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JEE Mains · Physics · STD 11 - Kinetic theory of gases
  • JEE Mains 2019
  • Medium
  • MCQ

For a given gas at \(1\,atm\) pressure, \(rms\) speed of the molecules is \(200\,m/s\) at \(127\,^oC.\) At \(2\,atm\) pressure and at \(227\,^oC,\) the \(rms\) speed of the molecules will be

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Answer & Solution

Correct Answer

(B) \(100\sqrt {5}\,m/s\)

Step-by-step Solution

Detailed explanation

\(V_{r m s}=\sqrt{\frac{3 R T}{M_{w}}}\)

\(\Rightarrow v_{m s} \propto \sqrt{T}\)

\(\text { Now, } \frac{v}{200}=\sqrt{\frac{500}{400}} \Rightarrow \frac{v}{200}=\frac{\sqrt{5}}{2}\)

\(\Rightarrow \mathrm{v}=100 \sqrt{5} \mathrm{m} / \mathrm{s}\)