Question

The vapour pressure of two pure liquids $A$ and $B$ which from an ideal solution are $500$ and $800$ torr respectively at $300K$. A liquid solution $A$ and $B$ for which the mole fraction of $A$ is $0.60$ is contained in a cylinder closed by a piston on which the pressure can be varied. What will be the pressure when 1 mol of the mixture has been vaporized?

Answer 1

Let $n_B$ moles of B is present in 1 mole of mixture that has been vapourized.

Thus,$y_B=\frac{n_B}{1}$

Mole fraction of B in the remaining liquid phase is given by:

${\partial }_B=\frac{1-n_B}{1}$

${\chi }_B=\frac{p-p^{0}r}{p_B^0-p_r^0}$

$[\because p^{0}r+(p_B^0-p_r\left){\chi }_B\right]$

$y_B=\frac{p_B}{p}\Rightarrow p_B^0-p_r^0$

After substituting of values of ${\chi }_B$ :

$1-n_B=\frac{p-p^{0}r}{p_B^0-p_r^0}$

and $n_B$=$\frac{(1-n_B)p^{0}B}{p}$

$n_B=\frac{p^{0}B}{p+p^{0}B}$

So, 1-$\frac{p^{0}B}{p+p^{0}B}$=$\frac{p-p^{0}r}{p_B^0-p_r^0}$

$\Rightarrow p=\sqrt{p_B^o.p_r^o=\sqrt{100X900}}$

$\Rightarrow p=300torr$ is the answer.