`4.0 mol` of an idela gas initially at `1.5atm` and `300K` is heated to `600K` where the pressure is `4.5atm`. Also, `C_(vm) = a+bT, a = 25 J K^(-1)mol^(-1), b = 0.03 J K^(-2) mol^(-1)`. Determine `DeltaU, DeltaH`.

Explanation

`DeltaU = int_(T_(1))^(T_(2))nC_(V)dT =nint_(T_(1))^(T_(2))(adT+bTdT)`
`=na (T_(2) - T_(1)) + (nb)/(2) (T_(2)^(2)-T_(1)^(2))`
`= 4(25 xx 300) +(4xx0.03)/(2) (600^(2) - 300)^(2) = 46.2KJ`
`DeltaH = int_(T_(1))^(T_(2)) C_(P) dT = n [(a +R) DeltaT +(b)/(2) (T_(2)^(2) - T_(1)^(2))] ( :' C_(P) = C_(V) +R]`
`= 4 [33.31 xx 300 + (0.03)/(2) (600^(2) - 300)^(2))] = 56.17 KJ`