Question

$N$ molecules each of mass $m$ of gas $A$ and $2N$ molecules each of mass $2m$ of gas $B$ are contained in the vessel which is maintained at a temperature $T$. The mean square of velocity of the 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 $w^2$. The ratio of $w^2 : v^2$ is

• A. $3 : 2$
• B. $1 : 3$
• C. $2 : 3$
• D. $1 : 1$

Answer 1

Answer: (C)

The mean square velocity of gas molecules is given by $ v^{2} = \frac{3kT}{m}$.
For gas $A$, $v^{2}_{A} = \frac{3kT}{m}\quad... \left(i\right)$
For a gas molecule,
$v^{2} = v^{2}_{x} +v^{2}_{y} + v^{2}_{z} = 3v^{2}_{x}\quad\left(\because v^{2}_{x} = v^{2}_{y} = v^{2}_{z}\right)$
or $\quad v^{2}_{x} = \frac{v^{2}}{3}$
From eqn. $\left(i\right)$, we get
$w^{2} = v^{2}_{x} =\left[\frac{\frac{3kT}{m}}{3}\right] = \frac{kT}{m}\quad...\left(ii\right)$
For gas $B$, $v^{2}_{B} = v^{2} = \frac{3kT}{m}\quad... \left(iii\right)$
Dividing eqn. $\left(ii\right)$ by eqn. $\left(iii\right)$, we get
$\frac{w^{2}}{v^{2}} = \frac{\frac{kT}{m}}{\frac{3kT}{2m}} = \frac{2}{3}$