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At what temperature, the rate of effusio...

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

A

110K

B

173K

C

373K

D

273K

Text Solution

Verified by Experts

The correct Answer is:
C
`(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`
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