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An ideal benzene toluene mixture contains 6 moles of benzene and 4 moles of toluene at...

An ideal benzene toluene mixture contains 6 moles of benzene and 4 moles of toluene at 25 ° C. The saturated vapor pressure of pure benzene at 25 ° C is 96 Torr, the saturated vapor pressure of pure toluene is 29 Torr. Initially, the mixture at 760 Torr pressure is only liquid. The total pressure is then slowly reduced to a pressure of 60 Torr. Calculate the composition of liquid phase and vapor phase, and calculate liquid and vapor mole fractions.

Solutions

Expert Solution

Given;

Moles of benzene = 6 mol

Moles of toluene = 4 mol

Total number of moles ; F = 6 + 4 = 10 mol

Composition of benzene in feed ; z = 6 / 10 = 0.4

It is given that the solution behaves ideally ; therefore Raoult's Law is valid.

Accoding to Raoult's Law ;

P = x Pb* + (1 - x) Pt*

where;

P = total pressure = 60 torr

Pb* = vapor pressure of benzene = 96 torr

Pt* = vapor pressure of toluene = 29 torr

x = composition of benzene in liquid phase

Then ;

60 = 96 x + (1 - x) 29

60 - 29 = 96 x - 29 x

x = 0.46

Composition of benzene in liquid phase = 0.46

Composition of toluene in liquid phase = 0.54

Also ;

y P = x Pb*

where

y = composition of benzene in vapor phase

y X 60 = 0.46 X 96

y = 0.46 X 96 / 60

y = 0.74

Composition of benzene in vapor phase = 0.74

Composition of toluene in vapor phase = 0.26

It is given that the initial mixture on reducing the pressure gets separated into liquid and vapor phase.

Therefore ;

Overall Mass Balance :

F = L + V

10 = L + V

Benzene Mass Balance :

z F = x L + y V

0.6 X 10 = 0.46 L + 0.74 V

6 = 0.46 L + 0.74 V

Solving the two equations simultaneously we get ;

L = 5 mol

V = 5 mol

Therefore;

Liquid mole fraction = L / F = 5 / 10 = 0.5

Vapor mole fraction = V / F = 5 / 10 = 0.5


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