Question

An ideal mixture of two liquids A and B is put in a cylinder containing piston. Piston is pulled out isothermally so that the volume of liquid decreases but that of vapours increases. Negligibly small amount of liquid was left and mole-fraction of A in vapour is 0.4. If ${\text{P}}_{\text{A}}^{{}^{\circ }}=\text{0.4}$ and ${\text{P}}_{\text{B}}^{{}^{\circ }}=\text{1.2}$ atm at the experimental temperature, which of the following is the total pressure at which the liquid is almost evaporated?

• A. 0.22 atm
• B. 0.431 atm
• C. 0.667 atm
• D. 1 atm

Answer 1

Answer: (C)

Mole-fraction in vapour phase, Y for A

${\text{Y}}_{\text{A}}=\text{0.4}\text{,}{\text{Y}}_{\text{B}}=\text{0.6}$

$$\begin{gathered} {\text{P}}_{\text{A}}^{{}^{\circ }}=\text{0.4} \\ {\text{P}}_{\text{B}}^{{}^{\circ }}=\text{1.2}\}\text{given} \end{gathered}$$

${\text{P}}_{\text{T}}={\text{P}}_{\text{A}}^{{}^{\circ }}{\text{x}}_{\text{A}}+{\text{P}}_{\text{B}}^{{}^{\circ }}{\text{x}}_{\text{B}}$ ...(i)

${\text{Y}}_{\text{A}}=\text{0.4}=\frac{{\text{P}}_{\text{A}}^{{}^{\circ }}{\text{x}}_{\text{A}}}{{\text{P}}_{\text{T}}}\Rightarrow \frac{\text{0.4}{\text{x}}_{\text{A}}}{{\text{P}}_{\text{T}}}=\text{0.4}$

or ${\text{P}}_{\text{T}}={\text{x}}_{\text{A}}$ ...(ii)

${\text{Y}}_{\text{B}}=\text{0.6}=\frac{{\text{P}}_{\text{B}}^{{}^{\circ }}{\text{x}}_{\text{B}}}{{\text{P}}_{\text{T}}}\Rightarrow \frac{\text{1.2}{\text{x}}_{\text{B}}}{{\text{P}}_{\text{T}}}=\text{0.6}$

or ${\text{P}}_{\text{T}}=2{\text{x}}_{\text{B}}$ ...(iii)

Comparing (ii) and (iii),

${\text{x}}_{\text{A}}=2{\text{x}}_{\text{B}}$

we know, ${\text{x}}_{\text{A}}+{\text{x}}_{\text{B}}=1$

Hence, ${\text{x}}_{\text{B}}=\frac{1}{3}$ and ${\text{x}}_{\text{A}}=\frac{2}{3}$

Putting in Equation (i)

${\text{P}}_{\text{T}}=\text{0.4}\times \frac{2}{3}+\text{1.2}\times \frac{1}{3}=\text{0.667 atm}$