Entropy: Clausius Inequality, Entropy Generation, Isentropic Devices and Second-Law Analysis
2. Entropy generation, the increase principle and the causes of irreversibility
Apply the balance to a system together with every body it exchanges heat with. The heat leaving one part enters another, so the ∫δQ/T terms cancel in the total and ΔS_total = ΔS_sys + ΔS_surr = S_gen ≥ 0. This is the principle of increase of entropy: the entropy of an isolated system (or of the universe) never decreases, and stays constant only for reversible processes. It decides which way a process can go, and the amount S_gen measures how irreversible it is. Entropy is not conserved. The entropy of a part may fall, but only by sending heat out, so ΔS_sys < 0 needs net heat rejection.
| Cause | Case | Entropy generated |
|---|---|---|
| Heat transfer across a finite ΔT | Q passes from a body at T_H to one at T_L | Q(1/T_L − 1/T_H); 600 kJ from 600 K to 300 K gives 1.0 kJ/K |
| Unrestrained expansion | n mol of ideal gas, volume V₁ to V₂ into vacuum | nR ln(V₂/V₁); 1 mol doubling gives 5.76 J/K |
| Friction, resistance | Work W dissipated as heat at T | W/T |
| Throttling | Ideal gas, p₁ to p₂, per unit mass | R ln(p₁/p₂) |
| Mixing | Different ideal gases at the same T and p | −R_u Σ n_i ln x_i, positive |
3. Entropy changes of gases, liquids, solids and reservoirs; T-s diagrams
From T ds = du + p dv = dh − v dp, an ideal gas with constant specific heats has, per unit mass,
- s₂ − s₁ = c_v ln(T₂/T₁) + R ln(v₂/v₁)
- s₂ − s₁ = c_p ln(T₂/T₁) − R ln(p₂/p₁)
Use molar c_p, c_v and R_u = 8.314 J/mol·K for a molar basis. An incompressible liquid or solid has c_p ≈ c_v = c and s₂ − s₁ = c ln(T₂/T₁). A reservoir at fixed T that receives heat Q has ΔS = Q/T. A phase change at T_sat has Δs = h_fg/T_sat. Worked example: 2 kg of air (c_p = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K) goes from 300 K, 100 kPa to 600 K, 400 kPa. Δs = 1.005 ln 2 − 0.287 ln 4 = 0.6966 − 0.3979 = 0.2987 kJ/kg·K, so ΔS = 0.597 kJ/K. Compression lowers entropy through the pressure term and heating raises it through the temperature term; the sign comes from the competition.
T-s diagram. For a reversible process δq = T ds, so the area under the path is the heat transferred per unit mass; for a reversible cycle the enclosed area is the net heat, equal to the net work. A reversible adiabatic (isentropic) process is a vertical line, a reversible isothermal one a horizontal line, and a Carnot cycle a rectangle. For an ideal gas (∂T/∂s)_v = T/c_v and (∂T/∂s)_p = T/c_p, so through the same state a constant-volume line is steeper than a constant-pressure line, since c_v < c_p. In the wet region isobars are horizontal. An irreversible adiabatic process moves to the right (s increases), which is why the real exit state of a turbine or compressor lies to the right of the isentropic one.
4. Isentropic processes and isentropic efficiency
Setting Δs = 0 in the two ideal-gas forms, for constant specific heats and γ = c_p/c_v, gives T₂/T₁ = (v₁/v₂)^(γ−1) = (p₂/p₁)^((γ−1)/γ) and p v^γ = constant. These apply only to a reversible adiabatic process of an ideal gas with constant specific heats. Worked example: air (γ = 1.4) compressed isentropically from 100 kPa and 300 K to 800 kPa gives T₂ = 300 × 8^(0.2857) = 543 K, much higher than the 300 K of an isothermal compression, so isentropic compression needs more work.
Real adiabatic machines are irreversible, so s₂ > s₁ and the exit state at the same exit pressure differs from the isentropic exit state 2s. The isentropic efficiency compares the actual device with the ideal one that has the same inlet state and exit pressure (steady flow, adiabatic, changes of kinetic and potential energy neglected except for the nozzle):
| Device | Definition | Expression |
|---|---|---|
| Turbine | Actual work / isentropic work | (h₁ − h₂)/(h₁ − h₂s) |
| Compressor | Isentropic work / actual work | (h₂s − h₁)/(h₂ − h₁) |
| Pump (liquid) | Isentropic work / actual work | v(p₂ − p₁)/(h₂ − h₁) |
| Nozzle | Actual exit KE / isentropic exit KE | (h₁ − h₂)/(h₁ − h₂s) |
Worked example (turbine). Air enters a turbine at 1000 K and expands to one-sixth of the inlet pressure, with c_p = 1.005 kJ/kg·K, γ = 1.4 and η_T = 0.85. T₂s = 1000 × (1/6)^0.2857 = 599.3 K. T₂ = 1000 − 0.85 × (1000 − 599.3) = 659.4 K. Actual work = 1.005 × 340.6 = 342.3 kJ/kg, against 402.7 kJ/kg isentropic. The entropy generated is c_p ln(T₂/T₂s) = 0.096 kJ/kg·K, because the exit pressure is the same.
5. Entropy balance: second-law analysis of systems and control volumes
Closed system: ΔS = Σ Q_k/T_k + S_gen, where T_k is the temperature of the boundary at which Q_k enters. Control volume (rate form): dS_cv/dt = Σ Q̇_k/T_k + Σ ṁ_i s_i − Σ ṁ_e s_e + Ṡ_gen, with Ṡ_gen ≥ 0. At steady state with one inlet and one outlet, Ṡ_gen = ṁ(s_e − s_i) − Σ Q̇_k/T_k. For an adiabatic steady device Ṡ_gen = ṁ(s_e − s_i) ≥ 0, so the exit entropy can never be less than the inlet entropy; the equality marks the reversible (isentropic) limit.
- Method. Choose the system and its boundary, write the mass, energy and entropy balances, fix the states from property data, and solve. Check that S_gen ≥ 0: if it comes out negative the data or the assumed direction is impossible.
- Feasibility test. An inventor claims an adiabatic air compressor takes air at 300 K, 100 kPa and delivers it at 400 K, 400 kPa. With c_p = 1.005 and R = 0.287 kJ/kg·K, s₂ − s₁ = 1.005 ln(400/300) − 0.287 ln 4 = 0.2891 − 0.3979 = −0.109 kJ/kg·K < 0, so the claim is impossible.
- Throttling of an ideal gas. Steady, adiabatic, no work: h₁ = h₂, so T₂ = T₁ and Ṡ_gen/ṁ = −R ln(p₂/p₁) = R ln(p₁/p₂). For 800 kPa to 200 kPa air, 0.287 ln 4 = 0.398 kJ/kg·K.
6. Second-law efficiency
The first-law (thermal) efficiency counts all energy equally, so it cannot tell a ruinous process from a good one. Second-law efficiency compares the performance with the best possible under the same conditions: η_II = η_th/η_th,rev for an engine, η_II = COP/COP_rev for a refrigerator or heat pump, w_actual/w_rev for a work-producing device and w_rev/w_actual for a work-absorbing one. It is at most 1 and equals 1 only for a reversible device. The work lost through irreversibility is W_lost = T₀ S_gen, where T₀ is the temperature of the surroundings, so lowering S_gen is the route to a better design.
Worked example. An electric resistance heater turns 1 kW of work into 1 kW of heat, a first-law efficiency of 100%. To keep a room at 295 K when the outside air is at 275 K, the reversible heat pump would have COP_rev = 295/20 = 14.75, while the heater has COP = 1. So η_II = 1/14.75 = 6.8%. The heater wastes the ability of high-grade electrical work to pump heat from outside. In the same way an engine that has η_th = 0.35 between 1000 K and 300 K has η_II = 0.35/0.70 = 0.50.
7. The third law of thermodynamics
The third law (Nernst's heat theorem, stated in the form due to Planck) says that the entropy of a pure, perfect crystalline substance is zero at absolute zero. The first two laws fix only entropy differences, which is why steam tables use an arbitrary reference state. The third law gives an absolute entropy, s(T) = ∫₀ᵀ (c_p/T′) dT′ plus h/T for each phase change on the way up, which is how absolute entropies of substances are tabulated at 298 K.
- For the integral ∫ c_p/T dT to stay finite near T = 0, the specific heats of a crystalline solid must tend to zero as T → 0.
- Absolute zero cannot be reached in a finite number of steps by any process; each step removes less entropy and heat as T falls. This agrees with the Kelvin scale argument that a reservoir at 0 K would give a 100% efficient engine.
Key takeaways
- Clausius inequality: ∮δQ/T ≤ 0, with equality for reversible cycles. Entropy is defined by dS = (δQ/T)_rev, is a property, and for any process ΔS = ∫δQ/T + S_gen with S_gen ≥ 0; the integral form alone holds only for reversible paths.
- Entropy of the universe never decreases (ΔS_sys + ΔS_surr = S_gen ≥ 0). A system's entropy falls only if heat leaves it. Causes of S_gen: finite-ΔT heat transfer, free expansion, friction, mixing, throttling, resistance, reaction.
- Ideal gas: Δs = c_p ln(T₂/T₁) − R ln(p₂/p₁) = c_v ln(T₂/T₁) + R ln(v₂/v₁); liquid or solid: c ln(T₂/T₁); reservoir: Q/T. On a T-s diagram area is heat, isentropes are vertical, and constant-v lines are steeper than constant-p lines.
- Isentropic ideal gas (constant specific heats): T₂/T₁ = (p₂/p₁)^((γ−1)/γ) and p v^γ = constant. Isentropic efficiency is actual/ideal for turbines and nozzles and ideal/actual for compressors and pumps; real exit entropy is higher than inlet.
- Entropy balance: Ṡ_gen = Σṁ_e s_e − Σṁ_i s_i − ΣQ̇/T_b ≥ 0. An adiabatic steady device needs s_exit ≥ s_inlet; a negative result means the claimed device is impossible. Second-law efficiency is the ratio of actual to reversible performance, and W_lost = T₀ S_gen.
- Third law: the entropy of a pure perfect crystal is zero at 0 K, so absolute entropies exist; specific heats of crystals tend to zero as T → 0, and absolute zero is unattainable in a finite number of steps.
Practice questions (17)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
For a cyclic process, the Clausius inequality states that
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Answer: C — ∮ δQ/T ≤ 0, with equality for a reversible cycle
The cyclic integral of δQ/T is zero for a reversible cycle and negative for an irreversible one, i.e. ∮ δQ/T ≤ 0. It cannot be zero for every cycle, because only the reversible form δQ_rev/T is the differential of a property. The statement ∮ δQ = 0 is wrong as well: for a cycle ∮ δQ equals the net work, not zero.In which of the following can the entropy of a closed system decrease?
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Answer: D — Cooling of the system by heat rejection to colder surroundings
For a closed system ΔS = ∫δQ/T + S_gen, and S_gen ≥ 0. A decrease needs ∫δQ/T < 0, which means net heat leaving the system. In the three adiabatic cases δQ = 0, so ΔS = S_gen ≥ 0 and the entropy cannot fall. The entropy of the surroundings rises by at least as much, so the universe still gains.One mole of an ideal gas expands from V to 2V at constant temperature in two ways: (X) reversibly while in contact with a reservoir at the gas temperature, and (Y) freely into an evacuated, thermally insulated chamber after a partition breaks. Which statement about entropy changes is correct?
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Answer: A — The gas has the same ΔS in X and Y, but the entropy of the universe changes only in Y
Entropy is a property and both paths join the same end states, so the gas has ΔS = R ln 2 = 5.76 J/K in both. In X the reservoir gives up the heat RT ln 2 and loses 5.76 J/K, so the universe changes by zero. In Y nothing outside changes, so the universe gains 5.76 J/K. The adiabatic argument for ΔS = 0 is the tempting error: δQ = 0 gives no information about ΔS for an irreversible process.On a T-s diagram of an ideal gas, two lines pass through the same state: a constant-volume line and a constant-pressure line. Which is correct?
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Answer: C — The constant-volume line is steeper, with slope T/c_v compared with T/c_p
Along a reversible constant-volume path T ds = c_v dT, so the slope is (∂T/∂s)_v = T/c_v. Along constant pressure it is T/c_p. Since c_v < c_p, the constant-volume line has the larger slope. At constant pressure more heat is needed per kelvin, so for the same ds the temperature rises less, which is the opposite of the first claim.Air, treated as an ideal gas with γ = 1.4, is compressed reversibly and adiabatically from 100 kPa to one-eighth of its initial volume. The final pressure is closest to
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Answer: B — 1840 kPa
For p v^γ = constant, p₂ = p₁ (v₁/v₂)^γ = 100 × 8^1.4 = 100 × 18.38 = 1838 kPa. The value 800 kPa applies to an isothermal compression, p₂ = p₁ × 8. The adiabatic pressure is larger because the gas also heats up as it is compressed.A gas expands adiabatically in a turbine from p₁ to p₂. Compared with a reversible adiabatic expansion between the same inlet state and the same p₂, the actual exit state has
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Answer: C — a higher temperature and a higher entropy, and the turbine delivers less work
Irreversibility generates entropy, so s₂ > s₂s. Along a constant-pressure line entropy and temperature (and enthalpy) rise together, so T₂ > T₂s. The work w = h₁ − h₂ is then smaller than the isentropic work h₁ − h₂s. A lower entropy at exit would contradict the second law for an adiabatic device, which excludes the other options with lower entropy.A steady adiabatic steam turbine receives steam with specific enthalpy 3200 kJ/kg. For expansion to the exit pressure the isentropic exit enthalpy would be 2400 kJ/kg, and the isentropic efficiency is 0.85. Kinetic and potential energy changes are negligible. The actual exit enthalpy is
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Answer: D — 2520 kJ/kg
η_T = (h₁ − h₂)/(h₁ − h₂s), so h₂ = 3200 − 0.85 × (3200 − 2400) = 3200 − 680 = 2520 kJ/kg. The value 2259 comes from dividing the isentropic work by the efficiency, which is the compressor formula, and 2824 from dividing h₂s by 0.85. The value 2400 would be the reversible case. Real exit enthalpy lies above the isentropic one.An electric resistance heater maintains a room at 295 K while the surroundings are at 275 K. It converts 1 kW of electrical work entirely into 1 kW of heat. Take the second-law efficiency of a heating device as its COP divided by the COP of a reversible heat pump between the same temperatures. The second-law efficiency of the heater is closest to
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Answer: D — 6.8%
The heater has COP = 1. A reversible heat pump between 295 K and 275 K has COP = 295/(295 − 275) = 14.75. So η_II = 1/14.75 = 0.068, i.e. 6.8%. The first-law efficiency of 100% is the tempting but irrelevant number. The value 93% is 1 − 0.068, and 20% is the temperature difference 20 K treated as a percentage.For a steady-flow adiabatic device with one inlet and one outlet, which condition must the specific entropies satisfy according to the second law?
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Answer: A — s_exit ≥ s_inlet, with equality for the reversible device
With Q̇ = 0 the entropy balance reads Ṡ_gen = ṁ(s_exit − s_inlet) ≥ 0, so s_exit ≥ s_inlet for any adiabatic device, whether it produces or absorbs work. Equality holds only if the device is reversible. The claim s_exit = s_inlet confuses adiabatic with isentropic: an irreversible adiabatic machine has s_exit > s_inlet.Which of the following processes generate entropy (S_gen > 0)?
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Answer: A — Adiabatic free expansion of an ideal gas into a vacuum; B — Conduction of heat from a body at 500 K to a body at 300 K; D — Steady adiabatic throttling of an ideal gas through a valve with a pressure drop
Free expansion, conduction across a finite temperature difference and throttling are spontaneous dissipative processes; their entropy generation is nR ln(V₂/V₁), Q(1/T_L − 1/T_H) and R ln(p₁/p₂) respectively, all positive. Reversible isothermal expansion against a reservoir at the gas temperature has no dissipation: the gas gains entropy, but the reservoir loses exactly as much, so S_gen = 0.Which of the following statements about the third law of thermodynamics are correct?
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Answer: A — The entropy of a pure, perfect crystalline substance is zero at absolute zero; C — The specific heat of a crystalline solid tends to zero as T → 0, so that ∫ c_p/T dT stays finite
The Planck form of the third law applies to pure perfect crystals, and finiteness of the absolute entropy integral forces c_p → 0 as T → 0. Substances with frozen disorder, glasses and mixtures keep a residual entropy, so the claim of zero entropy for every substance is false. The claim about reaching 0 K is false too: no process gets there in a finite number of steps.Which of the following statements about entropy of a closed system are correct?
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Answer: A — The change in entropy between two states is the same for every path joining them; B — The entropy of an adiabatic closed system never decreases; C — The entropy of the system can decrease only if heat is transferred out of it
Entropy is a property, so ΔS is path-independent. From ΔS = ∫δQ/T + S_gen with S_gen ≥ 0, an adiabatic system has ΔS ≥ 0, and a decrease needs net heat rejection. The integral form ΔS = ∫δQ/T holds only for reversible paths; for irreversible ones ΔS exceeds the integral.Two kilograms of air, an ideal gas with c_p = 1.005 kJ/kg·K and R = 0.287 kJ/kg·K (constant specific heats), are taken from 300 K and 100 kPa to 600 K and 400 kPa. The entropy change of the air is ____ kJ/K (two decimal places).
Numerical answer — type the value.
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Answer: 0.60
Δs = c_p ln(T₂/T₁) − R ln(p₂/p₁) = 1.005 ln 2 − 0.287 ln 4 = 0.6966 − 0.3979 = 0.2987 kJ/kg·K. For 2 kg, ΔS = 0.5975 kJ/K, i.e. 0.60. Forgetting the pressure term gives 1.39, and using c_v with the pressure term would be inconsistent.A metal rod conducts heat steadily at 5 kW from one end held at 500 K to the other held at 300 K. The rod is insulated along its length and its own state does not change with time. The rate of entropy generation in the rod is ____ W/K (two decimal places).
Numerical answer — type the value.
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Answer: 6.67
Taking the rod as the control volume at steady state, the entropy entering with heat is Q̇/T_hot and leaving is Q̇/T_cold, with no change of the rod's own entropy. Ṡ_gen = Q̇/T_cold − Q̇/T_hot = 5000 × (1/300 − 1/500) = 16.667 − 10.0 = 6.67 W/K.Air enters a steady adiabatic compressor at 300 K and is compressed to eight times the inlet pressure. Treat air as an ideal gas with c_p = 1.005 kJ/kg·K and γ = 1.4; kinetic and potential energy changes are negligible. If the isentropic efficiency is 0.80, the actual work input is ____ kJ/kg (nearest integer).
Numerical answer — type the value.
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Answer: 306
T₂s = 300 × 8^(0.4/1.4) = 300 × 1.8114 = 543.4 K. Isentropic work = 1.005 × (543.4 − 300) = 244.6 kJ/kg. For a compressor η_C = w_s/w_a, so w_a = 244.6/0.80 = 305.8 kJ/kg, i.e. 306. Multiplying by 0.80 instead gives 196, the commonest slip. The actual exit temperature is 300 + 243.4/0.8 = 604 K.Two kilograms of water at 80 °C are mixed adiabatically, at constant pressure and with no work, with three kilograms of water at 20 °C. Treat water as incompressible with c = 4.18 kJ/kg·K and take T(K) = t(°C) + 273. The entropy generated by the mixing is ____ J/K (nearest integer).
Numerical answer — type the value.
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Answer: 88
Energy balance: T_f = (2 × 353 + 3 × 293)/5 = 317 K. Entropy of the system: ΔS = 2 × 4.18 ln(317/353) + 3 × 4.18 ln(317/293) = −0.8995 + 0.9873 = 0.0880 kJ/K. The system is adiabatic, so S_gen = ΔS = 88 J/K. The hot water loses less entropy than the cold water gains, which is the signature of an irreversible process.A heat engine receives 500 kW of heat from a source at 800 K and rejects heat to the environment at 300 K, producing 200 kW of net power. Define the second-law efficiency as the thermal efficiency divided by the thermal efficiency of a reversible engine between the same two reservoirs. The second-law efficiency of the engine is ____ % (nearest integer).
Numerical answer — type the value.
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Answer: 64
Thermal efficiency η = 200/500 = 0.40. Reversible engine: 1 − 300/800 = 0.625. η_II = 0.40/0.625 = 0.64, i.e. 64%. Using Celsius temperatures would give a wrong reversible limit.