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

For adiabatic process ,
`T_(1)V_(1)^(gamma-1)= T_(2)V_(2)^(gamma-1) ` or `((V_(1))/(V_(2)))^(gamma-1) = (T_(2))/(T_(1))`
But `gamma=( C_(p))/(C_(v))`
`:gt gamma -1 = (C_(p))/(C_(v))-1= ( C_(p)-C_(v))/(C_(v))= (R)/(C_(v))` ,brgt Hence `((V_(1))/(V_(2)))^(R//C_(v))= (T_(2))/(T_(1))`
or `ln.(T_(2))/(T_(1))=(R)/(C_(v))ln.(V_(1))/(V_(2)) ` or `log.(T_(2))/(T_(1))= (R)/(C_(v)) log. (V_(1))/(V_(2))`
i.e., `log. (T_(2))/(300) = ( 8.314)/( 12.48) log. ( 12.5)/(25)`
or ` logT_(2) = log 300 + 0.666 log . ( 1)/(2)`
`= 2.4771 + 0.666 ( - 0.3010) = 2.2766`
or `T_(2) =` Antilog 2.2766 `=189 K`
`C_(p)=(DeltaH)/( DeltaT)`
or `DeltaH = C_(p) Delta T ` or for n moles `Delta H = n C_(p) Delta T`.
Taking the gas as ideal.
`PV = nRT ` or ` 1 xx 1.25 = n xx 0.0821 xx 300` or `n = 0.05` mole
Further, `C_(p) = C_(v) + R = 12.48 + 8.314 = 20.794J`
`:. Delta H = 0.05 xx 20.794 xx ( 189 - 300) = - 115 . 4 J`