A what temperature, the rate of effusion...

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

135K

373K

546 K

303K

Text Solution

Verified by Experts

The correct Answer is:

`( r ( N_(2)))/( r(SO_(2))) = 1.625 = sqrt(((3RT( N_(2)))/(28))/((3Rxx 323)/(64)))`
`= 1.625 = sqrt(( 16 xx T ( N_(2)))/( 7 xx 323))`
`:. T ( N_(2)) = 373 K`

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