Answer
At $0^{\circ}C$, $v_{mean}\approx 4.54\times10^{4}$ $cm/s$
At $40^{\circ}C$, $v_{mean}\approx 4.86\times10^{4}$ $cm/s$
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 $N_2$ molecules is $M=28$ $g$
For $T=(0+273)$ $K$ $=273$ $K$
$v_{mean}=\sqrt {\frac{8\times 8.3145\times 10^{7}\times 273}{\pi\times 28}}$ $cm/s$
$v_{mean}\approx4.54\times10^{4}$ $cm/s$
For $T=(40+273)$ $K$ $=313$ $K$
$v_{mean}=\sqrt {\frac{8\times 8.3145\times 10^{7}\times 313}{\pi\times 28}}$ $cm/s$
$v_{mean}\approx 4.86\times10^{4}$ $cm/s$
