One moles of an ideal gas which `C_(V) = 3//2 R` is heated at a constant pressure of `1 atm` from `25^(@)C` to `100^(@)C`. Calculate `DeltaU, DeltaH` and the entropy change during the process.

Explanation

As `(C_(P) - C_(V)) = R rArr C_(P) = C_(V) +R`
`rArr C_(P) = (3)/(2)R+R =(5)/(2)R`
Heat given at constnat pressure
`(DeltaH) = nC_(P) DeltaT =1 xx (5)/(2)R xx (373-298)`
`rArr (DeltaH) = 1xx (5)/(2) xx 1.987 xx 75 = 372.56 cal`
Work done in the process `=- P DeltaV = - P (V_(2)-V_(1))`
`=- P ((nRT_(2))/(P)-(nRT_(1))/(P))` as `(PV = nRT)`
`=nRT (T_(2)-T_(1)) =- 1 xx 1.987 xx (373-298) =- 149.05 cal`
From the first law of thermodynamics,
`DeltaU = q +w = 372.56 - 149.05 = 223.51 cal`