Answer to Question #97034 in Physical Chemistry for Abhishek

Question

The vapour pressures of pure liquids . A and B at 300 K are 4.5×10⁴ Pa and 6.5× 10⁴ Pa respectively. calculate the mole fractions of of A and B in vapour and liquid phases of a solution when the liquid equilibrium total vapour pressure of the binary liquid solution is 5.5× 10⁴ Pa at 300 K . Assume that the liquid and the vapour are ideal .

Answer 1

Let mole fraction of A in liquid phase is $x_a$ and B in liquid phase is $x_b=1-x_a$

Then according to question,

$$0.55=0.45x_a+0.65(1-x_a)$$

$$0.2x_a=0.1\\or\\x_a=0.5$$

$$x_b=1-x_a=0.5$$

Composition in vapour phase will be-

$p_a=0.45x_a=0.225\ bar\\p_b=0.65x_b=0.325\ bar$

Mole fraction of A in vapour phase , $y_a=\frac{0.225}{0.55}=0.4090$

Mole fraction of B in vapour phase,$y_b=\frac{0.325}{0.550}=0.5909$