Solution thermodynamics theory—Part IV
Chapter 11
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
)ln()( PyRTT iiig
i
iig
iigi
iig
i
yRTPTG
PRTTG
ln),(
ln)(
iiidi xRTPTG ln),(
This eqn. is obtained by combining
Component i in a mixtureof ideal gases
Now we define
Ideal solution model
Other thermodynamic propertiesfor the ideal solution: partial molar volume
iii
idi
ii
id
i
T
i
xT
idiid
i
VxVxV
VP
G
P
GV
,
iiid
i xRTPTGG ln),(
partial molar entropy in the ideal solution
ii i
iiiid
ii
iid
iii
P
i
xP
idiid
i
xxRSxSxS
xRSxRT
G
T
GS
ln
lnln,
iiid
i xRTPTGG ln),(
partial molar enthalpy in the ideal solution
ii
iidi
ii
id
iiiiiid
iid
iidi
HxHxH
HxRTTSxRTGSTGH
lnln
iiid
i xRTPTGG ln),(
Chemical potential ideal solution
i
iii f
fRTG
ˆln
iii fRTTG ln)(
iii fRTT ˆln)( Chemical potential component i in a Real solution
Chemical potential Pure component i
Subtracting:
For the ideal solution
i
idi
iid
i f
fRTG
ˆln
Lewis-Randall rule
iid
i
iiid
i fxf
ˆ
ˆi
idi
iid
i
iiid
i
f
fRTG
xRTG
ˆln
ln
Lewis-Randall rule
(Dividing by Pxi each side of the equation)
When is the ideal solution valid?
• Mixtures of molecules of similar size and similar chemical nature
• Mixtures of isomers• Adjacent members of homologous series
Virial EOS applied to mixtures
RT
BPZ 1
ijji j
i ByyB
How to obtain the cross coefficients Bij
10ˆ BBB ijij
cij
cijij
ij T
PBB ˆ
Mixing rules for Pcij, Tcij, ij, 11-70 to 11.73
Fugacity coefficient from virial EOS
i
22111212
1222111
2
ˆln
BBB
yBRT
P
For a multicomponent mixture, see eqn. 11.64
problem• For the system methane (1)/ethane (2)/propane (3)
as a gas, estimate
at T = 100oC, P = 35 bar, y1 =0.21, and y2 =0.43
321321ˆ,ˆ,ˆˆ,ˆ,ˆ and fff
Pyf
BBT
P
kid
kid
k
kkkkrk
rkidk
ˆ
)(exp 10
Assume that the mixture is an ideal solution
Obtain reduced pressures, reduced temperatures, and calculate
Results: methane (1) ethane (2) propane (3)
barfff
barfff
ididid 57.9ˆ;25.13ˆ;18.7ˆ
76.0;88.0;98.0
76.9ˆ;25.13ˆ;49.7ˆ
78.0ˆ;88.0ˆ;02.1ˆ
321
321
321
321
Virial model
Ideal solution
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
idE MMM The most important excess function is the excess Gibbs free energy GE
Excess entropy can be calculated from the derivative of GE wrt T
Excess volume can be calculated from the derivative of GE wrt P
And we also define partial molar excess properties
:
ln)(
ˆln)(
gsubtractin
fxRTTG
fRTTG
iiiid
i
iii
ii
ii fx
f Definition of activity coefficient
Summary
iR
i
iE
i
RTG
RTG
ˆln
ln
Summary
iiii
iiidi
iigi
igi
xRTG
xRTG
yRTG
ln
ln
ln
Note that:
ii
iiiiiiii
idii
iiiiiidi
iiiigi
igi
fRTT
fxRTTxRTG
fRTT
fxRTTxRTG
PyRTTyRTG
ˆln)(
ln)(ln
ˆln)(
ln)(ln
ln)(ln
problem• The excess Gibbs energy of a binary liquid mixture
at T and P is given by
2121 )8.16.2(/ xxxxRTGE
a) Find expressions for ln 1 and ln 2 at T and P
Using x2 =1 – x1
GE/RT= x12 -1.8 x1 +0.8 x1
3
Since i comes from
iE
i RTG lnWe can use eqns 11.15 and 11.16
21
12
11
21
ln
ln
RTdx
dGxGG
RTdx
dGxGG
EEE
EEE
then
211
1
4.228.1/
xxdx
RTdGE
31
212
31
2111
6.1ln
6.14.128.1ln
xx
xxx
And we obtain
If we apply the additivity rule and the Gibbs-Duhem equation
0ln
ln
ii
i
ii
i
E
dx
xRT
G
At T and P
(b and c) Show that the ln i expressions satisfy these equations Note: to apply Gibbs-
Duhem, divide the equation by dx1 first
Plot the functions and show their values
-3
-2.5
-2
-1.5
-1
-0.5
0
0 0.2 0.4 0.6 0.8 1
x1
GE/RT
ln g2
ln g1
ln 2
ln 1
GE/RT