Gas at a pressure P₀ 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

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$\frac{P_{0}}{2}$

$2 P_{0}$

$4 P_{0}$

$P_{0}$

answer is B.

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Detailed Solution

$P = \frac{1}{3} \frac{mN}{V} v_{rms}^{2} ∴ P \propto {mv}_{rms}^{2}$

$\Rightarrow \frac{P_{2}}{P_{1}} = \frac{m_{2}}{m_{2}} \times {(\frac{v_{2}}{v_{1}})}^{2} \Rightarrow \frac{P_{2}}{P_{1}} = \frac{(\frac{m_{1}}{2})}{m_{1}} {(\frac{2 v_{1}}{v_{1}})}^{2} \Rightarrow \frac{P_{2}}{P_{1}} = 2 \Rightarrow P = 2 P_{0}$

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