Carnot cycle

Example

Take $n$ moles of an ideal gas with constant heat capacity ratio $\gamma > 1$, and two reservoirs at temperatures $T_H > T_C$. The Carnot cycle consists of four reversible steps, through volumes $V_1 < V_2 < V_3$ and $V_1 < V_4 < V_3$:

1. isothermal expansion at $T_H$ from $V_1$ to $V_2$, absorbing heat $Q_H = nRT_H \ln(V_2/V_1)$;
2. isentropic expansion from $V_2$ to $V_3$, during which the gas cools from $T_H$ to $T_C$;
3. isothermal compression at $T_C$ from $V_3$ to $V_4$, rejecting heat $Q_C = nRT_C \ln(V_3/V_4)$;
4. isentropic compression from $V_4$ back to $V_1$, during which the gas warms from $T_C$ to $T_H$.

Check that it is a thermodynamic cycle and compute its efficiency. The gas ends at volume $V_1$ and temperature $T_H$, its initial state. On an isentrope of an ideal gas, $TV^{\gamma - 1}$ is constant, so $T_H V_2^{\gamma-1} = T_C V_3^{\gamma-1}$ and $T_H V_1^{\gamma-1} = T_C V_4^{\gamma-1}$. Dividing gives $V_2/V_1 = V_3/V_4$. The two isentropic steps exchange no heat. On an isotherm of an ideal gas the internal energy is constant, so the heat equals the work $nRT\ln(V_{\text{end}}/V_{\text{start}})$; this gives $Q_H$ and $Q_C$. Hence the thermal efficiency is $$\eta = 1 - \frac{Q_C}{Q_H} = 1 - \frac{T_C}{T_H},$$ the largest value allowed by Carnot's theorem. For $T_H = 500\,\mathrm{K}$ and $T_C = 300\,\mathrm{K}$, $\eta = 0.4$.