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

`:. C_(V) = 3//2 R :. C_(P) = C_(V) +R = (3)/(2)R +R =(5)/(2)R`
Therefore, Heat given at constant pressure `= m. C_(P). DeltaT`
or `DeltaH` or `dq = 1 xx (5)/(2) xxR xx (373 -298)`
or `DeltaH =1 xx(5)/(2) xx 1.987 xx 75 = 373. 56 cal`
Now work done in the process `=- P DeltaV`
`W =- P (V_(2) - V_(1))`
`=- P ((nRT_(2))/(P)-(nRT_(1))/(P)) ( :' PV =nRT)`
`=- nR (T_(2)-T_(1)) = - 1 xx 1.987 xx (373 - 298)`
`=- 149. 025 cal`
Therefore, from the first law of thermodynamics
`DeltaU = q +W = 372.56 - 149.05 rArr DeltaU = 233.51 cal`
Also, `dq_(rev) = nC_(P).dT`
`ds = (dq_(rev))/(T) :. dS =(nC_(P).dT)/(T)`
or `DeltaS = int_(T_(1))^(T_(2)) (nC_(P).dT)/(T) = nC_(P) log_(e)"(T_(2))/(T_(1))`
`:.DeltaS = 2.303 nC_(P) log_(10).(T_(2))/(T_(1))`
`=2.303 xx 1xx (5)/(2) xxR xx log_(10).(373)/(298)`
`= 1.122 cal K^(-1) mol^(-1)`