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The temperature of anideal gas is increased from $120 \mathrm{~K}$ to $480 \mathrm{~K}$. If at $120 \mathrm{~K}$, the root mean square speed of gas molecules is $v$, then at $480 \mathrm{~K}$ it will be
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The correct answer is:
$2 v$
$$
\begin{aligned}
& \text { Hints: } \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}} \\
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\sqrt{\frac{120}{480}}-=\sqrt{\frac{1}{4}}=\frac{1}{2} \\
& \mathrm{~V}_2=2 v
\end{aligned}
$$
\begin{aligned}
& \text { Hints: } \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}} \\
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\sqrt{\frac{120}{480}}-=\sqrt{\frac{1}{4}}=\frac{1}{2} \\
& \mathrm{~V}_2=2 v
\end{aligned}
$$
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