A diatomic gas consisting of rigid molecules is at a temperature of $87^{\circ} \mathrm{C}$. If the moment…

A diatomic gas consisting of rigid molecules is at a temperature of $87^{\circ} \mathrm{C}$. If the moment of inertia of the rotating diatomic rigid molecule is $2.76 \times 10^{-39} \mathrm{gcm}^2$, then the rms angular speed of the molecule is (Boltzmann constant $=1.38 \times 10^{-23} \mathrm{JK}^{-1}$ )
  1. $6 \times 10^{12}$ rads $^{-1}$
  2. $3 \times 10^{12} \mathrm{rads}^{-1}$
  3. $6 \times 10^{13} \mathrm{rads}^{-1}$
  4. $3 \times 10^{13}$ rads $^{-1}$

Solution

Moment of inertia of diatomic molecule, $ I=\frac{2}{3} M r^2 $
rms speed of molecule, $ v=\sqrt{\frac{3 k T}{M}} $ rms angular speed, $ \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 \times 10^{12} \mathrm{rad} / \mathrm{s} \end{aligned} $

Asked in: AP EAMCET 2018 (23 Apr Shift 1)

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