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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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Verified Answer
The correct answer is:
$2 \mathrm{P}$
We know,
$\begin{aligned}
& \mathrm{v}_{\mathrm{ms}}^2=\frac{3 \mathrm{PV}}{\mathrm{Nm}} \\
& \Rightarrow \mathrm{P}=\frac{1}{3} \frac{\mathrm{mN}}{\mathrm{V}} \mathrm{v}_{\mathrm{ms}}^2 \\
& \Rightarrow \mathrm{P} \propto \mathrm{v}_{\mathrm{ms}}^2 \\
& \therefore \quad \frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{\mathrm{m}_2}{\mathrm{~m}_1} \times\left[\frac{\mathrm{v}_2}{\mathrm{v}_1}\right]^2 \\
&=\frac{\left(\frac{\mathrm{m}_1}{2}\right)}{\mathrm{m}_1}\left[\frac{2 \mathrm{v}_1}{\mathrm{v}_1}\right]^2 \\
& \ldots . .\left(\because \text { given } \mathrm{m}_2=\frac{\mathrm{m}_1}{2} \text { and } \mathrm{v}_2=2 \mathrm{v}_1\right)
\end{aligned}$
$\frac{\mathrm{P}_2}{\mathrm{P}_1}=2$
$\therefore \quad P_2=2 P_1$
$=2 \mathrm{P} \quad \cdots\left(\right.$ given $\left.\mathrm{P}_1=\mathrm{P}\right)$
$\begin{aligned}
& \mathrm{v}_{\mathrm{ms}}^2=\frac{3 \mathrm{PV}}{\mathrm{Nm}} \\
& \Rightarrow \mathrm{P}=\frac{1}{3} \frac{\mathrm{mN}}{\mathrm{V}} \mathrm{v}_{\mathrm{ms}}^2 \\
& \Rightarrow \mathrm{P} \propto \mathrm{v}_{\mathrm{ms}}^2 \\
& \therefore \quad \frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{\mathrm{m}_2}{\mathrm{~m}_1} \times\left[\frac{\mathrm{v}_2}{\mathrm{v}_1}\right]^2 \\
&=\frac{\left(\frac{\mathrm{m}_1}{2}\right)}{\mathrm{m}_1}\left[\frac{2 \mathrm{v}_1}{\mathrm{v}_1}\right]^2 \\
& \ldots . .\left(\because \text { given } \mathrm{m}_2=\frac{\mathrm{m}_1}{2} \text { and } \mathrm{v}_2=2 \mathrm{v}_1\right)
\end{aligned}$
$\frac{\mathrm{P}_2}{\mathrm{P}_1}=2$
$\therefore \quad P_2=2 P_1$
$=2 \mathrm{P} \quad \cdots\left(\right.$ given $\left.\mathrm{P}_1=\mathrm{P}\right)$
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