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JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases
The root mean square speed of molecules of a given mass of a gas at \(27^{\circ} C\) and \(1\) atmosphere pressure is \(200\, ms ^{-1}\). The root mean square speed of molecules of the gas at \(127^{\circ} C\) and \(2\) atmosphere pressure is \(\frac{ x }{\sqrt{3}}\, ms ^{-1} .\) The value of \(x\) will be ......\(ms ^{-1} .\)
- A \(200\)
- B \(300\)
- C \(400\)
- D \(500\)
Answer & Solution
Correct Answer
(C) \(400\)
Step-by-step Solution
Detailed explanation
\(v _{ rms }=\sqrt{\frac{3 RT }{ M }}\) \(v _{ rms } \propto \sqrt{ T }\) \(\frac{\left( v _{ rms }\right)_{2}}{\left( v _{ rms }\right)_{1}}=\sqrt{\frac{ T _{2}}{ T _{1}}}\) \(=\sqrt{\frac{400}{300}}\) \(=\frac{2}{\sqrt{3}}\)…
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