Problem — Monatomic Carnot cycle (400-250 K)

Problem

A reversible Carnot cycle is carried out by $1\;\text{mol}$ of monatomic gas between $T_H = 400\;\text{K}$ and $T_C = 250\;\text{K}$. During the hot isothermal expansion the volume doubles. Calculate: (a) the absorbed $Q_H$, (b) the released $|Q_C|$, (c) the net work produced, (d) the efficiency, comparing it with the Carnot value. Use $R = 8{,}31\;\text{J/(mol·K)}$, $\ln 2 \approx 0{,}693$.

Solution

(a) Heat absorbed (hot isotherm A→B, $V_B/V_A = 2$):

$Q_H = nRT_H\ln 2 = 1\cdot 8{,}31\cdot 400\cdot 0{,}693 = \ev{2304\;\text{J} \approx 2{,}3\cdot 10^3\;\text{J}}$

(b) Heat released. From the adiabats $V_C/V_D = V_B/V_A = 2$, so the cold isotherm also sees the volume double:

$|Q_C| = nRT_C\ln 2 = 1\cdot 8{,}31\cdot 250\cdot 0{,}693 = \ev{1440\;\text{J} \approx 1{,}4\cdot 10^3\;\text{J}}$

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

$L_\text{netto} = 2304 - 1440 = \ev{864\;\text{J} \approx 8{,}6\cdot 10^2\;\text{J}}$

(d) Efficiency and check.

$\eta = \frac{L_\text{netto}}{Q_H} = \frac{864}{2304} = 0{,}375 \approx \ev{38\%}$

$\eta_\text{Carnot} = 1 - \frac{T_C}{T_H} = 1 - \frac{250}{400} = 0{,}375 = 37{,}5\% \quad\checkmark$

The two values coincide: the Carnot cycle reaches exactly the maximum efficiency between the two temperatures.

Links

Topics: Macchine termiche Concepts: Ciclo di Carnot · Rendimento Skills: Risoluzione di un ciclo termodinamico Objects: Macchina termica · Gas ideale