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(B) The mean square velocity of a gas molecule is given by $\langle v^2 \rangle = \frac{3kT}{m}$.For gas $A$,the mean square of the $x$-component is $

Vedclass · Accepted Answer

$2: 3$

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(B) The mean square velocity of a gas molecule is given by $\langle v^2 angle = \frac{3kT}{m}$.For gas $A$,the mean square of the $x$-component is $\omega^2 = \langle v_x^2 angle$. Since $\langle v^2 angle = \langle v_x^2 angle + \langle v_y^2 angle + \langle v_z^2 angle$ and $\langle v_x^2 angle = \langle v_y^2 angle = \langle v_z^2 angle$,we have $\langle v^2 angle = 3\omega^2$.Thus,$3\omega^2 = \frac{3kT}{m} \implies \omega^2 = \frac{kT}{m}$...$(i)$For gas $B$,the mean square velocity is $V^2 = \frac{3kT}{2m}$...(ii)Dividing $(i)$ by (ii):$\frac{\omega^2}{V^2} = \frac{kT/m}{3kT/2m} = \frac{kT}{m} 	imes \frac{2m}{3kT} = \frac{2}{3}$.

$N$ molecules of gas $A$,each having mass $m$,and $2N$ molecules of gas $B$,each of mass $2m$,are contained in the same vessel at a constant temperature $T$. The mean square velocity of $B$ is $V^2$ and the mean square of the $x$-component of $A$ is $\omega^2$. The value of $\frac{\omega^2}{V^2}$ is

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