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VKC Notes · Liquid state · Civil Services P-I · IFoS P-I

4.3 Liquid State: Surface Tension, Kelvin Equation, Capillary Action, Viscosity and Liquid Crystals

★★★ 2021–2025: 9 Civil Services Mains parts, 75 marks, every year since 2021, plus 1 Geo-Scientist Prelims question. Almost all are short numericals on γ, Δp = 2γ/r and h = 2γ cos θ/ρgr, or a “why” on the Kelvin equation.

Syllabus lines. Civil Services (Liquid state): “Kelvin equation; surface tension and surface energy, wetting and contact angle, interfacial tension and capillary action”. IFoS (Solid state): “Elementary study of liquid crystals”. Viscosity of liquids is added because it sits next to these lines in every textbook and in the gaseous-state unit (4.2).

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

4.3.1 Surface tension and surface energy ★★★

(4.3.1)
  • A molecule inside the liquid is pulled equally on all sides; one at the surface feels a net inward pull. Bringing molecules to the surface costs work, so the liquid shrinks its area. γ is the work to make unit new area (surface Gibbs energy) or the force per unit length along the surface.SN
  • Stronger intermolecular forces give larger γ: Hg 0.485, water 0.072, ethanol 0.022 N m⁻¹ at 20–25 °C. Ethanol (H-bonded) > dimethyl ether: the two are isomers with similar dispersion forces, but only ethanol forms O–H···O bonds (b.p. 78 °C against −25 °C) CSE 2024 Q3(b)(i) 5m.SN
  • γ is intensive (energy per area), like the dielectric constant; energy and enthalpy are extensive Geo Pre 2024 Q49. γ falls as T rises and vanishes at the critical temperature (Add-on 4.3.8).
  • Breaking up a drop costs γΔσ at constant volume: , so . A drop of radius 0.4 cm split into 125 drops: r = 0.08 cm, Δσ = 10.053 − 2.011 = 8.042 cm², ΔE = 72 × 8.042 = 579 erg = 5.79 × 10⁻⁵ J CSE 2021 Q3(a) 10m.SN

Solved PYQ CSE 2025 Q3(b) 15m

10.0 J of work turns one mole of water (25 °C) into droplets of uniform size; γ = 0.072 J m⁻². (i) Radius of the droplets; (ii) molecules per droplet.

  1. Area made: σ = w/γ = 10.0/0.072 = 138.9 m². For N drops: N·(4/3)πr³ = V and N·4πr² = σ; divide: r = 3V/σ = 3(1.80 × 10⁻⁵)/138.9 = 3.9 × 10⁻⁷ m.
  2. One drop: (4/3)πr³ = 2.46 × 10⁻¹⁹ m³, so N = 7.31 × 10¹³ drops; molecules per drop = 6.022 × 10²³/7.31 × 10¹³ = 8.2 × 10⁹.
  3. The 10 J is stored as surface Gibbs energy (ΔG = +10 J): dispersion is not spontaneous; small drops coalesce.SN

4.3.2 Curved surfaces: the Laplace equation ★★★

(4.3.2)
  • Derivation: let a drop grow from r to r + dr. Work against the pressure difference Δp·4πr²dr equals the surface work γ·8πrdr, so Δp = 2γ/r. Pressure is always higher on the concave side.SN
  • Water drop r = 150 nm: Δp = 2(0.072)/1.50 × 10⁻⁷ = 9.6 × 10⁵ Pa (9.5 atm) CSE 2024 Q3(b)(ii) 5m.SN
  • Maximum bubble pressure: a bubble blown at the tip of a capillary in a liquid has one surface; the pressure is largest when it is a hemisphere of the tube radius. Δp = 2.16 torr = 288.0 Pa, γ = 0.072 N m⁻¹: r = 2γ/Δp = 5.0 × 10⁻⁴ m, diameter 0.10 cm CSE 2023 Q4(c) 10m.SN
  • Plain hydrostatics is asked in the same unit: a 15 cm straw at 45° lifts water a vertical h = 15 sin 45° = 10.6 cm, so Δp = ρgh = 1000 × 9.81 × 0.1061 = 1.04 × 10³ Pa = 7.8 torr (only the vertical height counts) CSE 2023 Q4(b) 10m.
liquidpin pout  (vapour)(a) dropletpin  − pout  = 2γ/rvapourpin liquid around(b) bubble in a liquidone surface: 2γ/rairpin film: two surfaces(c) soap bubble in airtwo surfaces: 4γ/r
Fig. 4.3.1 Excess pressure inside curved surfaces: one surface gives 2γ/r, a soap film (two surfaces) 4γ/r.
Draw it in 60 s: a circle with inward arrows (drop), a circle inside a shaded liquid (bubble), two concentric circles (soap bubble); write 2γ/r, 2γ/r, 4γ/r under them.

4.3.3 Wetting, contact angle and capillary action ★★★

(4.3.3)
  • Contact angle θ: angle inside the liquid between the solid and the liquid surface at the contact line. Young’s equation balances the three tensions along the solid. θ < 90°: the liquid wets (adhesion beats cohesion, water on clean glass, θ ≈ 0); θ > 90°: non-wetting (mercury on glass, about 140°).SN
  • From Young, : for the same solid, the liquid with the larger own surface tension has the larger θ. Liquid A (γ seven times B) has the higher contact angle; with gravity switched off A gathers into one nearly spherical drop touching the glass at a small patch, while B (wetting) creeps over the walls and coats the beaker CSE 2023 Q1(e)(ii) 5m.SN
  • Capillary rise: the vertical pull 2πrγ cos θ round the rim holds up the column πr²hρg, so h = 2γ cos θ/ρgr. Water in a tube of radius 0.1 mm rises 14.7 cm; mercury (θ ≈ 140°) is depressed. Method of measuring γ: capillary rise (θ ≈ 0).SN
  • Same tube: hB/hA = (γB/γA)(ρA/ρB). A has half the γ and twice the density of B; hA = 2.0 cm, so hB = 2.0 × 2 × 2 = 8.0 cm CSE 2022 Q4(c) 10m.SN
  • Other methods of measuring γ: drop weight (stalagmometer, γ1/γ2 = m1/m2 for equal drop sizes), maximum bubble pressure (4.3.2), du Noüy ring (force to pull a ring off the surface).
hwater in glass: θ < 90°, risesh < 0mercury in glass: θ ≈ 140°, fallsγLV γSV γSL θYoung:γSV  = γSL  + γLV  cos θ
Fig. 4.3.2 Capillary rise of water (concave meniscus) and depression of mercury (convex meniscus) in glass; Young’s balance at the contact line.
Draw it in 60 s: two beakers with a thin tube each: water higher inside with a ∪ meniscus, mercury lower with a ∩ meniscus; a drop on a line with γLV along the drop surface, γSV outward, γSL inward, θ inside the drop.

4.3.4 The Kelvin equation ★★★

(4.3.4)
  1. Derivation: the liquid in a drop is under extra pressure Δp = 2γ/r (4.3.2). Raising the pressure on a liquid by Δp raises its chemical potential by VmΔp; at equilibrium the vapour must match it: RT ln(p/p*) = Vm·2γ/r.SN
  2. Water at 25 °C (γ 0.072 N m⁻¹, Vm 18.07 cm³ mol⁻¹): p/p* = 2.86 at r = 1 nm, 1.11 at 10 nm, 1.01 at 100 nm, 1.001 at 1 μm (Fig. 4.3.3).SN
  • Consequences: small drops evaporate and large ones grow (Ostwald ripening); vapour can be supersaturated without condensing until dust or ions give large nuclei (cloud seeding, Wilson cloud chamber); a concave meniscus in a narrow pore has a lower vapour pressure, so vapour condenses in pores below p* (capillary condensation).SN
  • Superheating: a vapour bubble starts at molecular size. Inside a tiny cavity the vapour pressure is below p*, while the pressure needed to hold the bubble open is pext + 2γ/r (2.2 atm for r = 1 μm in boiling water). At the normal b.p. p* only equals pext, so the bubble collapses; the liquid must overshoot the b.p. before boiling starts, then boils violently (bumping). Boiling chips supply ready-made cavities CSE 2025 Q1(e) 5m.SN
11010010000.51.01.52.02.53.0radius r / nm (log scale)p / p*droplet, 1 nm: 2.8610 nm: 1.11cavity, 1 nm: 0.35convex (drop): p > p*concave (bubble, pore): p < p*
Fig. 4.3.3 Kelvin equation for water at 25 °C (computed): vapour pressure over a droplet and inside a cavity, relative to a flat surface.
Draw it in 60 s: log r on the x-axis, a dashed line at p/p* = 1; one curve falling from above onto the line (drop), its mirror rising from below (cavity); mark 1 nm and 10 nm.

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