In the case of gaseous nitrogen find :
(a) the temperature at which the velocities of the molecules `v_1 = 300 m//s` and `v_2 = 600 m//s` are associated with equal values of the Maxwell distribution friction `F (v)` ,
(b) the velocity of the molecules `v` at which the value of the Maxwell distribution function `F (v)` for the temperature `T_0` will be the same as that for the temperature `eta` times higher.

Explanation

(a) We have,
`(v_1^2)/(v_p^2) e^(-v_1^2//v_p^2) = (v_2^2)/(v_p^2) e^(-v_2^2//v_p^2)` or `((v_1)/(v_2))^2 = e^(v_1^2 - v_2^2//v_p^2)` or `v_p^2 = (2 kT)/(m) = (v_1^2 - v_2^2)/((1 n v_1^2//v_2^2))`
So `T = (m(v_1^2 - v_2^2))/(2 k 1n (v_1^2)/v_2^2) = 330 K`
(b) `F (v)=(4)/(sqrt(pi)) (v^2)/(v_p^2) e^(-v^2//v_p^2) xx (1)/(v_p) ((1)/(v_p)` comes from `F(v) dv = df (u), du = (dv)/(v_p))`
Thus `(v^2)/(v^3) e^(-v^2//v^2) p_1 = v^2//v_(p_2) e^(-v_2//v_(2_2)) v_(p_1)^2 = (2 kT_0)/(m), v_(p_2)^2 = (2 kT_0)/(m) eta ` now
`e -(mv^2)/(2 kT_0)(1 - (1)/(eta)) = (1)/(eta^(3//2))` or `(mv^2)/(2 kT_0)(1 - (1)/(eta)) = (3)/(2) 1n eta`
Thus `v = sqrt((3 kT_0)/(m)) sqrt((1 n eta)/(1 - 1//eta))`.