The Clausius inequality states that for any process, dS ≥ đQ/T, with equality holding for reversible processes and strict inequality for irreversible processes. Integrating over a complete cycle gives ∮(đQ/T) ≤ 0, with the integral being zero only for reversible cycles. The Clausius inequality provides a mathematical expression of the second law and establishes entropy as a measure of irreversibility and the spontaneity of processes.
Prove the Clausius inequality from the Carnot cycle and second law. Apply it to irreversible processes and cycles to verify sign changes.
You know the second law — heat flows spontaneously from hot to cold, and no engine converts all heat to work — and you know entropy as a state function measuring disorder or the number of accessible microstates. The Clausius inequality is the quantitative bridge between them: for any process, it tells you whether entropy has increased or decreased and by how much relative to the heat exchanged, giving the second law a precise mathematical form.
The physical reasoning begins with what we know about the most efficient possible cycle. A Carnot cycle operating reversibly between temperatures T_H and T_C has efficiency η = 1 − T_C/T_H, which means Q_C/Q_H = T_C/T_H. With careful sign convention (Q_H > 0 absorbed at T_H, Q_C > 0 rejected at T_C), this gives Q_H/T_H − Q_C/T_C = 0. For any *irreversible* cycle between the same temperatures, Carnot's theorem says the efficiency is strictly less: more heat is rejected per unit of heat absorbed, so Q_C/Q_H > T_C/T_H, giving Q_H/T_H − Q_C/T_C < 0. Any real cycle can be approximated by a sum of Carnot sub-cycles, leading to the general statement: ∮ đQ/T ≤ 0 for any cyclic process, with equality if and only if the cycle is entirely reversible.
For a non-cyclic process from state A to state B, combine the actual process with a reversible return path from B to A. The cycle inequality gives ∫_{A→B,actual} đQ/T + ∫_{B→A,rev} đQ/T ≤ 0. The second integral equals −(S_B − S_A) because entropy is a state function and the path is reversible (dS = đQ/T exactly on a reversible path). Rearranging: S_B − S_A ≥ ∫_A^B đQ/T, or in differential form dS ≥ đQ/T, with equality only on a reversible path. For an isolated system (đQ = 0), this gives dS ≥ 0 — the entropy of an isolated system never decreases. This is the second law in its sharpest form.
The difference σ = ΔS − ∫ đQ/T ≥ 0 is the entropy generated internally by irreversibility — friction, heat transfer across a finite temperature difference, turbulence, chemical reactions out of equilibrium. A process with σ = 0 is reversible; any σ > 0 marks irreversibility and represents lost work. In engineering thermodynamics, minimizing entropy generation is the route to maximum efficiency. Practical sources of irreversibility — heat transfer across finite ΔT in a heat exchanger, throttling through a valve, mixing of fluids at different temperatures — each have quantifiable σ values. The Clausius inequality thus converts the qualitative second law ("irreversible processes increase entropy") into a quantitative tool: compute ∫ đQ/T along the process, compare to ΔS, and the gap directly measures how much work was irreversibly destroyed.