N molecules each of mass m of gas A and 2 N molecules each of mass 2m of gas B are contained in the same vessel which is maintined at a temperature T. The mean square of the 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 tye is denoted by `omega^(2)`. What is the ratio of `omega^(2)//v^(2) = ?`

Options

A. 1
B. 2
C. `1//3`
D. `2//3`

Answer

D

Explanation

For any molecule, the mean square velocity `barV_(A)^2 = barV_x^2 + barV_y^2 + barV_z^2`
due to symmetry we have `(barV_x^2 = barV_y^2 = barV_x^2)`
`V_A^2 = V_x^2 + V_y^2 + V_z^2 = 3V_x^2`
`or V_x^2 = V_A^2`…(i)
Take gas A,
`V_A^2 = (3k_BT)/m`
From eq. (i) ,
`w^2 = V_x^2 = V_A^2/3 = 1/3 ((3k_BT)/m) = (k_BT)/(m)...(ii)`
Take gas , B
`V_B^2 = V^2 = (3k_BT)/(2m)`
From eq. (ii) and (iii)
`w^2/V^2 = (k_BT//m)/(3k_BT//2m) = 2/3`.