At what temperature, the rate of effusion of `N_2` would be 1.625 times than the rate of `SO_2` at `500^@C` ?

Options

A. 273 K
B. 830 K
C. 110 K
D. 173 K

Answer

B

Explanation

`(r_(SO_(1)))/(r_(N_(2))) = sqrt((M_(N_(2)))/(M_(SO_(2)))) xx (T_(SO_(2)))/(T_(N_(2)))`
or `(x)/(1.625 x) = sqrt((28)/(64)) xx (773)/(T_(N_(2)))`
or `T_(N_(2)) = sqrt((28)/(64)) xx 773 xx 1.625`
`sqrt7 xx (773 xx 1.625)/(4) = 830` K