Excess Pressure in Bubbles and Drops
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Direct answer
The pressure inside a curved liquid surface exceeds the outside pressure by an amount fixed by surface tension and the geometry. A liquid drop has one surface and excess pressure 2T/r; a soap bubble has two surfaces (inner and outer) and excess pressure 4T/r; an air bubble submerged in liquid has one liquid–gas surface and behaves like a drop at 2T/r. The number of surfaces, not the substance, is what students must count. Blowing a soap bubble from radius r1 to r2 demands work equal to the rise in surface energy, W = 8πT(r2² − r1²), because both surfaces grow. When two bubbles connect, air flows from the smaller (higher excess pressure) into the larger, and the common interface adopts a radius set by 1/r = 1/r1 − 1/r2.
What you must remember
- Drop (one surface): excess pressure 2T/r, derived by balancing surface tension 2πr T around the rim against the pressure difference on the great circle πr².
- Soap bubble (two surfaces): 4T/r; the film has an inner and an outer surface, each pulling with 2T/r — the single most tested distinction in this chapter.
- Air bubble in liquid: 2T/r, one interface; strictly the radius used is of the gas pocket, and the pressure outside it is the liquid pressure at that depth, P0 + hρg.
- Cylindrical film (soap film on a wire loop): excess pressure T/r, one curved surface of a cylinder.
- Surface energy: a drop carries T × 4πr²; a soap bubble carries 2 × T × 4πr² = 8πr²T, so doubling a drop's radius costs the difference in these energies — work that must come from outside.
- Coalescing bubbles: two soap bubbles of radii r1 < r2 connected by a tube share air until the interface radius is r = r1r2/(r2 − r1); the smaller bubble shrinks because 4T/r1 > 4T/r2.
- Capillary connection: the same 2T/r curvature logic drives capillary rise h = 2T cos θ/(rρg), with θ the angle of contact — JEE frequently mixes the two in one question.
Blowing a bubble, in joules
A soap solution with T = 0.025 N/m is blown from radius 2 cm to 4 cm. The bubble's surface energy is 8πr²T at each stage, so the work done is 8πT(r2² − r1²) = 8π × 0.025 × (16 − 4) × 10⁻⁴ = 8π × 0.025 × 1.2 × 10⁻³ ≈ 7.5 × 10⁻⁴ J. Part of this went into creating fresh surface; the rest pushed the atmosphere back as the bubble grew, which is why the full thermodynamic accounting includes a P dV term at the more advanced level, though JEE's standard treatment equates work with the change in surface energy.
Now the coalescence classic. Bubbles of radii 1 cm and 3 cm are connected. Excess pressures are 4T/1 and 4T/3, so the smaller is at higher internal pressure and pumps air into the larger until the pressure difference across their common wall balances: 4T/r = 4T/r1 − 4T/r2 gives 1/r = 1/1 − 1/3 = 2/3, so the interface curves with r = 1.5 cm. The smaller bubble need not vanish — it stops shrinking when its shrinking radius and the shared-wall geometry satisfy this relation, and JEE Advanced has asked precisely for the interface radius rather than the naive "smaller disappears" answer.
Where students slip
The dominant error is the surface count: quoting 4T/r for a mercury drop in vacuum (it is 2T/r — drops have one surface) or 2T/r for a soap bubble. The second slip is absolute versus excess pressure: a bubble at depth h in water has internal pressure P0 + hρg + 4T/r, and problems that ask "pressure inside" expect all three terms. Watch also the radius arithmetic in energy problems — 8πr²T has radius squared, so a factor-of-two radius error costs a factor of four, and options are spaced to catch exactly that. Finally, in the two-bubble problem, students who compare radii instead of pressures conclude the large bubble feeds the small one; pressure, not size, drives the flow.
Frequently asked questions
Why is a soap bubble's excess pressure double a drop's?
Because the film has two free surfaces, inner and outer, each contributing 2T/r, giving 4T/r in total.
What is the excess pressure inside an air bubble in water?
2T/r above the surrounding liquid pressure at that depth, since a submerged air bubble presents only one liquid–gas interface.
How much work is needed to blow a soap bubble to radius r?
The surface energy is 8πr²T (two surfaces), so blowing from nothing at constant surface tension requires that much work; expanding r1 to r2 costs 8πT(r2² − r1²).
What happens when two soap bubbles of different radii are connected?
Air flows from the smaller bubble to the larger because its excess pressure 4T/r is higher, until the common interface settles at radius r1r2/(r2 − r1).
Does atmospheric pressure affect the excess pressure formula?
No — the excess depends only on surface tension and radius; atmospheric pressure sets the baseline inside and outside equally and cancels in the difference.