = 7.1183.
(b) In air, pO2 = 0.21 atm. = 0.21 760
AggmpercckilitySo S 10093.896.1591183.776021.0lub
7629.4
)21.0,1(21.0
1)(
2
Oairair patmppc
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
At 15400C liquid iron dissolves 0.04% nitrogen in equilibrium with
nitrogen gas at one atmospheric gas pressure and 0.23% oxygen in
equilibrium with oxygen gas at one atmosphere gas pressure. At that
temperature nitrogen pentoxide gas was passed over liquid iron such
that equilibrium was attained with fully dissociated gas at a net pressure
of one atmosphere. What is the nitrogen and oxygen contents of the melt?
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
N2 gas : O2 gas
1 mol : (5/2) mols
2 mol : 5 mols.
2 vol : 5 vol.
ONgOgNgON 52)(2
5)()( 2252
atm7
5and atm
7
222 ON pp
04.0104.02
NNNNN KKpKS
pKOS NN )N ed(dissociat 52'
.%02138.07
204.0 Ans
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
23.0
123.0
O
O
OO
K
K
pKS
%19439.07523.0)N issociated( 52 OdSO
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Alternative Standard States
1. Infinitely dilute, atom fraction standard stateHenerian standard state is obtained from Henry’s law which, strictly being limiting law obeyed by the solute in the dilute solution is expressed by
If the solute obeys Henry’s law over a finite concentration range, then
tconslawsHenerytheisandstatedards
RoultiantrwAofactivitytheisawhere
xasx
a
A
A
AAA
A
tan'tan
..
0
0
0
AAA xa .0
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Alternative Standard States
Henerian standard state is obtained by the extending the Henry’s law line to xA = 1.
This state represents pure solute in the hypothetical, nonphysical state in which it would exist as a pure component if it obeyed Henry’s law over the entire composition range (i.e., as it does for a dilute solution)
Having defined the Henrian standard state, the activity of A in solution with respect to the Henrian standard state is given by:
activityheneriantheish
andstatedardsheneriantheisThis
1=xatγ=1=h
A
A0AA
tan
.
tcoefficienactivityheneriantheisfwhere
xfh
A
AAA .
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Alternative Standard States
In the range of composition in which the solute obeys Henry’s Law, fA =1 and solute exhibit the Henerian ideality
hA = xA
In the range of composition in which the solute obeys Henry’s Law, fA =1 and γA = γA
0
tconsxA
A
AA
AA
A
A
Affx
x
h
a
tan.
.
tconsx
AA
A
Ah
a
tan
0
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Alternative Standard States
The free energy change accompanying the transfer of one mole of solute A from pure substance standard state (Raoultian standard state) to Henerian standard state, that is:
A (in the Raoultian standard state) → A( in the Henerian standard state)
is given by
The partial molar free energy of the solute at constant concentration is independent of standard state. The value of ΔGA
0 remains unchanged if is added and is subtracted from the right hand side of the above equation
000
)()()(
RAHAGGHRGA
)(RAG )(HAG
0)(
0)(
)()(000
)()(
)()()(
HARA
RAHA
GGGG
GGGGHRG
HARA
HARAA
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Alternative Standard States
AAA aTRGGBut ln0
0
tan
0
ln
ln)(,
A
tconsxA
AA
RT
h
aRTHRGHence
A
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Alternative Standard States
2. Infinitely dilute, wt% standard state.
The use of this standard state eliminates the necessity of converting weight
percentages, obtained via chemical analysis, to mole fractions for the
purpose of thermodynamic calculations. This standard state is particularly
convenient to use in metallurgical systems containing dilute solutes. This
standard state can formally be defined as:
0.%.%
0.%1.%
AwtA
A
Awtaor
AwtasAwt
a
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Alternative Standard States
If the concentration up to 1 weight-percent of solute A, then aA
= 1 at wt%A =1 and this 1 weight-percent solution is then the
standard state
W.r. t 1 weight-percent standard state, the activity of solute A
is given by
Where fA(1wt%) is the 1 wt.% activity coefficient and in the
range of composition in which A obeys the Henry’s law fA(1wt%)
Awtfh wtAwtA .%.%)1(%)1(
Awth wtA .%%)1(
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Alternative Standard States
We can also write
In the range of composition in which the solute obeys Henry’s Law, fA(1wt%) =1 and γA = γA
0 , therefore,
ncompositiotconswtA
AA
wtA
A
Awtf
x
h
a
tan%)1(%)1( %.
.
ncompositiotcons
AA
wtA
A
Awt
x
h
a
tan
0
%)1( %
.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Alternative Standard States
We know that
Where MA and MB are the molecular weight of A and B. the first term in
the denominator is small compared to the second and the relation may be simplified as
BA
AA
M
Awt
M
AwtM
Awt
x.%100.%
.%
A
BA
A
B
B
AA
M
M
Awt
xTherefore
M
MAwt
M
M
Awt
x
.100.%,
.100
..%100
.%
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Alternative Standard States
The free energy change accompanying the transfer of one mole of solute A from pure substance standard state (Raoultian standard state) to 1 wt.% standard state, that is:
A (in the Raoultian standard state) → A( in the 1 wt.% standard state)
is given by
A
BA
constxWtA
AA
M
MRTRT
h
aRTwtRG
A
.100lnln
ln%).1(
0
.%)1(
0
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Alternative Standard States
The free energy change accompanying the transfer of one mole of solute A from Henerian standard state to 1 wt.% standard state, that is:
A (in the Henerian standard state) → A( in the 1 wt.% standard state)
is given by
A
B
AA
BA
AAA
M
MRT
RTM
MRTRT
HRGwtRGwtHG
.100ln
ln.100
lnln
)(%).1(%).1(
00
000
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
Calculate the free energy change when the standard state of manganese is
transferred from pure liquid state to infinitely dilute wt% solution of
manganese in iron at 16270C melting point of Mn =12450C, at wt of Mn =
54.94, Fe = 55.85. Assume ideal behaviour of the solution.
Solution:
T = 1627 + 273 = 1900, MMn = 54.94, MFe = 55.85
Mn (pure substance standard state) Mn (dilute wt% standard state)
)(10 idealMn
Mn
FeMn
wtMn
MnMn M
MRT
h
aRTHRG
100lnln)( 0
%)1(
0
molJ /72455
94.54100
85.551ln1900314.8
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
From experimental measurements of the equilibria between H2-H2O gas
mixtures pure silica and silicon dissolved in liquid iron, the free energy
accompanying the transfer of standard state from pure silicon to the
infinitely dilute, wt.% solution of silicon in iron i.e.:
Si (pure, 1) Si (wt% dil. in Fe) has been expressed as G0 = -28500 –
5.8T cal./mol. At 16000C, the activity coefficient of silicon in iron ,
relative to pure silicon as the standard state is 0.0014 at 1 atomic% Si.
Calculate the activity coefficient of silicon, relative to the wt% standard
state at this concentration.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
G0 = -28500 – 5.8T = -28500 -5.8 1873 = -39360 cal.
wt% Si = 0.50
Si (pure, 1) Si (1wt% in Fe)
This is the activity coefficient of Si at infinite dilution, relative to pure Si as the standard state.
85.55%100
09.28%
09.28%
01.0%SiwtSiwt
Siwtsiatom
3936007.28100
85.55.ln
100ln RT %)1(
000
Si
Si
FeSiSi RT
M
MwtRG
00128.00 Si
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
Hence the activity coefficient of Si at 1 at% Si, relative to pure Si as the
standard state is calculated as follows:
000014.001.00014.0.)( SiSiSi xpurea
Si
FeSi
wtSi
Si
M
M
h
a
100
0
%)1(
FeSi
SiSiSi M
Mawth
0
100.%)1(
55.0
85.5500128.0
09.28100000014.0
FeSi
SiSi
MSiwt
MSiwt
MSiwt
x)%100(%
%
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
50.0%85.55
)%100(09.28
%09.28
%01.00
Siwtor
SiwtSiwt
Siwtr
hSi = fSi . wt% Si
.1.150.0
55.0
%Ans
Siwt
hf Si
Si
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
The Raoultian activity coefficient of Al at infinite dilution, in
liquid Fe-Al alloys is reported to be 0.063 at 16000C. Calculate the
standard free energy of formation of Al2O3(s) at 16000C for each of the
three standard states for solution of Al in Fe.
MAl = 26.98 and MFe = 55.85
given 1. 2 Al(l) + 3/2O2(g) = Al2O3(s) G0 = -1682927 – 323.24TJ
2. 2 Al(l, H) + 3/2O2(g) = Al2O3(s)
3. 2 Al(l, wt %) + 3/2O2(g) = Al2O3(s)
0Al
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
(1)
G0 = -1082927 +323.24 1873 = -1077498.5 J
Total free energy change for reaction (2) =
)1()()(2
3)(2 322 sOAlgOlAl
)3(),()(
)2()()(2
3),(2 322
HlAllAl
sOAlgOHlAl
00 ln)(
)3(
AlAl
AlAl RT
h
aRTHRG
reactionFor
0)(
0)(
0)( 2 HRRH GGG
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
= -1077498 – 2 8.314 1873 ln (0.063)
= -1077498 – (-86102) = -991396 J.
Total fre energy change for the reaction (4) =
)5(%)1,()(
)4()()(2
3%)1,(2 322
wtlAllAl
sOAlgOwtlAl
00)( ln2 AlR RTG
Al
FeAl
wtAl
AlAl M
MRT
h
aRTwtRG
reactionFor
100ln%)1(
)5(0
%)1(
0
0%)1(
0)(
0%)1( 2 wtRRwt GGG
Al
FeAlR M
MRTG
100ln2
00
)(
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
= -1077498 – 2 8.314 1873 ln
= -1077498 – 2 8.314 1873(-6.6422)
= -1077498+206866.8
= -870631J Ans.
98.26100
85.55063.0
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Problem
Vanadiam melts at 1720oC. The raoultion activity coefficient of vanadian
at infinite dilution in liquid iron at 1620oC is 0.069. Calculate the free
energy change accompanying the transfer of standard state from pure solid
vanadium to the infinitely dilute wt% solution of V in pure iron at 1620oC.
Given: heat of fusion of V = 4800 cal/g atom
MFe = 55.85 and MV = 50.95.
Solution:
.1893069.01993 0 KatKT Vf
V
..deg/4084.21993
4800molcal
T
HS
f
ff
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solution
Free energy of fusion of vanadium (s l) at the operating temperature 1893K.
We calculate Gf for fusion because the melting point of V is more than the operating temperature (at which V is in solid state).
V (pure l) V (wt%)
= -27033 cal/mol
For energy change V (pure solid) V infinitely dilute wt% solution
= ΔGf + ΔGV0 (R →H) = 240.9 – 27033
= -26792.1 cal/mol.
./9.2404084.218934800 molecalSTHG fff
95.50100
85.55069.0ln1983987.1
100ln%)1(
00
V
FeVV M
MRTwtRG
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Chemical Potential
The general equation for the free energy change of a system with temperature and pressure dG = VdP – SdT, does not take into account any variation in free energy due to concentration changes.
We know
From the fundamentals of partial differentiation we have
The coefficient called the ‘chemical potential’ and is
denoted by hence
....),,,,( 21 nnPTGG
......2....,,2
1....,,1,,
2
1
1
2
dnn
Gdn
n
GdP
P
GdT
T
GGd
nexceptnPT
nexceptnPTniTniP
i
nexceptnPTi
dnn∂
G∂+dPV+dTS=
i
1
∑....,,
'''.
inexceptnTPin
G
.......,, 1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Chemical Potential
is an intensive variable
This gives a new sets of fundamental equations for the open systems.
i
nExceptnPTi
i
n
G
.....,, 1
ii dndPVdTSGd
ii dnVPddTSAd
ii dndPVSTdHd
ii dnVPdSTdUd
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Physical Meaning of Chemical potential Consider the change in free energy (dG/) of a system produced by the
addition of dnA mole of component A at constant pressure and temperature. The change in free energy of a system is given by is the partial molar free energy of component A in solution
Chemical potential of either 1 g mol or 1 g atom of a substance dissolved
in a solution of definite concentration is the partial molar free energy. Thus
AAAA dnGdnGd
solutionofquantitylargeafor,,
A
BnTPAA n
GG
solutionof mole onefor,,
A
BnTPAA n
GG
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Equality of chemical potentialamongst phases at equilibrium
We know:
At constant T and P:
Consider two phases(I and II) in the system. Then,
Consider moving an infinitesimal of quantity dn1 from phase
I to phase II. Then,
ii dndPVdTSGd
∑ ii dnμ=G′d
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Equality of chemical potentialamongst phases at equilibrium
Therefore, total free energy change of the system is
For equilibrium at constant temp and pressure
Hence,
It can be generalized for all components at constant T and
P when phase I and II are at equilibrium as
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Equality of chemical potentialamongst phases at equilibrium
Where P is the total no. of phases in the system
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Phase Rule
Phase(P)
A phase is defined as any homogeneous and physically distinct part of a
system which is separated from other part of the system by a bounding
surface. For example, at 273.15K, three phases ice, water and water vapour
can exist in equilibrium. When ice exists in more than one crystalline
form, each form will represent a separate phase because it is clearly
distinguishable from each other.
Components(C)
The number of components in a system at equilibrium is the smallest
number of independently variable constituents by means of which the
composition of each phase present can be expressed directly or in the form
of a chemical equation.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Phase Rule
As an example let us consider decomposition of calcium carbonate :
CaCO3 (s) = CaO(s) + CO2(g)
According to the above definition, at equilibrium this system will consist of two components since the third one is fixed by the equilibrium conditions.
Thus we have three phases – two solids (CaCO3 and CaO) and a gas (CO2) and the system has only two components.
If CaO and CO2 are taken, the composition of calcium carbonate
phase can be expressed as xCaO + xCO2 giving xCaCO3 (by the
chemical reaction).
The composition of the three phases could be expressed equally by taking CaCO3 and CaO or CaCO3 and CO2 as the components.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Phase Rule
The dissociation of any carbonate, oxide or similar compounds involves
two components; the same is true in the case of salt hydrate equilibria, for
example : CuSO4.5H2O(s) = CuSO4.3H2O(s) + 2H2O(g) when the simplest
components are evidently CuSO4 and H2O.
In the slightly more complicated equilibrium : Fe(s) + H2O(g) = FeO(s) +
H2(g) it is necessary to choose three components in order that the
composition of each of the three phases can be expressed.
The composition of the two solid phases could be given in terms of Fe and
O, but these alone are insufficient to define the gaseous phase which is a
mixture of hydrogen and water vapour, a third component, viz., H2O is
necessary.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Phase Rule
The water system for example consists of one component, viz., H2O each
of the phases in equilibrium i.e. solid, liquid and vapour may be regarded
as being made of this component only.
Degrees of freedom(F)
The number of degrees of freedom is the number of variable factors, such
as temperature, pressure and concentration that need to be fixed in order
that the condition of a system at equilibrium may be completely defined
when referring to its equilibrium phase diagrams.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Derivation of the PhaseRule Equation
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Derivation of the PhaseRule Equation
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Derivation of the PhaseRule Equation
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Application of phase ruleto single component system
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Application of phase ruleto two component system
For two component system: F = 3 – P as the pressure is kept
constant
P=1 F= 2 can vary T and composition - bivariant
P=2 F= 1 can vary either T or P – monovariant
P=3 F= 0 no free variables it is a fixed point. - invarient
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Two componentsEutectic system
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Phase rule in Reactive Components
Consider a system consisting of N chemical species and
there are P number of phases.
In this case the number of components differ from number
of species.
Let us consider there are three out of N species are
chemically active and participate in the following reaction:
AB(s) = A(g) + B(g).
The number of total variables = P(N-1) + 2
Total number of constraints due to phase equilibrium= N(P-
1).
There is another additional constraints: AB(s) = A(g) + B(g).
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Phase rule in Reactive Components Additional, in the absence of A(g) and B(g) in the starting
reactant mixtures, stoichiometric consideration requires that PA= PB.
Some times, special constraints are placed on the system. For example, the system under consideration, the partial pressure of A has been fixed at 2 atm.
So this way total no. of constraints are = N(P-1)+1+1+1. F = [P(N-1)+2] – [N(P-1)+1+1+1] = (N-2) – P +1 = C – P
+1 Generalizing, for a system in which there are ‘r’
independent chemical equilibria, ‘s’ stoichiometric relation and ‘t’ special constraints we have
F = (N – r – s - t) – P +2 = C- P + 2 –t
where C = N - r - s
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Application of Phase rulein Reactive ComponentsProblem: A system is composed of a solid phase CaCO3, a solid phase CaO,
and a gas phase CO2 . The following equilibrium occurs:
CaCO3(s) = CaO(s) + CO2(g)
How many components are there and what are the degrees of freedom?Solution:
Species: CaCO3(s) , CaO(s), CO2(g) : N =3, Phases : two solid and a gas phase P = 3. No. of independent reaction equilibria r = 1. There is no stoichiometric or special constraints.
So s = 0 and t = 0 C = N-r-s = 3-1-0= 2F = C-P+2-t = 2-3+2-0 = 1Either temperature or pressure must be specified.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Application of Phase rulein Reactive ComponentsProblem:A pure solid NH4Cl is introduced into an evacuated chamber. It is then allowed to decompose and equilibrium has been established by following reaction:
NH4Cl(s) = NH3(g) + HCl(g)
Calculate the number of components and degrees of freedom.Solution:
N = 3 (NH4Cl(s) , NH3(g) , HCl(g))
P = 1 solid (NH4Cl(s) ) + 1 gases (NH3(g) + HCl(g ) = 2r = 1
s = 1 as P NH3(g) = P HCl(g)
t = 0C = N – r – s = 3 – 1 -1 = 1F = C – P + 2 – t = 1 -2 + 2 – 0 = 1
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Application of Phase rulein Reactive Components
Problem:Show that the system in which the reaction
Mn(s) + 2/3 AlCl3(g, 1atm) = MnCl2(l) + 2/3 Al (l)
is at equilibrium is invariant.
Solution: N = 4 (Mn(s), AlCl3(g), MnCl2(l), Al (l))
P = 1 solid (Mn) + 2 liquids(MnCl2(l)and Al (l)) + 1 gas (AlCl3(g)) = 4
r = 1s = 0
t = 1 (1 atm of AlCl3(g)) C = N –r – s = 4 – 1 – 0 = 3F = C – P + 2 – t = 3 – 4 + 2 – 1 = 0This is an invariant system
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Application of Phase rulein Reactive ComponentsProblem: Consider reduction of FeO with CO under standard
conditions i.e. P = 1 atm. FeO(s) + CO(g) = Fe(s) + CO2(g).
Calculate the number of components and degrees of freedom.
Solution:In this system we have P = 3 (i.e. two solids FeO and Fe and a gaseous phase CO+CO2) and N = 4, r = 1, s = 0 and
t = 1 (PCO + PCO2 =1 atm)C = N –r –s = 4 -1 -0 = 3F = C - P + 2-t = 3 – 3 + 2 -1 = 1Thus the above system has only one degree of freedom,
either temperature or pressure.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Phase Diagram
Graphical representation of what phases are present
in materials systems at various temperatures, pressures and compositions are called phase
diagrams
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solid Solution
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Solid Solution
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Types of Solid Solubility
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Phase Diagram
Solidus
The phase boundary between solid and two phase
region.
Liquidus
The phase boundary between liquid and two phase
region.
Solvus
The solid state phase boundary between terminal solid
solution and two phase region.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Single – Component System – Variation of Free energy with Temperature
S(l) > S(s), So slope of the line
for liquid H2O is greater than
solid H2O
)()(
)()(
l
P
l
s
P
s
ST
G
ST
G
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Single – Component System – Variation of Free energy with Temperature
Where ΔS is the change in
molar entropy which occurs
as a result of the change of
state. The slope od the line is –ve
which shows that at all temp
SH2O(l) > SH2O(s)
ST
G
P
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Single – Component System – Variation of Free energy with Pressure
V(l) < S(s) for H2O So slope
of the line for solid H2O
is greater than liquid H2O
at all pressure
)()(
)()(
s
T
s
l
T
l
VP
G
VP
G
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Variation of Free energies of solid liquid and vapour H2O with Temperature and Pressure
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Binary Phase Diagrams
Isomorphous System: The system which exhibits
complete solid solubility and liquid solubility is
called an isomorphous sytem. The crystal
structure of both the components as well as solid
solution are same.
Eutectic system: The system which exhibit limited
solid solubility or terminal solid solution is called
an eutectic system.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Isomorphous system
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Eutectic system
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Experimental Determination of Liquidus and Solidus– Cooling Curve Method
T = f(t) cooling curves measurements for several compositions
Pure metal : melts /solidifies /allotropic transformation takes place at one temperature.
Binary solutions: melts /solidifies over a range of temperature.
The temperature at which the start of solidification takes place is called liquidus.
The temperature at which the end of solidification takes place is called solidus.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Experimental Determination of Liqidus and Solidus– Cooling Curve Method
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Experimental Determination of Solvus– Cooling Curve Method
The common methods for the determining the solvus are microscopic examination and X-ray diffraction methods.
A series of small ingots of alloys of different compositions are prepared and homogenized.
They are annealed at various temperatures for prolonged time (a few days) and then quenched.
High temperature phases may be retained on quenching.
Subsequent, metallographic studies and X-ray diffraction reveals the various phases present at that temperature for given alloys
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Experimental Determination of Solvus– Cooling Curve Method
A phase boundary is first bracketed between two compositions.
The exact location of the boundary is determined by studying a more alloys of closely varying compositions in the boundary region.
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Free energy – Composition Diagram
The intercepts of the two axes by the tangent of the Gibbs free energy curve of the α phase at the composition represent
similarly, for the β phase
2X
21 and
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Free – Energy Composition Diagram for Binary Systems
Isomorphous System
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics
Free – Energy Composition Diagram for Binary Systems
Eutectic System
D E P A R T M E N T O F
Metallurgical Engineering Institute of Technology Banaras Hindu University
Metallurgical Thermodynamics