JEE Main 2024 — Kinetic Theory of Gases Question with Solution

The r.m.s velocity of the molecule of a gas of molecular weight $M$ at Kelvin temperature $T$is given by, $V_{rms}=\sqrt{\frac{3RT}{M}}$.

Let $M_O$ and $M_H$ are molecular weight of oxygen and hydrogen and $T_O$ and $T_H$ the corresponding Kelvin temperature at which $V_{r.m.s}$ is same for both,

$V_{r.m.s}=\sqrt{\frac{3R{T}_O}{M_O}}=\sqrt{\frac{3R{T}_H}{M_H}}$. Hence $\frac{T_O}{M_O}=\frac{T_H}{M_H}$

$T_O=273°\mathrm{C}+47°\mathrm{C}=320 \mathrm{K}$, 

$M_O=32$, 

$$\begin{gathered} M_H=2 \\ \Rightarrow T_H=\frac{2}{32}\times 320=20 \mathrm{K} \end{gathered}$$