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

Explanation

The mean square velocity of gas molecule is given by
`V^(2) = (3 kT)/(m)`
For gas A,
`v_(A)^(2) = (3 kT)/(m)`
For a gas molecule, `v^(2) = v_(x)^(2) + v_(y)^(2) + v_(z)^(2)`
`= 3 v_(x)^(2) (v_(x)^(2) = v_(y)^(2) = v_(z)^(2))`
or `v_(x)^(2) = (v^(2))/(3)`
From Eq. (i) we get
`v_(x)^(2) = (v_(A)^(2))/(3) = [((3 kT//m))/(3)] = (kT)/(m)`
For gas b,
`v_(B)^(2) = v^(2) = (3 kT)/(m_(B)) = (3 kT)/(m)`
Dividing Eq. (ii) by Eq. (iii) we get
`(omega^(2))/(v^(2)) = (kT // m)/(3 kT // 2 m) = (2)/(3)`