At what temperature, the rate of effusion of `N_(2)` would be 1.625 times that of `SO_(2)` at `50^(@)C` ?

Options

A. 110K
B. 173K
C. 373K
D. 273K

Answer

C

Explanation

`(r_(N_2))/(r_(SO_2)) = (V_(rms) N_2)/(V_(rms) SO_2) = sqrt( (T_(N_2))/(T_(SO_2)) . (M_(SO_2))/(M_(N_2)) ) = sqrt((T_(N_2))/(323) xx 64/28)`
`1.625 = sqrt((T_(N_2))/(323) . 16/7)`
`T_(N_2) = ((1.625)^2 xx 323 xx 7)/(16) = 373K`