Solution Thermodynamics Theory-Ch 11

8/18/2019 Solution Thermodynamics Theory-Ch 11

1/50


2/50

topics

• Fundamental equations for mixtures

• Chemical potential

• Properties of individual species insolution (partial properties)

• Mixtures of real gases

• Mixtures of real liquids


3/50

A few equations

dT nS dP nV nGd

PV U H

from

dT nS nS Td nH d nGd TS H G

)()()(

d(nH)obtain

)()()()(

−=

+≡

−−=−≡ For a closed system

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?


4/50

G in a mixture (opensystem)

+

∂

∂+

∂

∂= dT

T

nGdP

P

nGnGd

n P nT ,,

)()()(


5/50

G in a mixture of components at !and P

i

k

iidndT nS dP nV nGd

∑=+−= 1)()()( µ

ow is this e!"ation red"ced if n #$


6/50

" phases (each at ! and P) in a closeds#stem

i

k

i

idndT nS dP nV nGd ∑=+−= 1)()()( µ

β α

)()()( nM nM nM +=

dT nS dP nV nGd )()()( −=

Apply this equation to each phase

Sum the equations for each phase, take into account that

In a closed system:


7/50

$e end up with

0=+ ∑∑ β β α α µ µ ii

ii

i

i dndn

How are dni and dni

β related at constant n?


8/50

For " phases% components atequili&rium

β α

β α

β α

µ µ ii

P P

T T

=

=

=

%or all i # $, &,'k

Thermal e!"ilibri"m

echanical e!"ilibri"m

hemical e!"ilibri"m


9/50

'n order to solve the Ppro&lem

• eed models for µi in each phase

• xamples of models of µi in the vaporphase

• xamples of models of µi in the liquidphase


10/50

ow we are going to learna&out

• Partial molar properties

• *ecause the chemical potential is a partial

molar propert#

• At the end of this section thin a&out this * $hat is the chemical potential in ph#sical terms

* $hat are the units of the chemical potential

* +ow do we use the chemical potential to solve aP (phase equili&rium) pro&lem


11/50

Partial molar propert#

i jnT P i

i n

nM M

≠

∂

∂=

,,

)(

Solution property

Partial property

Pure-species property


12/50

example

i jnT P i

innV V

≠

∂∂= ,,)(

0lim

~

)(

)(

→∆

∆=∆

∆=∆

wn

ww

ww

nV nV

nV nV

pen !eaker: ethanol " water, equimolar

#otal $olume n%

# and P

Add a drop of pure water, nw

&i', allow for heat e'chan(e, until temp #

)han(e in $olume ?


13/50

!otal vs, partial properties

i

i

i

i

i

i

M nnM

M x M

∑

∑=

=

See deri$ation pa(e *+


14/50

-erivation of Gi&&s.-uhemequation

ii

i

i

i

i

x P xT

M x M

dx M dT T

M dP

P

M dM

∑

∑

=

+

∂∂

+

∂∂

=,,


15/50

Gi&&s.-uhem at constant !/P

P&Tconstant0=∑ ii

i M d x

seful for thermodynamic consistency tests


16/50

*inar# solutions

1

12

1

21

dxdM x M M

dx

dM

x M M

−=

+=

See deri$ation pa(e *+.


17/50

!tain d&/d'0 from 1a2


18/50

xample 11,0

• $e need "% cm0 of antifree2esolution3 0 mol4 methanol in water,

• $hat volumes of methanol and water

(at "5oC) need to &e mixed to o&tain"% cm0 of antifree2e solution at "5oC

• -ata3

water /07.18 /77.17

methanol /7.!0 /".8

1

2

1

1

mol cmV mol cmV

mol cmV mol cmV

==

==


19/50

solution

• Calculate total molar volume

• $e now the total volume% calculate thenum&er of moles required% n

• Calculate n1 and n"

• Calculate the volume of each pure species


20/50

3ote cur$es for partial molar $olumes


21/50

From Gi&&s.-uhem3

P&Tconstant0=∑ ii

i M d x

02211 =+ V d xV d x

4i$ide !y d'0, what do you conclude respect to the slopes?


22/50

6ead and wor example11,7

• Given +87x19:x"9x1x"(7x19"x")determine partial molar enthalpies asfunctions of x1% numerical values for pure.

species enthalpies% and numerical valuesfor partial enthalpies at in;nite dilution

• Also show that the expressions for the

partial molar enthalpies satisf# Gi&&s.-uhem equation% and the# result in thesame expression given for total +,


23/50

ow we are going to start looingat models for the chemical

potential of a given component in amixture• !he ;rst model is the ideal gas mixture

• !he second model is the ideal solution

• As #ou stud# this% thin a&out thedi


24/50

The ideal-gas mixturemodel

• => for an ideal gas

• Calculate the partial molar volumefor an ideal gas in an ideal gas

mixture


25/50

or an# par a mo ar proper # o er


26/50

or an# par a mo ar proper # o erthan volume% in an ideal gas

mixture3

P to p fromintegrate

T constant at P Rd dS

pT H P T H

pT M P T M

i

ig

i

i

ig

i

ig

i

i

ig

i

ig

i

ln

),(),(

),(),(

−=

==

=


27/50

Partial molar entrop# (igm)

i

ig

ii

ig

i

ig

i

i

ig

ii

ig

i

y R P T S pT S P T S

y R P T S pT S

ln),(),(),(

ln),(),(

−==

−=


28/50

Partial molar Gi&&s energ#

P y RT T y RT P RT T

P RT T G P and 1atm P etween g integratin

T constant at P RTd dP P

RT dP V dG

y RT GG

y RT TS H S T H G

iiii

ig

i

i

ig

i

ig

i

ig

i

i

ig

i

ig

i

ig

i

i

ig

i

ig

i

ig

i

ig

i

ig

i

ln)(lnln)(

ln)(

ln

ln

ln

+Γ =++Γ =

+Γ = =

===

+==

+−=−=

µ

µ )hemical potential of

component i in an

ideal (as mi'ture

#his is µ for a pure component 555

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++


29/50

Pro&lem

• $hat is the change in entrop# when ,?m0 of C=" and ,0 m0 of "% each at 1 &arand "5oC &lend to form a gas mixture atthe same conditions@ Assume ideal gases,

i

i

i

i

ig

ii

ig

i

i i

i

ig

ii

ig

i

i

i

ig

i

ig

i

ig

i

y y RS yS

y y RS yS yS

y RS S

ln

ln

ln

∑∑∑ ∑∑−=−

−==

−=

We showed that


30/50

solution

∑−=∆i

ii y ynRS ln

n # -V./T# $ bar $ m0. 1/ 2 &34 56

∆S # &7894: ;.5


31/50

Pro&lem• $hat is the ideal wor for the

separation of an equimolar mixtureof methane and ethane at 1?5oC and0 &ar in a stead#.ow process into

product streams of the pure gases at05oC and 1 &ar if the surroundingstemperature !σ 8 0B@02 6ead section 78+ 1calculation of ideal work292 #hink a!out the process: separation of (ases and chan(e of state

First calculate H and S for methane and for ethane chan(in( their state from

P0, #0, to P9#9

Second, calculate H for de-mixing and S for de-mixing from a mi'ture of ideal(ases


32/50

solution

ig

mixdei

i

i

ig

mixdei

i

i

i

i

i

ig

mixde

ig

mixde

ig T

T

ig

i

ig T

T

ig

i

S S yS

H H y H

y y RS H

P

P

T

dT T !pS

dT T !p H

−

−

−

−

∆+∆=∆

∆+∆=∆

=∆=∆

−=∆=∆

∑

∑

∑

∫ ∫

ln0

ln)(

)(

1

22

1

2

1

#


33/50

ow we introduce a new concept3fugacit#

• $hen we tr# to model realDs#stems% the expression for thechemical potential that we used for

ideal s#stems is no longer valid

• $e introduce the concept of fugacit#that for a pure component is the

analogous (&ut is not equal) to thepressure


34/50

$e showed that3

)ln()( P y RT T iiig

i +Γ = µ

P RT T G iig

i ln)( +Γ =

iii f RT T G ln)( +Γ =

Pure component i , ideal (as

)omponent i in a mixture

of ideal (ases

et;s define:

For a real fluid, we define

Fu(acity of pure species i


35/50

6esidual Gi&&s free energ#

i

R

i

iig

ii

R

i

RT G

P

f RT GGG

φ ln

ln

=

=−=

%alid for species iin any phase and

any condition


36/50

>ince we now how to calculateresidual propertiesE

P

dP " RT

G

RT G

P

ii

Ri

i

R

i

)1(ln

ln

0−==

=

∫ φ φ

aals, etc


37/50

examples

• From irial =>

• From van der $aals=>

RT

P #iii =φ ln

i

iiiii

" T R

P a

RT

P " "

22ln1ln −

−−−=φ


38/50

Fugacities of a ".phases#stem

l

ii

l

i

$ii

$i

f RT T G

f RT T G

ln)(

ln)(

+Γ =

+Γ =

ne component, two phases:

saturated liquid and saturated $apor at Pisat and #i

sat

>hat are the equili!rium conditions for a pure component?


39/50

Fugacit# of a pure liquid at Pand !

sat

i sat

i

l

i

l

i

sat

i

$

i

sat

i

l

i

sat

i

sat

i

$

il

i P

P f

P f

P f

P f

P

P f P f

)(

)(

)(

)()()( =


40/50

Fugacit# of a pure liquid at Pand !

dP V

RT

P P f P

P

l

i

sat

i

sat

i

l

i sat i

∫ =1

e#$)( φ


41/50

example• For water at 0oC and for P up to 1% Pa

(1 &ar) calculate values of f i and φ i fromdata in the steam ta&les and plot them vs, P

−−

−=−= )(1

)(1

ln%

%%

% iiii

ii

i

i

S S T

H H

RGG RT f

f

et Hi@ and Si

@ from the steam ta!les at *o) and the lowest P, 0 kPa

At low P, steam is an ideal (as BC f i@ BP@

#hen (et $alues of Hi and Si at *o) and at other pressure P and calculate f i 1P2


42/50


43/50

Pro&lem

• For >=" at : B and 0 &ar% determine goodestimates of the fugacit# and of G66!,

>=" is a gas% what equations can we use to

calculate f 8 φP

Find !c% Pc% and acentric factor% ω, !a&le *1% p, :H

Calculate reduced properties3 !r% Pr

!r81,0I0 and Pr80,H5


44/50


45/50

+igh P% high !% gas3 use Jee.Bessler correlation

• From ta&les 15 and 1: ;nd φ and φ1

• φ 8 ,:?"K φ1 8 1,057

• φ 8 φ φ1ω = 0.724

• f 8 φ P 8 ,?"7 x 0 &ar 8 "1?,17 &ar

• G66! 8 lnφ = −0.323


46/50

Pro&lem

• stimate the fugacit# of c#clopentane at11oC and "?5 &ar, At 11 oC the vaporpressure of c#clopentane is 5,":? &ar,

• At those conditions% c#clopentane is a high P liquid

dP V RT

P P f P

P

l

i

sat

i

sat

i

l

i sat i∫ =

1e#$)( φ

%ind Tc, -c, >c, Vc and acentric factor, ω, Table $, p9


47/50

%ind Tc, -c, >c,, Vc and acentric factor, ω, Table $, p9@47

alc"late red"ced properties Tr , -r sat

Tr # 79384@ and -r sat # 79$$3

At - B -r sat we can "se the virial CDS to calc"late φisat

2.!

1

".1

0

10

172.01&.0'

!22.008.0

)(e#$

r r

r

r sat

i

T #

T #

# #

T

P

−=−=

+= ω φ


48/50

φisat # 79:

-


49/50

coeLcient

ω

φ φ φ

φ ω φ φ

ω φ

φ

φ

φ

)(

lnlnln

)1(ln

)1(ln

)1(ln

ln

10

10

0

1

0

0

0

0

=

+=+−=

−=

=

−==

=

∫ ∫

∫

∫

i

r

r P

r

r P

i

r

r P

ii

r c

P ii

R

i

i

R

i

P

dP

" P

dP

"

P dP "

P P P

P dP "

RT G

RT G

r r

r

#a!les =0* to =0.

ee-Eessler

+$ 0 -ue Monda#


50/50

+$ 0% -ue Monda#%>eptem&er 1?

• Problems 11.! 11."! 11.#! 11.1!11.1$

W E 8, F"e onday, September &8

• Pro!lems 0080+, 0080 1!2, 00890, 008991a2, 00891a2, 00897

Solution Thermodynamics Theory-Ch 11

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