Gas at a pressure $P_{0}$ is contained in a vessel masses of all the molecules are halved and their speeds are doubled, the resulting pressure ‘P’ will be equal to

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$4 P_{0}$

$2 P_{0}$

$P_{0}$

$P_{0} / 2$

answer is B.

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Detailed Solution

$\begin{array}{l} root mean square velocity of molecules of gas is V_{rms} = \sqrt{\frac{3 P V}{m}} \\ h e r e P i s p r e s s u r e; V i s v o l u m e; m = m a s s o f g a s \\ \frac{V_{1}}{V_{2}} = \sqrt{\frac{P_{1}}{P_{2}} \times \frac{m_{2}}{m_{1}}} \\ g i v e n V_{2} = 2 V_{1}; m_{2} = \frac{m}{2}; P_{1} = P_{0}; s u b s t i t u t e i n a b o v e e q u a t i o n \\ \frac{V_{1}}{2 V_{1}} = \sqrt{\frac{P_{0}}{P_{2}} \times \frac{\frac{m}{2}}{m}} \\ \frac{1}{2} = \sqrt{\frac{P_{0}}{P_{2}} \times \frac{1}{2}} \\ \frac{1}{4} = \frac{P_{0}}{P_{2}} \times \frac{1}{2} \\ P_{2} = 2 P_{0} \end{array}$