Thermodynamic

2. Open
Close
Isolated
Energy &
Matter
Surrounding
System types.mp4

3. • According to the nature of the changes occurs in the system, it may be reversible and
irreversible, system can be classified as reversible and irreversible system.
• “Chemical changes in which driving and opposing forces are differ by large amount in
which driving force large than opposing force, if opposing force increased by small amount,
reaction does not reverse is called as irreversible process.”
• “A reaction proceeds in only one direction is called as irreversible reaction while the reaction
proceeds in both directions is called as reversible reaction”.
“A chemical reaction in which driving and opposing forces
are differ by infinitely small amount and proceeds till
attends equilibrium state, if opposing force is increased by
small amount, reaction become reverse is called as
reversible reaction.

4. The properties of the system are classified into two groups as-
a) Extensive properties:- An extensive properties of the system is any property
whose magnitude is depends on the amount of present in the system.
e.g.- pressure, volume, entropy, free energy, etc.
b) Intensive properties:- Intensive properties of a system is any property whose
amount is independent on the total amount but depends on the concentration of
the substance or substances in a system.
e.g.- density, refractive index, chemical potential, viscosity, surface tension, etc.
Properties of system
SystemSystem
Properties depends
on amount
Properties depends
on amount
Properties
independs on
amount
Properties
independs on
amount
Extensive
properties
Extensive
properties
Intensive
properties
Intensive
properties

5. AccordingAccording
to processto process
occurred inoccurred in
systemsystem
Isothermal
process
Isobaric
process
Isochronic
process
Cyclic
process
Adiabatic
process

6. • According to the process occurring in the system or according to the fundamental parameter change in a
system, it can be classified as-
• 1) Isothermal process:- A process occurring at constant temperature is called as isothermal process and a
system is called as isothermal system.
Δ T = 0
• 2) Isobaric process:- A process occurring at constant pressure is called as isothermal process and a
system is called as isobaric system.
Δ P = 0
• 3) Isochronic process:- A process occurring at constant volume is called as isothermal process and a
system is called as isochronic system.
Δ V = 0
• 4) Cyclic process:- A process in which a change in any state function such as internal energy is equal to
zero or a process in system having initial and final state are same is called as cyclic process and system is
called as cyclic system.
Change in state function = 0
• 5) Adiabatic process:- A process in which system does not exchange heat with it’s surrounding is called
as adiabatic process and a system is adiabatic system.
q = 0.
• All fundamental parameters or properties (P, V, T, H, E, G, A, density, surface tension, K, etc) are
classified into two types as-
• a) State function:- A parameter or properties of the system is depends on only initial and final state of the
system but not on the path through which change occurs is called as path function.
• e.g.- free energy, entropy, pressure, volume, etc.
• b) Path function:- A parameter or properties of the system is depends on the path through which change
occurs but not on the initial and final state of the system is called as path function.

8. First law thermodynamics

9. Electrolysis

10. First Law of ThermodynamicsFirst Law of Thermodynamics
This law is called as law of conservation of energy. “The energy canThis law is called as law of conservation of energy. “The energy can
neither be created nor destroyed, but it can be converted from one formneither be created nor destroyed, but it can be converted from one form
to another form, hence total energy of universe remains constant.”to another form, hence total energy of universe remains constant.”
q = W + ΔE (1)
or ΔE = q + W
The magnitude of ‘q’ and ‘W’ is depends on the manner in which the
work is performed in passing from the initial and final state.
The value of ΔE and W is explain by two different functions called as
entropy and free energy.

11. • According to the physical significance of entropy-
In Carnot cycle, only some part of heat is
converted into work and remaining part of
heat is given to sink.
The part of heat which is not available to perform work is a
measure of entropy change accompanying the process. Hence the
fraction of energy unavailable to perform the work is T.ΔS.S = q/T = ∆H/T

12. *** Free Energy and Equilibrium ***
•
How much quantity of heat is used to performed work done is
determined by mew thermodynamic state function is called as free
energy. The heat energy is supplied in the form of internal energy
or enthalpy.
[Heat energy (E or H)] = (Heat available to perform work done) +
(Heat unavailable to perform work done)
If heat energy is supplied in the form of internal energy (E), the part of internal
energy available to perform work is called Helmholtz free energy. It is denoted by
the latter A. The internal energy unavailable to perform the work done is TS, S-is
entropy at temperature T.
[Heat energy (E)] = (E available to perform work done) +
(E unavailable to perform work done)
E = A + TS
A = E - TS (1)

14. Helmholtz free energy
Helmholtz free energy of any system A is defined as -
A = E - TS
Where, A, E, S and T are state functions.
The change in A during any process is given as-
ΔA = A2 - A1
= (E2 – T2S2) – (E1 – T1S1)
= (E2 - E1) – (T2S2 – T1S1)
= ΔE – ΔTS (3)
Equation (3) is the general definition of ΔA. For isothermal process T2 = T1, then
equation (3) becomes-
ΔA = ΔE – TΔS (4)
Equation (4) explain physical interpretation (significance) of ΔA, under isothermal
conditions – TΔS = qrev (according to the definition of change in entropy).
ΔA = ΔE – qrev. (5)
According to the first law of thermodynamics for reversible and isothermal process-
qrev = ΔE + Wm or Wm = qrev - ΔE (a)
From equation (5) and (a)-
ΔA = -Wmax (6)
The equation (6) indicates that, at constant temperature the maximum work done by the
system is due to expense of (decrease in) Helmholtz free energy of the system.
Therefore, A is called as Work function of a system.

16. • An alternative equation for the variation of A with T can be obtained
as follows-
• Differentiate A/T with respect to T at constant V gives-
• [δ(A/T)/δT]v = [T(δA/δT)v – A]/T2
• = [T(-S) – A]/T2
• = - [A + TS]/T2
• = - E/T2
(13)
• Equation (9) shows that the dependence of A for a pure substance on
both T and V, these can be shown by (12) and (13) for T and (11) for
V.

17. *** ΔA for reaction ***
For any reaction such as-
ΔA for such reaction is sum of the A’s for the products minus a similar sum of
for the reactants.
ΔA = [cAC + dAD + ……] – [aAA + bAB + ……..]
Where, cAC, dAD, aAA, and bAB are molar Helmholtz free energies of products
and reactants respectively. At any temperature T,
ΔA = [Ep – ER] – T[Sp - SR]
ΔA = ΔE - TΔS (4)
Differentiate equation (12) with respect to temperature at constant volume as -
(δΔA/δT)v = [c(δAC/δT)v + d(δAD/δT)v + ……] – [a(δAA/δT)v + b(δAB/δT)v +
……..]
(δΔA/δT)v = - [cSC + dSD + ……] + [aSA+ bSB + …] = - ΔS (14)
Form equation (14), equation (4) becomes-
ΔA = ΔE + T(δΔA/δT)v (15)
Alternative expression for the variation of ΔA with T is-
Differentiate ΔA/T with respect to temperature at constant volume as -
[δ(ΔA/T)/δT]v = [T(δΔA/δT)v – ΔA]/T2
= [T(-ΔS) – ΔA]/T2
= - [ΔA + TΔS]/T2
2
aA + bB + ............ cC + dD + ..................

18. Gibb’s free Energy
If heat energy is supplied in the form of enthalpy, the part of enthalpy available to perform workIf heat energy is supplied in the form of enthalpy, the part of enthalpy available to perform work
is calledis called Gibbs free energyGibbs free energy. It is denoted by the latter G or F. The enthalpy unavailable to. It is denoted by the latter G or F. The enthalpy unavailable to
perform the work done is TS, S-is entropy at temperature T.perform the work done is TS, S-is entropy at temperature T.
[Heat energy (H)][Heat energy (H)] == (H available to perform work done)(H available to perform work done)
++ (H unavailable to perform work done)(H unavailable to perform work done)
H = G + TS
G = H - TS (1)
Where, H and S are enthalpy and entropy of the system respectively. G, H, S and T are state
functions.
Change in G during any process is given as-
ΔG = G2 - G1
= (H2 – T2S2) – (H1 – T1S1)
= (H2 - H1) – (T2S2 – T1S1)
= ΔH – ΔTS (2)
Equation (2) is the general definition of ΔG. For isothermal process T2 = T1 = T, then equation
(2) becomes-
ΔG = ΔH – TΔS (3)

19. Relationship between G and A
Consider equation (3) is- ΔG = ΔH – TΔS
But H is also state function- ΔH = ΔE + PΔV (at constant P)
Therefore,
ΔG = ΔE + PΔV – TΔS
= (ΔE – TΔS) + PΔV (ΔA = ΔE – TΔS)
ΔG = ΔA + PΔV (4)
In general- G = A + PV (5)
The equation (4) and (5) shows the relationship between G and A.

20. Physical significance of G
At constant temperature, and at constant pressure,
TΔS = qr ΔH = ΔE + PΔV.
Put these values into equation (3) ΔG = ΔH – TΔS as-
ΔG = (ΔE + PΔV) – qr
= - (qr - ΔE - PΔV)
But according to the first law of thermodynamics,
qr = ΔE + Wmax or Wmax = qr - ΔE
Therefore, ΔG = - (Wmax - PΔV) (6a)
Or -ΔG = (Wmax - PΔV) (6b)
Conclusion: Equation (6b) shows that ΔG is the available free energy for the work
done other than P-V type at constant T and P. the work is done due to expense of
Gibb’s free energy.

21. Variation of G with P and T
According to the definition of G [i.e. from equation (1)]-
G = H - TS
Differentiate it completely as-
dG = dH - TdS - SdT (7)
According to the definition of entropy and first law of thermodynamics-
dqr = TdS = dE + PdV (for reversible process) (8)
According to the definition of enthalpy- H = E + PV
Differentiate it completely- dH = dE + PdV + VdP (9)
Therefore equation (8) and (9) dH = VdP + TdS (10)
From equation (7) and (10)-
dG = VdP + TdS -TdS -SdT
dG = VdP - SdT
(11)
Equation (11) indicates that G is function of a T and P as the independent variables, we have-
G = f (T,P)

22. G = f (T,P)
Differentiate it completely-
dG = (δG/δT)P dT + (δG/δP)T dP (12)
Comparing equation (11) and (12) we have-
(δG /δP)T = V
(δG/δT)P = -S (13)
An alternative equation for the variation of G with T can be obtained as follows-
Differentiate G/T with respect to T at constant P gives-
[δ(G/T)/δT]P = [T(δG/δT)P – G]/T2
= [T(-S) – G]/T2
= - [G + TS]/T2
= -H/T2
(14)
Equation (11) shows that the dependence of G for a pure substance on both T and
P, these can be shown by (14) and (13) for T.

23. ΔG for reaction
For any reaction such as-
ΔG = ΣGp – ΣGR (1)
ΔG = [Hp – HR] – T[Sp - SR]
ΔG = ΔH - TΔS at constant T (2)
ΔH and ΔS are change in enthalpy and entropy of the reaction respectively at constant
temperature T and pressure P.
Also ΔG is function of T and P as -
d(ΔG) = (δΔG/δT)p dT + (δΔG/δP)T dP (3)
These derivatives are also obtained by differentiate equation (2) with respect to T at
constant P and with respect to P at constant T.
(δΔG/δT)p = (δΔH/δT)p - T(δΔS/δT)p – ΔS
(δΔG/δT)p = ΔCp - TΔCp/T – ΔS
= –ΔS (4)
Also (δΔG/δP)T = Vp – VR = ΔV (5)
Where, ΔV is change in molar volume during the reaction.
From equation (3), (4) and (5), we have-
d(ΔG) = ΔVdP –ΔSdT (6)
From equation (4) and (2)-
ΔG = ΔH + T[δ(ΔG)/δT]p (7)
The equation (7) is called as Gibb’s-Helmholtz equation where (δΔG/δT)p is called
aA + bB + ............ cC + dD + ..................
But (δH/δT)p = Cp;
(δΔH/δT)p = ΔCp and
(δΔS/δT)p = ΔCp/T

24. Significance of Gibb’s-Helmholtz equation:
Gibb’s-Helmholtz equation is-
ΔG = ΔH + T[δ(ΔG)/δT]p (7)
(i) Consider the reaction-
Zn + CuSO4 (sol) gives ZnSO4 (sol) + Cu e.m.f. = 1.10 Volt
For the reaction, heat of reaction is equal to ΔH. If same reaction is carried out reversibly in a cell by applying
the external voltage which is infinitely greater than that of cell emf. The work is done denoted by the later
ΔG instead if heat evolved. The difference between these two i.e. (ΔG - ΔH) is equal to a either -TΔS or
T[δ(ΔG)/δT]p.
(ii) From equation (2) and (7)-
- TΔS = T[δ(ΔG)/δT]p. (8)
But by the definition of the entropy- TΔS = qr.
Therefore, -qr = T[δ(ΔG)/δT]p. (9)
And equation (7) becomes- ΔG = ΔH -qr
Or qr = ΔH – ΔG (10)
The equation (10) represents the heat interchange between the system and when
process is conducted isothermally and reversibly.
Case-I): when ΔH > ΔG, qr is positive; heat is absorbed from the surrounding.
Case-II): when ΔH < ΔG, qr is negative; heat is evolved to the surrounding.
Case-III): when ΔH = ΔG, heat is neither absorbed nor evolved with the
surrounding.

25. *** Properties and significance of ΔG ***
i)Free energy is state function i.e. it is depends on the initial and final state of the system.
ii)Free energy is a definite quantity of any substance at any given temperature and pressure and
is changed when these two variables are changed. G = f(T,P).
iii)Absolute values of the free energies of substances are not known but its change can be
calculated.
iv)The free energy change can be expressed in simple equations similar to thermo chemical
equations which can easily added or subtracted.
v)The sign of the free energy is very important-
(a) ΔG is negative, indicate that energy is evolved or emitted during the chemical
reaction i.e. it is spontaneous reaction.
(b) ΔG is positive, indicate that energy is absorbed during the chemical reaction i.e. it
is non-spontaneous reaction.
(c) ΔG is zero indicate that system is in equilibrium state.
A negative sign of the free energy change for a process does not mean that the process will be
takes place. It only provides the information about the change.
E.g. the O2 and H2 can coexist at room temperature without combining, although ΔG for such
change is -56.690 cal at 250
C. In presence of the catalyst such as- platinized asbestos reaction
proceeds with explosive violence.

26. *** Calculation of free energy change ***
The variation of G with pressure at constant temperature can be given by simple equation as-
(δG /δP)T = V
Or dG = VdP (at constant T)
The total free energy change can be obtained by integration with proper limits -
G2 P2
∫ dG = ∫VdP
G1 P1
P2
ΔG = G2 – G1 = ∫ VdP (1)
P1
In equation (1), volume is depends on the pressure for n moles of the ideal gas as -
PV = nRT or V = nRT/P
P2
ΔG = G2 – G1 = ∫ (nRT) dP/P
P1
= nRT ln (P2/P1)
= 2.303 nRT log (P2/P1)
(2)
For real gases, obtain a suitable equation for the free energy change from equation (2).
In solids and liquids only volume can be constant over the appropriate range of pressure.
P2
ΔG = G2 – G1 = ∫VdP = V(P2 – P1) = VΔP
(3)
P1
For an ideal gas, according to the Boyle’s law, P2/P1 = V1/V2, therefore equation (2) becomes-
ΔG = 2.303 nRT log (V1/V2) (4)
In case of solids and liquids, the free energy changes may be considered to be constant
over a fairly wide range of pressure at given temperature. The ΔG change in gases can be
calculated.

27. The Reaction isotherm (Van’t Hoff’s Isotherm)
Van’t Hoff’s isotherm gives the net work done that can be obtained from gaseous reactants at
constant temperature, when both reactants and products are at suitable orbitary pressure.
Consider a general reaction-
If aA, aB, aC and aD are activities of reactants A, B and products C, D
respectively.
The free energies of each of these substances per mole at T is-
GA = Go
A + RT lnaA (1)
GB = Go
B + RT lnaB (2)
GC = Go
C + RT lnaC (3)
GD = Go
D + RT lnaD (4)
Where, Go
A, Go
B, Go
C and Go
D are the free energies at unit activity of the A, B,
C and D respectively or are the standard free energies of A, B, C and D
respectively.
aA + bB + ............ cC + dD + ..................
A quantitative relation of the free energy change for the chemical reaction has been developed
which is known as Reaction isotherm or Van’t Hoff’s isotherm.

28. Therefore the free energy change of the chemical reaction -
ΔG = ΣGp – ΣGR
= [cGC + dGD + …….] – [aGA +bGB + ……..]
= [cGo
C + RT lnaC
c
+ dGo
D + RT lnaD
d
+ …….] –
[aGo
A + RT lnaA
a
+ bGo
B + RT lnaB
b
+ ……..]
= [(cGo
C + dGo
D + ……) – (aGo
A + bGo
B +……)] +
RT[(lnaC
c
+ lnaD
d
+ …….) – (lnaA
a
+ RT lnaB
b
+ ……..)]
ΔG = ΔGo
+ RT ln{(aC
c
.aD
d
…….)/(aA
a
.aB
b
……..)} (5)
The equation (5) is called as the Van’t Hoff’s isotherm or reaction isotherm. When activities of
all reactants and products are equal to one then ΔG = ΔGo
.
At equilibrium step of the reaction - ΔG = 0
0 = ΔGo
+ RT ln{(ac
C x ad
D…….)/(aa
A x ab
B……..)}
At equilibrium, according to the law of mass action, Ka = (ac
C ad
D…….)/(aa
Aab
B……..)
Therefore, 0 = ΔGo
+ RT ln Ka
Or ΔGo
= - RT ln Ka (6)
Ka = e -ΔGo/RT
(7)
Form equation (5) and (6)-
ΔG = RT ln{(ac
C ad
D…….)/(aa
Aab
B……..)} - RT ln Ka
= RT ln Qa - RT ln Kaa where Qa = {(ac
C ad
D…….)/(aa
Aab
B……..)}
= RT ln (Qa/Ka) (8)

29. For gases activities are proportional to the partial pressures of the components of reaction
mixture hence the activities of the equation (5) can be replaced by the partial pressures,
therefore-
ΔG = ΔGo
+ RT ln{(PC
c
.PD
d
…….)/(PA
a
.PB
b
……..)} (9)
= RT ln Qp - RT ln Kp (10)
Where Kp is the equilibrium constant and
Qp = (Pc
C.Pd
D…….)/(Pa
A.Pb
B……..)
At equilibrium state, ΔG = ΔGo
and Qp = 1.
ΔGo
= - RT ln Kp = -W or Kp = e -ΔGo/RT
(11)
i.e. net work done of the reaction is equal to the decrease in free energy of the system and
it is calculated by using expression -
RT ln Kp or 2.303RT log Kp.
It will be observed that, ΔG is positive when Kp is less than unity.
If ΔG is negative when Kp is greater than unity and is zero when Kp = 0 or 1.

30. *** Van’t Hoff Isochore ***
The Van’t Hoff isochore is obtained by comparing the Van’t Hoff isotherm with Gibbs
-Helmeholtz equation.
The Gibbs-Helmeholtz equation is ΔG = ΔH + T[d(ΔG)/dT]p
Or -ΔH = T[d(ΔG)/dT]p - ΔG
Dividing both side by T2
gives:- -ΔH/T2
= {T[d(ΔG)/dT]p - ΔG}/T2
The right hand side of the above equation is obtained by differentiating ΔG/T w.r.t. T at constant P-
[d(ΔG/T)/dT]p = {T[d(ΔG)/dT]p - ΔG}/ T2
-ΔH/T2
= [d(ΔG/T)/dT]p (12)
•According to Van’t Hoff isotherm- ΔG = -RT ln Kp (11)
•Dividing both side by T and differentiating it w.r.t. T at constant P-
• -[d(ΔG/T)/dT]p = R d(lnKp)/dT (13)
•Comparing equation (13) and (12), we have-
• ΔH/T2
= R d(lnKp)/dT
•Or (ΔH/R).dT/T2
= d(lnKp) (14)
•The above equation is known as Van’t Hoff isochore. If ΔH remains same over the range of
temperature, we have on integration -
• lnKp = (ΔH/R)∫dT/T2
• = -ΔH/RT + constant of integration.
•Appling the limits T1 and T2 at the equilibrium constants Kp1 and Kp2 respectively, we have-
• lnKp2 - lnKp1 = -(ΔH/R)[1/T2 - 1/T1]
• ln[Kp2/Kp1] = (ΔH/R)[1/T1 - 1/T2]
• log[Kp2/Kp1] = (ΔH/2.303R)[(T2 – T1)/T1T2] (15)
•By using this equation, equilibrium constant at any temperature or heat of reaction is calculated.

31. • In reversible reaction, the sign of ΔGo
indicates whether the forward or reverse
reaction is spontaneous as- ΔGo
= -
2.303RT logKp
• i) If ΔGo
is negative; logK must be positive &
K is greater than one and reaction proceeds
spontaneously in forward direction.
• ii) If ΔGo
is positive; logK must be negative
& K is less than one and reaction proceeds
spontaneously in reverse direction.
• iii) If ΔGo
is zero; logK must be zero & K is
equal to one and reaction is in equilibrium
state.

32. • *** Van’t-Hoff equation in terms of Kc***
• We known that equilibrium constant in terms of partial
pressure Kp and in terms of concentration Kc are related to
each other by the equation as-
• Kp = Kc (RT)Δn
• Taking logarithm of both side-
• lnKp = lnKc + Δn ln(RT)
• Differentiate it with respect to temperature, we get-
• d(lnKp)/dT = d(lnKc)/dT + Δn/T (16)
• From equation (16) and (14)-
• ΔH/RT2
= d(lnKc)/dT + Δn/T
• d(lnKc)/dT = ΔH/RT2
- Δn/T (17)
• = ΔH/RT2
- ΔnRT/RT2
(18)
• = (ΔH – ΔnRT)/RT2
• But (ΔH – ΔnRT) = ΔE,
• d(lnKc)/dT = ΔE/RT2
(19)
• Where, ΔE is the heat of reaction at constant volume.

33. Thermodynamics of open system
Apply thermodynamics to mixtures
of variable composition
Thermodynamic properties of
single component
Thermodynamic properties of
single component
Partial properties
The equilibrium
thermodynamic state of a
simple one-component open
system can be specified
n, the amount of the
single component
Pressure
Temperatur
e
Open system
Properties
of
Partial Molar quantities
(property per unit mole)

34. Thermodynamics of open system
Differential relation for a general
extensive quantity, X, in a one-
component system
Explain by T, P and n
Molar quantity
Xm
Molar quantity
Xm
The molar quantity Xm is an intensive quantity. Because an intensive quantity cannot
depend on an amount of matter, Xm depends only on T and P.
Partial molar
quantity
Similarly define S, G, H, V, etc

35. Molar quantity
Partial molar Gibbs free energy or
chemical potential µ
,
For n moles of ideal gas
From this derive equation for S, H, V, A, etc
Hints

36. Partial molar quantity
A partial molar property is the contribution (per mole) that a substance makes to an overall
property of a mixture. Dependence of one variable (property) with respect to particular
variable by keeping other variables constants is called as partial molal quantities.
Let X is any extensive property of the system is said to be function of P and T and the
number of moles of the various constituents present in it.
Multi – component system
X = f (P, T, n1, n2, n3, …….,nj )
Differentiate X completely -
dX = (dX∕dT)P,ni dT + (dX∕dP)T,ni dP +
(dX∕dn1)P,T,ni dn1 + (dX∕dn2)P,T,ni dn2 + ……..
+ (dX∕dnj)P,T,ni dnj
ni represents in each instance all the n’s
except the one with respect to which X
is differentiated
partial molal
quantity (Xj)

37. Partial molar quantity
Therefore, X1
= (dX∕dn1
)P,T,nj
, X2
= (dX∕dn2
)P,T,nj
,…….., Xj
= (dX∕dnj
)P,T,ni
dX = (dX∕dT)P,ni dT + (dX∕dP)T,ni dP + X1 dn1 + X2 dn2+…….. + Xj dnj
At constant temperature and pressure -
(dX)T,P
= X1
dn1
+ X2
dn2
+…….. + Xj
dnj
On integration
X = X1 n1 + X2 n2+…….. + Xj nj
Equation shows that any extensive property of a mixture at constant T and P, is
expressed as a sum of X x n for individual components of solution. In each product, the
‘n’ represents a capacity factor and X represents a partial molal quantity represent an
intensive factor.

38. X = X1 n1 + X2 n2+…….. + Xj nj
Partial molar quantity
Differentiate
it completely
(dX)T,P
= X1
dn1
+ X2
dn2
+…….. + Xj
dnj
+ n1
dX1
+ n2
dX2
+…….. + nj
dXj
(dX)T,P
= X1
dn1
+ X2
dn2
+…….. + Xj
dnj
But Put it into
n1 dX1 + n2 dX2+…….. + nj dXj = 0
For two component
system
n1 dX1 + n2 dX2 = 0
Gibb’s - Duhem
equation (put X = G/μ)

39. Variation of chemical potential with temperature and pressure
The general expression for the differential change of the Gibbs energy, G, is given by:
dG = - S dT + V dP
Relation between the change of G with respect
to a change of pressure at constant
temperature
On
integration
PV =
nRT
= = nRT ln
OROR

40. Concept of activity
chemical potential
The activity is a dimensionless quantity that is equal to unity if the substance is in its
standard state.
We define the activity coefficient γi of a non-ideal gas as –
So that the activity coefficient of a gas equals the ratio of the fugacity to the pressure:
The activity coefficient of an ideal gas equals unity. The chemical potential of a non-ideal
gas can be written –