AP EAMCET · PHYSICS · Kinetic Theory of Gases
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
- A \(1.25\)
- B \(0.74\)
- C \(0.52\)
- D \(0.26\)
Answer & Solution
Correct Answer
(C) \(0.52\)
Step-by-step Solution
Detailed explanation
As \(V_1=1.5 \mathrm{~V}_2\) \(\begin{aligned} & \sqrt{\frac{3 \mathrm{RT}}{\mathrm{m}_1}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{m}_2}} \times 1.5 \\ & \frac{3}{\mathrm{~m}_1}=\frac{8}{\pi} \frac{1}{\mathrm{~m}_2} \times 2.25\end{aligned}\)…
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