The density of hydrogen gas at standard atmospheric pressure $(P = 1.01 \times 10^{5} \ Pa)$ is $0.09 \ kg/m^{3}$. Find the root mean square velocity $(v_{rms})$ and the average kinetic energy of $1 \ mole$ of the gas.

(N/A) According to the kinetic theory of gases,the pressure $P$ is given by $P = \frac{1}{3} \rho v_{rms}^{2}$.
Therefore,$v_{rms} = \sqrt{\frac{3P}{\rho}}$.
Substituting the values $P = 1.01 \times 10^{5} \ Pa$ and $\rho = 0.09 \ kg/m^{3}$:
$v_{rms} = \sqrt{\frac{3 \times 1.01 \times 10^{5}}{0.09}} = \sqrt{\frac{3.03 \times 10^{5}}{0.09}} = \sqrt{33.67 \times 10^{5}} = \sqrt{3.367 \times 10^{6}} \approx 1834.9 \ m/s$.
The average kinetic energy of $1 \ mole$ of an ideal gas is given by $E = \frac{3}{2} RT$.
Using the relation $PV = RT$ for $1 \ mole$,we have $E = \frac{3}{2} PV$.
Since $\rho = \frac{M}{V}$,where $M$ is the molar mass of $H_{2}$ $(2 \times 10^{-3} \ kg/mol)$,the volume of $1 \ mole$ is $V = \frac{M}{\rho} = \frac{2 \times 10^{-3}}{0.09} \approx 0.0222 \ m^{3}$.
Thus,$E = \frac{3}{2} \times (1.01 \times 10^{5}) \times (0.0222) \approx 3363.3 \ J$.