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Gas at a pressure $P_{0}$ in contained as 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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${4P}_{0}$

$2 P_{0}$

$P_{0}$

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

answer is B.

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Detailed Solution

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

so $\frac{P_{2}}{P_{1}} = \frac{m_{2}}{m_{1}} \times {(\frac{v_{2}}{v_{1}})}^{2} = \frac{\frac{m_{1}}{2}}{m_{1}} (\frac{2 v_{1}}{v_{1}}) = 2$

$\Rightarrow P_{2} = 2 P_{1} = 2 P_{0}$

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