Two boxes are at the same temperature. The first box contains gas with molecular mass $m_1$ and rms speed…

Two boxes are at the same temperature. The first box contains gas with molecular mass $m_1$ and rms speed $v_1$. The second box contains gas with molecular mass $\mathrm{m}_2$ and average speed $\mathrm{v}_2$. If $\mathrm{v}_1=1.5 \mathrm{v}_2$, then $\frac{\mathrm{m}_1}{\mathrm{~m}_2}$ is
  1. $1.25$
  2. $0.74$
  3. $0.52$
  4. $0.26$

Solution

As $V_1=1.5 \mathrm{~V}_2$ $\begin{aligned} & \sqrt{\frac{3 \mathrm{RT}}{\mathrm{m}_1}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{m}_2}} \times 1.5 \\ & \frac{3}{\mathrm{~m}_1}=\frac{8}{\pi} \frac{1}{\mathrm{~m}_2} \times 2.25\end{aligned}$ $\frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{3 \pi}{2.25 \times 8}=0.52$ ~

Asked in: MHT CET Full Test 13

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