The temperature of an ideal gas is increased from $140 \mathrm{~K}$ to $560 \mathrm{~K}$. If the r.m.s.…
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
$4 v$
$\frac{v}{4}$
$\frac{v}{2}$
$2 v$
Solution
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, $v^{\prime}=2 v$