Consider a binary solution of two volatile liquid components 1 and 2. and are the mole fractions of component 1 in liquid and vapor phase, respectively. The slope and intercept of the linear plot of vs are given respectively as: [2025]

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Solution

Q.
Consider a binary solution of two volatile liquid components 1 and 2. $x_{1}$ and $x_{2}$ are the mole fractions of component 1 in liquid and vapor phase, respectively. The slope and intercept of the linear plot of $\frac{1}{x_{1}}$ vs $\frac{1}{y_{1}}$ are given respectively as: [2025]

1 $\frac{P_{1}^{0}}{P_{2}^{0}}, \frac{P_{2}^{0} - P_{1}^{0}}{P_{2}^{0}}$

2 $\frac{P_{1}^{0}}{P_{2}^{0}}, \frac{P_{1}^{0} - P_{2}^{0}}{P_{2}^{0}}$

3 $\frac{P_{2}^{0}}{P_{1}^{0}}, \frac{P_{1}^{0} - P_{2}^{0}}{P_{2}^{0}}$

4 $\frac{P_{2}^{0}}{P_{1}^{0}}, \frac{P_{2}^{0} - P_{1}^{0}}{P_{2}^{0}}$

Ans.
(1)

Vapour pressure of component 1 in solution $(P_{1})$ $= P_{1}^{°} x_{1} - (A) [Raoult’s law]$

Vapour pressure of component 2 in solution $(P_{2})$ $= P_{2}^{°} x_{2} [Raoult’s law]$

Total vapour pressure of solution $(P_{T})$ $= P_{1} + P_{2} = P_{1}^{°} x_{1} + P_{2}^{°} x_{2} [Dalton’s law]$

$P_{T} = P_{1}^{°} x_{1} + P_{2}^{°} x_{2} = P_{1}^{°} x_{1} + P_{2}^{°} (1 - x_{1}) = (P_{1}^{°} - P_{2}^{°}) x_{1} + P_{2}^{°}$

Also, vapour pressure of component 1 in solution $(P_{1})$
${=P}_{T} y_{1} = [(P_{1}^{°} - P_{2}^{°}) x_{1} + P_{2}^{°}] y_{1} - (B) [Dalton’slaw]$

Equating (A) and (B)

$P_{1}^{°} x_{1} = [(P_{1}^{°} - P_{2}^{°}) x_{1} + P_{2}^{°}] y_{1}$

$\frac{P_{1}^{°}}{y_{1}} = \frac{[(P_{1}^{°} - P_{2}^{°}) x_{1} + P_{2}^{°}]}{x_{1}}$

$\frac{P_{1}^{°}}{y_{1}} = (P_{1}^{°} - P_{2}^{°}) + \frac{P_{2}^{°}}{x_{1}}$

$\frac{P_{2}^{°}}{x_{1}} = \frac{P_{1}^{°}}{y_{1}} + (P_{2}^{°} - P_{1}^{°})$

$\frac{1}{x_{1}} = \frac{P_{1}^{°}}{P_{2}^{°}} \times \frac{1}{y_{1}} + \frac{(P_{2}^{°} - P_{1}^{°})}{P_{2}^{°}}$

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