Correct Answer - C
Here, `W= -146KJ= -146xx10^(3)J`
`T_(2)-T_(1)= 7^(@)C`
`R= 8.3J "mole"^(-1)K^(-1)`
`=8.3xx10^(3)"J kilo mole"^(-1)K^(-1)`
As `W=(R(T_(2)-T_(1)))/(1-gamma)`
`:. -146xx10^(3)= (8.3xx10^(3)xx7)/(1-gamma)`
`gamma-1=(8.3xx10^(3)xx7)/(146xx10^(3))= 0.40`
`gamma= 1.40`
`:. The gas must be diatomic.