Question

Mixture of two liquids $A$ and $B$ is placed in cylinder containing piston. Piston is pulled out isothermally so that the,volume of liquid decreases but that of vapour increases. When negligibly small amount of liquid was remaining, the mole fraction of $A$ in vapour is $0.4$. Given $P_A^{\circ }=0.4atm$ and $P_B^{\circ }=1.2atm$ at the experimental temperature. Calculate the total pressure at which the liquid has almost evaporated. (Assume ideal behaviour) (write the value to the nearest integer)

Answer 1

$P_T=P_A^{\circ }{X}_A+P_B^{\circ }{X}_B=P_A^{\circ }{X}_A+P_B^{\circ }(1-X_A)$

$P_T=P_B^{\circ }+X_A(P_A^{\circ }-P_B^{\circ })$

$Y_A=\frac{P_A^{\circ }\times X_A}{P_T}=\frac{P_A^{\circ }\times X_A}{P_B^{\circ }+X_A(P_A^{\circ }-P_B^{\circ })}$

$0.4=\frac{0.4X_A}{1.2-0.8X_A}$

$1.2=1.8\times X_A$

$X_A=\frac{2}{3}$

So$X_B=\frac{1}{3}$

$P_T=0.4\times \frac{2}{3}+1.2\times \frac{1}{3}=\frac{2}{3}=0.66atm$