Entropy and the entropy statement of the second law

The Carnot topic gave η_C = 1 − T_C/T_H for a reversible engine, which is the statement Q_C/Q_H = T_C/T_H, or Q_H/T_H = Q_C/T_C. Now fix one sign convention and keep it: heat counts positive when it enters the working substance. The engine receives +Q_H at T_H and −Q_C at T_C, and the ratio becomes a sum: Q_H/T_H + (−Q_C)/T_C = 0. Put numbers in. An engine taking 1000 J at 600 K and rejecting 500 J at 300 K has 1000/600 − 500/300 = 1.667 − 1.667 = 0. Notice what did not cancel: the heats themselves sum to +500 J, the work delivered. Dividing each heat by the temperature at which it flowed is what makes the terms cancel. That is the seed of everything here: for a reversible two-reservoir cycle, the sum of Q/T round the loop vanishes. The rest of the topic is the generalisation of that one sentence to any cycle, and then to any process.

A general reversible cycle exchanges heat at continuously varying temperatures, so the sum becomes an integral, ∮dQ/T, with T the temperature at which dQ enters. Slice the cycle with a family of closely spaced adiabats. Between two neighbouring adiabats the upper and lower arcs of the cycle are short enough to replace by isotherms, so each strip is a small Carnot cycle and contributes dQ_H/T_H − dQ_C/T_C = 0. The interior adiabats are traversed once in each direction by neighbouring strips and cancel, leaving only the original boundary. Summing the strips: ∮dQᵣₑᵥ/T = 0 for every reversible cycle. Now take two reversible paths A and B from state 1 to state 2. A forwards and B backwards form a reversible loop, so ∫_A dQ/T − ∫_B dQ/T = 0. The integral is the same along A, along B, and along every other reversible route: it depends only on the end states. A path-independent integral defines a function of state, and this one is named entropy: S₂ − S₁ = ∫₁² dQᵣₑᵥ/T. Heat is not a state function; heat divided by the temperature at which it flows, summed along a reversible path, is.

Test the claim on a case you can integrate. One mole of a monatomic ideal gas (C_V = 3R/2 = 12.47 J mol⁻¹ K⁻¹) goes from (300 K, 10 L) to (600 K, 20 L). Path A: heat it at constant volume, Q = nC_VΔT = 12.47 × 300 = 3741 J, then expand it isothermally at 600 K, Q = nRT ln 2 = 8.314 × 600 × 0.6931 = 3458 J; total 7199 J. Path B: expand isothermally at 300 K first, Q = 8.314 × 300 × 0.6931 = 1729 J, then heat at constant volume, 3741 J; total 5470 J. The heats differ by 1.7 kJ, as heat may, since the work differs. Now divide before summing. The isochoric leg contributes ∫nC_V dT/T = nC_V ln 2 = 8.64 J K⁻¹ on either path. The isothermal leg contributes Q/T = 3458/600 = 5.76 J K⁻¹ on A and 1729/300 = 5.76 J K⁻¹ on B: the heat was doubled by doubling T, and T divides it straight back out. Both paths give 8.64 + 5.76 = 14.4 J K⁻¹. Written generally, ΔS = nC_V ln(T₂/T₁) + nR ln(V₂/V₁), a formula with nothing in it but the end states.

A real engine between the same reservoirs has η < η_C, so it rejects more heat than the reversible one: Q_C/T_C > Q_H/T_H, and the loop sum goes negative. Take 1000 J from 600 K with η = 0.30: W = 300 J, Q_C = 700 J, and ∮dQ/T = 1000/600 − 700/300 = 1.667 − 2.333 = −0.667 J K⁻¹. This is the Clausius inequality, ∮dQ/T ≤ 0, and the T in it is the temperature of the reservoir supplying or receiving dQ, because during an irreversible step the substance itself may have no single temperature to quote. Read the sign as a test. A cycle with ∮dQ/T > 0 would reject less heat than Carnot allows and beat η_C, which the Kelvin statement of the previous topic forbids; equality marks the reversible cycle; the more negative the value, the more the cycle wastes. Its magnitude is the entropy the cycle hands to the reservoirs on balance: the hot one lost 1.667 J K⁻¹, the cold one gained 2.333 J K⁻¹, and 0.667 J K⁻¹ appeared from nowhere.

Let a system go from 1 to 2 by any process at all, then bring it back reversibly. The loop obeys Clausius: ∫₁² dQ/T (actual path) + ∫₂¹ dQᵣₑᵥ/T ≤ 0. The second integral is S₁ − S₂, so ∫₁² dQ/T ≤ S₂ − S₁: along an irreversible path the heat-over-temperature integral underestimates the entropy change. Each dQ came from a reservoir at that same T, whose entropy fell by dQ/T, so the surroundings changed by exactly −∫dQ/T ≥ −ΔS_sys. Rearranged: ΔS_sys + ΔSₛᵤᵣᵣ ≥ 0. That is the entropy statement of the second law, and for an isolated system, where dQ = 0, it collapses to ΔS ≥ 0. Try it on conduction. Let 1000 J leak from 600 K to 300 K through a rod: ΔSᵤₙᵢᵥ = 1000/300 − 1000/600 = +1.667 J K⁻¹. The reverse leak would give −1.667 J K⁻¹ and never happens. Nothing was destroyed and nothing was conserved; energy was, entropy was created. That is why this is a directional constraint and not a conservation law: it sorts processes into allowed and forbidden, and it does so with an inequality, not an equals sign.

Boltzmann read the same quantity as S = k ln W, with W the number of microstates that look like the given macrostate. Logarithms turn a product of counts into a sum of entropies, which is why S is extensive. Test it on a free expansion: N molecules of ideal gas at fixed T double their volume into a vacuum. Each molecule has twice as many places to be, so W₂/W₁ = 2^N and ΔS = k ln 2^N = Nk ln 2 = nR ln 2, which is 5.76 J K⁻¹ per mole. That is exactly what ∫dQᵣₑᵥ/T gives along a reversible isothermal expansion between the same end states, nR ln(V₂/V₁). The actual expansion, though, had Q = 0: along the real path ∫dQ/T = 0, and 0 < 5.76 J K⁻¹ is the Clausius inequality once more. So the recipe stands: to compute ΔS you invent a reversible path with the same end states, because dQ/T has no meaning inside an irreversible lurch. Counting also shows what forbidden means. The chance of finding a mole back in its original half is 2^(−N) = 10^{(−1.8 × 10²³)}: not impossible, merely unobservable, and for ten molecules, 1 in 1024, it happens all the time.