Kinetic Theory

JEE Main level, single correct, medium

31 of 168

Question 31 of 168

The rmsrms speed of oxygen molecule in a vessel at particular temperature is (1+5x)12v,{\left(1+\dfrac{5}{x}\right)}^{\dfrac{1}{2}}v, when vv is the average speed of the molecule. The value of xx will be:
(take π=227\pi =\dfrac{22}{7})

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Answer: (C) 2828

The RMS speed is given by
vrms=3RTM{v}_{rms}=\sqrt{\dfrac{3RT}{M}}
The average speed is given by
vavg=8RTπM{v}_{avg}=\sqrt{\dfrac{8RT}{\pi M}}
The ratio is
⇒vrmsvavg=3RTM8RTπM=3π8=38×227⇒vrmsvavg=3328\Rightarrow \dfrac{{v}_{rms}}{{v}_{\text{avg}}}=\sqrt{\dfrac{\dfrac{3RT}{M}}{\dfrac{8RT}{\pi M}}=}\sqrt{\dfrac{3\pi }{8}}=\sqrt{\dfrac{3}{8}\times \dfrac{22}{7}} \Rightarrow \dfrac{{v}_{rms}}{{v}_{\text{avg}}}=\sqrt{\dfrac{33}{28}}
Comparing it with the given value (1+5x)12{\left(1+\dfrac{5}{x}\right)}^{\dfrac{1}{2}},
vrmsvavg=1+528⇒x=28\dfrac{{v}_{rms}}{{v}_{\text{avg}}}=\sqrt{1+\dfrac{5}{28}} \Rightarrow x=28

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