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)

• A. $\sqrt{\frac{2kT}{I}}$
• B. $\sqrt{\frac{kT}{I}}$
• C. $\sqrt{\frac{3kT}{I}}$
• D. $\sqrt{\frac{5kT}{I}}$

Answer 1

Answer: (A)

Given, Diatomic gas is at temperature $T$ .
Rotational kinetic energy = $2\times \left(\frac{kT}{2}\right)$
Because diatomic gas has two degrees of rotation
Rotational Kinetic energy,
$K{E}_{rot}=\frac{1}{2}I{\left(\omega \right)}^2\Rightarrow \frac{1}{2}I{\left(\omega \right)}^2=2\left(\frac{kT}{2}\right)\Rightarrow \frac{1}{2}I{\left(\omega \right)}^2=kT\Rightarrow I{\left(\omega \right)}^2=2kT\Rightarrow {\left(\omega \right)}^2=\frac{2kT}{I}$
$\therefore \omega =\sqrt{\frac{2kT}{I}}$