Compression in contact with one reservoir: reversible or sudden

Question 1. During the reversible compression, the gas remains in thermal equilibrium with the reservoir: its temperature is $T_0$ throughout the process. Its pressure increases gradually from $P_0$ to $P_1=xP_0$. In the final state,

$$P_{f,\mathrm{rev}}=xP_0,\qquad T_{f,\mathrm{rev}}=T_0,\qquad V_{f,\mathrm{rev}}=\frac{nRT_0}{xP_0}=\frac{V_0}{x},$$

where the initial equation of state $nRT_0=P_0V_0$ has been used.

For the sudden compression, the external pressure is set to $P_1$. At final mechanical equilibrium, the gas pressure equals this external pressure: $P_{f,\mathrm{irr}}=P_1=xP_0$. Final thermal equilibrium with the reservoir requires $T_{f,\mathrm{irr}}=T_0$. The equation of state then gives

$$V_{f,\mathrm{irr}}=\frac{nRT_{f,\mathrm{irr}}}{P_{f,\mathrm{irr}}} =\frac{nRT_0}{xP_0}=\frac{V_0}{x}.$$

Both processes therefore lead to the same final state $(P_f,T_f,V_f)=(xP_0,T_0,V_0/x)$.

Question 2. Entropy is a state function: its change depends only on the initial and final states, which are identical in both experiments. For the ideal gas,

$$\Delta S_{\mathrm{gas}} =nc_V\ln\frac{T_f}{T_0}+nR\ln\frac{V_f}{V_0} =nc_V\ln1+nR\ln\frac1x=-nR\ln x.$$

Thus $\Delta S_{\mathrm{gas,rev}}=\Delta S_{\mathrm{gas,irr}}=-nR\ln x<0$. This decrease in the gas's entropy does not determine the entropy produced: exchanges with the reservoir must also be included.

Question 3. During the reversible compression, $P_{\mathrm{ext}}=P=nRT_0/V$. The work received is

$$\begin{aligned} W_{\mathrm{rev}}&=-\int_{V_0}^{V_0/x}\frac{nRT_0}{V}\,\mathrm dV\\ &=-nRT_0[\ln V]_{V_0}^{V_0/x} =-nRT_0\ln(1/x)=nRT_0\ln x>0. \end{aligned}$$

The gas returns to its initial temperature, so $\Delta U=nc_V(T_0-T_0)=0$. The first law $\Delta U=Q+W$ gives $Q_{\mathrm{rev}}=-W_{\mathrm{rev}}$. The reservoir receives the opposite heat at temperature $T_0$:

$$\Delta S_{\mathrm{th,rev}}=\frac{-Q_{\mathrm{rev}}}{T_0}=nR\ln x.$$

The entropy received by the gas is $S_{\mathrm{e},\mathrm{rev}}=Q_{\mathrm{rev}}/T_0=-nR\ln x$, so $S_{\mathrm{i},\mathrm{rev}}=\Delta S_{\mathrm{gas}}-S_{\mathrm{e},\mathrm{rev}}=0$. The reservoir receives entropy $S_{\mathrm{e},\mathrm{th}}=-Q_{\mathrm{rev}}/T_0$, equal to its entropy change: it too produces no entropy. Adding gives $\Delta S_{\mathrm{gas}}+\Delta S_{\mathrm{th,rev}}=0$: no entropy is produced in the combined system.

Question 4. The sudden compression takes place under constant external pressure $P_1$. The work received is therefore

$$\begin{aligned} W_{\mathrm{irr}}&=-P_1(V_f-V_0) =-xP_0\left(\frac{V_0}{x}-V_0\right)\\ &=P_0V_0(x-1)=nRT_0(x-1)>0. \end{aligned}$$

Again $\Delta U=0$, so $Q_{\mathrm{irr}}=-nRT_0(x-1)$.

Question 5. The reservoir receives $-Q_{\mathrm{irr}}$, hence

$$\Delta S_{\mathrm{th,irr}}=\frac{-Q_{\mathrm{irr}}}{T_0}=nR(x-1).$$

The entropy balance for the gas is

$$\Delta S_{\mathrm{gas}}=\frac{Q_{\mathrm{irr}}}{T_0}+S_{\mathrm{i}}, \qquad S_{\mathrm{i}}=-nR\ln x+nR(x-1).$$

For the ideal reservoir, the entropy received is $-Q_{\mathrm{irr}}/T_0$, equal to its change: it produces no entropy. The combined gas—reservoir system receives no heat from any other body; the work received from outside carries no entropy. Its entropy increase is therefore produced internally:

$$S_{\mathrm{i}}=\Delta S_{\mathrm{gas}}+\Delta S_{\mathrm{th,irr}} =-nR\ln x+nR(x-1).$$

Set $f(x)=x-1-\ln x$. Then $f(1)=0$ and $f'(x)=1-1/x=(x-1)/x>0$ for $x>1$. The function increases from zero, so $S_{\mathrm{i}}=nRf(x)>0$. Moreover,

$$W_{\mathrm{irr}}-W_{\mathrm{rev}} =nRT_0(x-1)-nRT_0\ln x=T_0 S_{\mathrm{i}}.$$

Since $Q=-W$ in both cases, the additional work is rejected as heat to the reservoir. This is the cost of irreversibility.