At 300,K,the rms speed of oxygen molecules is | vedclass

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(B) The root mean square (rms) speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.The average speed of gas molecules is given by $v_{

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(B) The root mean square (rms) speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.The average speed of gas molecules is given by $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.According to the problem,$v_{rms} = \sqrt{\frac{\alpha+5}{\alpha}} 	imes v_{avg}$.Substituting the formulas: $\sqrt{\frac{3RT}{M}} = \sqrt{\frac{\alpha+5}{\alpha}} 	imes \sqrt{\frac{8RT}{\pi M}}$.Squaring both sides: $\frac{3RT}{M} = \left(\frac{\alpha+5}{\alpha}ight) 	imes \frac{8RT}{\pi M}$.Canceling $\frac{RT}{M}$ from both sides: $3 = \left(\frac{\alpha+5}{\alpha}ight) 	imes \frac{8}{\pi}$.Given $\pi = \frac{22}{7}$,so $\frac{8}{\pi} = \frac{8 	imes 7}{22} = \frac{56}{22} = \frac{28}{11}$.Thus,$3 = \left(\frac{\alpha+5}{\alpha}ight) 	imes \frac{28}{11}$.$33\alpha = 28(\alpha + 5)$.$33\alpha = 28\alpha + 140$.$5\alpha = 140$.$\alpha = 28$.

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

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