Entropy of a gas: when the heat capacity depends on $T$

Question 1. The heat capacity is molar: for $n$ moles, the total heat capacity is $nc_V(T)$. Since $U$ depends only on $T$,

$$\mathrm dU=nc_V(T)\,\mathrm dT=n(A+BT)\,\mathrm dT.$$

Integrate from $T_{\mathrm{ref}}$ to $T$, using $\tau$ as the integration variable:

$$\begin{aligned} U(T)-U_{\mathrm{ref}} &=n\int_{T_{\mathrm{ref}}}^{T}(A+B\tau)\,\mathrm d\tau\\ &=n\left[A\tau+\frac B2\tau^2\right]_{T_{\mathrm{ref}}}^{T}\\ &=nA(T-T_{\mathrm{ref}})+\frac{nB}{2}(T^2-T_{\mathrm{ref}}^2). \end{aligned}$$

Adding $U_{\mathrm{ref}}$ gives $U(T)$. When calculating $\Delta U=U(T_f)-U(T_i)$, the reference terms cancel:

$$\boxed{\Delta U=nA(T_f-T_i)+\frac{nB}{2}(T_f^2-T_i^2).}$$

We cannot write $\Delta U=nc_V(T_f)(T_f-T_i)$: the heat capacity varies during heating and must remain inside the integral.

Question 2. During the locally reversible process, the work received is $\delta W=-P\,\mathrm dV$. The first law gives

$$\mathrm dU=\delta Q_{\mathrm{rev}}-P\,\mathrm dV, \qquad \delta Q_{\mathrm{rev}}=\mathrm dU+P\,\mathrm dV.$$

Replace $\mathrm dU$ by the previous result and $P$ by $nRT/V$:

$$\delta Q_{\mathrm{rev}}=n(A+BT)\,\mathrm dT+\frac{nRT}{V}\,\mathrm dV.$$

Local reversibility allows us to write $\mathrm dS=\delta Q_{\mathrm{rev}}/T$. Dividing each term by $T$ gives

$$\boxed{\mathrm dS=n\left(\frac AT+B\right)\mathrm dT +\frac{nR}{V}\,\mathrm dV.}$$

Question 3. Choose a reversible path from $(T_i,V_i)$ to $(T_f,V_i)$ at constant volume, followed by a path from $(T_f,V_i)$ to $(T_f,V_f)$ at constant temperature. In the first step, $\mathrm dV=0$; in the second, $\mathrm dT=0$. Adding the two entropy changes gives

$$\begin{aligned} \Delta S &=n\int_{T_i}^{T_f}\left(\frac AT+B\right)\mathrm dT +nR\int_{V_i}^{V_f}\frac{\mathrm dV}{V}\\ &=nA\left[\ln T\right]_{T_i}^{T_f} +nB\left[T\right]_{T_i}^{T_f} +nR\left[\ln V\right]_{V_i}^{V_f}\\ &=nA(\ln T_f-\ln T_i)+nB(T_f-T_i) +nR(\ln V_f-\ln V_i). \end{aligned}$$

Using $\ln a-\ln b=\ln(a/b)$, we obtain

$$\boxed{\Delta S=nA\ln\frac{T_f}{T_i}+nB(T_f-T_i) +nR\ln\frac{V_f}{V_i}.}$$

Entropy is a state function: its change depends only on the initial and final states. The reversible path is used to calculate this change, but the result also applies to an actual irreversible process between the same states.

Question 4. The volume is constant: the gas receives no work. The first law and question 1 therefore give

$$Q_{\mathrm{gas}}=\Delta U =nA(T_f-T_i)+\frac{nB}{2}(T_f^2-T_i^2).$$

The combined gas—reservoir system is isolated, so $\Delta U_{\mathrm{gas}}+\Delta U_{\mathrm{th}}=0$. With no work exchanged, this implies $Q_{\mathrm{th}}=-Q_{\mathrm{gas}}$. For the gas, $V_f=V_i$ makes the volume term from question 3 vanish:

$$\Delta S_{\mathrm{gas}}=nA\ln\frac{T_f}{T_i}+nB(T_f-T_i).$$

The reservoir remains at $T_f$, hence

$$\Delta S_{\mathrm{th}}=\frac{Q_{\mathrm{th}}}{T_f} =-\frac{Q_{\mathrm{gas}}}{T_f}.$$

The isolated combined system exchanges no entropy with the rest of the universe: the entropy produced is the sum of the two changes,

$$\begin{aligned} S_{\mathrm{i}} &=\Delta S_{\mathrm{gas}}+\Delta S_{\mathrm{th}}\\ &=nA\ln\frac{T_f}{T_i}+nB(T_f-T_i)\\ &\quad-\frac{nA(T_f-T_i)}{T_f}-\frac{nB(T_f^2-T_i^2)}{2T_f}. \end{aligned}$$

To determine the sign, return to the integral expressions for the two terms:

$$\Delta S_{\mathrm{gas}}=n\int_{T_i}^{T_f}\frac{A+BT}{T}\,\mathrm dT, \qquad \frac{Q_{\mathrm{gas}}}{T_f} =n\int_{T_i}^{T_f}\frac{A+BT}{T_f}\,\mathrm dT.$$

The limits are identical, so the integrands can be subtracted:

$$S_{\mathrm{i}}=n\int_{T_i}^{T_f}(A+BT) \left(\frac1T-\frac1{T_f}\right)\mathrm dT>0.$$

Indeed, $A+BT>0$ and $1/T-1/T_f>0$ for $T_i\leq T<T_f$. The integral is therefore strictly positive when $T_f>T_i$: heating by contact with this reservoir is irreversible.