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A container contains certain gas of mass ‘m’ of high pressure. Some of the gas has been allowed to escape from the container and after some time the pressure of the gas becomes half and its absolute temperature 2/3 rd. The amount of the gas escaped is

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$\frac{2}{3} m$

$\frac{1}{2} m$

$\frac{1}{4} m$

$\frac{1}{6} m$

answer is C.

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Detailed Solution

From ideal gas equation

PV=nRT

$v = \frac{nRT}{p}$

From combined gas equation

$\frac{n_{1} {RT}_{1}}{P_{1}} = \frac{n_{2} R T_{2}}{P_{2}}$

$\frac{W_{1} {RT}_{1}}{{MP}_{1}} = \frac{W_{2} R T_{2}}{M P_{2}}$

$\frac{mRT}{MP} = \frac{W_{2} R \frac{2 T}{3}}{M} \times \frac{2}{p}$

$W_{2} = \frac{3}{4} m$

Gas escaped

$m - x = \frac{3}{4} m$

$x = \frac{m}{1} - \frac{2 m}{4}$

$= \frac{4 m - 3 m}{4} = \frac{1 m}{4}$

∴ Hence the amount of gas escaped is $\frac{m}{4}$

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