Past exam of the quantum condensed matter course of the University of Cambridge / 2007 / Section A / A4

Creating a liquid surface changes the molecular environment and costs interfacial free energy. In the dispersion-only cleavage model, the attractive Van der Waals force per area has magnitude $A/(6\pi h^3)$. Separating two contacting halves from a molecular cutoff $h_0$ to infinity takes work per areaHere $A_H$ denotes the Hamaker constant, to distinguish it from an area. Cleavage creates two free surfaces, giving the Hamaker estimate of liquid surface tensionThe continuum attraction diverges at zero separation and neglects molecular impenetrability and short-range repulsion. The cutoff is therefore an effective contact separation of molecular size, commonly of order $0.1$–$0.3\mathrm{nm}$, not a limit that can be taken to zero.

An experimental limitation is the strong temperature dependence of surface tension. For water in equilibrium with its vapour, the recommended experimental values decrease from $71.98\mathrm{mN}{\mathrm{m}}^{-1}$ at $25{}^{\circ }\mathrm{C}$ to $58.92\mathrm{mN}{\mathrm{m}}^{-1}$ at $100{}^{\circ }\mathrm{C}$, and the fitted tension vanishes at the critical point. A fixed pairwise cleavage energy misses interfacial entropy, changing density and many-body correlations.^[ref] The dispersion-only model is unsuitable when cohesion is dominated by hydrogen bonds, ionic interactions or metallic bonding; liquid metals are a clear example of the latter failure.

For a smooth homogeneous solid and a partially wetting liquid at equilibrium, tangential balance at the contact line gives Young's equationThe angle is measured through the liquid. Replacing solid-vapour area by solid-liquid plus liquid-vapour area changes free energy by $-S$ per wetted area, where the spreading coefficient isFor partial wetting, $S={\gamma }_{LV}(\cos \theta -1)<0$. Positive $S$ favours a spreading film and zero macroscopic equilibrium contact angle; the finite-angle Young construction cannot be extrapolated to $\cos \theta >1$.

The intended chemically matched surface approximation is ${\gamma }_{SL}\simeq 0$. It gives $S\simeq 23-21.3=1.7\mathrm{mN}{\mathrm{m}}^{-1}>0$, so increasing the wetted area lowers free energy. The initially steep drop spreads and flattens toward complete wetting:The initial $90^{\circ }$ angle describes a nonequilibrium shape and must not be inserted into equilibrium Young's equation to fix ${\gamma }_{SL}$. More precisely, complete wetting requires ${\gamma }_{SL}\le 1.7\mathrm{mN}{\mathrm{m}}^{-1}$ in this surface-energy model; chemical similarity alone is not a measurement of that interfacial energy.

Water on such a low-energy, oil-like surface is expected to bead into a comparatively rounded, high-angle drop because the substrate does not compensate its cohesive interactions. Its numerical contact angle needs the water-solid interfacial energy, which is not supplied.

The liquid does not wet the glass because $\cos 140^{\circ }<0$. The liquid meniscus is convex, with the capillary mercury level depressed below the reservoir. Balancing vertical surface-tension force against the hydrostatic column gives Jurin's law, using the inner radius $r=0.5\mathrm{mm}$:Thus the depression is approximately $10.1\mathrm{mm}$. The simple height law neglects small meniscus-volume corrections. Mercury's high surface tension reflects strong metallic bonding, not merely dispersion between weakly interacting molecular units.

A thin soap bubble has two interfaces, so the Young–Laplace equation gives $\Delta p=4\gamma /R=8\gamma /d$. HenceAt equal surface tension, the smaller bubble has the higher internal pressure. Air diffuses through their separating film from the smaller to the larger bubble, so the smaller shrinks and the larger grows. The pressure difference increases as the smaller shrinks. This is gas transfer driven by curvature, not necessarily immediate merging of the interfaces; coalescence additionally requires film rupture.