At 300 K , the rms speed of oxygen molecules is α + 5 α times to that of its average speed in the gas. Then, the value of α will be (use π = 22 7 )

RMS speed is given by, v r m s = 3 R T M and average speed is given by, v a v g = 8 π R T M . As per question, v r m s = α + 5 α v a v g ⇒ 3 R T M = α + 5 α 8 π R T M ⇒ 3 = α + 5 α × 8 22 7 ⇒ α = 28

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

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Solution:

RMS speed is given by, $v_{r m s} = \sqrt{\frac{3 R T}{M}}$ and average speed is given by, $v_{a v g} = \sqrt{\frac{8}{π} \frac{R T}{M}}$ .

As per question,

$v_{r m s} = \sqrt{\frac{α + 5}{α}} v_{a v g}$

$\Rightarrow \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{α + 5}{α}} \sqrt{\frac{8}{π} \frac{R T}{M}}$

$\Rightarrow 3 = \frac{α + 5}{α} \times \frac{8}{\frac{22}{7}}$

$\Rightarrow α = 28$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:KTG

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