Question

At what temperature is the average velocity of $O_2$ molecule equal to the root mean square velocity at ${27}^{\circ }\text{C}$?

Answer 1

$u_{\text{avg}}=u_{\text{rms}}$
$\sqrt{\frac{8R{T}_1}{\pi M}}=\sqrt{\frac{3R{T}_2}{M}}$
$\frac{8R{T}_1}{\pi M}=\frac{3R\times 300}{M}$
$T_1=353.57\text{K}$
Therefore, $T_1={80.57}^{\circ }\text{C}$