Calculating entropy changes

Entropy is defined along a reversible path, and real processes are not reversible, so the first move in any calculation is to stop looking at what actually happened. Because S is a function of state, ΔS depends only on the initial and final equilibrium states. Drop a 350 K block into 290 K water: the real process is violent and irreversible, but the block still ends at a definite equilibrium temperature, and you may pretend it reached that temperature by being walked down through a continuous series of baths, each infinitesimally cooler than the last. That imagined path is reversible, dQᵣₑᵥ/T is legal along it, and the integral it returns is the true ΔS of the block. Keep the discipline in this order: name the two end states first, then choose the most convenient reversible path joining them. The path may be one the system could never actually follow.

Heat a mass m whose specific heat capacity c is effectively constant. On a reversible path the heat delivered between T and T + dT is dQᵣₑᵥ = mc dT, so dS = mc dT/T and ΔS = mc ln(T₂/T₁). The logarithm is not decoration. Take 0.500 kg of water from 293.15 K to 353.15 K: the heat is Q = 0.500 × 4186 × 60.0 = 1.256 × 10⁵ J, but ΔS = 2093 × ln(353.15/293.15) = 2093 × 0.1862 = 390 J K⁻¹ — not Q/T₁ = 428 J K⁻¹, and not Q/T₂ = 356 J K⁻¹. Dividing the whole heat by any one temperature fails because T moves while the heat is being delivered. Only when the interval is narrow enough that T barely shifts does Q/T become a fair approximation, and the two wrong answers above bracket the right one.

At a transition the temperature stands still while heat pours in, which makes the integral trivial: T comes out of ∫dQᵣₑᵥ/T and ΔS = Q/Tₜᵣ = ± mL/Tₜᵣ. Melting 1.00 kg of ice at 273.15 K needs L_f = 3.34 × 10⁵ J kg⁻¹, so ΔS = 3.34 × 10⁵ ÷ 273.15 = +1.22 × 10³ J K⁻¹; freezing the same kilogram at the same temperature gives −1.22 × 10³ J K⁻¹, the sign following the heat. These terms are large and they are easy to drop. Warming that kilogram of ice from 263 K to 273 K with c_ice = 2100 J kg⁻¹ K⁻¹ contributes only 2100 × ln(273/263) = 78 J K⁻¹, while the melt that follows contributes about sixteen times as much. Any calculation that crosses a transition must be cut into stages and the stages added.

For n moles of ideal gas the first law along a reversible path gives dQᵣₑᵥ = dU + p dV = n C_V dT + nRT dV/V. Divide through by T and the temperatures separate cleanly: dS = n C_V dT/T + nR dV/V, so ΔS = n C_V ln(T₂/T₁) + nR ln(V₂/V₁). Two independent logs, one per state variable, and either may vanish. An isothermal change leaves only nR ln(V₂/V₁); a constant-volume change leaves only n C_V ln(T₂/T₁); and a reversible adiabatic change, on which TV^(γ−1) stays constant, makes the two terms cancel exactly, which is why it is called isentropic. For 2.00 mol of a diatomic gas, C_V = 5R/2 = 20.79 J mol⁻¹ K⁻¹, going from 300 K and 0.0200 m³ to 450 K and 0.0500 m³: ΔS = 2.00 × 20.79 × ln 1.50 + 2.00 × 8.314 × ln 2.50 = 16.86 + 15.24 = 32.1 J K⁻¹.

The system is only half of the account. A reservoir — a room, the sea, a thermostat bath — is by definition so large that the exchange leaves its temperature untouched, so its own path is already reversible and its entropy change is just Q/Tᵣₑₛ, with Q signed as the reservoir receives it. That is what makes irreversibility visible in numbers. Let 1.00 kJ conduct straight down a metal bar from a 600 K furnace into a 300 K room. The furnace loses 1000/600 = 1.67 J K⁻¹, the room gains 1000/300 = 3.33 J K⁻¹, and the bar ends in the state it started in and contributes nothing. The total is +1.67 J K⁻¹, positive for one reason only: the same energy was divided by a smaller temperature on arrival. Reversibility is the limit Tₕ → T_c, where the total tends to zero.

The sharpest test of the method is a gas expanding freely into an evacuated, insulated chamber. No heat crosses the boundary and the gas pushes on nothing, so Q = 0, W = 0 and ΔU = 0, which for an ideal gas fixes T₂ = T₁. Integrating dQ/T along what actually happened returns zero, and that answer is wrong, because the gas was nowhere near equilibrium and the definition never applied. Take the legal route: the endpoints are (T, V₁) and (T, V₂), and a reversible isothermal expansion joins those same two states, so ΔS = nR ln(V₂/V₁). For 1.00 mol doubling its volume that is 8.314 × ln 2 = 5.76 J K⁻¹. The surroundings received no heat and no work, so ΔSₛᵤᵣᵣ = 0 and the universe gains the whole 5.76 J K⁻¹ — worth, against 290 K surroundings, 290 × 5.76 = 1.67 kJ of work that the expansion threw away.