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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 :-

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$(v_{rms})$ $= \sqrt{\frac{3}{2}} V_{p}$

$(v_{rms})$ $= V_{p}$

$(v_{rms})$ $= \sqrt{\frac{1}{3}} V_{p}$

$(v_{rms})$ $= \sqrt{\frac{2}{3}} V_{p}$

answer is B.

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Detailed Solution

$(v_{rms})$ $= \sqrt{\frac{3 RT}{M}} and V_{mp} = \sqrt{\frac{2 RT}{M}}$

Thus $(v_{rms})$ $= \sqrt{\frac{3}{2}} V_{mp}$

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