Thermodynamics,Gibs energy and phase equilibrium

1. Solution Thermodynamics
Partial molal property:
NF = f(P,T,𝑵𝟏,𝑵𝟐……………..) ..(1)
d(NF) =
𝝏(𝑵𝑭)
𝝏𝑷 𝑻,𝑵𝟏,𝑵𝟐,….
dP +
𝝏(𝑵𝑭)
𝝏𝑻 𝑷,𝑵𝟏,𝑵𝟐…….
dT + 𝒊
𝝏(𝑵𝑭)
𝝏𝑵𝒊 𝑷,𝑻,𝑵𝒋≠𝒊
d𝑵𝒊
NdF + FdN = N(
𝝏𝑭
𝝏𝑷
)𝑻,𝑵𝒊
dP + N(
𝝏𝑭
𝝏𝑻
)𝑷,𝑵𝒊
dT + 𝑭𝒊d𝑵𝒊
Where, 𝑭𝒊 =
𝝏(𝑵𝑭)
𝝏𝑵𝒊 𝑷,𝑻,𝑵𝒋≠𝒊
..(2)
This is called partial molal property.

2. 𝑵𝒊 = 𝒀𝒊 N
d𝑵𝒊 = 𝒀𝒊 dN + d𝒀𝒊 N
So NdF + FdN = N(
𝝏𝑭
𝝏𝑷
)𝑻,𝑵𝒊
dP + N(
𝝏𝑭
𝝏𝑻
)𝑷,𝑵𝒊
dT + 𝑭𝒊 (𝒀𝒊 dN + d𝒀𝒊 N)
N 𝒅𝑭 − (
𝝏𝑭
𝝏𝑷
)𝑻,𝑵𝒊
𝒅𝑷 − (
𝝏𝑭
𝝏𝑻
)𝑷,𝑵𝒊
𝒅𝑻 − 𝐅𝐢 𝐝𝐘𝐢 + 𝑭 − 𝒀𝒊𝑭𝒊 dN = 0
dF = (
𝝏𝑭
𝝏𝑷
)𝑻,𝑵𝒊
𝒅𝑷 + (
𝝏𝑭
𝝏𝑻
)𝑷,𝑵𝒊
𝐝𝐓 + 𝐅𝐢 𝐝𝐘𝐢 ..(3)
F = 𝒊 𝒀𝒊𝑭𝒊 ..(4)
(one may choose any N & dN,so they are independent & arbitrary)
NF = 𝒊 𝑵𝒊𝑭𝒊 ..(5)
dF = 𝒊 𝑭𝒊d𝒀𝒊 (at constant P,T) ..(6)

3. d(NG) =
𝜕(𝑁𝐺)
𝜕𝑃 𝑇,𝑁1,𝑁2,….
dP +
𝜕(𝑁𝐺)
𝜕𝑇 𝑃,𝑁1,𝑁2…….
dT + 𝑖
𝜕(𝑁𝐺)
𝜕𝑁𝑖 𝑃,𝑇,𝑁𝑗≠𝑖
d𝑁𝑖
 
i
j
N
T
P
i
i G 



 ,
,
i
]
N
NG
[

This is Partial Molal Gibb’s Free Energy or Chemical Potential
 
i
,
,
i
dN
]
N
NG
[
Pd(NV)
-
Td(NS)
d(NU)  



 i
j
N
v
s
 
i
,
,
i
dN
]
N
NH
[
vd(NP)
Td(NS)
d(NH)  




 i
j
N
P
s
 
i
,
,
i
dN
]
N
NA
[
Pd(NV)
-
d(NT)
d(NA)  




 i
j
N
v
T
S
 
i
,
,
i
dN
]
N
NG
[
vd(NP)
Sd(NT)
d(NG)  





 i
j
N
T
P

4. NvdP
-
Td(NS)
-
(NH)
NvdP
-
NSdT
d(NG)
NSdT
-
Td(NS)
-
(NH)
d(NG)
T(NS)
-
(NH)
(NG)
d
d




   
   
i
i
i
i
i
i
i
i
N
T
P
N
p
s
N
T
P
N
p
s
G
A
H
U
Similarly
G
H
i
j
ii
j
i
j
i
j





















 

,
]
N
NG
[
]
N
NH
[
dN
]
N
NG
[
dN
]
N
NH
[
,
,
i
,
,
i
i
,
,
i
i
,
,
i
i
i G


 
i
j
N
T
P 


 ,
,
i
i ]
N
NG
[

However, we use the definition,
As P,T are measurable variables and can be controlled
easily,

5. GIBBS DUHEM EQUATION
NG = 𝐺𝑖𝑁𝑖 ……………………..(5)
d(NG) = 𝐺𝑖𝑑𝑁𝑖 + 𝑁𝑖𝑑𝐺𝑖
d(NG) = 𝐺𝑖𝑑𝑁𝑖 - NSdT + N𝜗dP……………………..(11)
Subtracting 𝑁𝑖𝑑𝐺𝑖 + NSdT – N𝜗dP = 0……………………..(15)
This is Gibb’s – Duhem eqn.
At constant P, T, 𝑁𝑖𝑑𝐺𝑖 = 𝑁𝑖 𝑑𝜇𝑖 = 0 …………………….(16)
𝑌𝑖 𝑑𝜇𝑖 = 0 …………………….(17)

6. PHASE EQUILIBRIA:
Let us consider a closed multi-component system involving 2 phases a & b at P & T
They interact till they reach a state of eqbm.
d(𝑁𝐺)𝑎
= -(𝑁𝑆)𝑎
dT + (𝑁𝜗)𝑎
dP + 𝜇𝑖
𝑎
𝑑𝑁𝑖
𝑎
……………(18)
d(𝑁𝐺)𝑏 = -(𝑁𝑆)𝑏dT + (𝑁𝜗)𝑏dP + 𝜇𝑖
𝑏
𝑑𝑁𝑖
𝑏
………….(19)
d(NG) = d(𝑁𝐺)𝑎
+ d(𝑁𝐺)𝑏
At constant P & T,
d(NG) = 𝜇𝑖
𝑎
𝑑𝑁𝑖
𝑎
+ 𝜇𝑖
𝑏
𝑑𝑁𝑖
𝑏
……………..(20)

7. GIBB’S FREE ENERGY MINIMUM PRINCIPLE:
It states that in a state of eqbm. ,the Gibb’s Free Energy of a system at a constant P & T reaches a minimum value..
d(NG) = 0(at eqbm.) ……………….(21)
𝜇𝑖
𝑎
𝑑𝑁𝑖
𝑎
+ 𝜇𝑖
𝑏
𝑑𝑁𝑖
𝑏
= 0 ………………..(22)
For closed system, d𝑁𝑖 = d𝑁𝑖
−𝑎
+ d𝑁𝑖
𝑏
= 0 ……………….(23)
d𝑁𝑖
−𝑏
= -d𝑁𝑖
𝑎
(𝜇𝑖
𝑎
− 𝜇𝑖
𝑏
) 𝑑𝑁𝑖
𝑎
= 0 ………………..(24)
𝜇𝑖
𝑎
= 𝜇𝑖
𝑏
……………….(25)
This is the criterion for eqbm. between two phases. chemical potential for each of the components
is identical in both the phases.

8. Fugacity of a component in a solution
For a pure fluid (ideal gas) dGT =VdP=RTdlnP
For a pure fluid (real gas) dGT =VdP=RTdlnf
For a component in a solution, fugacity is defined as
i
 =partial molal volume of i-th species
dp
f
RTd
d
G
d i
i 
 


^
ln ………………(26)
For a pure fluid:
 



P
dP
P
RT
f
f
RT
G
G
0
0
0
)
(
ln 
For a multi-component system:
^
0
^
0
ln
f
f
RT

 
 ………………(27)

9. i
f
ˆ
0
ˆ
i
f
= Fugacity of i-th species in the multi-component system
= Fugacity of i-th species at standard state
For gas phase, standard state is chosen at a pressure P where the gas behaves ideally
= pi (partial pressure) = yi P …………………… (28)
0
ˆ
i
f
p
y
f
RT
i
^
0
ln

 
 …………………… (29)

 0
ˆ
ˆ
ˆ
i
i
i
f
f
 )
ˆ
(
P
y
f
i
i
………………....... (30)
i
i
i RT 

 ˆ
ln
0

 ……………………… (31)

10. IDEAL GAS SOLUTION
Each species is uninfluenced by the presence of the other species.
Gibb’s theorem: A partial molal property(other than volume) of a constituent species in an
ideal gas mixture is equal to the corresponding molar property of the species as a pure ideal
gas at the mixture temperature but at a pressure equal to its partial pressure in the mixture.
)
,
(
)
,
( i
ig
i
ig
i p
T
F
P
T
F 

 i
iF
y
F
…………………………………..(32)
…………………………………..(4)
a. Enthalpy, ig
H
)
,
(
)
,
( P
T
H
p
T
H ig
i
i
ig
i  ig
H
(
)
,
(
)
,
(
)
,
( P
T
H
p
T
H
P
T
H ig
i
i
ig
i
ig
i 

ig
i
ig
i H
H 
is a function of T only) ……………………..(33)
……………………………..(34)

11.  

 ig
i
i
ig
i
i
ig
H
y
H
y
H ……………………………..(34)
So, change in enthalpy due to mixing, 0

 mixing
H

12. b. Entropy,
P
Rd
T
dT
C
dS ig
p
ig
i ln
/ 

At constant T,
𝑝𝑢𝑟𝑒
𝑚𝑖𝑥
𝑑𝑆𝑖
𝑖𝑔 = −
𝑃
𝑝𝑖
𝑅𝑑𝑙𝑛𝑃
i
i
i
ig
i
i
ig
i y
R
P
P
y
R
P
p
R
P
T
S
p
T
S ln
ln
ln
)
,
(
)
,
( 





i
ig
i
i
ig
i y
R
P
T
S
p
T
S ln
)
,
(
)
,
( 
 ……………………………..(36)
From equation 32, )
,
(
)
,
( i
ig
i
ig
i p
T
S
P
T
S  ………………………………..(37)
Putting equation (36) into (37)
i
ig
i
ig
i y
R
P
T
S
P
T
S ln
)
,
(
)
,
( 
 ……………………………….(38)
i
ig
i
ig
i y
R
S
S ln


……………………………(39)

13. Putting equation (39) into (4)
  


 i
i
ig
i
i
ig
i
i
ig
y
y
R
S
y
S
y
S ln

 
 i
i
ig
i
i
ig
y
y
R
S
y
S ln ………………………………..(40)
i
i
mixing y
y
R
S 


 ln
.
1

i
y .
0

 mixing
S
So, Entropy increases for mixing.

14. Gibb’s F ree energy:
𝐺𝑖𝑔 = 𝐻𝑖𝑔 -T𝑆𝑖𝑔
𝐺𝑖
𝑖𝑔
=𝐻𝑖
𝑖𝑔
-T𝑆𝑖
𝑖𝑔
---------(41)
Using Eq.s 34 & 39
𝐺𝑖
𝑖𝑔
=𝐻𝑖
𝑖𝑔
-T𝑆𝑖
𝑖𝑔
+ RT𝑙𝑛𝑦𝑖
𝐺𝑖
𝑖𝑔
=µ𝑖𝑔=𝐺𝑖𝑔 + RT𝑙𝑛𝑦𝑖 -----------(42)
𝐺𝑖𝑔= ⅀ 𝑦𝑖𝐺𝑖
𝑖𝑔
+ RT⅀ 𝑦𝑖𝑙𝑛𝑦𝑖 ----------(43)
∆𝐺𝑚𝑖𝑥𝑖𝑛𝑔 = RT⅀ 𝑦𝑖𝑙𝑛𝑦𝑖
∆𝐺𝑚𝑖𝑥𝑖𝑛𝑔 is negative.

15. Problem: A chamber with a partition contains 0.1 1b mole of 𝑁2 and 𝑂2
At 70° F & 1 atm. The partition is removed and the gases allowed to mix. C alculate the total changeof
entropy of the system due to mixing.
Solution:
∆S = - R⅀ 𝑦𝑖𝑙𝑛𝑦𝑖
∆S= -R [ 0.5ln 0.5 + 0.5ln 0.5 ]
= R ln 0.5 Btu/mole R
Total change in entropy
= -0.2 R ln 0.5
= 0.2 x 1.98 x ln 2 = 0.274 Btu/R.

16. Vapour Pressure:
When there is an equality between the rate of vapourisation and condension between the
liquid and the vapour phase , a dynamic equilibrium is attained. At this eqbm. the pressure
exerted by the vapour is called the saturation vapour pressure of the liquid at that temp.
Antoine eqn : ………..(44a)
P in Torr , t in ᵒC T↑ Ps ↓
Saturated Vapour:
When the partial pressure of a vapour is equal to the sat. vap. pressure at the existing
temperature, it is called saturated vapour. PA = PA
S ……….(44b)
'
'
'
10
log
C
t
B
A
s
P




17. Dew Point:
The temp at which the vapour becomes saturated at the existing pressure, is called th Dew point.
Boiling Point:
The tmp at which the sat. vapour pressure of a liq. Becomes equal to th total pressure on the surface, is called the boiling
point. PA
S = P ……………….. (44c)
Vapour Pressure of Immiscible Liquids:
PS = PA
S + PB
S ……………….. (45)
Vapour pressure exerted by a mixture of a immiscible liq is always greater than the vapour pressures of either the liquids, and its
boiling point is lower. This principle is used in steam Distillation.
Vapour pressure of solutions:
Vapour pressure exerted by a component in a solution of two or more liquids is always less than that of the pure substance.
Raoult’s Law(ideal Solution): Pi = xiPi
S …………………(45)
Pi = Vap. pressure of i-th species in the solution.
Pi
S = Sat. vap pressure of I in pure state at the existing

18. Henry’s Law : (Dilute solutions)
Pi = Hi xi ……(46)
Hi=Henry’s constant
Xi = Mole fraction of the solute i in the solution

19. Vapour-Liquid Equilibrium

20. Vapour-Liquid Equilibrium (VLE)
Lewis-Randall rule (ideal solution):
i
i
id
i
f
x
f 
ˆ ………..(47)
id
i
f
ˆ =Fugacity of i-th component in an ideal solution.
i
x =Mole fraction of i-th component in the solution.
i
f =Fugacity of pure species i in the same physical state as that of
the solution as that of the solution & at the same P & T.
a) For ideal gases & ideal solution; L.R. rule is reduced to Dalton’s law, ………….(48)
b) For ideal liquid solution L.R. rule is reduced to Raoult’s law. …………….(49)
P
y
p i
i 
P
s
i
i
i x
p 
b
i
a
i µ
µ 
……………(25)
Criterion for equilibrium:
Both the phases are at the same P & T. b
i
a
i f
f ˆ
ˆ 
l
i
v
i f
f ˆ
ˆ 
For VLE:
…………….(50)
…………….(51)

21. Non-ideal Solution:
Activity & Activity Coefficient : For an ideal solution L.R. rule is not valid. Interactions between like & unlike
molecules are not identical
^
^
id
i
i
i
f
f

 …………….(52)
i
i
i
id
i
i
i f
x
f
f 
 
 ˆ
ˆ
…………..(53)
i
i
i
f
f
a
^
 ……………(54)
i
i
i f
a
f 
ˆ …………….(55)
i
i
i x
a 

……………(56)
i

i
a
i
f
ˆ = Fugacity of i-th component in the solution.
= Activity
= Activity coefficient

22. At equilibrium
l
i
v
i f
f ˆ
ˆ 
p
y
f i
i
v
i
^
^


From eq.(30) :
……………………(51)
……………………(51)
i
i
i
l
i f
x
f 

^
So, at equbm
i
i
i
i
i f
x
p
y 
 
^
[General Eqn. for VLE] ………………..(57)
VLE of Non-ideal soln at Low Pressure:
At low pressure , the vapour phase behaves ideally s
i
i p
f 
1
^

i
 and
Therefore, s
i
i
i
l
i
i
i
v
i p
x
f
p
y
p
f 




^
^
s
i
i
i
i p
x
p
y 
 Non-ideal soln
………………………(58)
………………………(58)

 
 s
i
i
i
i p
x
p
p
y 

23. For ideal soln 1

i

s
i
i
i p
x
p
y 

 
 s
i
i
i p
x
p
p
y
……………..(60)
……………..(61)
Liquid –phase properties from VLE data: ( , )
^
l
i
f i

Prob: From a set of experimental VLE data of P-X1-Y1 for methyl-ethyl ketone (1)/toluene(2) system at 50ᵒC.
Calculate the liquid-phase properties.
Solution:
From equation (58)
From equation (58)
s
i
i
i
l
i
i
i
v
i p
x
f
p
y
p
f 




^
^
s
i
i
i
i
p
x
p
y


………………………….(A)
………………………….(B)
Equations (A) and (B) may be used to calculate ^
1
l
f
^
2
l
f 1
 2

and

24. VLE Data :
P(kPa) x1 y1
12.30( ) 0.0000 0.0000 0.0000 12.300 _ 1.000 0.000 _
15.51 0.0895 0.2716 4.2120 11.298 1.304 1.009 0.032 0.389
18.61 0.1981 0.4565 8.4960 10.114 1.188 1.026 0.054 0.342
21.63 0.3195 0.5934 12.8350 8.795 1.114 1.050 0.068 0.312
24.01 0.4232 0.6815 16.3630 7.697 1.071 1.078 0.072 0.297
25.92 0.5119 0.7440 19.2840 6.636 1.044 1.105 0.071 0.283
27.96 0.6096 0.8050 22.5080 5.542 1.023 1.135 0.063 0.267
30.12 0.7135 0.8639 26.0210 4.099 1.010 1.163 0.051 0.248
31.75 0.7934 0.9048 28.7270 3.023 1.003 1.189 0.038 0.234
34.15 0.9102 0.9590 32.7500 1.400 0.997 1.268 0.019 0.227
36.09( ) 1.0000 1.0000 36.0900 0.000 1.000 _ 0.000 _
P
y
f l
1
1
ˆ  P
y
f l
2
2
ˆ  S
P
x
P
y
1
1
1
1 
 S
P
x
P
1
2
2
2

  RT
GE
2
1 x
RTx
GE
S
P2
S
P1

25. L.R Rule
L.R Rule
𝑓2
𝑓1
𝑓𝑖
𝑓𝑖 = 𝛾𝑖𝑥𝑖𝑓𝑖
𝑥1
1. Actual 𝑓𝑖 line becomes tangent to the L.R line at 𝑥𝑖=1
2. At 𝑥𝑖=0 , 𝑓𝑖 = 0
3. At 𝑥𝑖 0 , using L’ Hospital rule, lim
𝑥𝑖→0
𝑓𝑖
𝑥𝑖
= lim
𝑥𝑖→0
𝑑𝑓𝑖
𝑑𝑥𝑖
= 𝐻𝑖 ..................................... (62)

26. L.R Rule
Actual
Henry’s Law
𝑓𝑖
0 1
𝐻𝑖 is Henry’s constant.
This is the slope of the straight line drawn tangent to the curve
at 𝑥𝑖=0. Henry’s law is valid at 𝑥𝑖 0 or for very dilute solution

27. Prove that Henry’s law is related to L.R rule or Raoult’s law (for ideal solution) through Gibb’s-Duhem equation.
[at constant T,P ] ………………………………(16)
1
2
^
2
1
2
^
1
1
1
^
2
2
1
^
1
2
2
1
1
0
ln
ln
0
ln
ln
ln
0
dx
dx
f
dx
d
x
f
dx
d
x
f
d
x
f
d
x
f
RT
d
d
x
d
x
i
i












^
2
2
2
^
2
^
1
1
1
^
1
^
2
2
^
2
2
^
1
1
1
^
1
^
2
2
2
^
1
1
1 ln
ln
f
x
dx
f
d
f
x
dx
f
d
f
dx
d
f
x
f
dx
d
f
x
f
dx
d
x
f
dx
d
x



………………………………(26)
………………………………(63)
………………………………(64)

28. 2
2
2
2
0
1
1
1
1
1 ˆ
ˆ
lim
ˆ
ˆ
lim
1
1
x
f
dx
f
d
x
f
dx
f
d
x
x 


1
1 
x 1
1
ˆ f
f 
1
)
ˆ
(
1
1
1
1
1
1


x
dx
f
d
f
So when Henry’s law is valid for one species (#2 here)
L.R. rule is valid for other species(#1 here)

30. (1) If L.R. rule is valid
(2) The plot shows that at
(3)
1

i

1

i
 1
1 
x
i
i
i
i
f
x
f
^

 …………..(53)