Entropy, Clausius inequality and entropy generation

Why it matters

Entropy turns the second law into a number you can calculate. It tells you whether a process can happen, how far a real compressor or turbine falls short of ideal (isentropic efficiency), and how much potential work is destroyed by friction, mixing and heat transfer across temperature differences. T–s diagrams of every power and refrigeration cycle are built on it.

Key ideas

Clausius inequality. For any closed system executing a cycle, ∮ δQ/T ≤ 0, where T is the absolute temperature at the boundary where δQ crosses. The equality holds for a reversible cycle, the inequality for an irreversible one; a cycle with ∮δQ/T > 0 is impossible.

Entropy as a property. Because ∮(δQ/T)_rev = 0 for every reversible cycle, the integrand is the differential of a property, entropy S: dS = (δQ/T)_rev. ΔS between two states is the same for every path, but you must calculate it along a reversible path (any convenient one) between the end states.

Entropy change of a closed system. For any process S₂ − S₁ = ∫δQ/T + S_gen, where the integral is the entropy transfer with heat and S_gen ≥ 0 is the entropy generated by irreversibilities inside the boundary. Work carries no entropy. S_gen = 0 only for an internally reversible process; S_gen < 0 is impossible. Note that the system's entropy can fall if it rejects heat — only S_gen has a fixed sign.

Increase-of-entropy principle. For an isolated system (or system + surroundings) no heat crosses the outer boundary, so ΔS_isolated = S_gen ≥ 0. Real processes move isolated systems toward higher entropy; equilibrium is the state of maximum entropy.

Gibbs (T ds) relations. For a simple compressible substance: T·ds = du + p·dv and T·ds = dh − v·dp. They relate properties only, so they hold for any process (reversible or not), and they are the starting point for every Δs formula.

Special cases.

• Reversible adiabatic → isentropic (Δs = 0). An adiabatic process with friction has Δs > 0.
• Reversible isothermal heat transfer → ΔS = Q/T.
• Reservoir → ΔS = Q/T_reservoir (Q positive into the reservoir).
• Phase change at constant p and T → Δs = h_fg/T_sat.

T–s diagram. The area under an internally reversible path on a T–s diagram is the heat transferred, q = ∫T ds. A Carnot cycle is a rectangle; isentropic processes are vertical lines.

Isentropic efficiency compares a real adiabatic device with the isentropic one between the same pressures: turbine η_T = (h₁ − h₂)/(h₁ − h₂s), compressor η_C = (h₂s − h₁)/(h₂ − h₁).

Formulas

∮ δQ/T ≤ 0 dS = (δQ/T)_rev ΔS = ∫δQ/T + S_gen, with S_gen ≥ 0 Δs = cp·ln(T₂/T₁) − R·ln(p₂/p₁) (ideal gas, constant cp) Δs = cv·ln(T₂/T₁) + R·ln(v₂/v₁) (ideal gas, constant cv) Δs = c·ln(T₂/T₁) (incompressible solid or liquid) ΔS_reservoir = Q/T_res S_gen = ΔS_system + ΔS_surroundings ≥ 0

• S: entropy, kJ/K; s: specific entropy, kJ/(kg·K)
• δQ: heat transfer, kJ; T: absolute temperature at the boundary, K
• cp, cv, c, R: kJ/(kg·K); p: absolute pressure; v: specific volume
• S_gen: entropy generated, kJ/K (≥ 0)

Worked examples

Example 1 (standard). Air is taken from 300 K, 100 kPa to 600 K, 500 kPa. Find the change in specific entropy. cp = 1.005, R = 0.287 kJ/(kg·K).

1. Δs = cp·ln(T₂/T₁) − R·ln(p₂/p₁)
2. = 1.005 × ln 2 − 0.287 × ln 5 = 0.6966 − 0.4619
3. = 0.2347 kJ/(kg·K) — independent of how the change was made.

Example 2 (GATE level). A 2 kg copper block (c = 0.386 kJ/(kg·K)) at 500 K is dropped into 10 kg of water (c = 4.18 kJ/(kg·K)) at 300 K in an insulated, rigid container. Find the final temperature and the entropy generated.

1. Energy balance (no Q, no W): m_c·c_c·(T_f − 500) + m_w·c_w·(T_f − 300) = 0
2. T_f = (0.772 × 500 + 41.8 × 300)/(0.772 + 41.8) = (386 + 12 540)/42.572 = 303.63 K
3. Copper: ΔS = m·c·ln(T_f/T₁) = 0.772 × ln(303.63/500) = −0.3851 kJ/K
4. Water: 41.8 × ln(303.63/300) = +0.5023 kJ/K
5. Isolated combination: S_gen = ΣΔS = −0.3851 + 0.5023 = +0.1172 kJ/K (positive, so the process is irreversible, as expected for heat transfer across a finite ΔT).

Common mistakes

• Using ΔS = Q/T for an irreversible process or a varying temperature; integrate along a reversible path instead.
• Believing the entropy of a system can never decrease. Only the entropy of an isolated system (or S_gen) cannot.
• Using °C inside ln(T₂/T₁) — it must be kelvin.
• Using the pressure term with the wrong sign: raising pressure at constant T lowers entropy.
• Treating every adiabatic process as isentropic.
• Forgetting the surroundings when asked for total entropy change.

For GATE AE

Expect entropy-change numericals for ideal gases (often with a pressure and temperature change), mixing or quenching problems asking for S_gen, heat transfer between reservoirs, and isentropic-efficiency problems for compressors and turbines in propulsion contexts. Conceptual one-markers test the Clausius inequality, when S_gen is zero, and the shape of processes on a T–s diagram. Practise checking the sign of S_gen as a sanity test for every answer.

Quick check

1. 500 J of heat is added reversibly at a constant 250 K. Find ΔS.
2. ∮δQ/T for a cycle is +0.2 kJ/K. Is the cycle possible?
3. Can the entropy of a closed system decrease? How?
4. What is the entropy change in a reversible adiabatic process?
5. 5 kg of water (c = 4.18 kJ/(kg·K)) is heated from 25 °C to 75 °C. Find ΔS.

Answers: 1. 2 J/K. 2. No — it violates the Clausius inequality. 3. Yes, by rejecting heat (entropy leaves with the heat). 4. Zero. 5. About 3.24 kJ/K.