JEE Main 2025 Jan 23 Shift 2, Chemistry Q18: Vapour Pressure Of A Solution

JEE Main2025Jan 23, Shift 2Chemistry

Consider a binary solution of two volatile liquid components 1 and 2. $x_{1}$ and $y_{1}$ are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of $\frac{1}{x _{1}}$ vs $\frac{1}{y _{1}}$ are given respectively as :

A
$\frac{P _{1}^{0}}{P _{2}^{0}}, \frac{P _{2}^{0} - P _{1}^{0}}{P _{2}^{0}}$
B
$\frac{P _{2}^{0}}{P _{1}^{0}}, \frac{P _{1}^{0} - P _{2}^{0}}{P _{2}^{0}}$
C
$\frac{P _{1}^{0}}{P _{2}^{0}}, \frac{P _{1}^{0} - P _{2}^{0}}{P _{2}^{0}}$
D
$\frac{P _{2}^{0}}{P _{1}^{0}}, \frac{P _{2}^{0} - P _{1}^{0}}{P _{2}^{0}}$

Solution

For an ideal binary mixture, the partial pressure of component 1 equals both $P_{1}^{\circ} x_{1}$ (Raoult's law) and $P_{T} y_{1}$ (Dalton): $P_{T} y_{1} = P_{1}^{\circ} x_{1}$ .

The total pressure is $P_{T} = P_{2}^{\circ} + x_{1} (P_{1}^{\circ} - P_{2}^{\circ})$ .

Substituting and rearranging:

$\frac{P _{2}^{\circ} + x _{1} ( P _{1}^{\circ} - P _{2}^{\circ} )}{x _{1}} = \frac{P _{1}^{\circ}}{y _{1}}$

$\frac{P _{2}^{\circ}}{x _{1}} + (P_{1}^{\circ} - P_{2}^{\circ}) = \frac{P _{1}^{\circ}}{y _{1}}$

Dividing throughout by $P_{2}^{\circ}$ :

$\frac{1}{x _{1}} = \frac{P _{1}^{\circ}}{P _{2}^{\circ}} \cdot \frac{1}{y _{1}} + \frac{P _{2}^{\circ} - P _{1}^{\circ}}{P _{2}^{\circ}}$

$∴$ slope $= \frac{P _{1}^{\circ}}{P _{2}^{\circ}}$ and intercept $= \frac{P _{2}^{\circ} - P _{1}^{\circ}}{P _{2}^{\circ}}$ .

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JEE Main 2025 Jan 23 Shift 2, Chemistry Q18: Vapour Pressure Of A Solution

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