JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases
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
- A \(80\,m/s\)
- B \(100\sqrt {5}\,m/s\)
- C \(100\,m/s\)
- D \(80\sqrt {5}\,m/s\)
Answer & Solution
Correct Answer
(B) \(100\sqrt {5}\,m/s\)
Step-by-step Solution
Detailed explanation
\(V_{r m s}=\sqrt{\frac{3 R T}{M_{w}}}\) \(\Rightarrow v_{m s} \propto \sqrt{T}\) \(\text { Now, } \frac{v}{200}=\sqrt{\frac{500}{400}} \Rightarrow \frac{v}{200}=\frac{\sqrt{5}}{2}\) \(\Rightarrow \mathrm{v}=100 \sqrt{5} \mathrm{m} / \mathrm{s}\)
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