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)`.

Answer

`-115.96` joules

Explanation

For adiabatic expansion of a gas, we have `ln. (T_(1))/(T_(2)) = (R)/(C_(V)) " or "ln.(300)/(T_(2)) = (8.31)/(12.48) ln. (2.50)/(1.25)`
Solving the above equation, we get `T_(2) = 188.5 K` Number of moles of argon gas,
`n = (PV)/(RT) = (1 xx 1.25)/(0.082 xx 300) = 0.05`
Now, we know that `Delta H = n.C_(P).Delta T`
`:. Delta H = 0.05 xx 20.8 (188.5 - 300)`
`[C_(P) = C_(V) + R = (12.48 + 8.314) ~~ 20.8] = - 115.96J`