Question

A mixture of two gasses $A$ and $B$ is kept in a vessel which is maintained at a temperature $T$ . In the vessel there are $N$ molecules each of mass $m$ , of the gas $A$ and $2N$ molecules each of mass $2m$ of the gas $B$ . The mean square velocity of the molecules of the gas $B$ is $v^{2}$ and the mean square velocity of the molecules of the gas $A$ along the $x$ direction is $w^{2}$ . The value of $\frac{w^{2}}{v^{2}}$ is

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

Answer 1

Answer: (D)

Mean kinetic energy of the two types of molecules should be equal. The mean square velocity of A type molecules $=\omega ^{2}+\omega ^{2}+\omega ^{2}=3\omega ^{2}$
Therefore, $\frac{1}{2}m\left(3 \left(\omega \right)^{2}\right)=\frac{1}{2}\left(2 m\right)v^{2}$
This gives $\frac{\omega ^{2}}{v^{2}}=\frac{2}{3}$