If the r.m.s. velocity of a gas at a given te | vedclass

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(D) The formula for $r.m.s.$ velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.From this,we can see that $v_{rms} \propto \sqrt{\frac{T}{M}}$.Let

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$150$

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(D) The formula for $r.m.s.$ velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.From this,we can see that $v_{rms} \propto \sqrt{\frac{T}{M}}$.Let the initial state be $v_1, T_1, M_1$ and the final state be $v_2, T_2, M_2$.Given: $v_1 = 300 \ m/s$,$M_2 = 2M_1$,and $T_2 = \frac{T_1}{2}$.Using the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{M_1}{M_2} 	imes \frac{T_2}{T_1}}$.Substituting the values: $\frac{v_2}{300} = \sqrt{\frac{M_1}{2M_1} 	imes \frac{T_1/2}{T_1}} = \sqrt{\frac{1}{2} 	imes \frac{1}{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore,$v_2 = \frac{300}{2} = 150 \ m/s$.

If the $r.m.s.$ velocity of a gas at a given temperature (Kelvin scale) is $300 \ m/s$,what will be the $r.m.s.$ velocity of a gas having twice the molecular weight and half the temperature on the Kelvin scale?

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