Question

Ideal mixture of two miscible liquids A and B is placed in a cylinder containing piston. Piston is pulled out isothermally so that volume of liquid decreases but that of vapours increases. When negligibly small amount of liquid was left, mole fraction of A in vapour phase is 0.4. If $P_A^0=0.4atm$ and $P_B^0=1.2atm$ at the experimental temperature, the total pressure at which the liquid is almost evaporated in atm is (nearest integer).

Answer 1

Let $X_A$ and $X_B$ mole fraction of A and B be present in solution.
From Raoult's law
$P_T=P_A^0\cdot X_A+P_B^0\cdot X_B$
$P_T=\left(0.4\right)X_A+1.2\left(X_B\right)$ ....(i)
Also, $P_A^{\prime }=P_A^0\cdot X_A=P_T\times X_A^{\prime }$
where, $X_A^{\prime }$ are mole fraction of A in gaseous phase
$\therefore \left(0.4\right)X_A=P_T\cdot \left(0.4\right)$
$\therefore P_T=X_A$ ....(ii)
Similarly, for $X_B^{\prime }$ mole fraction of B in gaseous phase
$\left(1.2\right)X_B=P_T\times 0.6(\because X_A^{\prime }+X_B^{\prime }=1)$
$\therefore P_T=2X_B$ .....(iii)
By eqs. (i) and (ii)
$\left(0.6\right)X_A=\left(1.2\right)X_B$
or $X_A=2X_B$ ....(iv)
$\because X_A+X_B=1$ ..... (v)
$\therefore X_A=\frac{2}{3}$ and $X_B=\frac{1}{3}$
From eq. (i) $\therefore P_T=0.4\times \frac{2}{3}+1.2\times \frac{1}{3}=0.667atm$.