Excess pressure in drops

Blow a soap bubble and let it sit in still air. It does not collapse, and it does not keep growing — it holds a stable spherical shape with the air pressure trapped inside it slightly higher than the air pressure outside. That tiny pressure difference is not a curiosity; it is a direct, measurable consequence of surface tension acting on a curved surface, and it explains everything from why small bubbles connected to large ones shrink instead of grow, to why mist droplets are perfectly round, to why a blocked capillary in your lungs' alveoli would need pressure support. This article builds the excess-pressure relationship from a simple force balance, applies it to both a liquid drop (one curved surface) and a soap bubble (two curved surfaces), and works through the numbers you are expected to handle in ISC numericals. You are assumed to already know the definition of surface tension $T$ as force per unit length along a surface, and surface energy as energy per unit area — this article picks up exactly where that leaves off, asking: what pressure difference does a curved surface under tension actually sustain?

The core insight

A curved liquid surface under tension always pushes toward its centre of curvature, so the pressure just inside a drop exceeds the pressure outside by $P_{\text{excess}} = 2T/r$ for one surface (a drop) or $4T/r$ for two surfaces (a soap bubble), with excess pressure growing larger as the radius shrinks.

The core idea

A flat liquid surface has no tendency to push in any particular direction — the surface tension forces pulling along it are all in balance. A curved surface is different. Because surface tension acts like a stretched elastic skin trying to minimise its area, a curved surface is always pulled taut toward its centre of curvature, and this net inward pull must be balanced by a higher pressure on the concave (inner) side than on the convex (outer) side. This pressure difference is the excess pressure.

One surface — a liquid drop. A spherical drop of radius $r$ has a single liquid–air interface. Surface tension continually tries to shrink the drop's surface area (and hence its surface energy), and this tendency is held in check by the drop's own internal pressure pushing outward. The equilibrium excess pressure is

P_{\text{excess}} = P_{\text{inside}} - P_{\text{outside}} = \frac{2T}{r}

where: $T$ is the surface tension of the liquid (N/m) and $r$ is the radius of the drop (m). The same formula applies to an air bubble trapped inside a liquid, since that too has only one liquid surface (liquid outside, air inside).

Two surfaces — a soap bubble. A soap bubble is a thin film of liquid with air on both sides, so it has two curved surfaces — an outer film face and an inner film face — each independently contributing an excess pressure of $2T/r$ . The contributions add, doubling the result:

P_{\text{excess}} = \frac{4T}{r}

Why smaller means higher. Both formulas show excess pressure varying as $1/r$ — inversely with radius. A tiny drop or bubble must sustain a larger pressure difference than a large one to stay in equilibrium against the same surface tension. This inverse relationship is the single most exam-tested fact in this topic, and it drives a genuinely counter-intuitive result: connect a small soap bubble and a large one by a tube, and the small bubble (higher internal pressure) empties its air into the large one (lower internal pressure) and shrinks further, rather than growing to match it.

Figure

A small drop and a large drop side by side showing thinner arrows for the large drop and thicker arrows for the small drop to represent higher excess pressure in the smaller drop

Derivation AExcess pressure inside a drop from a force balance

Show that balancing the inward pull of surface tension along the rim of an imaginary cut against the outward push of the pressure difference across that cut's cross-section gives exactly $P_{\text{excess}} = 2T/r$ for a spherical liquid drop of radius $r$ and surface tension $T$ , without requiring calculus or knowledge of surface curvature beyond the sphere's own geometry.

Step 1. Imagine slicing a spherical liquid drop of radius $r$ exactly in half through its centre, so you are left with two identical hemispherical shells held together (in reality) only by the internal pressure and the surface tension acting along the circular rim where the cut was made.

Step 2. The surface tension acts along the entire circular rim of the cut, pulling the two hemispheres toward each other. The rim is a circle of radius $r$ , so its circumference is $2\pi r$ , and since surface tension $T$ is force per unit length, the total inward pulling force along the rim is $F_{\text{tension}} = T \times 2\pi r$ .

Step 3. This inward pull must be exactly balanced by the outward push of the excess pressure $P_{\text{excess}}$ , which acts over the flat circular cross-section exposed by the cut. That cross-section has area $\pi r^2$ , so the outward force is $F_{\text{pressure}} = P_{\text{excess}} \times \pi r^2$ .

Step 4. For the hemisphere to be in equilibrium (neither half flying apart nor collapsing inward), these two forces must be equal: $P_{\text{excess}} \times \pi r^2 = T \times 2\pi r$ .

Step 5. Cancelling $\pi r$ from both sides gives $P_{\text{excess}} \times r = 2T$ , which rearranges directly to $P_{\text{excess}} = 2T/r$ — the excess pressure formula for a single curved surface, obtained purely from a static force balance and the sphere's own geometry, with no calculus needed. For a soap bubble, the identical argument applies to each of its two surfaces, so the tension force along the rim doubles to $T \times 2 \times 2\pi r$ , giving $P_{\text{excess}} = 4T/r$ .

Worked example

A water drop of radius $1.0\ \text{mm}$ sits in air, with the surface tension of water $T = 0.072\ \text{N/m}$ . A soap bubble of radius $2.0\ \text{cm}$ is blown from a solution with $T = 0.030\ \text{N/m}$ . Find the excess pressure inside each, and state which one has the larger excess pressure.

Step 1. For the water drop (one surface), convert the radius to metres: $r = 1.0\ \text{mm} = 1.0\times10^{-3}\ \text{m}$ . Apply $P_{\text{excess}} = 2T/r$ :

P_{\text{drop}} = \frac{2 \times 0.072}{1.0\times10^{-3}} = \frac{0.144}{1.0\times10^{-3}} = 144\ \text{Pa}

Step 2. For the soap bubble (two surfaces), convert the radius: $r = 2.0\ \text{cm} = 2.0\times10^{-2}\ \text{m}$ . Apply $P_{\text{excess}} = 4T/r$ :

P_{\text{bubble}} = \frac{4 \times 0.030}{2.0\times10^{-2}} = \frac{0.12}{0.02} = 6.0\ \text{Pa}

Step 3. Compare the two results directly. Even though the soap bubble formula has an extra factor of $2$ compared with the drop formula, the bubble's radius ( $2.0\ \text{cm} = 20\ \text{mm}$ ) is $20$ times larger than the drop's radius ( $1.0\ \text{mm}$ ), and the $1/r$ dependence dominates: $144\ \text{Pa}$ for the tiny drop versus only $6.0\ \text{Pa}$ for the much larger bubble.

The water drop has the larger excess pressure ( $144\ \text{Pa}$ versus $6.0\ \text{Pa}$ for the bubble), purely because its radius is so much smaller — a direct illustration of the $1/r$ trend, not because a drop's single-surface formula gives a smaller pressure than a bubble's double-surface formula in general.

Competitive stretch

If the same soap bubble were instead blown to a smaller radius of $1.0\ \text{cm}$ using the same solution, its excess pressure doubles to $P = 4(0.030)/(1.0\times10^{-2}) = 12\ \text{Pa}$ , exactly twice the $6.0\ \text{Pa}$ found for the $2.0\ \text{cm}$ bubble — confirming $P \propto 1/r$ , and showing why, if the two bubbles were connected by a tube, the $1.0\ \text{cm}$ bubble (higher pressure) would empty air into the $2.0\ \text{cm}$ bubble and shrink further rather than grow.

Common traps

Using $2T/r$ for a soap bubble instead of $4T/r$ — the extra factor of two comes from the bubble having two liquid surfaces (inner and outer film faces), whereas a liquid drop or an air bubble inside a liquid has only one.
Assuming a bigger drop or bubble means a bigger excess pressure — it is the opposite: excess pressure is inversely proportional to $r$ , so smaller drops and bubbles have the higher excess pressure.
Treating $P_{\text{excess}}$ as the absolute pressure inside the drop — it is only the difference, $P_{\text{inside}} - P_{\text{outside}}$ ; the actual internal pressure is atmospheric (or surrounding) pressure plus this excess.
Applying the bubble formula $4T/r$ to an air bubble trapped inside a liquid — that configuration has only one liquid surface (liquid on the outside, air inside), so it uses $2T/r$ , the same as an ordinary drop, not the two-surface soap-bubble case.

Ask the tutor: Want to see how this excess-pressure result connects to capillary rise in a tube, or need a few more mixed drop-versus-bubble numericals to practice the $1/r$ trend?

Why does a soap bubble have two surfaces but a water drop has only one, if both are round?

A water drop is a solid ball of liquid — air only touches its outer surface, so there is exactly one liquid–air interface. A soap bubble is a thin hollow film of soapy water enclosing trapped air, with air both inside the shell and outside it. That means the film has two faces exposed to air, an outer one and an inner one, and each independently contributes its own $2T/r$ excess pressure, which is why the two add up to $4T/r$ for the whole bubble.

If excess pressure is higher for smaller drops, why don't tiny raindrops instantly burst apart?

The excess pressure inside a small drop is indeed higher, but "higher" here means higher than the surrounding air pressure, not so high that the drop becomes unstable. The internal pressure and the inward pull of surface tension are in equilibrium by definition — that is exactly what the formula $P_{\text{excess}} = 2T/r$ describes: the internal pressure needed to exactly balance surface tension's tendency to shrink the drop, not a runaway pressure that would rupture it.

In the two-connected-bubbles case, does the smaller bubble ever stop shrinking?

As the smaller bubble loses air and its radius decreases further, its excess pressure ( $\propto 1/r$ ) actually increases even more, which keeps driving air out toward the larger, lower-pressure bubble. In practice the smaller bubble keeps shrinking until it empties into the larger one entirely (or the tube is closed), rather than settling into a stable intermediate size — this is precisely why the situation feels counter-intuitive compared to everyday expectation.

Excess pressure in drops

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads