Entropy and irreversibility

Take a reversible Carnot engine between 600 K and 300 K that absorbs 6000 J from the boiler. It rejects Q_c = 6000 × 300/600 = 3000 J, so the net heat around the cycle is 3000 J, not zero: heat is not a state function and there is no such thing as the heat content of a gas. Now weight each transfer by its temperature. The boiler term is 6000/600 = 10.0 J K⁻¹ in, the condenser term is 3000/300 = 10.0 J K⁻¹ out, and they cancel exactly. That cancellation is not a coincidence of these numbers; it is Carnot's Qₕ/Tₕ = Q_c/T_c. Any reversible cycle can be covered by a mesh of narrow Carnot cycles whose shared adiabatic edges are traversed twice in opposite senses and cancel, leaving ∮ δQᵣₑᵥ/T = 0 for the cycle as a whole. A quantity whose loop integral always vanishes has a path-independent integral between any two points, so ∫ δQᵣₑᵥ/T defines the difference of a state function. Clausius named it entropy.

A rigid insulated vessel holds gas in one chamber and vacuum in the other; the membrane bursts. Audit it: Q = 0 because the walls are insulated, W = 0 because the gas pushes on nothing, so ΔU = 0 and an ideal gas ends at the temperature it started at. It is tempting to write ΔS = Q/T = 0, and it is wrong — the definition carries a reversible subscript, and a bursting membrane passes through no equilibrium states at all. What survives is that the endpoints, (T, V₁) and (T, V₂), are both proper equilibrium states. So invent a route between them that the formula does accept: a slow isothermal expansion against a piston, in contact with a bath at T. On that route ΔU = 0 again, so Qᵣₑᵥ = W = nRT ln(V₂/V₁), and ΔS = Qᵣₑᵥ/T = nR ln(V₂/V₁). For one mole doubling its volume that is 8.314 × ln 2 = 5.76 J K⁻¹. The gas keeps no memory of which route it took; only the endpoints entered the answer.

Label four gas molecules and ask which half of a box each occupies. There are 2⁴ = 16 equally likely arrangements. Exactly one puts all four on the left; six put two on each side. The even split is not favoured by any force — it is favoured six to one by sheer count. Write Ω for the number of microscopic arrangements consistent with a macroscopic description; Boltzmann's S = k_B ln Ω makes entropy the logarithm of that count. The logarithm is doing real work: microstate counts multiply when two systems are put side by side, Ω = Ω₁Ω₂, while entropies must add, and only a logarithm converts one into the other. Counting also reproduces the last section. Each molecule free to roam volume V has a position count proportional to V, so N of them give Ω ∝ V^N; doubling V multiplies Ω by 2^N and ΔS = k_B ln 2^N = N k_B ln 2 = nR ln 2, the identical 5.76 J K⁻¹ per mole. For a mole the all-on-one-side state has probability 2^{(−6 × 10²³)}: not rare so much as never.

Apply the Clausius inequality to a cycle made of a real, irreversible leg from state 1 to state 2 and a reversible return, and the reversible leg contributes exactly −(S₂ − S₁). What is left is ΔS ≥ ∫ δQ/Tᵣₑₛ for the real leg, with Tᵣₑₛ the temperature of whatever supplied the heat, not the temperature inside the churning system. Define the shortfall as entropy generated, S_gen = ΔS − ∫ δQ/Tᵣₑₛ ≥ 0. This is the sharpest difference between the two laws. Energy is conserved: the joules are only moved and relabelled. Entropy is not conserved, it is manufactured — by friction, by conduction across a finite temperature gap, by unrestrained expansion, by mixing, by current through a resistor, by any finite-rate process at all. Nothing in nature destroys it. A reversible process is simply the idealised limit in which S_gen is driven to zero, which costs infinite time.

A freezer pulls 500 J out of its contents at 270 K and dumps 600 J into a 300 K kitchen, the extra 100 J being the work its compressor did. The contents lose entropy: ΔS = −500/270 = −1.85 J K⁻¹. The kitchen gains ΔS = +600/300 = +2.00 J K⁻¹. The total is +0.15 J K⁻¹, positive, so nothing illegal has happened even though one object's entropy plainly fell. Water freezing does the same, shedding 334 kJ per kilogram to its surroundings; so does a crystal growing, and so does every living thing. Only ΔSₜₒₜₐₗ is signed by the second law. The familiar statement that entropy never decreases is the special case of an isolated system, where there are no surroundings to take up the slack, so ΔS_sys and ΔSₜₒₜₐₗ are the same number. Over-generalising that case is the commonest way to conclude that refrigerators and biology are impossible.

Let a hot body at Tₕ and a cold one at T_c exchange a small amount of heat dQ. The hot one loses dQ/Tₕ of entropy, the cold one gains dQ/T_c, and the total change is dQ(1/T_c − 1/Tₕ), which is positive while Tₕ > T_c, shrinks as the gap closes, and reaches zero when the temperatures are equal. So the second law is not assuming the direction of heat flow, it is deriving it: the common temperature is the state where total entropy stops rising, which is exactly what equilibrium means. The same picture answers the unsettling question about time. Newton's laws and Maxwell's equations run equally well backwards, and a reversed film of a free expansion violates none of them; it merely asks the gas to land in one particular microstate out of 2^N. For ten particles that reversal is watched routinely in simulation, which is why entropy increase is a statement about probability. At 10²³ particles the probability is so small that the statement becomes a law.