example ch 9

8/12/2019 Example Ch 9

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Use the van Laar activity coefficient expression to predict the compositions of co-existing

liquid phases (I and II) comprised of two partially miscible liquids (1) and (2) at 50

Example 9.1

o

12 211 2 12 21

2 212 1 21 2

21 2 12 1

ln ; ln ; 2.5; 3.5

[1 ] [1 ]

A A A A

A x A x

A x A x

γ γ = = = =+ +

C and 4

bar. At these conditions the van Laar equations are given by:

Solution:

This system is a binary; however, there are 2 phases since ‘1’ and ‘2’ are partially miscible.

One of the phases (I) is rich in component ‘1’ (with some ‘2’ dissolved in it); while the

second phase (II) is rich in component ‘2’ (with some ‘1’ dissolved in it). Estimates of the

composition of both phases are required. We use two equations of type 9.34 as there are 2components.

For component 1:

1 1 1 1 1 1

2 21 1

1 1

exp( ) exp( )

[1 ] [1 ](1 ) (1 )

I I I II II II

I II

I II

x x x x x x

x x

α α γ γ

α α

β β

= = =

+ +− −

..(a)

For component 2:

2 2 2 2 2 22 22 2

2 2

exp( ) exp( )

[1 ] [1 ](1 ) (1 )

I I I II II II

I II

I II

x x x x x x

x x

β β γ γ β β

α α

= = =

+ +− −

..(b)

Lastly1 2

1 I I x x+ = ..(c)

And:1 2 1 II II x x+ = ..(d)

Equations I – IV need to be solved simultaneously using a suitable algorithm to obtain the

final solution:1 10.12, 0.85 I II x x= = ; and

2 20.88, 0.15 I II x x= =



Example 9.2

Estimate solubility of a solid A in a liquid B at 300o

( ) ( )1/2 1/23 3

100 / ; 125 / ; 9.5 cal / cc ; 7.5 cal / cc . L L

A B A BV cm mol V cm mol δ δ = = = =

K, using (i) ideal solution

assumption, (ii) regular solution model for liquid-phase. The following data are

available:

Heat of fusion for A: 17.5 kJ/mol. Melting point for A = 350o

K.

Assuming that solid-phase is pure naphthalene (which is in equilibrium with

solution of naphthalene in hexane), z1 = 1, γ1S

,

,

1 1ln ln m

fus

T A A A

m A

H x

R T T γ

∆= − − −

= 1, we start with the simplified

equation:

For ideal solution the above equation reduces to:

,

,

1 1ln m A

fus

T

A

m A

H x

R T T

∆ = − −

Using the data provided:17500 1 1

ln8.314 300 350

A x

= − −

Thus, the ideal solubility x A

By Regular solution theory:

= 0.38.

2 2 ln ( ) L

A A A A B RT V γ δ δ = − Φ

∴ ( )

22

ln 1

A

L fus A B A B

A

m

V H T x

RT RT T

δ δ Φ − ∆= − − −

..(1)

Also L

B b B L L

A A B B

x V

x V x V Φ =

+ ..(2)

T = 300 K,1mT = 350o

( ) ( )1/2 1/23 3100 / ; 125 / ; 9.5 cal / cc ; 7.5 cal / cc . L L

A B A BV cm mol V cm mol δ δ = = = =

K

Solution

algorithm:

(1) Assume x A (to start with assume x A

(2) Calculate Φ

= 0)

B

(3) Use equation (1) to calculate new x

from eqn 2

(4) If x

A

A,i+1 – xA,i < 0.01, xA,i+1 is the solution or else, return to step ‘1’.



The final converged value for 0.08 A

x ≈

Note that the result differs significantly from that obtained by assuming ideal

solution behaviour for the liquid phase.

Example 9.3

Compute the eutectic composition and temperature for a mixture of two substances A and B

using the following data:

Property A B Normal Tm (o 180K) 181

fusH∆ (J/mol) 6600 9075

We use the ideal solution behaviour for the liquid phase. The following equation then holds at

the eutectic point:

, ,

, ,

exp exp 1 fus fus

m A m B A B

m A m B

T T T T H H

RT T RT T

− − ∆ ∆+ =

On substituting all relevant data:

6600 180 9075 180exp exp 1

8.314 180 8.314 181

T T

x T x T

− − + =

On solving by trial and error, ( ) 0150T eutectic K ≈

The eutectic composition is found from the following equation:

exp A

A

fusm A

A

m

T T H x

RT T − ∆=

Substituting all the available data with T = 1500

0.5. A x ≈

K, the eutectic composition is found to be:

Example 9.4



A certain solid A has a vapour pressure of 0.01 bar at 3000

We start with the following equation:

K. Compute its solubility at the

same temperature in a gas B at a pressure of 1.0bar. The molar volume of the solid is

125cc/mol.

( ) ˆexp

S sat

A Asat sat

A A A A

V P PP y P

RT φ φ

− =

Since 210 , 1.0sat sat

A AP bar φ −= ≈

Further as the total system pressure is 1.0bar, it follows that

ˆ 1.0 A

φ ≅

Thus the solubility of the solid at the system pressure is given by:

( )( / ) exp

S sat

A Asat

A A

V P P y P P

RT

− =

Substituting all relevant data the solubility is:

21.05 10 . A

y x −=

example ch 9

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