Excess of pressure inside a liquid drop, a soap bubble, and an air bubble

The free surface of a liquid becomes curved on contact with a solid, and the degree of curvature depends on the nature of the liquid-air or liquid-gas interface involved, through the interfacial surface tension. Because a curved surface has energy associated with it, and a given volume of liquid minimizes that surface energy by minimizing its surface area, small liquid drops naturally become spherical (the shape of least surface area for a given volume). Whenever the free surface of a liquid is curved, there is always a PRESSURE DIFFERENCE between the concave (inner) side and the convex (outer) side of that surface: for a PLANE (flat) liquid surface, the tangential surface-tension forces on either side of any surface element act in exactly opposite directions and cancel out, so the pressure is the same on the liquid side as on the vapour side. But for a CURVED liquid surface, the surface-tension forces no longer cancel -- their horizontal components cancel, but their vertical components add up -- producing a net force directed normal to the surface: inward, toward the centre of curvature, for a convex surface, and outward, away from the centre of curvature, for a concave surface. The upshot is that, for a curved liquid surface in equilibrium, the pressure on its CONCAVE side is always GREATER than the pressure on its CONVEX side. EXCESS PRESSURE FORMULAS. (1) AIR BUBBLE inside a liquid of surface tension T and radius R: considering the forces on a hemispherical half of the bubble -- the surface-tension force $F_T=2\pi R T$ acting around the rim, the outside-pressure force $F_{P1}=P_1\pi R^2$, and the inside-pressure force $F_{P2}=P_2\pi R^2$ -- equilibrium requires $F_{P2}=F_T+F_{P1}$, i.e. $(P_2-P_1)\pi R^2=2\pi R T$, giving the excess pressure $\Delta P=P_2-P_1=\dfrac{2T}{R}$. (2) SOAP BUBBLE in air, radius R: a soap bubble has TWO liquid surfaces (an inner one and an outer one, both in contact with air), so the surface-tension force is DOUBLE that of a single-surface bubble, $F_T=2\times2\pi RT=4\pi RT$; the identical equilibrium argument then gives $\Delta P=\dfrac{4T}{R}$, exactly TWICE the excess pressure of an equivalent air bubble or drop. (3) LIQUID DROP of radius R, surface tension T (a single free surface, like the air bubble): the identical derivation gives $\Delta P=\dfrac{2T}{R}$, but now with the HIGHER pressure on the INSIDE of the drop (unlike the air bubble, where the higher pressure P2 is the pressure of the gas trapped inside the bubble surrou …