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A gas at pressure $\mathrm{P}_0$ is contained in a vessel. If the masses of all the molecules are halved and their velocities are doubled, the resulting pressure would be equal to
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The correct answer is:
$2 \mathrm{P}_0$
The pressure is given as $\mathrm{P}=\frac{1}{3} \frac{\mathrm{mN}}{\mathrm{V}} \mathrm{v}_{\mathrm{rms}}^2$
$\Rightarrow \mathrm{P} \propto \mathrm{mv}_{\mathrm{ms}}^2$
$\therefore \quad \frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{\mathrm{m}_2 \mathrm{v}_2^2}{\mathrm{~m}_1 \mathrm{v}_1^2}$
$\therefore \quad \frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{\frac{\mathrm{m}_1}{2}\left(2 \mathrm{v}_1\right)^2}{\mathrm{~m}_1 \mathrm{v}_1^2}=2$
$\therefore \quad \mathrm{P}_2=2 \mathrm{P}_0$ $\ldots\left(\because \mathrm{P}_1=\mathrm{P}_0\right)$
$\Rightarrow \mathrm{P} \propto \mathrm{mv}_{\mathrm{ms}}^2$
$\therefore \quad \frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{\mathrm{m}_2 \mathrm{v}_2^2}{\mathrm{~m}_1 \mathrm{v}_1^2}$
$\therefore \quad \frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{\frac{\mathrm{m}_1}{2}\left(2 \mathrm{v}_1\right)^2}{\mathrm{~m}_1 \mathrm{v}_1^2}=2$
$\therefore \quad \mathrm{P}_2=2 \mathrm{P}_0$ $\ldots\left(\because \mathrm{P}_1=\mathrm{P}_0\right)$
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