6.1 Second and Third Laws: Entropy, Gibbs and Helmholtz Energies, Maxwell Relations
★★★ 2021–2025: 15 Mains parts, 139 marks (Civil Services 7 parts / 75 m · IFoS 7 / 59 m · Geo-Scientist 1 / 5 m) and 30 Geo-Scientist Prelims questions (2021–2026), four to six every year: the most-asked Prelims chemistry topic.
Syllabus lines. Civil Services: “Second law of thermodynamics; entropy as a state function, entropy changes in various processes, entropy–reversibility and irreversibility, free energy functions; thermodynamic equation of state; Maxwell relations; temperature, volume and pressure dependence of U, H, A, G, Cp and Cv, α and β; … criteria for equilibrium …; Nernst heat theorem, introductory idea of third law of thermodynamics”. IFoS: the same up to “Nernst heat theorem and third law of thermodynamics” (plus statistical thermodynamics). Geo-Scientist: “Chemical potential in terms of Gibbs energy and other thermodynamic state functions and its variation with temperature and pressure” (Mains) and the second law, entropy and spontaneity in Prelims.
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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes
6.1.1 Second law, heat engines and the Carnot cycle ★★★ Prelims every year
- Kelvin–Planck: no cyclic engine can turn heat from one reservoir wholly into work. Clausius: heat does not flow by itself from cold to hot. Both say the same: some heat must always be rejected.SN
- A Carnot engine runs a reversible cycle of two isotherms and two adiabats. Its efficiency depends only on the two temperatures, not on the working substance, and no engine between the same reservoirs can beat it Geo Pre 2022 Q56.SN
- Steam turbine, 400 °C in, 150 °C out: η = 1 − 423/673 = 0.371, so wmax = 0.371 q, reached only if every step is reversible (no friction, heat exchanged across zero temperature difference, ΔSuniv = 0) CSE 2025 Q1(j) 5m.SN
- Prelims numbers: 127 °C/27 °C → 1 − 300/400 = 25 % Geo Pre 2024 Q55; 300 K/200 K, 1000 J of work → qh = w/η = 1000/(1/3) = 3000 J Geo Pre 2023 Q55; 90 % with Tc = 12 °C → Th = 285.15/0.1 = 2851.5 K = 2578 °C Geo Pre 2026 Q57.
6.1.2 Entropy, the Clausius inequality and entropy changes ★★★
- Because ∮ đqrev/T = 0 for every reversible cycle, S is a state function: ΔS depends only on the end states. For an irreversible cycle the integral is negative (Clausius inequality) Geo Pre 2022 Q58.SN
- Isolated system: ΔS ≥ 0 (equality only at equilibrium or for a reversible change). For the universe: ΔSuniv > 0 spontaneous, = 0 reversible. An adiabatic change has ΔS ≥ 0; a closed system’s S can fall if heat leaves it Geo Pre 2025 Q62 Geo Pre 2026 Q61.SN
| Process (ideal gas, n mol) | ΔS |
|---|---|
| Isothermal, V1 → V2 | nR ln(V2/V1) = −nR ln(p2/p1) |
| Heating at constant p (or V) | nCp,m ln(T2/T1) (or nCV,m ln(T2/T1)) |
| General change (T, p both change) | nCp,m ln(T2/T1) − nR ln(p2/p1) |
| Phase change at Ttrs | ΔtrsH/Ttrs |
- Two paths, one answer: isobaric expansion then isochoric cooling gives nCp ln(T1/T0) + nCV ln(T0/T1) = nR ln(V1/V0), the same as the direct isotherm: entropy is a state function IFoS 2025 Q3(b)(ii) 10m.SN
- He, 300 K, 1.20 atm → 100 K, 12.0 atm: 21 ln(1/3) − 8.3 ln 10 = −42.3 J K⁻¹ Geo Pre 2023 Q56; 100 g water heated 27 → 127 °C (cp = 4.2 J g⁻¹ K⁻¹): 420 ln(400/300) = 121 J K⁻¹ ≈ 122 Geo Pre 2025 Q61; positive ΔS for heating water and boiling it, negative for freezing Geo Pre 2026 Q58.
Solved PYQ CSE 2022 Q3(b) 10m
A 74.6 g ice cube melts in the sea at 0 °C and 1 atm (ΔfusH = 6.01 kJ mol⁻¹). Find ΔSsys, ΔSsurr, ΔSuniv.
- n = 74.6/18.02 = 4.14 mol; q = 4.14 × 6010 = 24.88 kJ, taken in at 273.15 K.
- ΔSsys = +24 880/273.15 = +91.1 J K⁻¹; the sea gives the same heat at the same T: ΔSsurr = −91.1 J K⁻¹.
- ΔSuniv = 0: ice and water are at equilibrium, the melting is reversible (and ΔG = 0). In a sea at 1 °C, ΔSuniv > 0 and melting is spontaneous.SN
6.1.3 Entropy of phase change and Trouton’s rule ★★
- Melting and boiling: ΔH > 0 and ΔS > 0 Geo Pre 2021 Q59. Freezing 2.0 g water at 0 °C: −(2.0/18.0)(6000)/273 = −2.4 J K⁻¹ Geo Pre 2025 Q60.
- Trouton’s rule holds for liquids with no special order: toluene (33.2 kJ/383.8 K = 86.5 J K⁻¹ mol⁻¹) obeys it. Methanol is H-bonded in the liquid, so vaporising it gains extra disorder: 104 J K⁻¹ mol⁻¹. HF stays associated (chains, rings) in the vapour too, so it gains little: about 26 J K⁻¹ mol⁻¹ IFoS 2022 Q3(b)(ii) 5m.SN
Solved PYQ CSE 2024 Q1(c) 10m
Hg(l): ΔfH° = 0, S° = 77.4; Hg(g): ΔfH° = 60.78 kJ mol⁻¹, S° = 174.7 J K⁻¹ mol⁻¹. Normal boiling point?
- At Tb liquid and vapour are at equilibrium at 1 atm: ΔG° = ΔH° − TΔS° = 0.
- ΔvapS° = 174.7 − 77.4 = 97.3 J K⁻¹ mol⁻¹; Tb = 60 780/97.3 = 625 K (352 °C); measured 630 K.
- Assumptions: ΔH° and ΔS° keep their 298 K values (ΔCp = 0); vapour ideal; 1 atm ≈ 1 bar.SN
6.1.4 Third law and absolute entropy ★★
- Nernst heat theorem: ΔS → 0 for any change as T → 0. Third law: the entropy of a perfect crystal is zero at 0 K, so S can be measured absolutely: limT→0 S = 0 and ST = ∫0T Cp dT/T Geo Pre 2021 Q56.SN
- Residual entropy: CO and N₂O freeze with molecules pointing either way, S(0) = R ln 2 = 5.8 J K⁻¹ mol⁻¹ ideally (5 measured for CO).
- H₂ + ½O₂ → H₂O(g): ΔrS°(298) = 188.825 − 130.684 − ½(205.138) = −44.43; ΔrCp = 33.58 − 28.84 − ½(29.37) = −9.945; ΔrS°(373) = −44.43 − 9.945 ln(373/298) = −46.66 J K⁻¹ mol⁻¹ IFoS 2024 Q1(b) 8m.SN
- Ne vs Ar: both monatomic, only translation counts. The Sackur–Tetrode equation gives Sm(Ar) − Sm(Ne) = (3/2)R ln(39.95/20.18) = +8.5 J K⁻¹ mol⁻¹: heavier atoms have more closely spaced translational levels, so more states are reached IFoS 2024 Q8(c) 10m.
- Heating curve: the two vertical jumps are ΔfusS° and ΔvapS° (BC and DE in the Prelims figure) Geo Pre 2021 Q51.
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