a. The entropy balance for the 1 mol of gas contained in the piston and cylinder is
\underline{S}_{f}-\underline{S}_{i}=\frac{Q}{T}+S_{\text {gen }}
where T is the constant temperature of this isothermal system and Q is the total heat flow (from both the thermostatic bath and the cylinder walls) to the gas. From Eq. g of Illustration 3.4-7, we have for the 1 mol of gas
Q=R T \ln \frac{\underline{V}_{f}}{\underline{V}_{i}}
and from Eq. 4.4-2, we have
\begin{aligned}\underline{S}\left(T_{2}, \underline{V}_{2}\right)-\underline{S}\left(T_{1}, \underline{V}_{1}\right) &=C_{ V }^{*} \int_{T_{1}}^{T_{2}} \frac{d T}{T}+R \int_{\underline{V}_{1}}^{\underline{V}_{2}} \frac{d \underline{V}}{\underline{V}} \\&=C_{ V }^{*} \ln \left(\frac{T_{2}}{T_{1}}\right)+R \ln \left(\frac{\underline{V}_{2}}{\underline{V}_{1}}\right)\end{aligned} (4.4-2)
\underline{S}_{f}-\underline{S}_{i}=R \ln \frac{\underline{V}_{f}}{\underline{V}_{i}}
since the temperature of the gas is constant. Thus
S_{ gen }=\underline{S}_{f}-\underline{S}_{i}-\frac{Q}{T}=R \ln \frac{\underline{V}_{f}}{\underline{V}_{i}}-\frac{1}{T}\left\{R T \ln \frac{\underline{V}_{f}}{\underline{V}_{i}}\right\}=0
so that the gas undergoes a reversible expansion in all four processes.
b. The entropy balance for the isothermal system consisting of 1 mol of gas and the piston and cylinder is
S_{f}-S_{i}=\frac{Q}{T}+S_{ gen }
where Q is the heat flow to the piston, cylinder, and gas (Q^{ NET } of Illustration 3.4-7) and S_{f}-S_{i} is the entropy change for that composite system:
S_{f}-S_{i}=\left(\underline{S}_{f}-\underline{S}_{i}\right)_{ gas }+\left(S_{f}-S_{i}\right)_{\text {piston-cylinder }}
Since the system is isothermal,
\left(S_{f}-\underline{S}_{i}\right)_{ gas }=R \ln \frac{\underline{V}_{f}}{\underline{V}_{i}} (see Eq. 4.4-2)
and
\left(S_{f}-S_{i}\right)_{\text {piston-cylinder }}=0 see Eq. 4.4-6)
\underline{S}\left(T_{2}\right)-\underline{S}\left(T_{1}\right)=\int_{T_{1}}^{T_{2}} C_{ P } \frac{d T}{T} (4.4-6)
Consequently,
S_{\text {gen }}=R \ln \frac{V_{f}}{\underline{V}_{i}}-\frac{Q}{T}=\frac{1622.5-Q^{\text {NET }}}{298.15} J / K
so we find, using the entries in Table 1 of Illustration 3.4-7, that
S_{\text {gen }}= \begin{cases}1.4473 J / K & \text { for process a } \\ 0.8177 J / K & \text { for process b } \\ 0.4343 J / K & \text { for process c } \\ 0 & \text { for process d }\end{cases}
Thus, we conclude that for the piston, cylinder, and gas system, processes a, b, and c are not reversible, whereas process d is reversible.
Comment
From the results of part (a) we find that for the gas all expansion processes are reversible (i.e., there are no dissipative mechanisms within the gas). However, from part (b), we see that when the piston, cylinder, and gas are taken to be the system, the expansion process is irreversible unless the expansion occurs in differential steps. The conclusion, then, is that the irreversibility, or the dissipation of mechanical energy to thermal energy, occurs between the piston and the cylinder. This is, of course, obvious from the fact that the only source of dissipation in this problem is the friction between the piston and the cylinder wall.