The relation between root mean square speed $$v_{rms}$$ and most probable speed $$v_p$$ for the molar mass $$M$$ of oxygen gas molecule at the temperature of $$300$$ K will be

We need to find the relation between the root mean square speed $$v_{rms}$$ and the most probable speed $$v_p$$ for oxygen gas at $$300$$ K.

Recall the expressions for $$v_{rms}$$ and $$v_p$$: $$v_{rms} = \sqrt{\frac{3RT}{M}}$$ and $$v_p = \sqrt{\frac{2RT}{M}}$$, where $$R$$ is the universal gas constant, $$T$$ is the temperature, and $$M$$ is the molar mass.

Taking the ratio gives $$\frac{v_{rms}}{v_p} = \frac{\sqrt{\frac{3RT}{M}}}{\sqrt{\frac{2RT}{M}}} = \sqrt{\frac{3}{2}}$$, so that $$v_{rms} = \sqrt{\frac{3}{2}} \, v_p$$.

This relation holds for any ideal gas at any temperature, as the ratio depends only on the numerical constants 3 and 2 in the respective formulas.

The correct answer is Option C: $$v_{rms} = \sqrt{\frac{3}{2}}v_p$$.