Solution thermodynamics theory—Part I
Chapter 11
topics
• Fundamental
Fundamental equations for mixtures
equations for mixtures
• Chemical potential
• Properties of individual species in solution i
f i di id l
i i
l i
(partial properties)
• Mixtures of real gases • Mixtures of real liquids q
A few equations
For a closed system
G  H  TS
d (nG )  d (nH )  Td (nS )  (nS )dT
f
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)
G in a mixture (open system)
  (nG ) 
  (nG ) 
d (nG )  
dP  
dT 


 P  T ,n
 T  P ,n
G a
G in a mixture of k components at T and P
tu e o co po e ts at a d
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 ) 
I a closed
In
l
d system:
t
d (nG )  (nV )dP  (nS )dT
We end up with
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
For 2 phases, k components at equilibrium






T T
P P
i  i
For all i = 1, 2,…k
Thermal equilibrium
Mechanical equilibrium
q
Chemical equilibrium
In order to solve the VLE problem
In order to solve the VLE problem
• Need models for 
Need models for i in each phase
in each phase
• Examples of models of 
l
f
d l f i in the vapor phase
h
h
• Examples of models of i in the liquid phase
Now we are going to learn:
• Partial molar properties
p p
• Because the chemical potential is a partial molar p
p
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
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
Partial molar property
  (nM ) 
Mi  


n
i

 P ,T ,n j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 nnw 0
Open beaker: ethanol + water
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
Total vs. partial properties
M   xi M i
i
nM   ni M i
i
See derivation page 384
Derivation of Gibbs‐Duhem
Derivation of Gibbs
Duhem equation
equation
 M 
 M 
dM  
 dP  
 dT   M i dxi
 P T , x
 T  P , x
i
M   xi M i
i
Gibbs‐Duhem
Gibbs
Duhem at constant T&P
at constant T&P
 x dM
i
i
0
constant T & P
i
Useful for thermodynamic consistency tests
Binary solutions
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
Example 11.3
• We
We need 2,000 cm
need 2,000 cm3 of antifreeze solution: 30 of antifreeze solution: 30
mol% methanol in water.
oC) • What volumes of methanol and water (at 25
(
)
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  17.77cm3 / mol
V2  18.07cm3 / mol water
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:
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
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
values for partial enthalpies at infinite dilution
• Also
Also show that the expressions for the partial molar show that the expressions for the partial molar
enthalpies satisfy Gibbs‐Duhem equation, and they result in the same expression given for total H.