Question

Vapour pressure of $A$ and $B$ mixture at $80^\circ C$ is given by $P\left(\right.inmmHg\left.\right)=100+200x_{A},$ where $x_{A}$ is mole fraction of $A.$ Solution is prepared by mixing $2mol$ of $A$ and $3mol$ of $B$ and if the vapours are removed and condensed into liquid and again brought to the temperature of $-80^\circ C,$ If the ratio of mole fraction of $A$ in vapour state and mole fraction of $B$ in vapour state above the condensate is $x:1,$ find value of $'x'$ is -

Answer 1

$P_{T}=100+200\times 0.4=180$
$P_{A}^{0}=300,P_{B}^{0}=100,y_{A}=\frac{300 \times 0 . 4}{180}=\frac{2}{3}$
$y_{A}^{'}=\frac{\frac{2}{3} \times 300}{\frac{2}{3} \times 300 + \frac{1}{3} \times 100}=\frac{6}{7}$
$y_{B}^{'}=\frac{1}{7},\frac{y '_{A}}{y '_{B}}=\frac{6}{1}=6$