Solution thermodynamics
theory—Part I
Chapter 11
topics
• Fundamental equations for mixtures
• Chemical potential
• Properties of individual species in solution
(partial properties)
• Mixtures of real gases
• Mixtures of real liquids
A few equations
G  H  TS
For a closed system
d (nG )  d (nH )  Td (nS )  (nS )dT
from
H  U  PV
obtain d(nH)
d (nG )  (nV )dP  (nS )dT
Total differential form, what are (nV) and (nS)
Which are the main variables for G??
What are the main variables for G in an open system of
k components?
G in a mixture (open system)
  (nG ) 
  (nG ) 
d (nG )  
dP  
dT 


 P  T ,n
 T  P ,n
G in a mixture of k components at T and P
k
d (nG )  (nV )dP  (nS )dT   i dni
i 1
How is this equation reduced if n =1?
2 phases (each at T and P) in a closed system
Apply this equation to each phase
k
d (nG )  (nV )dP  (nS )dT   i dni
i 1
Sum the equations of both phases, take into account that
(nM )  (nM )  (nM ) 
In a closed system:
d (nG )  (nV )dP  (nS )dT
We end up with

i

i
dni   i dni  0



i
How are dni and dni related at constant n?
For 2 phases, k components at equilibrium






T T
P P
i  i
For all i = 1, 2,…k
Thermal equilibrium
Mechanical equilibrium
Chemical equilibrium
In order to solve the VLE problem
• Need models for i in each phase
• Examples of models of i in the vapor phase
• Examples of models of i in the liquid phase
Now we are going to learn:
• Partial molar properties
• Because the chemical potential is a partial molar
property
• At the end of this section think about this
– What is the chemical potential in physical terms
– What are the units of the chemical potential
– How do we use the chemical potential to solve a VLE
(vapor-liquid equilibrium) problem
Partial molar property
 (nM ) 
Mi  


n
i

 P ,T ,n ji
Solution property
Partial property
Pure-species property
example
 (nV ) 
Vi  

 ni  P ,T ,n ji
 (nV )  Vw nw
~
 (nV )  Vw nw
lim nw 0
Open beaker: ethanol + water, equimolar
Total volume nV
T and P
Add a drop of pure water, nw
Mix, allow for heat exchange, until temp T
Change in volume ?
Total vs. partial properties
M   xi M i
i
nM   ni M i
i
See derivation page 384
Derivation of Gibbs-Duhem equation
 M 
 M 
dM  
 dP  
 dT   M i dxi
 P T , x
 T  P , x
i
M   xi M i
i
Gibbs-Duhem at constant T&P
 x dM
i
i
0
constant T & P
i
Useful for thermodynamic consistency tests
Binary solutions
See derivation page 386
dM
M 1  M  x2
dx1
dM
M 2  M  x1
dx1
Obtain dM/dx1 from (a)
Example 11.3
• We need 2,000 cm3 of antifreeze solution: 30
mol% methanol in water.
• What volumes of methanol and water (at 25oC)
need to be mixed to obtain 2,000 cm3 of
antifreeze solution at 25oC
• Data:
V1  38.63cm / mol
V1  40.73cm / mol methanol
V2  17.77cm / mol
V2  18.07cm / mol water
3
3
3
3
solution
• Calculate total molar volume of the 30% mixture
• We know the total volume, calculate the number of
moles required, n
• Calculate n1 and n2
• Calculate the total volume of each pure species needed
to make that mixture
Note curves for partial molar volumes
From Gibbs-Duhem:
 x dM
i
i
0
constant T & P
i
x1dV1  x2 dV2  0
Divide by dx1, what do you conclude respect to the slopes?
Example 11.4
• Given H=400x1+600x2+x1x2(40x1+20x2) determine
partial molar enthalpies as functions of x1, numerical
values for pure-species enthalpies, and numerical
values for partial enthalpies at infinite dilution
• Also show that the expressions for the partial molar
enthalpies satisfy Gibbs-Duhem equation, and they
result in the same expression given for total H.