The rate of effusion of two gases 'a' and 'b' under identical conditions of temperature and pressure are in the ratio of 2:1. What is the ratio of rms velocity of their molecules if T_a and T_bare in the ratio of 2:1

The rate of effusion of two gases 'a' and 'b' under identical conditio

1.	The rate of effusion of two gases 'a' and 'b' under identical conditions of temperature and pressure are in the ratio of 2:1. What is the ratio of rms velocity of their molecules if T_a and T_bare in the ratio of 2:1
Answer» `2:1` `SQRT(2)`:1 `2sqrt(2):1` `1:sqrt(2)` Solution :`(gamma_a)/(gamma_b) = sqrt((M_b)/(M_a)) = 2/1 implies M_b = 4M_a`. `((U_(rms))_(a))/((U_(rms))_b) = sqrt((T_a )/(T_b XX M_a)) = sqrt((2)/(1) xx 4/1) = 2sqrt(2) :1`.