`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 maintained at a temperature `T`. The mean square 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 `omega^(2)`. What is the ratio of `omega^(2)//v^(2) = ?`

Options

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

Answer

C

Explanation

The mean sequare velocity of gas molecules is giben by `v^(2)=(3kT)/(m).`
For gas A, `v_(A)^(2)=(3kT)/(m)" "...(i)`
For a gas molecule.
`v^(2)-v_(x)^(2)+v_(y)^(2)+v_(z)^(2)=3v_(x)^(2)" "(thereforev_(x)^(2)=v_(y)^(2)=v_(z)^(2))`
or `v_(x)^(2)=(v^(2))/(3)`
From eqn. (i), we get
`w^(2)=v_(x)^(2)=[(3kT)/(m/(3))]=(kT)/(m)" "...(ii)`
For gas B, `v_(B)^(2)=v^(2)=(3kT)/(2m)" "...(iii)`
Dividing eqn. (ii) by eqn. (iii), we get
`(w^(2))/(v^(2))=(KT)/((m/(3kT))/(2m))=2/3`