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The temperature of a gas is raised from $27^{\circ} \mathrm{C}$ to $927^{\circ} \mathrm{C}$. The root mean square speed
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gets doubled
Given, $\begin{aligned} T_1 & =27^{\circ} \mathrm{C}=300 \mathrm{~K} \\ T_2 & =927^{\circ} \mathrm{C}=1200 \mathrm{~K}\end{aligned}$
$v_{\mathrm{rms}}=\sqrt{\left(\frac{3 k T}{m}\right)}$
$\begin{aligned} & \Rightarrow \quad v_{\mathrm{rms}} \propto \sqrt{T} \\ & \therefore \quad \frac{\left(v_{\text {rms }}\right)_1}{\left(v_{\text {rms }}\right)_2}=\sqrt{\frac{T_1}{T_2}} \\ & \frac{\left(v_{\text {rms }}\right)_1}{\left(v_{\text {rms }}\right)_2}=\sqrt{\frac{300}{1200}}=\frac{1}{2} \\ & \Rightarrow \quad\left(v_{\mathrm{rms}}\right)_2=2\left(v_{\mathrm{rms}}\right)_1 \\ & \end{aligned}$
$v_{\mathrm{rms}}=\sqrt{\left(\frac{3 k T}{m}\right)}$
$\begin{aligned} & \Rightarrow \quad v_{\mathrm{rms}} \propto \sqrt{T} \\ & \therefore \quad \frac{\left(v_{\text {rms }}\right)_1}{\left(v_{\text {rms }}\right)_2}=\sqrt{\frac{T_1}{T_2}} \\ & \frac{\left(v_{\text {rms }}\right)_1}{\left(v_{\text {rms }}\right)_2}=\sqrt{\frac{300}{1200}}=\frac{1}{2} \\ & \Rightarrow \quad\left(v_{\mathrm{rms}}\right)_2=2\left(v_{\mathrm{rms}}\right)_1 \\ & \end{aligned}$
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