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Question

Pressure over an ideal binary liquid solution containing 10 moles each of liquid A and B is gradually decreased isothermally. At what pressure, half of the total amount of liquid will get converted into vapour? ($P_{A}^{o}$= 200 torr, $P_{B}^{o}$= 100 torr)

Options

A.150 torr
B.166.5 torr
C.133.3 torr
D.141.4 torr

Solution

Let the final composition:

Liquid (10 mole): A = x mole, B = (10 – x) mole

Vapour (10 mole): A = (10 – x) mole, B = x mole

Now, $\frac{Y_{A}}{Y_{B}} = \frac{X_{A}}{X_{B}}.\frac{P_{A}^{o}}{P_{B}^{o}} \Rightarrow \frac{x}{10 - x}.\frac{200}{100}$

$\therefore x = 4.14$

Now, $P = X_{A}.P_{A}^{o} + X_{B}.P_{B}^{o} = \frac{x}{10} \times 200 + \frac{10 - x}{10} \times 100 = 141.4\text{ torr}$

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