19. (a) Work is done only for the ab portion of the process. This portion is at constant pressure, so the
work done by the gas is

4V0
W = p0 dV = p0(4V0 - V0) =3p0V0 .
V0
(b) We use the first law: "Eint = Q - W . Since the process is at constant volume, the work done by
the gas is zero and Eint = Q. The energy Q absorbed by the gas as heat is Q = nCV "T , where
CV is the molar specific heat at constant volume and "T is the change in temperature. Since the
3
gas is a monatomic ideal gas, CV = R. Use the ideal gas law to find that the initial temperature
2
is Tb = pbVb/nR =4p0V0/nR and that the final temperature is Tc = pcVc/nR =(2p0)(4V0)/nR =
8p0V0/nR. Thus,

3 8p0V0 4p0V0
Q = nR - =6p0V0 .
2 nR nR
The change in the internal energy is "Eint = 6p0V0. Since n = 1 mol, this can also be written
Q =6RT0. Since the process is at constant volume, use dQ = nCV dT to obtain

Tc
dQ dT Tc
"S = = nCV = nCV ln .
T T Tb
Tb
3
Substituting CV = R and using the ideal gas law, we write
2
Tc pcVc (2p0)(4V0)
= = =2 .
Tb pbVb p0(4V0)
3 3
Thus, "S = nR ln 2. Since n =1, this is "S = R ln 2.
2 2
(c) For a complete cycle, "Eint =0 and "S =0.