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At temperature $T, N$ molecules of gas A each having mass $m$ and at the same temperature $2N$ molecules of gas $B$ each having mass $2\,m$ are filled in a container. The mean sqaure velocity of molecules of gas $B$ is $v^2$ and mean square of $x$ component of velocity of molecules of gas$ A$ is $w^2$. The ratio of $w^2/v^2$ is :
$1$
$2$
$1/3$
$2/3$
Solution
Total $K.E.$ of $A$ type molecules $=\frac{3}{2} m \omega^{2}$
Total $K.E.$ of $A$ type of molecule is
$K . E_{A}=\frac{1}{2} m\left[\left(\nu_{r . m s}\right)_{x}^{2}+\left(\nu_{r . m s}\right)_{y}^{2}+\left(\nu_{r . m s}\right)_{z}^{2}\right]$
But $\left(\nu_{r. m s}\right)_{x}=\omega$
So $\left(\nu_{r. m s}\right)_{y}=\left(\nu_{r. m s}\right)_{z}=\omega$
Total $K.E.$ of $B$ type molecules
$=\frac{1}{2} \times 2 m \nu^{2}=m \nu^{2}$
Now, $\frac{3}{2} \times m \omega^{2}=m \nu^{2}\left(\omega^{2} / \nu^{2}\right)=\frac{2}{3}$
