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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main · Physics · Kinetic Theory of Gases

A diatomic gas consisting of rigid molecules is at a temperature of...

A diatomic gas consisting of rigid molecules is at a temperature of 87∘C87^{{}^{{}^{\circ}}} \mathrm{C}. If the moment of inertia of the rotating diatomic rigid molecule is 2.76×10−39gcm22.76 \times 10^{-39} \mathrm{gcm}^2, then the rms angular speed of the molecule is (Boltzmann constant =1.38×10−23JK−1=1.38 \times 10^{-23} \mathrm{JK}^{-1} )A diatomic gas consisting of rigid molecules is at a temperature of 87^ ^ ^ C . If the moment of inertia of the

Options

  1. A6×10126 \times 10^{12} rads ^{-1}
  2. B3×1012rads−13 \times 10^{12} \mathrm{rads}^{-1}
  3. C6×1013rads−16 \times 10^{13} \mathrm{rads}^{-1}
  4. D3×10133 \times 10^{13} rads ^{-1}

Correct answer

A. 6×10126 \times 10^{12} rads ^{-1}

Step-by-step solution

Moment of inertia of diatomic molecule, I=23\frac{2}{3}M r² A diatomic gas consisting of rigid molecules is at a temperature of 87^ ^ ^ C . If the moment of inertia of the
rms speed of molecule, v=3kTM\sqrt{\frac{3 k T}{M}}rms angular speed, ω=vr=3kT/M3I/2M=2kTI=2×1.38×10−23×(87+273)2.76×10−46=360×1023=36×1024=6×1012rad/s\begin{aligned}\omega=\frac{v}{r}& =\frac{\sqrt{3 k T / M}}{\sqrt{3 I / 2 M}}=\sqrt{\frac{2 k T}{I}}\\ & =\sqrt{\frac{2 \times 1.38 \times 10^{-23} \times(87+273)}{2.76 \times 10^{-46}}}\\ & =\sqrt{360 \times 10^{23}}=\sqrt{36 \times 10^{24}}\\ & =6 \times10^{12} \mathrm{rad}/ \mathrm{s} \end{aligned}
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