Question

The vapour pressure of pure liquid $A\text{and}B\text{are}450\text{and}700mmHg$respectively, at $350K$. Find out the composition of the liquid mixture if total vapour pressure is $600mmHg$. Also find the composition of the vapour phase.

Answer 1

Given:
vapour pressure of pure liquid,
$P_A^0=450mmHg$
vapour pressure of pure liquid,
$P_B^0=700mmHg$
total vapour pressure,
$P_{total}=600mmHg$
According to Raoult's law,
$P_{gas}=P_{gas}^0\times {\chi }_{gas}$
And $P_{total}=P_A+P_B$
$P_{total}=P_A^0\times {\chi }_A+P_B^0\times {\chi }_B$
$P_{total}=P_A^0\times {\chi }_A+P_B^0\times (1-{\chi }_A)$
{Since, ${\chi }_A+{\chi }_B=1$}
$\Rightarrow 600=450\times {\chi }_A+700\times (1-{\chi }_A)$
$\Rightarrow 600=700-250\times {\chi }_A$
$\Rightarrow 250\times {\chi }_A=100$
$\Rightarrow {\chi }_A=\frac{100}{250}=0.4$
$\Rightarrow {\chi }_B=1-{\chi }_A$
$\Rightarrow {\chi }_B=1-0.4=0.6$
Thus, composition of A and B in liquid phase is $0.4\text{and}0.6$ respectively.

Vapour pressure of compounds
We know that $P_A=P_A^0\times {\chi }_A$
$P_A=450\times 0.4=180mmHg$
And $P_B=P_B^0\times {\chi }_B$
$P_B=700\times 0.6=420mmHg$

Mole fraction in vapour phase

We know that $y_{gas}=\frac{P_{gas}}{P_{toatl}}$
In vapour phase, mole fraction of liquid $A$

$y_A=\frac{P_A}{P_A+P_B}=\frac{180}{180+420}=0.3$
And $y_B=1-y_A=0.7$

Thus, composition of A and B in vapour phase is $0.3\text{and}0.7$ respectively.