Kinetic Theory of Gases: PYQ Walkthrough (2)

Question

The rms speed of oxygen molecules ($O_2$, $M = 32$ g/mol) at $300$ K is $v_1$. At what temperature would hydrogen molecules ($H_2$, $M = 2$ g/mol) have the same rms speed? (JEE Main 2022)

Solution — Step by Step

Final answer: $T \approx 18.75$ K.

Why This Works

The rms speed depends only on $T/M$. For two gases to have the same $v_{\text{rms}}$, the ratio $T/M$ must be equal. Since hydrogen is $16$ times lighter than oxygen, it needs $16$ times less temperature to match the same rms speed.

Physically, lighter molecules move faster at the same temperature. To slow hydrogen down to the rms speed of oxygen at $300$ K, we cool it heavily — which is why the answer comes out so low.

Alternative Method

Use ratios directly. $\frac{v_{\text{rms}, O_2}}{v_{\text{rms}, H_2}} = \sqrt{\frac{T_1 M_2}{T_2 M_1}} = 1$. So $T_1 M_2 = T_2 M_1 \Rightarrow T_2 = \frac{T_1 M_2}{M_1} = \frac{300 \times 2}{32} = 18.75$ K.

Common Mistake

Plugging molar mass in g/mol but $R = 8.314$ J/(mol·K) — that gives wrong units. If you carry the ratio approach, units cancel and this trap disappears. If you compute $v_{\text{rms}}$ explicitly, convert $M$ to kg/mol ($32 \times 10^{-3}$ kg/mol).