The RMS speeds of the molecules of two gases A and B are in the ratio 0.866: 1 at 27°C for A and 127 — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

RMS speed $v \propto \sqrt{T/M}$. The ratio is $\frac{v_A}{v_B} = \sqrt{\frac{T_A / M_A}{T_B / M_B}} = 0.866 \approx \frac{\sqrt{3}}{2}$. Substituting the temperatures $T_A = 300 \text{ K}$ and $T_B = 400 \text{ K}$: $\sqrt{\frac{300}{M_A} \times \frac{M_B}{400}} = \frac{\sqrt{3}}{2}$. Squaring both sides yields $\frac{3}{4} \frac{M_B}{M_A} = \frac{3}{4}$, which simplifies to $M_A = M_B$. The gases must have identical molar masses. Pairs (a), (b), and (c) have equal molar masses (28, 32, 44 respectively). Pair (d) has unequal molar masses (64 and 80). Therefore, correct answer is A,B,C.