The most probable speed of ${\mathrm{O}}_2$ molecules at$\mathrm{T}\left(\mathrm{K}\right)$ is

Formula Used: v= $\sqrt{\frac{2\mathrm{RT}}{\mathrm{M}}}$

Detailed Solution: We know that most probable speed 

v= $\sqrt{\frac{2\mathrm{RT}}{\mathrm{M}}}$

Where M= Molecular weight

Now,  ${\mathrm{v}}_{{\mathrm{O}}_2}=\sqrt{\frac{2\mathrm{RT}}{32}}$= $\sqrt{\frac{\mathrm{RT}}{16}}$

Therefore, option (C) is the answer.