A sample of argon gas at 1 atm 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_(V), m` for argon is `12.48 J. K^(-1). " mol"^(-1))`.

Explanation

Relation between volume & absolute temperature for an ideal gas involved in a reversible adiabatic process :
`T_(1)V_(1)^(gamma-1)= T_(2)V_(2)^(gamma-1)`
Given : `C_(V_(m))= 12.48 J. K^(-1) " mol"^(-1)`
`therefore C_(P_(m))=R+C_(V_(m))=(8.314+12.48)`
`= 20.794 J. K^(-1)" mol"^(-1)`
`therefore gamma= (C_(P,m))/(C_(V,m))=(20.794)/(12.48)=1.66`.
It is also given that `T_(1)=(273+27)= 300 K`.
`V_(1)= 1.25 dm^(3) and V_(2)= 2.50 dm^(2)`.
`therefore (T_(1))/(T_(2))= ((V_(2))/(V_(1)))^(gamma-1)= ((2.5)/(1.25))^(1.66-1)= 1.58`
or, `T_(2)=(300)/(1.58)=189.87 K`.
Number of moles of argon gas in the sample,
`n= (PV)/(RT)=(1xx1.25)/(0.0821xx300)= 0.05 ` mol
So, change in enthalpy in the process,
`Delta H = nC_(P, m)(T_(2)-T_(1))`
`= 0.05xx20.794(189.87-300) J = -114.5 J`