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A container consists of 1 mole of liquid water and 1 mole of $N_{2} (g)$ , at a temperature of 300 K If aqueous tension of water at 300 K is 38 mm of Hg and the container is fitted with a piston operating at 0.1321 atm then calculate volume occupied by $N_{2}$ gas:
$(R = 0.0821 atm L / mol k)$

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300 L

600 L

20 L

80 L

answer is A.

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Detailed Solution

The pressure on water molecule;

$\begin{array}{l} P_{H_{2} O} = \frac{38}{760} = 0.05 atm \\ P_{N_{2}} = 0.1321 - 0.05 = 0.0821 \\ V = \frac{n R T}{P} = 300 L \end{array}$

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