AdvanceThermodynamics_Materi_6

8/10/2019 AdvanceThermodynamics_Materi_6

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ACTIVITY MODELS


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Modified Raoult's Law and Excess GibbsFree Energy

Equilibrium criteria:For vapor phase:For liquid phase, we may use an activity coefficient( i), giving:

Poynting method is used to calculate the purecomponent liquid phase fugacities:Combining:

f iV = f i

L

f iV = yi i P

f i L= xi i f i

L

f i L= i

sat P isat exp

V i L P P i

sat

RT yi i P = xi i i

sat P isat exp

V i L P P i

sat

RT


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K-ratio ==> Ki=yi/xi

At low pressure: Poynting Correction and ratio offor the component approach unity ==> MODIFIED

RAOULT'S LAW

K i =i L P i

sat

P [isat exp

V i L P P i

sat / RT

i

] isat / iK i =

i L P i

sat

Patau y i P= xi i P i

sat


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Excess Gibbs free energy: G E=G-G is

dst diperoleh:

Activity Coefficient and excess Gibbs free energyare coupled.Excess Gibbs energy is zero for an ideal solution

Activity Coefficients as derivatives:

G E = RT i

xi ln i

G E

n i T , P, n j i= RT ln i


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Comparison with Equation of State

Methods


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Determination of G E from

Experimental DataModified Raoult's Law:Excess Gibbs energy from activity coefficient (for

binary system):

plot G E/RT vs x 1Example (1): system of 2-propanol (1) + water (2)from citation data: T=30 oC, P 1

sat = 60.7 mmHg, P 2sat = 32.1

mmHg, y 1 = 0.6462, when x 1 = 0.6369 at P=66.9 mmHg

==> determine activity coefficient and relate to excess Gibbsenergy

i= yi P

xi P isat

G E

RT = x1 ln 1 x2 ln 2

1= y1 P

x1 P 1sat =

0.6462 66.90.6369 60.7

= 1.118 2 = y2 P

x2 P 2sat =

0.3538 66.90.3631 32.1

= 2.031

G E

RT = x1 ln 1 x2 ln 2 =0.6369 ln 1.118 0.3631ln 2.031 =0.328


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plot to G E/RT vs x 1


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One-Parameter MARGULES

Equationthe simplest expression for Gibbs excess energyfunction

Parameter A is a constant which is not associated withthe other uses of the variable (equation of stateparameters, Helmholtz energy, Antoine coefficients).

Example: derive the expression for the activitycoefficients for the one-parameter Margules equation

==>

G E

RT

= A x1 x2

G E

RT = A n2

n1n

1 RT G E n i T, P, n j i= An 2[ 1n n 1n 2]= A n 2n [1 n1n ]= Ax2 1 x1


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==>Dengan cara yg sama ==>

Example (2): continued from example (1), at 30o

Cshow x 1=0.1168 and y 1=0.5316 at P=60.3 mmHg.What are the pressure and vapor phase compositionspredicted by the one-parameter Margules Equation.at x 1=0.6369, we found G

E/RT = 0.328.

Fitting the Margules equation:

at the new composition:

ln 1= Ax22

ln 2= Ax12

G E

RT = A x1 x2 A= 0.328 /[ 0.6369 0.3631 ]=1.42

x1= 0.1168, we find : ln 1= Ax22= 1.42 0.8832 2= 1.107, 1 = 3.03

ln 2= Ax12= 1.42 0.1168 2= 0.0194, 2= 1.02


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Substituting into Modified Raoult's Law: y1 P = x1 1 P 1

sat = 0.1168 3.03 60.7 =21.48 mmHg y

2P = x

1 2P

2

sat = 0.8832 1.02 32.1 =28.92 mmHg

P = y1 P y2 P = 50.4 mmHg y1= y1 P /P = 21.48 /50.4 = 0.426


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Two-Parameters MARGULES

Equationrequires differentiation of nG E/RT with respect to a

mole number the equation is multiplied by n (mole numbers)where x 1=n 1/(n 1+n 2), and x 2=n 2/(n 1+n 2):

differentiation with respect to n 1:

Reconversion of n i to x i:

G E

x1 x2 RT = A21 x1 A12 x2 alternatively

G E

RT = A21 x1 A12 x2 x1 x2

nG E

RT = A

21n

1 A

12n

2

n1 n 2

n1 n2 2 nG E / RT n1 T , P, n 2= ln 1

ln 1= n2

[ A21 n 1 A12 n2

1n1 n2

2 2n 1n1 n 2

3

n1 A21n 1 n 2

2

]ln 1= x2[ A21 x1 A12 x2 1 2x 1 A21 x1]


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note: x 2=1-x 1, so:

For the limiting conditions of infinite dilution:

ln 1= x22[ A12 2 A21 A12 x1]

ln 2= x12[ A21 2 A12 A21 x2]Margules Equation

ln 1= A12 x1= 0 andd ln 2

= A21 x2= 0


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Example of Margules Equation

Following is set of VLE data for the systemmethanol(1)/water(2) at 333.15 K:


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calculate i ==> modified Raoult's law

calculate ln iplot ln i vs x i

A12 and A 21 are values of the intercepts at x 1=0 and x 1=1 ofthe straight line drawn to represent the G E/x1x2RT:

we get: A 12 =0.372 and A 21 =0.198

calculate G E/RT and G E/x1x2RT

We have equation:or ==>

i= yi P

xi P isat

ln 1= A12 x1= 0 andd ln 2= A21 x2= 0

G E

RT = 0.198 x1 0.372 x2 x1 x2

G E

RT = x1 ln 1 x2 ln 2


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Use 1 and 2 to calculate VLE predictions:

Another method:by fitting ln i vs x i for the following Margules equation:

or to obtain A 12 and A 21

i = yi P

xi P isat P = x1 1 P 1

sat x2 2 P 2sat

y1= x1 1 P 1sat

x1 1 P 1sat x2 2 P 2

sat

ln 1= x22[ A12 2 A21 A12 x1] ln 2= x12[ A21 2 A12 A21 x2]


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van Laar Equation for Activity

Coefficient Model

by infinite solutions method:

or rearranging to:

ln 1= A12

[1

A12 x1 A21 x2

]

2

van Laar Equation ln 2= 21

[1 A21 x1 A12 x2]2

ln 1= A12 x1= 0 andd ln 2

= A21 x2= 0

A12= ln 1 [1 x2 ln 2 x1 ln 1]2

A21= ln 2 [1 x1 ln 1 x2 ln 2 ]2


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Example of van Laar Equation

Consider benzene(1) + ethanol(2) system which exhibits anazeotrope at 760 mmHg and 68.24 oC containing 44.8 mole% ethanol. Calculate the composition of the vapor in

equilibrum with an equimolar liquid solution at 760 mmHggiven the Antoine constants:log P 1

sat =6.87987-1196.76/(T+219.161)

log P2

sat =8.1122-1592.86/(T+226.18)

Solution:

at T=68.24 oC, P 1sat =519.7 mmHg; P 2

sat =503.5 mmHg

at azeotrope: x 1=y 1 ==> 1=P/P 1sat

; =P/P 2sat

1=760/519.7=1.4624; =760/503.5=1.5094

where x 1=0.552; x 2=0.448


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Wilson's Equation for Activity

Coefficient

Vi, V j : molar volume of pure liquid i, j, at temperature T

ln 1= ln x1 x2 12 x2 12

x1 x2 12

21 x1 21 x2

Wilson's Equation

ln 2= ln x1 21 x2 x1 12

x1 x2 12 21

x1 21 x2

12=

V 2

V 1exp

A12 RT

and. 21

=V 1

V 2exp

A21 RT

G E

RT = x1 ln x1 x2 12 x2 ln x2 x1 21


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NRTL Equation for Activity

Coefficient (3 parameters)

, b 12 , and b 21 are parameters specific to a particular pair ofspecies, and are independent of composition and temperatureNRTL: Non Random Two Liquid

ln 1= x22

[21 G21

x1 x2 G 21

2

G12 12 x2 x1 G 12

2]

NRTL EquationG E

x1 x2 RT =

G 21 21 x1 x2 G 21

G 12 12 x2 x1 G 12

ln 2= x12

[12

G 12

x2 x1 G 12

2 G 21 21

x1 x2 G 212

]G 12= exp 12 G21= exp 2112=

b12 RT

21= b21 RT


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For the infinite-dilution values of the activity coefficient: ln 1

= 21 12 exp 12

ln 2

= 12 21 exp 21


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UNIQUAC Equation for Activity

CoefficientFor multicomponent solution:

UNIQUAC equation requires two adjustable parameters

characterized from experimental data for each binarysystem ==>

G E

RT =

j x j ln j/ x j 5

jq j x j ln j/ j

jq j x j ln

ii ij

ln k = ln k COMB ln k

RES

ln k COMB = ln k / xk 1 k / xk 5q k [ln k / k 1 k / k ]

ln k RES = q k [1 ln i i ik j j kj

i

i ij]


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UNIQUAC for Binary Solution

==>

==>

where ==>

Q, R, v ???

ln 1 = ln1

x11 1

x15q 1 [ln

1

1

1 11

]

q1

[1 ln 1 2 21

11 2 21

212

1 12 2

]ln 2= ln

2

x21 2

x25q 2 [ln

2

2

1 22

]

q 2[1 ln 1 12 2

1 21

1 2 21

2

1 12 2] j

x j r j

i

xi r i j

x j q j

i

xi q i

r j= k

vk j Rk ; q j=

k vk

j Q k


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R k parameter ==> group volume

Q k parameter ==> group surface area

Molecule size (r j) and molecule shape (q j) may becalculated by multiplying the group parameters by thenumber of times each group appears in the molecule,and summing all the groups in the moleculewhere v k

(j) is the number of groups of the k th type in the j th molecule.


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T h i f Fitti M d l t


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Technique for Fitting Model to

Experimental dataTools in Matlab: lsqcurvefit%Assume you det er mi ned xdat a and ydat a exper i ment al l yxdat a = [ 0. 9 1. 5 13. 8 19. 8 24. 1 28. 2 35. 2 60. 3 74. 6 81. 3] ;ydat a = [ 455. 2 428. 6 124. 1 67. 3 43. 2 28. 1 13. 1 - 0. 4 - 1. 3 - 1. 5] ;x0 = [ 100; - 1] % St ar t i ng guess[ x, r esnor m] = l sqcur vef i t ( @myf un, x0, xdat a, ydat a)

function F = myf un( x, xdat a)

F = x( 1) *exp( x( 2) *xdat a) ;

Result:

x =498. 8309 - 0. 1013

r esnor m =9. 5049

or using GUI of Matlab: >>cftool

AdvanceThermodynamics_Materi_6

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