Answer
For $Hg_2$, $v_{mean}\approx 1.25\times10^{4}$ $cm/s$
For $H_2$, $v_{mean}\approx 1.78\times10^{5}$ $cm/s$
Work Step by Step
According to the Maxwell–Boltzmann distribution, the average speed $(v_{mean})$ of the gas molecules is expressed as
$v_{mean}=\sqrt {\frac{8RT}{\pi M}}$
where $T$ is the temperature and $M$ is the molar mass of the molecules.
As the temperature is not specified here, we have to calculate $v_{mean}$ at room temperature $T=298$ $K$.
The molar mass of $Hg_{2}$ molecules is $M=401.18$ $g$
For $T=298$ $K$
$v_{mean}=\sqrt {\frac{8\times 8.3145\times 10^{7}\times 298}{\pi\times 401.18}}$ $cm/s$
$v_{mean}\approx 1.25\times10^{4}$ $cm/s$
The molar mass of $H_2$ molecules is $M=2$ $g$
For $T=298$ $K$
$v_{mean}=\sqrt {\frac{8\times 8.3145\times 10^{7}\times 298}{\pi\times 2}}$ $cm/s$
$v_{mean}\approx 1.78\times10^{5}$ $cm/s$
