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JEE Mains · Physics · STD 11 - Kinetic theory of gases
  • JEE Mains 2021
  • Hard
  • Numerical

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 ......\(ms ^{-1} .\)

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Answer & Solution

Correct Answer

(C) \(400\)

Step-by-step Solution

Detailed explanation

\(v _{ rms }=\sqrt{\frac{3 RT }{ M }}\)

\(v _{ rms } \propto \sqrt{ T }\)

\(\frac{\left( v _{ rms }\right)_{2}}{\left( v _{ rms }\right)_{1}}=\sqrt{\frac{ T _{2}}{ T _{1}}}\)

\(=\sqrt{\frac{400}{300}}\)

\(=\frac{2}{\sqrt{3}}\)

\(\left( v _{ rms }\right)_{2}=\frac{2}{\sqrt{3}}\left( v _{ rms }\right)_{1}\)

\(=\frac{2}{\sqrt{3}} \times 200\)

\(\left( v _{ rms }\right)_{2}=\frac{400}{\sqrt{3}} m / s\)

\(=400\)

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