Kinetic Theory

JEE Main level, single correct, medium

45 of 168

Question 45 of 168

The relation between root mean square speed (vrms)\left({v}_{rms}\right) and most probable speed (vp)\left({v}_{p}\right) for the molar mass MM of oxygen gas molecule at the temperature of 300K300K will be

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Answer: (C) vrms=32vp{v}_{rms}=\sqrt{\dfrac{3}{2}}{v}_{p}

RMS speed of O2{O}_{2} gas is vrms=3RTM=3RT32{v}_{rms}=\sqrt{\dfrac{3RT}{M}}=\sqrt{\dfrac{3RT}{32}} and most probable speed of O2{O}_{2} gas is vmostprobable=2RTM=2RT32{v}_{mostprobable}=\sqrt{\dfrac{2RT}{M}}=\sqrt{\dfrac{2RT}{32}}.
Here temperature is same.
Thus, vrmsvmostprobable=3RT322RT32=32\dfrac{{v}_{rms}}{{v}_{mostprobable}}=\dfrac{\sqrt{\dfrac{3RT}{32}}}{\sqrt{\dfrac{2RT}{32}}}=\sqrt{\dfrac{3}{2}}
Hence, vrms=32vp{v}_{rms}=\sqrt{\dfrac{3}{2}}{v}_{p}

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