A sample of argon gas at `1atm` pressure and `27^(@)C` expands reversibly and adiabatically from `1.25 dm^(3)` to `2.50 dm^(3)`. Calculate the enthalpy change in this process. `C_(vm)` for orgon is `12.48J K^(-1) mol^(-1)`.

Explanation

Number of moles of argon present in the sample
`n =(PV)/(RT) = (1.25 xx 1)/(0.0821xx300) = 0.05075`
For adiabatic expansion,
`(T_(1))/(T_(2)) = ((V_(1))/(V_(1)))^(gamma-1)`
or `(300)/(T_(1)) = ((2.50)/(1.25))^(1.66-1)`
or `T_(2) = 188.55K`
`C_(P) = C_(V) +R = 12.48 +8.314 = 20.794 J K^(-1)mol^(-1)`
`DeltaH = n xx C_(P) xx DeltaT`
`= 0.05075 xx 20.794 xx (300 - 188.55) = 117.6 J`