Solution thermodynamics theory—Part IV Chapter 11.
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Transcript of Solution thermodynamics theory—Part IV Chapter 11.
![Page 1: Solution thermodynamics theory—Part IV Chapter 11.](https://reader031.fdocuments.us/reader031/viewer/2022012306/56649d6e5503460f94a4f0c9/html5/thumbnails/1.jpg)
Solution thermodynamics theory—Part IV
Chapter 11
![Page 2: Solution thermodynamics theory—Part IV Chapter 11.](https://reader031.fdocuments.us/reader031/viewer/2022012306/56649d6e5503460f94a4f0c9/html5/thumbnails/2.jpg)
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
![Page 3: Solution thermodynamics theory—Part IV Chapter 11.](https://reader031.fdocuments.us/reader031/viewer/2022012306/56649d6e5503460f94a4f0c9/html5/thumbnails/3.jpg)
)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
![Page 4: Solution thermodynamics theory—Part IV Chapter 11.](https://reader031.fdocuments.us/reader031/viewer/2022012306/56649d6e5503460f94a4f0c9/html5/thumbnails/4.jpg)
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),(
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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),(
![Page 6: Solution thermodynamics theory—Part IV Chapter 11.](https://reader031.fdocuments.us/reader031/viewer/2022012306/56649d6e5503460f94a4f0c9/html5/thumbnails/6.jpg)
partial molar enthalpy in the ideal solution
ii
iidi
ii
id
iiiiiid
iid
iidi
HxHxH
HxRTTSxRTGSTGH
lnln
iiid
i xRTPTGG ln),(
![Page 7: Solution thermodynamics theory—Part IV Chapter 11.](https://reader031.fdocuments.us/reader031/viewer/2022012306/56649d6e5503460f94a4f0c9/html5/thumbnails/7.jpg)
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
![Page 8: Solution thermodynamics theory—Part IV Chapter 11.](https://reader031.fdocuments.us/reader031/viewer/2022012306/56649d6e5503460f94a4f0c9/html5/thumbnails/8.jpg)
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)
![Page 9: Solution thermodynamics theory—Part IV Chapter 11.](https://reader031.fdocuments.us/reader031/viewer/2022012306/56649d6e5503460f94a4f0c9/html5/thumbnails/9.jpg)
![Page 10: Solution thermodynamics theory—Part IV Chapter 11.](https://reader031.fdocuments.us/reader031/viewer/2022012306/56649d6e5503460f94a4f0c9/html5/thumbnails/10.jpg)
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When is the ideal solution valid?
• Mixtures of molecules of similar size and similar chemical nature
• Mixtures of isomers• Adjacent members of homologous series
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Virial EOS applied to mixtures
RT
BPZ 1
ijji j
i ByyB
![Page 14: Solution thermodynamics theory—Part IV Chapter 11.](https://reader031.fdocuments.us/reader031/viewer/2022012306/56649d6e5503460f94a4f0c9/html5/thumbnails/14.jpg)
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
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Fugacity coefficient from virial EOS
i
22111212
1222111
2
ˆln
BBB
yBRT
P
For a multicomponent mixture, see eqn. 11.64
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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
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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
![Page 18: Solution thermodynamics theory—Part IV Chapter 11.](https://reader031.fdocuments.us/reader031/viewer/2022012306/56649d6e5503460f94a4f0c9/html5/thumbnails/18.jpg)
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
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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
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:
ln)(
ˆln)(
gsubtractin
fxRTTG
fRTTG
iiiid
i
iii
ii
ii fx
f Definition of activity coefficient
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Summary
iR
i
iE
i
RTG
RTG
ˆln
ln
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Summary
iiii
iiidi
iigi
igi
xRTG
xRTG
yRTG
ln
ln
ln
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Note that:
ii
iiiiiiii
idii
iiiiiidi
iiiigi
igi
fRTT
fxRTTxRTG
fRTT
fxRTTxRTG
PyRTTyRTG
ˆln)(
ln)(ln
ˆln)(
ln)(ln
ln)(ln
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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
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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
![Page 27: Solution thermodynamics theory—Part IV Chapter 11.](https://reader031.fdocuments.us/reader031/viewer/2022012306/56649d6e5503460f94a4f0c9/html5/thumbnails/27.jpg)
then
211
1
4.228.1/
xxdx
RTdGE
31
212
31
2111
6.1ln
6.14.128.1ln
xx
xxx
And we obtain
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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
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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
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