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

4 P_{0}

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2 P_{0}

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P_{0}

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\frac{P_{0}}{2}

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Solution

The correct option is B $2 P_{0}$
$\begin{matrix} P = \frac{1}{3} \frac{m N}{V} v_{r m s}^{2} ∴ p \propto m v_{r m s}^{2} s o \frac{P_{2}}{P_{1}} = \frac{m_{2}}{m_{2}} \times {(\frac{v_{2}}{v_{1}})}^{2} = \frac{m_{1} / 2}{m_{1}} {(\frac{2 v_{1}}{v_{1}})}^{2} = 2 P_{2} = 2 P_{1} = 2 P_{0} \end{matrix}$
Hence,
option B is correct answer.

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