Solution thermodynamics
theory—Part IV
Chapter 11
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