A gas at temperature T consists of only rigid | vedclass

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(A) According to the equipartition theorem,each rotational degree of freedom contributes $\frac{1}{2}kT$ to the average energy of a molecule.For a rig

Vedclass · Accepted Answer

$\sqrt{\frac{2kT}{I}}$

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(A) According to the equipartition theorem,each rotational degree of freedom contributes $\frac{1}{2}kT$ to the average energy of a molecule.For a rigid diatomic molecule,there are $2$ rotational degrees of freedom.Therefore,the average rotational kinetic energy is given by: $K.E._{rot} = 2 	imes (\frac{1}{2}kT) = kT$.The rotational kinetic energy is also expressed as: $K.E._{rot} = \frac{1}{2}I\omega_{rms}^2$.Equating the two expressions: $\frac{1}{2}I\omega_{rms}^2 = kT$.Solving for $\omega_{rms}$: $\omega_{rms}^2 = \frac{2kT}{I}$.Thus,$\omega_{rms} = \sqrt{\frac{2kT}{I}}$.

$A$ gas at temperature $T$ consists of only rigid diatomic molecules. If the moment of inertia of a molecule is $I$,what is the root mean square angular velocity of the rotating molecule? ($k$ = Boltzmann constant)

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