The root mean square velocity of `SO_(2)` gas becomes the same as that of methane at `27^(@)C` when the temperature is `:`

Options

A. `327^(@)C`
B. `127^(@)C`
C. `54^(@)C`
D. `927^(@)C`

Answer

D

Explanation

r.m.s. velocity `= sqrt(( 3RT)/( M))`
r.m.s. velocity of `SO_(2) = sqrt((3RT)/( 64))`
r.m.s. velocity of `CH_(4)` at 300K `= sqrt(( 3R xx 300)/( 16))`
Now `sqrt(( 3RT)/( 64)) = sqrt(( 3R xx 300)/(16))`
or `( T )/( 64) = ( 300)/( 16)`
or `T = ( 300 xx 64)/( 16) = 1200K`
or ` T = 1200 - 273 = 927^(@)C`
At constant pressure,
`( V_(1))/( T_(1)) = ( V_(2))/( T_(2))`
or `T_(2) = ( V_(2) T_(1))/( V_(1))`
`= ( 100 xx 300)/( 250) = 1200K`
or `T_(2) = 1200 - 273 = 927^(@)C`