Worked exercise — simplified Carnot cycle

Problem

$1\;\text{mol}$ of monatomic gas performs a Carnot cycle between $T_H = 600\;\text{K}$ and $T_C = 300\;\text{K}$. In the hot isothermal expansion the volume doubles. Calculate the efficiency and compare it with $\eta_\text{Carnot}$. Use $R = 8{,}31\;\text{J/(mol·K)}$, $\ln 2 \approx 0{,}693$.

Solution

Heat absorbed (A→B, isothermal at $T_H$). In the reversible isothermal expansion $Q_H = nRT_H\ln(V_B/V_A)$, with $V_B/V_A = 2$:

$Q_H = nRT_H\ln 2 = 1\cdot 8{,}31\cdot 600\cdot 0{,}693 = 3456\;\text{J}$

Volume ratio on the cold isotherm. From the two adiabatics of the cycle we get $V_C/V_D = V_B/V_A = 2$: the cold isotherm also sees the volume double.

Heat released (C→D, isothermal at $T_C$).

$|Q_C| = nRT_C\ln 2 = 1\cdot 8{,}31\cdot 300\cdot 0{,}693 = 1728\;\text{J}$

Net work. In a cycle $\Delta E_\text{int} = 0$, so $L_\text{net} = Q_H - |Q_C|$:

$L_\text{net} = 3456 - 1728 = 1728\;\text{J}$

Efficiency.

$\eta = \frac{L_\text{net}}{Q_H} = \frac{1728}{3456} = \ev{0{,}5 = 50\%}$

Check with the Carnot formula.

$\eta_\text{Carnot} = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 0{,}5 = 50\% \quad\checkmark$

The two values coincide: the Carnot cycle reaches exactly the maximum efficiency allowed between the two temperatures, and this does not depend on the type of gas.

Collegamenti

Argomenti: Heat engines Concetti: Carnot cycle · Efficiency · Thermodynamic transformations Competenze: Solving a thermodynamic cycle Oggetti: Heat engine · Ideal gas