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If temperature of gas molecules is raised from $127^{\circ} \mathrm{C}$ to $527^{\circ} \mathrm{C}$, the ratio of r.m.s. speed of the molecules is respectively
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Verified Answer
The correct answer is:
$1: \sqrt{2}$
We know
$V_{\mathrm{rms}}=\sqrt{\frac{3 R T}{m}}$
The temperature of the same gas molecule is raised.
$\mathrm{V}_{\mathrm{mm}} \propto \sqrt{\mathrm{T}}$
$\therefore \quad$ The ratio of the velocities is
$\begin{aligned}
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\sqrt{\mathrm{T}_1}}{\sqrt{\mathrm{T}_2}} \\
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\sqrt{400}}{\sqrt{800}} \\
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{1}{\sqrt{2}}
\end{aligned}$
$V_{\mathrm{rms}}=\sqrt{\frac{3 R T}{m}}$
The temperature of the same gas molecule is raised.
$\mathrm{V}_{\mathrm{mm}} \propto \sqrt{\mathrm{T}}$
$\therefore \quad$ The ratio of the velocities is
$\begin{aligned}
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\sqrt{\mathrm{T}_1}}{\sqrt{\mathrm{T}_2}} \\
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\sqrt{400}}{\sqrt{800}} \\
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{1}{\sqrt{2}}
\end{aligned}$
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