A system can be taken from state A to state B where SA = SB either by (a) directly along the adiabat S = constant, or (b) first along an isochore A to C and then along the isobar C to B. The difference in the work done by the system is simply the area enclosed between the two paths in a P–V diagram. Does this contradict the statement that the work delivered to a reversible work source (RWS) is the same for every reversible process? Explain.
The Attempt at a Solution
I don't think it contradicts the statement because although process b seems to involve a greater amount of work, it would also have a dQ which cannot be seen on the P-V diagram. If the path is taken along the adiabat, dQ=0. I think(and I could be wrong here) that the same amount of work is being delivered to the renewable work source, and the "extra" work of traveling along the isochor and isobar is actually in the form of heat.
I'm a bit shaky on this topic, and Callen's book can be very hard to follow at some points. I would really appreciate it if someone could help me out with understanding this. Thank you