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An ideal gas expands adiabatically. $(\gamma=1 \cdot 5)$ To reduce the r.m.s. velocity of the molecules 3 times, the gas has to be expanded
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Verified Answer
The correct answer is:
81 times
We know,
$\begin{aligned}
& V_{\text {rms }}=\sqrt{\frac{3 R T}{M_0}} \\
& \Rightarrow \mathrm{T} \propto V^2
\end{aligned}$
$\frac{\mathrm{T}_2}{\mathrm{~T}_1}=\frac{\mathrm{V}_2^2}{\mathrm{~V}_1^2}=\frac{\left(\frac{\mathrm{V}_1}{3}\right)^2}{\mathrm{~V}_1^2}=\frac{1}{9}.. (i)$
Also, $\mathrm{TV}^{\gamma-1}=$ Constant
$\begin{aligned}
& \Rightarrow \frac{V_2}{V_1}=\left(\frac{T_1}{T_2}\right)^{\frac{1}{\gamma-1}} \\
& \frac{V_2}{V_1}=(9)^2=81
\end{aligned}$
$\begin{aligned}
& V_{\text {rms }}=\sqrt{\frac{3 R T}{M_0}} \\
& \Rightarrow \mathrm{T} \propto V^2
\end{aligned}$
$\frac{\mathrm{T}_2}{\mathrm{~T}_1}=\frac{\mathrm{V}_2^2}{\mathrm{~V}_1^2}=\frac{\left(\frac{\mathrm{V}_1}{3}\right)^2}{\mathrm{~V}_1^2}=\frac{1}{9}.. (i)$
Also, $\mathrm{TV}^{\gamma-1}=$ Constant
$\begin{aligned}
& \Rightarrow \frac{V_2}{V_1}=\left(\frac{T_1}{T_2}\right)^{\frac{1}{\gamma-1}} \\
& \frac{V_2}{V_1}=(9)^2=81
\end{aligned}$
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