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

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

$\text{P}_{\text{A}}^{^\circ }=\text{0.4} \\ \text{P}_{\text{B}}^{^\circ }=\text{1.2} \left.\right\} \text{given}$

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

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