Question

$N$ molecules each of mass $m$ of gas A and $2N$ molecules each of mass $2m$ of gas B are contained in the same vessel at temperature T. The mean square of the velocity of molecules of gas B is $v^2$ and the mean square of $x$ component of the velocity of molecules of gas A is $w^2$. The ratio $\frac{w^2}{v^2}$ is

• A. $1$
• B. $2$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is D $\frac{2}{3}$

Total KE of A type of molecules = $\frac{3}{2}m{w}^2$
Total KE of A type of molecules is
$K.E{.}_A=\frac{1}{2}\left[\right(V_{r.m.s}{)}_x^2+(V_{r.m.s}{)}_y^2+(V_{r.m.s.}{)}_z^2\left)\right]$
But $(V_{rms}{)}_x=w$
So, $(V_{rms}{)}_y=(V_{rms}{)}_z=w$
Total KE of B type of molecules is
$=\frac{1}{2}\times 2m{v}^2=m.v^2$
Now as KE is dependent on temperature $T$
$\frac{3}{2}\times m{w}^2=m{v}^2\left(w^2/v^2\right)=\frac{2}{3}$