In a vessel, the ideal gas is at a pressure $\mathrm{P}$. If the mass of all the molecules is halved and their speed is doubled, then resultant pressure of the gas will be

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Aicharya · Accepted Answer

We know, $\begin{aligned} & \mathrm{v}_{\mathrm{ms}}^2=\frac{3 \mathrm{PV}}{\mathrm{Nm}} \ & \Rightarrow \mathrm{P}=\frac{1}{3} \frac{\mathrm{mN}}{\mathrm{V}} \mathrm{v}_{\mathrm{ms}}^2 \ & \Rightarrow \mathrm{P} \propto \mathrm{v}_{\mathrm{ms}}^2 \ & 	herefore \quad \frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{\mathrm{m}_2}{\mathrm{~m}_1} 	imes\left[\frac{\mathrm{v}_2}{\mathrm{v}_1}ight]^2 \ &=\frac{\left(\frac{\mathrm{m}_1}{2}ight)}{\mathrm{m}_1}\left[\frac{2 \mathrm{v}_1}{\mathrm{v}_1}ight]^2 \ & \ldots . .\left(\because 	ext { given } \mathrm{m}_2=\frac{\mathrm{m}_1}{2} 	ext { and } \mathrm{v}_2=2 \mathrm{v}_1ight) \end{aligned}$ $\frac{\mathrm{P}_2}{\mathrm{P}_1}=2$ $	herefore \quad P_2=2 P_1$ $=2 \mathrm{P} \quad \cdots\left(ight.$ given $\left.\mathrm{P}_1=\mathrm{P}ight)$

In a vessel, the ideal gas is at a pressure $\mathrm{P}$. If the mass of all the molecules is halved and…

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