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VKC Notes · Phase equilibria · Civil Services P-I · IFoS P-I

7.1 Phase Rule, Phase Diagrams, Clapeyron and Clausius–Clapeyron Equations, Eutectics and Partially Miscible Liquids

★★★ 2021–2025: 15 Mains parts, 176 marks (Civil Services 8 parts / 105 m · IFoS 7 / 71 m) and 1 Geo-Scientist Prelims question. Civil Services sets a 10–20-mark diagram almost every year: the diagram, labelled, is most of the marks.

Syllabus lines. Civil Services: “Clausius–Clapeyron equation; phase diagram for a pure substance; phase equilibria in binary systems, partially miscible liquids: upper and lower critical solution temperatures”. IFoS: “Phase equilibria in pure substances; Clausius–Clapeyron equation; phase diagram for a pure substance; phase equilibria in binary systems, partially miscible liquids: upper and lower critical solution temperatures”.

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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes

7.1.1 The phase rule ★★★

(7.1.1)
  • Phase (P): a uniform part separated by boundaries (all gases make one phase; each solid is a phase). Components (C): the least number of independent species that fix the composition of every phase, C = S − R − (extra restrictions), S species, R independent reactions. Degrees of freedom (F): intensive variables (T, p, compositions) that can be changed without changing P.SN
  • Derivation: each phase needs C − 1 mole fractions, plus T and p: P(C − 1) + 2 variables. Equilibrium makes μ of each component equal in all phases: C(P − 1) equations. F = P(C − 1) + 2 − C(P − 1) = C − P + 2.SN
  • NH₄Cl(s) ⇌ NH₃(g) + HCl(g): 3 species, 1 reaction. From pure NH₄Cl (or equal NH₃ and HCl) pNH₃ = pHCl is one more restriction: C = 3 − 1 − 1 = 1. Add excess NH₃ or HCl and that restriction goes: C = 2 IFoS 2024 Q1(d) 8m. Similarly CaCO₃ ⇌ CaO + CO₂: C = 2 (two solid phases, no composition link).SN

7.1.2 Clapeyron and Clausius–Clapeyron equations ★★★

  1. On a phase boundary μα(T, p) = μβ(T, p). Move along it: dμα = dμβ, and dμ = −SmdT + Vmdp gives (Vβ − Vα)dp = (Sβ − Sα)dT IFoS 2021 Q4(c) 15m.SN
  2. At a reversible transition ΔS = ΔH/T: the Clapeyron equation, exact for any two phases (Eq. 7.1.2).
  3. Liquid ⇌ vapour: Vm(l) ≪ Vm(g) = RT/p (ideal vapour); then d ln p/dT = ΔvapH/RT² (Clausius–Clapeyron); integrate with ΔvapH constant.SN
(7.1.2)
  • Slopes: vaporisation and sublimation curves always rise (ΔH, ΔV > 0); sublimation is steeper at the triple point (ΔsubH = ΔfusH + ΔvapH). Ice → water: ΔVm = −1.63 cm³ mol⁻¹, so dp/dT = 6010/(273.15 × −1.63 × 10⁻⁶) = −133 atm K⁻¹: pressure lowers the melting point of ice; for almost all other solids it raises it.SN
  • Water at 95 °C, 634 mm, ΔvapH = 40 593 J mol⁻¹: ln(p₂/p₁) = 4882.5 × 3.640 × 10⁻⁵ = 0.1777; p(100 °C) = 634 × 1.1945 = 757 mm, 3 mm short of 760 CSE 2021 Q1(d) 10m.SN
  • 529 torr at 363.2 K and 760 torr at 373.2 K: ΔvapH = 8.314 × ln(760/529)/(7.378 × 10⁻⁵) = 40.8 kJ mol⁻¹; ΔvapS = 109 J K⁻¹ mol⁻¹, above Trouton’s 88 (H-bonding) CSE 2022 Q4(b) 10m.
  • Mountain, 380 torr: 1/T = 1/373.15 + (8.314 ln 2)/40 650, T = 354.4 K (81 °C) IFoS 2025 Q4(a)(i) 5m. At the normal boiling point ΔG = 0, so ΔH = TΔS Geo Pre 2021 Q60.SN

7.1.3 Enthalpy of vaporisation up to the critical point ★

273373473573647010203040T / KΔvap H / kJ mol−1 triple point: ≈ 45 kJ (= ΔHsub  − ΔHfus )40.7 at 100 °CTc : ΔH = 0falls ever faster;vertical at Tc 
Fig. 7.1.1 ΔvapH of water against T (Watson’s relation, computed, from the 100 °C value).
Draw it in 60 s: a curve starting high at the left, falling slowly, then plunging to zero and meeting the T axis vertically at Tc.
  • At the triple point ΔvapH = ΔsubH − ΔfusH (water 51.06 − 6.01 = 45.05 kJ mol⁻¹). It falls slowly at first (Kirchhoff: dΔvapH/dT = ΔCp < 0), then faster as the vapour becomes dense; empirically ΔvapH ∝ (1 − T/Tc)0.38 (Watson). At Tc liquid and vapour become identical: ΔvapH = 0, ΔV = 0, ΔS = 0 CSE 2021 Q3(c)(i) 10m.SN
  • So the integrated Clausius–Clapeyron equation (constant ΔvapH) holds only over a short range well below Tc.

7.1.4 One-component systems: water and reading a diagram ★★★

  • C = 1: F = 3 − P. Areas (one phase) F = 2, lines (two phases) F = 1, triple point F = 0. Four phases can never coexist.SN
Tp1 atmm.p. 273.15 Kb.p. 373.15 Kiceliquid watervapourOBOAOCOC leans left:slope < 0O: triple point, F = 0273.16 K, 4.58 torrA: critical point647 K, 218 atmround A: vapour → liquid, no meniscus
Fig. 7.1.2 Phase diagram of water (not to scale; labelled values are real).
Draw it in 60 s: three curves meeting at O: OB low on the left, OA rising to the right and ending at A, OC nearly vertical leaning left; write ice, water, vapour in the areas; dashed 1 atm line.

Solved PYQ CSE 2022 Q3(d) 10m

Construct the phase diagram of water and explain the three curves; significance of critical pressure, critical temperature and triple point.

  1. OA (water ⇌ vapour, vapour-pressure curve, F = 1) rises and ends at A, the critical point (647 K, 218 atm): above Tc no pressure liquefies the vapour; pc is the pressure needed to liquefy it at Tc.SN
  2. OB (ice ⇌ vapour, sublimation curve) runs down to 0 K; steeper than OA at O. Supercooled water continues OA below O as a metastable dashed line, lying above OB.
  3. OC (ice ⇌ water) is nearly vertical and leans to the left because ice is less dense than water (Clapeyron slope −133 atm K⁻¹).SN
  4. O, the triple point (273.16 K, 4.58 torr, F = 0): the only point where ice, water and vapour coexist; the fixed point of the kelvin scale.SN
  • Reading a given diagram CSE 2023 Q3(b) 20m: one curve system and one triple point → one component; normal melting and boiling points (F = 1) are where the 1 atm line cuts the fusion and vaporisation curves; critical and triple points have F = 0. Isobaric heating across a curve: a sharp change, temperature constant until one phase is used up. A path round the critical point (vapour → supercritical → liquid) gives liquid with no meniscus at any stage.SN
  • Butane (Tm −138 °C, Tb 0 °C, Tc 152 °C, pc 38 atm; triple point just below −138 °C at 3 × 10⁻⁵ torr): compressing at 140 °C (below Tc) meets the vapour-pressure curve at about 31 atm (Clausius–Clapeyron through (273 K, 1 atm) and (425 K, 38 atm)): two phases appear, then all liquid. At 200 °C (above Tc) one phase throughout CSE 2025 Q3(c) 10m.
  • New drug: ln pl = −3010/T + 13.2, ln ps = −3820/T + 16.1 (bar). Triple point: 810/T = 2.9, T = 279.3 K, p = 11.3 bar. 1 bar is below the triple-point pressure, so no liquid at 1 bar; the solid sublimes at 3820/16.1 = 237 K, so at 298 K, 1 bar it is a gas. ΔsubH = 3820R = 31.8 kJ mol⁻¹ (ΔvapH = 25.0, ΔfusH = 6.7) CSE 2024 Q4(a) 15m.SN

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