Derive the equation of $v_{rms}$ in terms of the mass of a molecule.

(N/A) The average kinetic energy of one molecule of an ideal gas is given by:
$<\frac{1}{2} m v^{2}> = \frac{3}{2} k_{B} T$
Where $m$ is the mass of the molecule,$v$ is the velocity,$k_{B}$ is the Boltzmann constant,and $T$ is the absolute temperature.
Rearranging the equation for the mean square velocity $\langle v^{2} \rangle$:
$\langle v^{2} \rangle = \frac{3 k_{B} T}{m}$
The root mean square velocity $v_{rms}$ is defined as the square root of the mean square velocity:
$v_{rms} = \sqrt{\langle v^{2} \rangle} = \sqrt{\frac{3 k_{B} T}{m}}$
From this expression,it is evident that for a given temperature $T$:
$v_{rms} \propto \frac{1}{\sqrt{m}}$
Thus,at a constant temperature,lighter molecules possess higher root mean square speeds compared to heavier molecules.