A fixed mass of gas at constant pressure occupies a volume ' $V$ '. The gas undergoes a rise in temperature…
A fixed mass of gas at constant pressure occupies a volume ' $V$ '. The gas undergoes a rise in temperature so that the r.m.s. velocity of the molecules is doubled. The new volume will be
$\frac{\mathrm{V}}{2}$
$\frac{\mathrm{V}}{\sqrt{2}}$
2 V
4 V
Solution
$\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{KT}}{\mathrm{M}}} \Rightarrow \mathrm{~V}_{\mathrm{rms}}^2 \propto \mathrm{~T}$
$\therefore \quad$ When r.m.s. velocity is doubled, $\mathrm{T}_2=4 \mathrm{~T}$ ... (i)
At constant pressure, Volume $\propto$ Temperature
$\begin{aligned}
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\mathrm{T}_1}{\mathrm{~T}_2} \\
\therefore \frac{\mathrm{~V}}{\mathrm{~V}_2} & =\frac{\mathrm{T}}{4 \mathrm{~T}} \\
\mathrm{~V}_2 & =4 \mathrm{~V}
\end{aligned}$
...[From(i)]
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