Ideal gas process internal energy change

[arili]

Homework Statement

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.

Homework Equations

PV^gamma=PV^gamma
U = Q + W

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?

 

[TSny]

Your work for both approaches looks good to me. Hope I'm not overlooking something.

I would say that the numbers given in the problem are inconsistent. For example, you can use your value of $\gamma$ to find $C_P$ and then calculate the heat added going from B to C using $Q = nC_p\Delta T$. It doesn't agree with the value for the heat given in the problem.

 

[Chestermiller]

The discrepency is this: Who says that the amounts of heat are 100 and 150? This is inconsistent with your calculation using method 2. (Of course, this all assumes that that the process paths are irreversible).

Chet