Carnot cycle: energy and entropy along each branch

Question 1. The diagram shows adiabats $AB$ and $CD$, $BC$ on the hot isotherm and $DA$ on the cold isotherm. The cycle runs clockwise. Along adiabat $AB$, the adiabatic relation gives

$$T_{\mathrm{C}}V_A^{\gamma-1}=T_{\mathrm{H}}V_B^{\gamma-1},\qquad \left(\frac{V_A}{V_B}\right)^{\gamma-1}=\frac{T_{\mathrm{H}}}{T_{\mathrm{C}}}.$$

Raising to the power $1/(\gamma-1)$ and setting $b=(T_{\mathrm{H}}/T_{\mathrm{C}})^{1/(\gamma-1)}$ gives $V_A=bV_B$. Similarly, along $CD$,

$$T_{\mathrm{H}}V_C^{\gamma-1}=T_{\mathrm{C}}V_D^{\gamma-1} \quad\Longrightarrow\quad V_D=bV_C=brV_B.$$

Thus $V_D/V_A=(brV_B)/(bV_B)=r$: the isotherms have the same volume ratio, traversed in opposite directions.

Question 2. Along each reversible adiabat, $\delta Q_{\mathrm{rev}}=0$ and $\mathrm dS=\delta Q_{\mathrm{rev}}/T=0$:

$$Q_{AB}=Q_{CD}=0,\qquad \Delta S_{AB}=\Delta S_{CD}=0.$$

Along an ideal-gas isotherm, $\mathrm dU=0$, so $\delta Q_{\mathrm{rev}}=P\,\mathrm dV=nRT\,\mathrm dV/V$. For the expansion at the hot temperature,

$$\begin{aligned} Q_{BC}&=nRT_{\mathrm{H}}\int_{V_B}^{rV_B}\frac{\mathrm dV}{V} =nRT_{\mathrm{H}}[\ln V]_{V_B}^{rV_B}=nRT_{\mathrm{H}}\ln r>0,\\ \Delta S_{BC}&=\frac{Q_{BC}}{T_{\mathrm{H}}}=nR\ln r. \end{aligned}$$

For the compression at the cold temperature, $V_A/V_D=1/r$:

$$\begin{aligned} Q_{DA}&=nRT_{\mathrm{C}}\int_{V_D}^{V_A}\frac{\mathrm dV}{V} =nRT_{\mathrm{C}}[\ln V]_{V_D}^{V_A}\\ &=nRT_{\mathrm{C}}\ln\frac{V_A}{V_D}=-nRT_{\mathrm{C}}\ln r<0,\\ \Delta S_{DA}&=\frac{Q_{DA}}{T_{\mathrm{C}}}=-nR\ln r. \end{aligned}$$

In the $(S,T)$ diagram, entropy is constant along $AB$ and $CD$: these branches are vertical. Isotherms $BC$ and $DA$ are horizontal. Taking $S_A$ as the relative origin, the coordinates are

$$\begin{aligned} \begin{array}{c|cc} \text{State}&S-S_A&T\\ \hline A&0&T_{\mathrm{C}}\\ B&0&T_{\mathrm{H}}\\ C&nR\ln r&T_{\mathrm{H}}\\ D&nR\ln r&T_{\mathrm{C}} \end{array} \end{aligned}$$

This gives a clockwise rectangle of width $nR\ln r$ and height $T_{\mathrm{H}}-T_{\mathrm{C}}$.

Question 3. Over a cycle, $\Delta U=0$. Counting received energy as positive, the first law gives

$$W=-(Q_{AB}+Q_{BC}+Q_{CD}+Q_{DA}) =-nR(T_{\mathrm{H}}-T_{\mathrm{C}})\ln r<0.$$

The work delivered is $-W>0$ and the heat received from the hot reservoir is $Q_{BC}>0$. The efficiency is therefore

$$\eta=\frac{-W}{Q_{BC}} =\frac{nR(T_{\mathrm{H}}-T_{\mathrm{C}})\ln r}{nRT_{\mathrm{H}}\ln r} =\frac{T_{\mathrm{H}}-T_{\mathrm{C}}}{T_{\mathrm{H}}}=1-\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}.$$

Each reservoir receives the opposite of the heat it supplies to the gas:

$$\Delta S_{\mathrm{th,H}}=\frac{-Q_{BC}}{T_{\mathrm{H}}}=-nR\ln r,\qquad \Delta S_{\mathrm{th,C}}=\frac{-Q_{DA}}{T_{\mathrm{C}}}=nR\ln r.$$

The gas returns to its initial state, so $\Delta S_{\mathrm{gas}}=0$. The three changes sum to zero: no entropy is produced. Work delivered to the surroundings carries no entropy.

Question 4. Retain the received-energy sign convention for the machine: $Q_{\mathrm{H}}>0$ is the heat received from the hot reservoir, $Q_{\mathrm{C}}<0$ that received from the cold reservoir, and $W<0$ the work received over a cycle. The reservoirs receive the opposite heats, $-Q_{\mathrm{H}}$ and $-Q_{\mathrm{C}}$. Their total entropy change is the entropy produced, since the machine returns to its initial state and only work is transferred to the surroundings:

$$S_{\mathrm{i}}=-\frac{Q_{\mathrm{H}}}{T_{\mathrm{H}}}-\frac{Q_{\mathrm{C}}}{T_{\mathrm{C}}}.$$

Multiplying by $T_{\mathrm{C}}$ and isolating $Q_{\mathrm{C}}$ gives

$$T_{\mathrm{C}}S_{\mathrm{i}}=-\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}Q_{\mathrm{H}}-Q_{\mathrm{C}}, \qquad Q_{\mathrm{C}}=-\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}Q_{\mathrm{H}}-T_{\mathrm{C}}S_{\mathrm{i}}.$$

The first law gives $0=Q_{\mathrm{H}}+Q_{\mathrm{C}}+W$. The recovered work is $|W|=-W=Q_{\mathrm{H}}+Q_{\mathrm{C}}$, hence

$$\eta=\frac{|W|}{Q_{\mathrm{H}}}=1+\frac{Q_{\mathrm{C}}}{Q_{\mathrm{H}}} =1-\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}-\frac{T_{\mathrm{C}}S_{\mathrm{i}}}{Q_{\mathrm{H}}}.$$

Since $T_{\mathrm{C}}>0$ and $Q_{\mathrm{H}}>0$, positive entropy production reduces the efficiency.

Question 5. The heat received from the hot reservoir is the same for both engines. From question 2,

$$Q_{\mathrm{H}}=nRT_{\mathrm{H}}\ln r=1{,}0\times8{,}314\times500\times\ln3{,}4 \simeq5087\ \mathrm J.$$

The reversible engine's efficiency is

$$\eta_{\mathrm{Carnot}}=1-\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}} =1-\frac{300}{500}=0{,}40=40\,\%.$$

Question 4 gives the difference between the two efficiencies:

$$\Delta\eta=\eta_{\mathrm{Carnot}}-\eta_{\mathrm{irr}} =\frac{T_{\mathrm{C}}S_{\mathrm{i}}}{Q_{\mathrm{H}}} \simeq\frac{300\times2{,}0}{5087}=0{,}118.$$

The numerator $T_{\mathrm{C}}S_{\mathrm{i}}=600\ \mathrm J$ is an energy, so the ratio is dimensionless. The reduction is $100\Delta\eta\simeq11{,}8$ percentage points. Thus,

$$\eta_{\mathrm{irr}}=0{,}40-0{,}118\simeq0{,}282=28{,}2\,\%.$$

The efficiency falls from $40\,\%$ to $28{,}2\,\%$; a reduction of $11{,}8$ points is not a relative reduction of $11{,}8\,\%$.