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For a binary ideal solution of two volatile liquids A and B (with vapour pressures P_A⁰ and P_B⁰), when a plot of 1/P (reciprocal of total pressure) versus y_A (mole fraction of A in vapour phase) is drawn, which of the following correctly describes the plot?

1. A straight line with slope = (1/P_B⁰ - 1/P_A⁰)
2. A straight line with slope = (1/P_A⁰ - 1/P_B⁰)
3. A straight line with slope = 1/P_B⁰
4. A straight line with slope = P_A⁰ * P_B⁰

Correct answer: A straight line with slope = (1/P_B⁰ - 1/P_A⁰)

Solution

From Raoult's law: P = P_A⁰ * x_A + P_B⁰ * (1 - x_A). From Dalton's law: y_A = P_A⁰ * x_A / P, so x_A = y_A * P / P_A⁰. Substituting: P = P_A⁰ * (y_A * P / P_A⁰) + P_B⁰ * (1 - y_A * P / P_A⁰) = y_A * P + P_B⁰ - P_B⁰ * y_A * P / P_A⁰. So P - y_A * P + P_B⁰ * y_A * P / P_A⁰ = P_B⁰. P(1 - y_A + P_B⁰ * y_A / P_A⁰) = P_B⁰. 1/P = (1/P_B⁰)(1 - y_A + P_B⁰ * y_A / P_A⁰) = 1/P_B⁰ - y_A/P_B⁰ + y_A/P_A⁰ = 1/P_B⁰ + y_A * (1/P_A⁰ - 1/P_B⁰). This is a straight line: y-intercept = 1/P_B⁰, slope = (1/P_A⁰ - 1/P_B⁰). But option (A) says slope = (1/P_B⁰ - 1/P_A⁰) which is the negative. The correct slope is (1/P_A⁰ - 1/P_B⁰) matching option (B). However, conventionally if P_B⁰ > P_A⁰ then 1/P_A⁰ > 1/P_B⁰, so slope is positive. The standard derivation gives slope = (1/P_A⁰ - 1/P_B⁰), which is option (B).

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