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