Answer
At $20^{\circ}C$, $v_{mean}\approx 3.75\times10^{4}$ $cm/s$
At $30^{\circ}C$, $v_{mean}\approx 3.82\times10^{4}$ $cm/s$
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.
The molar mass of $CO_2$ molecules is $M=44$ $g$
For $T=(20+273)$ $K$ $=293$ $K$
$v_{mean}=\sqrt {\frac{8\times 8.3145\times 10^{7}\times 293}{\pi\times 44}}$ $cm/s$
$v_{mean}\approx 3.75\times10^{4}$ $cm/s$
For $T=(30+273)$ $K$ $=303$ $K$
$v_{mean}=\sqrt {\frac{8\times 8.3145\times 10^{7}\times 303}{\pi\times 44}}$ $cm/s$
$v_{mean}\approx 3.82\times10^{4}$ $cm/s$
