The r.m.s. speed of the molecules of a gas in | vedclass

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(C) The root mean square $(r.m.s.)$ speed of gas molecules is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M}}$.Here,$R$ is the universal gas con

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(C) The root mean square $(r.m.s.)$ speed of gas molecules is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M}}$.Here,$R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.From the formula,it is evident that $V_{rms}$ depends only on the temperature $T$ and the nature of the gas (molar mass $M$).It does not depend on the number of molecules or the density of the gas.Since the temperature remains constant and the gas type does not change,the $r.m.s.$ speed remains unchanged regardless of the amount of gas leaking out.Therefore,the $r.m.s.$ speed of the remaining molecules will be $400 \; m/s$.

The $r.m.s.$ speed of the molecules of a gas in a vessel is $400 \; m/s$. If half of the gas leaks out,at constant temperature,the $r.m.s.$ speed of the remaining molecules will be ..... $m/s$.

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