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

`DeltaH=nxxC_(p)xxDeltaT`
and `C_(p)=C_(v)+R=12.48+8.314`
`=20.794 jK^(-1)` (R in J)
For a given sample of argon gas, mole `(n)`
`n=(PV)/(RT)=(1xx1.25)/(0.821xx300)=0.05`
Also for reversible adiabatic change `(TV)^(gamma-1)` =constant
or `T_(2)V_(2)^(gamma-1)=T_(1)V_(1)^(gamma-1)`
or `T_(2)=T_(1)xx((V_(1))/(V_(2)))^(gamma-1)` (`gamma=1.66` for argon)
`300xx((1.25)/(2.50))^(1.66-1)=300xx(1/2)^(0.66)`
`T_(2)=189.85 K`
`:. DeltaT=T_(2)-T_(1)=189.85-300=-110.15`
Thus, `DeltaH=0.05xx20.794xx(-110.15)=114.52 J`