On the straight sloping leg,
p=p0(3−V/V0) and
dp=−p0dV/V0. Put
v=V/V0, which increases from
1 to
2. The reversible
heat transfer is
Thus
heat is absorbed for 1<v<15/8 and emitted for 15/8<v<2, with zero instantaneous
heat at
v=15/8. The switching point has
(p,V)=(9p0/8,15V0/8). In particular the net
heat on this leg,
23p0V0, is not the total positive
heat on it.