The velocity of the molecules of a gas at tem | vedclass

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(C) The root mean square velocity of gas molecules is given by the relation $v_{rms} \propto \sqrt{T}$.Given that the initial velocity is $v_1 = v$ at

Vedclass · Accepted Answer

$480$

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(C) The root mean square velocity of gas molecules is given by the relation $v_{rms} \propto \sqrt{T}$.Given that the initial velocity is $v_1 = v$ at temperature $T_1 = 120\,K$,and the final velocity is $v_2 = 2v$ at temperature $T_2$.Using the proportionality,we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.Substituting the values: $\frac{2v}{v} = \sqrt{\frac{T_2}{120}}$.$2 = \sqrt{\frac{T_2}{120}}$.Squaring both sides: $4 = \frac{T_2}{120}$.Therefore,$T_2 = 4 	imes 120 = 480\,K$.

The velocity of the molecules of a gas at temperature $120\,K$ is $v$. At what temperature in $K$ will the velocity be $2v$?

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