Solution thermodynamics
theory
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 for each phase, 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 PE 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
about
• 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 PE (phase
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
V1  18.07cm / mol water
3
3
3
3
solution
• Calculate total molar volume
• We know the total volume, calculate the number of
moles required, n
• Calculate n1 and n2
• Calculate the volume of each pure species
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?
Read and work 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.
Now we are going to start looking at models
for the chemical potential of a given
component in a mixture
• The first model is the ideal gas mixture
• The second model is the ideal solution
• As you study this, think about the differences,
not only mathematical but also the physical
differences of these models
The ideal-gas mixture model
• EOS for an ideal gas
• Calculate the partial molar volume for an ideal
gas in an ideal gas mixture
For an ideal gas mixture
Vi (T , P)  Vi (T , P)  V (T , P)
ig
ig
and
RT
pi  yi P  yi ig
V
ig
For any partial molar property other than
volume, in an ideal gas mixture:
M (T , P)  M (T , pi )
ig
i
ig
i
H (T , P)  H (T , pi ) 
ig
i
ig
i
dSiig   Rd ln P
at constant T
integrate from pi to P
Partial molar entropy (igm)
S (T , pi )  S (T , P)  R ln yi
ig
i
ig
i
S (T , P)  S (T , pi )  S (T , P)  R ln yi
ig
i
ig
i
ig
i
Partial molar Gibbs energy
Gi ig  H iig  TSiig  H iig  TSiig  RT ln yi
iig  Gi ig  Giig  RT ln yi
Chemical potential of
component i in an
ideal gas mixture
*******************************************************************************
RT
dG  Vi dP 
dP  RTd ln P at constant T
P
integrating between P  1atm and P
ig
i
ig
Giig  i (T )  RT ln P
This is  for a pure component !!!
iig  i (T )  RT ln P  RT ln yi  i (T )  RT ln yi P
Problem
• What is the change in entropy when 0.7 m3 of CO2
and 0.3 m3 of N2, each at 1 bar and 25oC blend to
form a gas mixture at the same conditions? Assume
ideal gases.
We showed that:
Siig  Siig  R ln yi
S   yi S
ig
ig
i
i
  yi S  R  yi ln yi
ig
i
i
S   yi S   R  yi ln yi
ig
ig
i
i
i
i
solution
S  nR yi ln yi
i
n = PV/RT= 1 bar 1 m3/ [R x 278 K]
S = 204.89 J/K
Problem
• What is the ideal work for the separation of an
equimolar mixture of methane and ethane at
175oC and 3 bar in a steady-flow process into
product streams of the pure gases at 35oC and
1 bar if the surroundings temperature Ts =
300K?
1) Read section 5.8 (calculation of ideal work)
2) Think about the process: separation of gases and change of state
First calculate H and S for methane and for ethane changing their state from
P1, T1, to P2T2
Second, calculate H for de-mixing and S for de-mixing from a mixture of ideal
gases
solution
ig
T2
H i   Cp (T )dT
ig
T1
Si  
ig
T2
T1
ig
dT
P2
Cp (T )
 ln
T
P1
ig
H de
 mix  0
ig
S de
 mix  R  yi ln yi
i
ig
H   yi H i  H de
 mix
= -7228 J/mol
i
ig
S   yi S i  S de
 mix
= -15,813 J/mol K
i
Wideal = H – Ts S = -2484 J/mol
Now we introduce a new concept: fugacity
• When we try to model “real” systems, the
expression for the chemical potential that we
used for ideal systems is no longer valid
• We introduce the concept of fugacity that for
a pure component is the analogous (but is not
equal) to the pressure
We showed that:
Giig  i (T )  RT ln P
i  i (T )  RT ln( yi P)
ig
Pure component i, ideal gas
Component i in a mixture
of ideal gases
Let’s define:
Gi  i (T )  RT ln f i
For a real fluid, we define
Fugacity of pure species i
Residual Gibbs free energy
fi
G  Gi  G  RT ln
P
R
i
ig
i
G  RT ln i
R
i
Valid for species i
in any phase and
any condition
Since we know how to calculate residual
properties…
G  RT ln i
R
i
R
i
P
G
dP
 ln i   ( Z i  1)
0
RT
P
Zi from an EOS, Virial, van der Waals, etc
examples
• From Virial EOS
Bii P
ln i 
RT
• From van der Waals EOS
bi P 
ai P

ln i  Z i  1  ln  Z i 
 2 2
RT  R T Z i

Fugacities of a 2-phase system
G  i (T )  RT ln f i
v
i
G  i (T )  RT ln f i
l
i
v
l
One component, two phases:
saturated liquid and saturated vapor at Pisat and Tisat
What are the equilibrium conditions for a pure component?
Fugacity of a pure liquid at P and T
v
sat
l
sat
l
f i ( Pi ) f i ( Pi ) f i ( P) sat
f i ( P) 
Pi
sat
v
sat
l
sat
Pi
f i ( Pi ) f i ( Pi )
l
Fugacity of a pure liquid at P and T
f i ( P)   P
l
sat sat
i
i
1
exp
RT

P
Pi
l
V
dP
i
sa t
example
• For water at 300oC and for P up to 10,000 kPa (100 bar)
calculate values of fi and i from data in the steam tables
and plot them vs. P
*

fi
1
1 Hi  Hi
*
* 
ln * 
(Gi  Gi )  
 ( Si  Si ) 
RT
R T
fi

At low P, steam is an ideal gas => fi* =P*
Get Hi* and Si* from the steam tables at 300oC and the lowest P, 1 kPa
Then get values of Hi and Si at 300oC and at other pressure P and calculate fi (P)
Problem
• For SO2 at 600 K and 300 bar, determine good
estimates of the fugacity and of GR/RT.
SO2 is a gas, what equations can we use to calculate f =
/P
Find Tc, Pc, and acentric factor, w, Table B1, p. 680
Calculate reduced properties: Tr, Pr
Tr=1.393 and Pr=3.805
High P, high T, gas: use Lee-Kessler
correlation
• From tables E15 and E16 find 0 and 1
• 0 = 0.672; 1 = 1.354
•  = 0 1w  0.724
• f =  P = 0.724 x 300 bar = 217.14 bar
• GR/RT = ln   0.323
Problem
• Estimate the fugacity of cyclopentane at 110oC and
275 bar. At 110 oC the vapor pressure of
cyclopentane is 5.267 bar.
• At those conditions, cyclopentane is a high P liquid
1
l
sat sat
f i ( P)  i Pi exp
RT

P
Pi
sa t
l
Vi dP
Find Tc, Pc, Zc,, Vc and acentric factor, w, Table B1, p.
680
Calculate reduced properties: Tr, Prsat
Tr = 0.7486 and Prsat = 0.117
At P < Prsat we can use the virial EOS to calculate isat
 Pr 0
1 
  exp  ( B  wB )
 Tr

0.422 1
0.172
0
B  0.083  1.6 ; B  0.139  4.2
Tr
Tr
sat
i
isat = 0.9
P-correction term:
Get the volume of the saturated liquid phase, Rackett equation
V
sat
 Vc Z
(1Tr )
c
2/7
Vsat = 107.55 cm3/mol
1
l
sat sat
f i ( P)  i Pi exp
RT
f = 11.78 bar

P
Pi
sa t
l
Vi dP
Generalized correlations: fugacity coefficient
GiR  RT ln i
P
GiR
dP
 ln i   ( Z i  1)
0
RT
P
P  Pc Pr
ln i  
Pr
ln i  
Pr
0
0
dPr
( Z i  1)
Pr
Pr
dPr
1 dPr
( Z  1)
 w Z
0
Pr
Pr
0
ln i  ln   w ln 
0
   0 ( 1 )w
1
Tables E13 to E16
Lee-Kessler
Now we introduce the concept of fugacity for
a given component in a mixture
• Fugacity of component i in a mixture plays an
analogous role to the partial pressure of the
same component i in an ideal mixture
• At low pressure, the fugacity of i in the
mixture tends to be the partial pressure of i.
• This means that the fugacity coefficient of i in
the mixture tends to 1 at low pressures
We showed that:
Giig  i (T )  RT ln P
i  i (T )  RT ln( yi P)
ig
Pure component i, ideal gas
Component i in a mixture
of ideal gases
Let’s define:
Gi  i (T )  RT ln f i
For a real fluid, we define
Fugacity of pure species i
i  i (T )  RT ln( yi P)
ig
Component i in a mixture
of ideal gases
Now lets consider component i in solution
i  i (T )  RT ln fˆi
Component i in a real
solution
Fugacity of component i in solution
We can also define a fugacity coefficient of i in solution
We can also relate fugacity (in solution) to
a residual property
M  M M
R
ig
M  Mi  M
R
i
ig
i
Residual partial Gibbs free energy
Gi  Gi  Gi
R
ig
ˆ
f
ig
i
i  i  RT ln
yi P
How to calculate fugacity coefficients in
solution
R


(
nG
/ RT ) 
ˆ
ln i  

ni

 P ,T ,n j
dP
nG / RT   (nZ  n)
0
P
P   ( nZ  n)  dP
P
dP
ˆ
ln i   
  ( Z i  1)

0
0

n
P
P
i


R
P
How to calculate fugacity coefficients in
solution
BP
Z  1
RT
Valid for mixtures of gases at
low and
moderated pressures
B   yi y j Bij
i
j
for a binary
B  y B11  2 y1 y2 B12  y B22
2
1
2
2
How to calculate fugacity coefficients in
solution
 (nZ i ) 
Zi  

 ni  P ,T ,n j
From the Virial EOS (truncated after the second term)
P   (nB ) 
Z1  1 


RT  n1  T ,n
2
You can show that
P
2
ˆ
ln 1 
( B11  y2 (2 B12  B11  B22 ))
RT
P
2
ˆ
ln 2 
( B22  y1 (2 B12  B11  B22 ))
RT
Derivation is in page 406, but try to do it by yourself first
problem
• For the system methane (1)/ethane (2)/propane (3)
as a gas, estimate
fˆ1 , fˆ2 , fˆ3 and ˆ1 , ˆ2 , ˆ3
at T = 100oC, P = 35 bar, y1 =0.21, and y2 =0.43


1
 Bkk   yi y j (2 ik   ij )
2 i j


 ik  2 Bik  Bii  Bkk
P
ˆ
ln k 
RT
 ii  0
 ik   ki
solution
1) Get Tc, Pc, and Vc for each of the components
2) Calculate mixture properties:
wij, TCij, PCij, Zcij and Vcij equations 11.70 to 11.74
For example we will have T11, T22, T33, T12, T13, T23
Bˆ ij  B 0  wij B1
0.422 1
0.172
B  0.083  1.6 ; B  0.139  4.2
Tr
Tr
0
Bij Pcij
ˆ
Bij 
RTcij
From this equation get Bij that we need to
calculate ik
For example
13  2B13  B11  B33
Results of  in cm3/mol
11 = 0, 22 =0, 33 =0
12 = 30.442, 13 = 107.809, 23 =23.482

P 
1
ˆ
ln k 
 Bkk   yi y j (2 ik   ij )
RT 
2 i j

 ik  2 Bik  Bii  Bkk
 ii  0
 ik   ki
fˆk  ˆk yk P
results
ˆ1  1.02; ˆ2  0.88; ˆ3  0.78
fˆ1  7.49; fˆ2  13.25; fˆ3  9.76bar
When we deal with mixtures of liquids or
solids
• We define the ideal solution model
• Compare it to the ideal gas mixture, analyze
its similarities and differences
The ideal solution model
i  i (T )  RT ln( yi P)
ig
Component i in a mixture
of ideal gases
That is obtained by using
Gi  i (T )  RT ln P
ig
In the first term
of this equation
  Gi (T , P)  RT ln yi
ig
i
ig
Now we define
  Gi (T , P)  RT ln xi
id
i
id
Ideal solution model
Other thermodynamic properties
for the ideal solution: partial molar volume
Gi  Gi (T , P)  RT ln xi
id
id
 Gi
 
 P
id
Vi
id
V
id

 G
  
T , x  P
id
i
  xiVi   xiVi
id
i
i

  Vi
T
partial molar entropy in the ideal solution
Gi  Gi (T , P)  RT ln xi
id
S
id
i
 Gi id
 
 T
id

 Giid
  
 P, x
 T

  R ln xi  Si  R ln xi
P
S id   xi Siid   xi Si  R xi ln xi
i
i
i
partial molar enthalpy in the ideal solution
Gi  Gi (T , P)  RT ln xi
id
id
H iid  Gi id  TSiid  Gi  RT ln xi  TSi  RT ln xi  H i
H id   xi H iid   xi H i
i
i
Chemical potential ideal solution
i  i (T )  RT ln fˆi
Chemical potential component i
in a Real solution
Gi  i (T )  RT ln f i
Chemical potential
Pure component i
Subtracting:
For the ideal solution
fˆi
i  Gi  RT ln
fi
i
id
ˆf id
 Gi  RT ln i
fi
Lewis-Randall rule
i  Gi  RT ln xi
id
i
id
ˆf id
 Gi  RT ln i
fi
ˆf id  x f
i
i i
id
ˆ
i  i
Lewis-Randall rule
When is the ideal solution valid?
• Mixtures of molecules of similar size and
similar chemical nature
• Mixtures of isomers
• Adjacent members of homologous series
problem
• For the system methane (1)/ethane (2)/propane (3)
as a gas, estimate
fˆ , fˆ , fˆ and ˆ , ˆ , ˆ
1
2
3
1
2
3
at T = 100oC, P = 35 bar, y1 =0.21, and y2 =0.43
Assume that the mixture is an ideal solution
Obtain reduced pressures, reduced temperatures, and calculate
k
id
 Prk 0
1 
 exp  ( Bkk  wBkk )
 Trk

ˆf id   id y P
k
k
k
Results: methane (1) ethane (2) propane
(3)
ˆ1  1.02; ˆ2  0.88; ˆ3  0.78
Virial model
ˆf  7.49; fˆ  13.25; fˆ  9.76bar
1
2
3
1  0.98; 2  0.88; 3  0.76
Ideal solution
ˆf id  7.18; fˆ id  13.25; fˆ id  9.57bar
1
2
3
Now we want to define a new type of
residual properties
• Instead of using the ideal gas as the reference,
we use the ideal solution
Excess properties
M  M M
E
id
The most important excess function is the excess Gibbs free energy GE
Excess entropy can be calculated from the derivative of GE with respect to T
Excess volume can be calculated from the derivative of GE with respect to P
And we can also define partial molar excess properties
ˆ
Gi  i (t )  RT ln f i
Gi  i (t )  RT ln xi f i
id
subtracting :
ˆf
i
i 
xi f i
Definition of activity coefficient
Summary
Gi  RT ln  i
R
ˆ
G  RT ln 
E
i
i
Summary
  G  RT ln yi
ig
i
ig
i
  Gi  RT ln xi
id
i
i  Gi  RT ln  i xi
HW # 3, Due Monday, September 17
• Problems 11.2, 11.5, 11.8, 11.12, 11.13
HW # 4, Due Monday, September 24
• Problems 11.18, 11.19 (b), 11.21, 11.22
(a), 11.24(a), 11.25