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The temperature of an ideal gas is increased from $120 \mathrm{~K}$ to $480 \mathrm{~K}$. If at $120 \mathrm{~K}$, the root mean square velocity of the gas molecules is $v$, at $480 \mathrm{~K}$ it becomes
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$2 u$
\(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}}}}{\sqrt{\frac{3 \mathrm{RT}_{2}}{\mathrm{M}}}}=\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}} \therefore \mathrm{v}_{2}=2 \mathrm{v}_{1}\)
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