Question

Even if surface tension is most important only for liquids, we shall here consider a simple regular "solid" surrounded by vacuum. The molecules are placed in a cubic grid (see the figure) with grid length equal to the molecular separation length L_mol. Each molecule in the interior has six bonds to its neighbors whereas a surface molecule has only five. If the total binding energy of a molecule in the bulk is ∈, a surface molecule will only be bound by $\frac{5}{6}\in$. The missing binding energy corresponds to adding an extra positive energy $\frac{1}{6}\in$ for each surface molecule.
The binding energy may be estimated as ∈ ≈ hm where h is the latent heat of evaportion and $\mathrm{m}=\frac{{\mathrm{M}}_{\mathrm{mol}}}{{\mathrm{N}}_{\mathrm{A}}}={\mathrm{\rho L}}_{\mathrm{mol}}^3$ is the mass of a single molecule. Dividing the molecular surface energy $\frac{1}{6}\in$ with the molecular area scale ${\mathrm{L}}_{\mathrm{mol}}^2$, we arrive at the following estimate of the surface energy density.

$\mathrm{\alpha }\approx \frac{\frac{1}{6}\in }{{\mathrm{L}}_{\mathrm{mol}}^2}\approx \frac{1}{6}\frac{\mathrm{hm}}{{\mathrm{L}}_{\mathrm{mol}}^2}\approx \frac{1}{6}{\mathrm{h\rho L}}_{\mathrm{mol}}$
The appearance of the molecular separation scale is quite understandable because ρL_mol represents the effective surface mass density of layer of thickness L_mol. One might expect that the smallness of the molecular scale would make the surface energy density insignificant in practice, but the small thickness is offset by the fairly large values of ρ and h in normal liquids, for example water.

Consider 3 liquids
(P)    h = 10⁵ J/kg, ρ = 10⁴ kg/m³, L_mol = 10^–7 m
(Q)    h = 4 × 10⁵ J/kg, ρ = 10³ kg/m³, L_mol = 3 × 10^–7 m
(R)     h = 8 × 10⁴ kg, ρ = 4000 kg/m³, L_mol = 1.5 × 10^–6 m