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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.
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