MHT CET · Physics · Kinetic Theory of Gases
If temperature of gas molecules is raised from \(127^{\circ} \mathrm{C}\) to \(527^{\circ} \mathrm{C}\), the ratio of r.m.s. speed of the molecules is respectively
- A \(1: 2\)
- B \(2: 1\)
- C \(1: \sqrt{2}\)
- D \(2: \sqrt{2}\)
Answer & Solution
Correct Answer
(C) \(1: \sqrt{2}\)
Step-by-step Solution
Detailed explanation
We know
\(V_{\mathrm{rms}}=\sqrt{\frac{3 R T}{m}}\)
The temperature of the same gas molecule is raised.
\(\mathrm{V}_{\mathrm{mm}} \propto \sqrt{\mathrm{T}}\)
\(\therefore \quad\) The ratio of the velocities is
\(\begin{aligned}
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\sqrt{\mathrm{T}_1}}{\sqrt{\mathrm{T}_2}} \\
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\sqrt{400}}{\sqrt{800}} \\
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{1}{\sqrt{2}}
\end{aligned}\)
\(V_{\mathrm{rms}}=\sqrt{\frac{3 R T}{m}}\)
The temperature of the same gas molecule is raised.
\(\mathrm{V}_{\mathrm{mm}} \propto \sqrt{\mathrm{T}}\)
\(\therefore \quad\) The ratio of the velocities is
\(\begin{aligned}
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\sqrt{\mathrm{T}_1}}{\sqrt{\mathrm{T}_2}} \\
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\sqrt{400}}{\sqrt{800}} \\
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{1}{\sqrt{2}}
\end{aligned}\)
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