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

(N/A) The average kinetic energy of one molecule of an ideal gas is given by:
$<\frac{1}{2} m v^{2}> = \frac{3}{2} k_{B} T$
Here,$m$ is the mass of one molecule,$v$ is the velocity,$k_{B}$ is the Boltzmann constant,and $T$ is the absolute temperature.
We know that the Boltzmann constant $k_{B} = \frac{R}{N_{A}}$,where $R$ is the universal gas constant and $N_{A}$ is the Avogadro number.
Substituting $k_{B}$ into the equation:
$<\frac{1}{2} m v^{2}> = \frac{3}{2} (\frac{R}{N_{A}}) T$
Multiplying both sides by $2$:
$m = \frac{3 R T}{N_{A}}$
Since the molar mass $M_{0} = m \times N_{A}$,we can write $m = \frac{M_{0}}{N_{A}}$. Substituting this:
$(\frac{M_{0}}{N_{A}}) = \frac{3 R T}{N_{A}}$
Canceling $N_{A}$ from both sides:
$M_{0} = 3 R T$
$ = \frac{3 R T}{M_{0}}$
By definition,the root mean square velocity $v_{rms} = \sqrt{}$.
Therefore,$v_{rms} = \sqrt{\frac{3 R T}{M_{0}}}$.