1. The problem statement, all variables and given/known data A sample of an ideal gas goes through the process shown in Figure P20.32. From A to B, the process is adiabatic; from B to C, it is isobaric with 100 kJ of energy entering the system by heat. From C to D, the process is isothermal; from D to A, it is isobaric with 150 kJ of energy leaving the system by heat. Determine the difference in internal energy Eint,B – Eint,A. 2. Relevant equations PV^gamma=PV^gamma U = Q + W 3. The attempt at a solution I know two ways to solve this, and I am just wondering why the two ways give different answers. Way 1: Uc-Ub = Qbc + Wbc = Qbc - Pb(Vc-Vb) = 5.79 kJ Ud-Uc = 0 kJ Ua-Ud = Qad + Wad = Qad - Pa(Va-Vd) = -48.7 kJ Ub - Ua = -((Uc-Ub)+(Ud-Uc)+(Ua-Ud)) = 42.9 kJ Way 2: PV^gamma = constant in adiabatic processes. So we can use Pa, Pb, Va, Vb to solve for gamma. I got 1.375. Thus Cv = R/(gamma-1). But when I substitute Ub - Ua = nCvT = (1/(gamma-1))(PbVb-PaVa) I get 19kJ. What's the reason for this discrepancy?