Q. 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)
NTA AbhyasNTA Abhyas 2020
Solution:
Given, Diatomic gas is at temperature $T$ .
Rotational kinetic energy = $2\times \left(\frac{k T}{2}\right)$
Because diatomic gas has two degrees of rotation
Rotational Kinetic energy,
$KE_{rot}=\frac{1}{2}I\left(\omega \right)^{2}\Rightarrow \frac{1}{2}I\left(\omega \right)^{2}=2\left(\frac{k T}{2}\right)\Rightarrow \frac{1}{2}I\left(\omega \right)^{2}=kT\Rightarrow I\left(\omega \right)^{2}=2kT\Rightarrow \left(\omega \right)^{2}=\frac{2 k T}{I}$
$\therefore \omega =\sqrt{\frac{2 k T}{I}}$
Rotational kinetic energy = $2\times \left(\frac{k T}{2}\right)$
Because diatomic gas has two degrees of rotation
Rotational Kinetic energy,
$KE_{rot}=\frac{1}{2}I\left(\omega \right)^{2}\Rightarrow \frac{1}{2}I\left(\omega \right)^{2}=2\left(\frac{k T}{2}\right)\Rightarrow \frac{1}{2}I\left(\omega \right)^{2}=kT\Rightarrow I\left(\omega \right)^{2}=2kT\Rightarrow \left(\omega \right)^{2}=\frac{2 k T}{I}$
$\therefore \omega =\sqrt{\frac{2 k T}{I}}$