Kinetic Theory

JEE Main level, single correct, medium

37 of 168

Question 37 of 168

At 300K300K, the rms speed of oxygen molecules is α+5α\sqrt{\dfrac{\alpha +5}{\alpha }} times to that of its average speed in the gas. Then, the value of α\alpha will be (use π=227\pi =\dfrac{22}{7})

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Answer: (B) 2828

RMS speed is given by, vrms=3RTM{v}_{rms}=\sqrt{\dfrac{3RT}{M}} and average speed is given by, vavg=8πRTM{v}_{avg}=\sqrt{\dfrac{8}{\pi }\dfrac{RT}{M}}.
As per question,
vrms=α+5αvavg{v}_{rms}=\sqrt{\dfrac{\alpha +5}{\alpha }}{v}_{avg}
⇒3RTM=α+5α8πRTM\Rightarrow \sqrt{\dfrac{3RT}{M}}=\sqrt{\dfrac{\alpha +5}{\alpha }}\sqrt{\dfrac{8}{\pi }\dfrac{RT}{M}}
⇒3=α+5α×8227\Rightarrow 3=\dfrac{\alpha +5}{\alpha }\times \dfrac{8}{\dfrac{22}{7}}
⇒α=28\Rightarrow \alpha =28

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