Problem — Derivation of Carnot efficiency

Problem

Consider a Carnot cycle between $T_H$ and $T_C$ carried out by $n$ moles of ideal gas. Starting from the formula $Q = nRT\ln(V_B/V_A)$ for the isotherms and from $TV^{\gamma-1} = \text{const}$ for the adiabats, show that the efficiency is $\eta = 1 - T_C/T_H$, i.e. it does not depend on the type of gas.

Solution

Vertices of the cycle. A→B hot isotherm at $T_H$; B→C adiabat; C→D cold isotherm at $T_C$; D→A adiabat.

Heats exchanged along the isotherms. For a reversible isotherm $\Delta E_\text{int} = 0$, so $Q = L_\text{gas} = nRT\ln(V_\text{fin}/V_\text{in})$:

$Q_H = nRT_H\ln\frac{V_B}{V_A} \qquad |Q_C| = nRT_C\ln\frac{V_C}{V_D}$

Constraint from the adiabats. Applying $TV^{\gamma-1} = \text{const}$ to the two adiabatic segments:

$T_H V_B^{\gamma-1} = T_C V_C^{\gamma-1} \qquad T_C V_D^{\gamma-1} = T_H V_A^{\gamma-1}$

Dividing the first by the second side by side, $T_H$ and $T_C$ cancel:

$\frac{V_B^{\gamma-1}}{V_A^{\gamma-1}} = \frac{V_C^{\gamma-1}}{V_D^{\gamma-1}} \quad\Rightarrow\quad \frac{V_B}{V_A} = \frac{V_C}{V_D}$

Ratio of the heats. The logarithms of the volume ratios are therefore equal, and cancel:

$\frac{|Q_C|}{Q_H} = \frac{nRT_C\ln(V_C/V_D)}{nRT_H\ln(V_B/V_A)} = \frac{T_C}{T_H}$

Efficiency. From $\eta = 1 - |Q_C|/Q_H$:

$\ev{\eta = 1 - \frac{T_C}{T_H}}$

Conclusion. In the derivation, $n$, $R$, $\gamma$ and the volumes have all disappeared: the efficiency depends only on the two temperatures, not on the working fluid. This is the core of the second law identified by Carnot: no real engine operating between the same temperatures can exceed this value.

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Topics: Macchine termiche Concepts: Ciclo di Carnot · Rendimento · Trasformazioni termodinamiche Skills: Impostazione simbolica Objects: Gas ideale