Q.Prove that root mean square velocity of gas molecule is directly proportional to the square root of its absolute temperature.

$v_{rms}$ is obtained from the kinetic theory expression for pressure of a gas, $PV=\tfrac13 Nm\overline{v^2}$, combined with the ideal gas equation.

From the kinetic theory of gases, the pressure exerted by $N$ molecules of mass $m$ each, in a container of volume $V$, is

$PV = \frac{1}{3}Nm\overline{v^2}$

where $\overline{v^2}$ is the mean square speed, so $v_{rms}=\sqrt{\overline{v^2}}$.

For one mole of gas, $N = N_A$ (Avogadro's number) and $Nm = M$ (molar mass), so

$PV = \frac{1}{3}M v_{rms}^2 \quad\Rightarrow\quad v_{rms}^2 = \frac{3PV}{M}$

Using the ideal gas equation $PV = RT$ for one mole,

$v_{rms}^2 = \frac{3RT}{M} \quad\Rightarrow\quad v_{rms} = \sqrt{\frac{3RT}{M}}$

Since $R$ and $M$ (molar mass of the gas) are constants for a given gas,

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