JEE Main 2022 — Kinetic Theory of Gases Question with Solution

The RMS speed of a gas at a given temperature is given by, $v_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}$.

Therefore, to double the RMS speed, the temperature should be increased to four times the initial temperature. Therefore,

$T_f=1200 \mathrm{K}, T_i=300 \mathrm{K}, \text{and} n=\frac{14}{28}=\frac{1}{2}$

The volume of the vessel will remain the same therefore, the process is an isochoric process. Therefore,

$Q=n{C}_v\Delta T=\frac{1}{2}\times \frac{5R}{2}\times 900$

$\Rightarrow Q=9360 \mathrm{J}$