Interfaces and interfacial energy

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1 Interfaces and interfacial energy 1/14 kinds: l/g }{{ l/l } mobile s/g s/l s/s Example. Estimate the percetage of water molecules on the surface of a fog droplet of diameter (i) 0.1 mm (naked eye visibility limit) (ii) 200 nm (optical microscope visibility limit)? (i) %, (ii) 1 % The smaller particle, the more pronounced surface phenomena Interfacial energy γ = ( ) G A T,p liquids: interfacial energy = surface tension dg = dw interface = γda Often denoted σ. Units: J m 2 = N m 1 CGS: dyn cm 1 = mn m 1 l/l, l/g = γldx surface molecules have higher energy Interfacial energy of a crystal depends on the direction (crystal plane)

Surface energy and vaporization enthalpy Order-of-magnitude estimates: ( ) Typical molecule molecule separation = r = Vm 1/3 NA Energy of neighboring molecules: u Number of neighbors in the bulk: N

2 Surface energy and vaporization enthalpy Order-of-magnitude estimates: ( ) Typical molecule molecule separation = r = Vm 1/3 NA Energy of neighboring molecules: u Number of neighbors in the bulk: N bulk Number of neighbors at surface: N surf 2/14 Vaporization intenal energy: vap U m = N bulk un A /2 Area per one surface molecule: A = r 2 Surface energy of one molecule: u p = (N bulk N surf )u/2 Surface tension: γ = u p /A = (N bulk N surf )u/(2a) γ vapu m (N bulk N surf ) Vm 2/3 N 1/3 A N bulk (Stefan s rule) Example. Water (25 C): N bulk 4, N surf 3, vap H m = kj mol 1, V m = 18 cm 3 mol 1 ( ) J mol 1 (4 3) γ ( m 3 mol 1 ) 2/3 ( mol 1 ) 1/3 4 Experiment: γ = N m 1 = N m 1

3 Temperature dependence 3/14 At saturated pressure, or (at lower T) at constant pressure In the critical point: vap H(T c ) = 0, approx.: vap H T c T γ = const Tc T V 2/3 m (Eötvös) The surface tension decreases with increasing temperature. It limits to zero at the critical point. credit: wikipedia Empirical correction (Ramay and Shields) γ = const Tc 6 K T V 2/3 m Using the critical exponent (Guggenheim-Katayama, van der Waals) γ = const (T T c ) 11/9

4 Laplace pressure 4/14 Soap bubble has Pressure in a droplet of radius r (Young Laplace): two surfaces! p = p inside p outside = 2γ ( generally 1 = γ + 1 ) where Rx a R y are the r R x R y main radii of curvature Derivation 1 from the surface energy vs. volume dependence: work needed to increase the surface by da is dw surf = γda sphere surface work needed to swell the drop by dv is dw vol = pdv 2 = 4πr dw vol = dw surf p = γda dv = γd(4πr2 ) d( 4 3 πr3 ) γ8πr dr = 4πr 2 dr = 2γ r Derivation 2 from the force F acting on the cross section area A : circumference = l = 2πr, F = lγ, A = πr 2, p = F/A

< 90 (water glass) hydrophobic (lyophobic)

5 Capillary action Capillary action in a capillary of radius r h = cos θ 2πr γ πr 2 ρg = θ = contact angle 2γ cos θ rρg wetting 5/14 non-wetting hydrophilic (lyophilic) surface: θ < 90 (water glass) hydrophobic (lyophobic) surface: θ > 90 (mercury glass, water-teflon, water lotus leave)

6 Young equation and spreading 6/14 On a solid: wetting non-wetting spreading Young equation: vector sum of interfacial tensions = 0 γ sg = γ ls + γ lg cos θ Spreading: γ sg > γ ls + γ lg (γ sg γ ls γ lg > 0) Liquid droplet on liquid: wetting spreading Droplet: γ BC < γ AB + γ AC, spreading: γ BC > γ AB + γ AC

Surface tension 7/14 Methods of calculation: balance of forces, energy minimum Example.

stoma) in the leaves of 10 m high trees. Water surface tension is γ = 72 mn m 1.

Calculate the thickness of a mercury puddle on a flat non-wetting surface. γ = 0.

7 Surface tension 7/14 Methods of calculation: balance of forces, energy minimum Example. Calculate the maximum size of pores (stomata, sg. stoma) in the leaves of 10 m high trees. Water surface tension is γ = 72 mn m 1. (Ignore osmotic pressure.) d = 2.9 µm Example. Calculate the thickness of a mercury puddle on a flat non-wetting surface. γ = N m 1, ρ = 13.6 g cm 3. credit: wikipedia [SEM image] 3.8 mm puddle of spilled crude oil on water credit:

8 Using units 8/14 Example. Estimate the typical size (volume) for which the surface forces are of the same order as gravitational forces. [ρ] = kg m 3, [γ] = N m 1 = kg s 2, [g] = m s 2 kg s 2 γ m = m s 2 kg m 3 l gρ, V ( ) γ 3/2 gρ for water V 0.02 cm 3 droplet volume ( cm 3 ) Slowest surface: λ = 2π γgρ l is called capillary length

Spreading: interface s/l created at the expense of l/l: Cohesion work l/l = W k = 2γ lg Adhesion work s/l = W a = γ sg + γ lg γ ls

9 Cohesion and adhesion work 9/14 Cohesion work (energy) W k (per unit area of the interface, here l/l) Adhesion work (energy) W a (per unit area of the interface, here l/s) W k = 2γ lg W a = γ sg + γ lg γ ls the same for s/s the same for l 1 /l 2, s 1 /s 2 Spreading: interface s/l created at the expense of l/l: Cohesion work l/l = W k = 2γ lg Adhesion work s/l = W a = γ sg + γ lg γ ls Harkins spreading coefficient: S l/s = W a W k = γ sg γ ls γ lg S l/s > 0 energy is gained liquid spreads NB sign: W a = energy needed to unstick

10 Chemical potential of a droplet Transfer (1 mol) of liquid from the bulk (below a flat interface, r = ) to droplets of diameter r. Pressure increases by p = 2γ/r and the chemical potential by µ = V m p = V m 2γ r Liquid is in equilibrium with vapor: 10/14 Kelvin equation ln ps r p s µ = µ + RT ln ps p st µ + µ = µ + RT ln ps r p st = + 2γ V(l) m RTr (also Gibbs Thomson or Ostwald Freundlich) saturated vapor pressure higher above a droplet / lower in a cavity Example. Saturated vapor pressure of water at 25 C is 3.15 kpa. Calculate the partial pressure of water above a membrane of pore diameters 100 nm. γ water = 72 mn m 1. complete wetting: 3.08 kpa, non-wetting: 3.22 kpa

11 Nucleation [tchem/showisi.sh] 11/14 Supersaturated vapor (p > p s or T < T boil ), supersaturated solution (c > c s ), superheated liquid (T > T boil ), etc., are metastable, beyond the spinodal unstable (cf. spinodal decomposition) Nucleation = creation of nuclei of a new (stable) phase in a metastable region. A Gibbs energy barrier must be crossed. Mechanism of nucleation: Saturation homogeneous (wet air: S > S = p/p s 4) heterogeneous on dirt, surface (wet air: S > 1.02) on ions (wet air: S > 1.25) Homogeneous nucleation by the Kelvin equation (CNT, classical nucleation theory): A nucleus grows for p > p s r minimum radius of the nucleus: r = 2γV m RT 1 ln S distillation boiling chips (stones) to prevent overheting bubble chambers to detect ionized radiation (obsolete) Spinodal decomposition = spontaneous split of an unstable phase into two phases. There is no barrier.

12 Example: critical nucleus size 12/14 Calculate the critical nucleus size for humid (150% rel. humidity) air at 25 C. γ = 72 mn m 1. r = m (2400 molecules)

Ostwald ripening 13/14 Higher saturated pressure above small droplets small droplets evaporate, big ones grow. Higher concentration above small crystals small crystals dissolve, large crystals grow.

13 Ostwald ripening 13/14 Higher saturated pressure above small droplets small droplets evaporate, big ones grow. Higher concentration above small crystals small crystals dissolve, large crystals grow. precipitate ageing (digestion) bigger crystals can be filtered snow quality change ice cream becomes crunchy fog drizzle credit: index.php/ostwald ripening, Clarke(2003)

14 Measuring surface tension of liquids [mz html/surftens.html] 14/14

8.2 Surface phenomenon of liquid. Out-class reading: Levine p Curved interfaces

8.2 Surface phenomenon of liquid. Out-class reading: Levine p Curved interfaces Out-class reading: Levine p. 387-390 13.2 Curved interfaces https://news.cnblogs.com/n/559867/ 8.2.1 Some interesting phenomena 8.2.1 Some interesting phenomena Provided by Prof. Yu-Peng GUO of Jilin