First slide

Law of equipartition of energy

Question

If I is the moment of inertia of a rigid diatomic molecule, then calculate the angular mean square velocity of a rotating molecule in the gas sample, if the temperature is T. (k = Boltzmann constant)

Difficult

A

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

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

$\sqrt{\frac{3 kT}{I}}$
D

$\sqrt{\frac{5 kT}{I}}$

Solution

Given, Diatomic gas is at temperature T.
Rotational kinetic energy = $2 \times (\frac{kT}{2})$ Because diatomic gas has two degrees of rotation
Rotational Kinetic energy
${KE}_{rot} = \frac{1}{2} {Iω}^{2}$
$\Rightarrow \frac{1}{2} {Iω}^{2} = 2 (\frac{kT}{2})$
$\Rightarrow \frac{1}{2} I ω^{2} = k T$
$\Rightarrow {Iω}^{2} = 2 kT$
$\Rightarrow ω^{2} = \frac{2 kT}{I}$
$∴ ω = \sqrt{\frac{2 kT}{I}}$

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