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) = ?`

Answer

0.67

Explanation

The mean square velocity of the gas molecules is given by
`C^2=3kT//m`
For gas `A:C_A^2=3kT//m` and
For gas `B:C_B^2=3kT//2m=v^2` …(i)
The molecule A has equal probability of motion in all directions, therefore
`C_x^2=C_9^2=C_x^2=omega^2`(given)
`therefore C_A^2=C_x^2+C_y^2=3C_x^2=3 omega^2`
or `omega^2=(C_A^2)/(3)=1/3((3kT)/(m))=(KT)/(m)` ...(ii)
Dividing (ii) by (i), we get : `(omega^2)/(v^2)=(kT//m)/(3kT//2m)=2/3 ~~ 0.67`