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 v² and the mean square of x component of the velocity of molecules of gas A is w². The ratio $\frac{w^{2}}{v^{2}}$ is

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

Answer 1

Answer: (D)

$\frac{2}{3}$

Answer 2

Mean square velocity of molecule $= \frac{3kT}{m}$

For gas A, x component of mean square velocity of molecule $= w^{2}$

∴ Mean square velocity $= 3w^{2} = \frac{3kT}{m}$ …..(i)

For B gas mean square velocity $= v^{2} = \frac{3kT}{2m}$ …..(ii)

From (i) and (ii) $\frac{3w^{2}}{v^{2}} = \frac{2}{1}$ so $\frac{w^{2}}{v^{2}} = \frac{2}{3}$.