Training The Carnot Cycle — Thermodynamic Limits & Irreversibility Proof of Carnot's Theorem & The Clausius Inequality
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Proof of Carnot's Theorem & The Clausius Inequality

30 min The Carnot Cycle — Thermodynamic Limits & Irreversibility

Carnot's theorem — that no engine operating between two fixed temperature reservoirs can be more efficient than a reversible one — is one of the most profound results in physics. This lesson presents the complete proof by contradiction (the Kelvin-Planck and Clausius statements of the second law), derives the Clausius inequality $\oint \delta Q/T \leq 0$, and shows how this leads inevitably to the existence of entropy as a state function.

Carnot's Theorem and the Clausius Inequality

Carnot's Theorem (1824)

For any engine operating between a hot reservoir at $T_H$ and a cold reservoir at $T_L$:

$$\eta_{\text{any engine}} \leq \eta_{\text{Carnot}} = 1 - \frac{T_L}{T_H}$$

with equality if and only if the engine is internally reversible.

Proof by Contradiction (Kelvin-Planck)

Suppose engine X has $\eta_X > \eta_C$. Use X to drive a reversed Carnot (a refrigerator) between the same reservoirs. Let X absorb $Q_H$ from the hot reservoir and produce work $W_X = \eta_X Q_H$. Feed this work to the refrigerator, which pumps $Q_L' = W_X / (\eta_C/(1-\eta_C)) = W_X T_L/(T_H - T_L)$ from the cold reservoir back to the hot reservoir. The net effect: the hot reservoir gains heat with no external work input — violating the Kelvin-Planck statement of the second law. Therefore $\eta_X \leq \eta_C$.

Clausius Inequality

For any thermodynamic cycle, real or ideal:

$$\oint \frac{\delta Q}{T} \leq 0$$

with equality ($= 0$) for internally reversible cycles. For an irreversible cycle the integral is strictly negative — the deficit equals the total entropy generated.

Entropy as a State Function

Because $\oint (\delta Q/T)_{\text{rev}} = 0$ for every reversible cycle, the integrand must be an exact differential. Define:

$$dS \equiv \frac{\delta Q_{\text{rev}}}{T}$$

$S$ is the entropy — a thermodynamic property of the system, independent of path, measured in J/K. For irreversible processes between the same end states:

$$\Delta S_{\text{system}} \geq \int_1^2 \frac{\delta Q}{T}$$

The excess is the entropy generated by irreversibilities (friction, free expansion, heat across a finite $\Delta T$).

Example 1 — Clausius Integral for an Irreversible Engine

An engine runs between $T_H = 800$ K and $T_L = 300$ K. It absorbs $Q_H = 8000$ J and exhausts $Q_L = 3500$ J. Evaluate $\oint \delta Q/T$ and find the entropy generated.

  1. $\oint \delta Q/T = Q_H/T_H - Q_L/T_L = 8000/800 - 3500/300 = 10 - 11.67 = -1.67$ J/K
  2. Since the integral is negative, the cycle is irreversible (as expected for a real engine).
  3. The entropy generated is $\dot{S}_{\text{gen}} = |\oint \delta Q/T| = 1.67$ J/K per cycle.
  4. Check efficiency: $\eta = (8000-3500)/8000 = 56.25\%$ vs.
  5. Carnot $62.5\%$ ✓.
Example 2 — The Thermodynamic Temperature Scale

Show that the Carnot efficiency ratio $Q_L/Q_H$ for a reversible engine defines an absolute temperature scale independent of the working fluid.

  1. For any two reversible engines cascaded between $T_1 > T_2 > T_3$, the efficiency product rule gives:
  2. $\eta_{13} = 1 - Q_3/Q_1$ and $\eta_{12}\cdot\eta_{23}$ must be consistent.
  3. The only function $f(T_1, T_3) = Q_1/Q_3$ that satisfies this cascade property is $f = T_1/T_3$.
  4. This means $Q_H/Q_L = T_H/T_L$ — the ratio of heat transfers equals the ratio of thermodynamic temperatures, which Kelvin used to define the absolute temperature scale.
  5. The ideal gas scale and the thermodynamic scale coincide.

Practice Problems

1. An engine absorbs 10 kJ at 600 K and rejects 4 kJ at 250 K. Is it possible? Calculate $\oint \delta Q/T$.
2. State the Kelvin-Planck and Clausius statements of the second law and show they are equivalent.
3. Why can entropy be defined only via reversible heat transfer, even though it is a property of irreversible states?
4. A Carnot heat pump operates between $-10$°C and $35$°C. Find its COP for heating ($\text{COP}_{HP} = Q_H/W$).
5. Engine A: $Q_H = 5$ kJ, $Q_L = 2$ kJ, $T_H = 500$ K, $T_L = 200$ K. Compute $\oint \delta Q/T$. How much entropy was generated?
6. Can the Clausius inequality equal zero for an irreversible process? Why?
Show Answer Key

1. $\oint \delta Q/T = 10000/600 - 4000/250 = 16.67 - 16 = +0.67$ J/K. Since $> 0$, this violates the Clausius inequality — the engine is impossible.

2. KP: "No device converts heat from a single reservoir entirely to work." Clausius: "Heat cannot spontaneously flow from cold to hot." Equivalence: assume KP violated → construct a device violating Clausius, and vice versa.

3. Entropy is a state function defined by connecting any two states via a reversible path and integrating $\delta Q/T$ along that path. Real processes may be irreversible, but entropy change depends only on end states, so we always compute it via the reversible surrogate.

4. $T_H = 308$ K, $T_L = 263$ K. $\text{COP}_{HP} = T_H/(T_H - T_L) = 308/45 = 6.84$.

5. $\oint \delta Q/T = 5000/500 - 2000/200 = 10 - 10 = 0$. Entropy generated = 0 — the engine is reversible (Carnot). Verify: $\eta = 3/5 = 60\%$ and $1 - 200/500 = 60\%$ ✓.

6. No. For an irreversible cycle the Clausius inequality is strictly $< 0$. Zero is attained only by reversible cycles.

Exact State-Point Analysis of the Carnot Cycle Irreversibility, Lost Work & the Gouy-Stodola Theorem