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. Which is maintained at a temperature T. The mean square of the velocity of molecules of B type is denoted by $V^2$ and the mean square of the X component of the velocity of A type is denoted by ${\omega }^2$, $\frac{{\omega }^2}{V^2}=$

• A. 2
• B. 1
• C. 1/3
• D. 2/3

Answer 1

The correct option is D 2/3

Total KE of A type of molecules =
$\frac{3}{2}m{\omega }^2$
Total KE of A type of molecule is
$K.E_A$ = $\frac{1}{2}\left[\right(V_{r.ms}{)}_x^2+(V_{rms}{)}_y^2+(V_{rms}{)}_z^2]$
but $(V_{rms}{)}_x$ = $\omega$
So $(V_{rms}{)}_y$ = $(V_{rms}{)}_z$ = $\omega$
Total KE of B type molecules
$=\frac{1}{2}\times 2m{v}^2$ = $m.v^2$
Now, $\frac{3}{2}\times m{\omega }^2$ = $m{v}^2$
or $\left({\omega }^2/v^2\right)$ = $\frac{2}{3}$