Answer
For $N_2$, $v_{mean}\approx 4.75\times10^{4}$ $cm/s$
For $CO_2$, $v_{mean}\approx 3.79\times10^{4}$ $cm/s$
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 $N_2$ molecules is $M=28$ $g$
For $T=298$ $K$
$v_{mean}=\sqrt {\frac{8\times 8.3145\times 10^{7}\times 298}{\pi\times 28}}$ $cm/s$
$v_{mean}\approx 4.75\times10^{4}$ $cm/s$
The molar mass of $CO_2$ molecules is $M=44$ $g$
For $T=298$ $K$
$v_{mean}=\sqrt {\frac{8\times 8.3145\times 10^{7}\times 298}{\pi\times 44}}$ $cm/s$
$v_{mean}\approx 3.79\times10^{4}$ $cm/s$
