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The temperature of an ideal gas is increased from $140 \mathrm{~K}$ to $560 \mathrm{~K}$. If the r.m.s. speed of gas molecules is $v$ at $140 \mathrm{~K}$, then at $560 \mathrm{~K}$, r.m.s. speed becomes
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$2 v$
The r.m.s. speed $v$ is directly proportional to the square root of temperature $T$ in kelvin.
$v \propto \sqrt{T}$
The new temperature is $T^{\prime}=4 \mathrm{~T}$
Hence, $v^{\prime}=2 v$
$v \propto \sqrt{T}$
The new temperature is $T^{\prime}=4 \mathrm{~T}$
Hence, $v^{\prime}=2 v$
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