Two compounds A and B in when mixed form nearly ideal solutions. At room temperature, $P_{A}^{0} = 40 m m H g$ and $P_{B}^{0} = 120 m m H g$ .
A liquid mixture is composed of $1 m o l$ of A and $3 m o l$ of B.
If the pressure over the mixture at room temperature is reduced. At what pressure does the first vapour form?

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Solution

Initially
$x_{A} = \frac{1}{1 + 3} = 0.25 x_{B} = 1 - 0.25 = 0.75$
Given:
$P_{A}^{0} = 40 m m H g P_{B}^{0} = 120 m m H g$
The pressure at which first vapour is formed is the bubble point pressure.
At bubble point:
When the first vapour is formed, the composition of A and B in the liquid mixture will be the same.
So, at bubble point
$x_{A} = 0.25$
$x_{B} = 0.75$
$P = x_{A} P_{A}^{0} + x_{B} P_{B}^{0} P = 0.25 \times 40 + 0.75 \times 120 P = 100 m m H g$

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Two compounds A and B in when mixed form nearly ideal solutions. At room temperature, P0A=40 mm Hg and P0B=120 mm Hg. A liquid mixture is composed of 1 mol of A and 3 mol of B. If the pressure over the mixture at room temperature is reduced. At what pressure does the first vapour form?