Correct Answer - A
`KE= (3)/(2)RT, T= -123+273= + 150K`
`(3)/(2)xxRxx150= 3xx8.314xx75= xJ`
`=225xx 8.314=x`
At `27^(@)C= 27+223= 300K`
`KE` for `=2xJ=(3)/(2)xx8.314xx300`
`N` molecules
`:. x "Joule" =3xx8.314xx75`
In both the cases `x J` corresponds to `N` molecules.