Surface energy
In a liquid, each molecule attracts its neighbours. A molecule deep inside is pulled equally from all sides, so its net pull is zero. A molecule at the surface has neighbours only below and beside it, so it feels a net inward pull.
To bring a molecule from inside to the surface, work must be done against this pull. So molecules at the surface have extra energy. This extra energy of the surface is the surface energy.
Because extra surface costs energy, a liquid always tries to make its surface area as small as possible. That is why free drops are round: a sphere has the least surface for a given volume.
Surface tension
Surface tension S is the force per unit length acting along the surface, at right angles to any line drawn on it: S = F/l. Unit: N/m.
Take a U-shaped wire frame with a sliding wire of length l and a soap film on it. The film has two surfaces (front and back), so it pulls the wire with force F = 2Sl. To pull the wire out by distance d, work done = F d = 2Sld. The new area made is 2ld (two faces). So:
Surface energy per unit area = work / new area = 2Sld / 2ld = S
So surface tension (N/m) and surface energy per area (J/m²) are the same quantity. Water: S ≈ 0.072 N/m at 20 °C. S falls as temperature rises, and soap lowers it a lot.
Angle of contact
Where a liquid surface meets a solid, the angle of contact θ is the angle between the solid surface and the tangent to the liquid surface, measured inside the liquid.
- θ < 90°: the liquid wets the solid. Attraction to the solid (adhesion) is stronger than attraction within the liquid (cohesion). Water on clean glass: θ ≈ 0°–10°.
- θ > 90°: the liquid does not wet the solid. Cohesion is stronger. Mercury on glass: θ ≈ 140°; water on a waxy lotus leaf: θ is large, so water rolls off as beads.
Soaps and detergents lower θ and S so water can enter fine gaps in cloth. Waterproofing agents raise θ so water stays out.
Excess pressure inside drops and bubbles
A curved surface pulls toward its centre, so pressure on the concave (inner) side is higher.
Liquid drop of radius r: grow the radius by Δr. Extra surface energy = S × 8πrΔr. Work by the extra pressure = ΔP × 4πr²Δr. Setting them equal:
ΔP = P(in) − P(out) = 2S/r (also true for an air bubble inside a liquid)
Soap bubble in air: it has two surfaces, inner and outer. So:
ΔP = 4S/r
Smaller drops and bubbles have more excess pressure. If you join a small and a big soap bubble by a tube, the small one shrinks and the big one grows.
Capillary rise
Dip a thin glass tube (a capillary) in water. Water rises inside it, and its top surface (meniscus) is curved like a cup. For mercury the level falls below the outside level, and the meniscus bulges up.
Formula (from excess pressure): for water with θ ≈ 0°, the meniscus is a hemisphere of radius a (tube radius). Pressure just under the meniscus is lower than outside by 2S/a. Water climbs until the column's hydrostatic pressure fills this gap: hρg = 2S/a. In general, with angle θ:
h = 2S cos θ / (r ρ g)
- h ∝ 1/r: thinner tube, higher rise.
- θ < 90° ⇒ cos θ positive ⇒ rise. θ > 90° ⇒ cos θ negative ⇒ fall (mercury).
- This is capillarity: it helps water move up soil and through paper, cloth and wicks.
Try it: the floating pin
Place a steel pin on a fork and lower it gently onto water in a bowl. It floats, held by the surface skin, even though steel is 8 times denser than water. Now touch the water at the edge with a drop of dish soap. The pin sinks at once: soap lowers S. In the 3D, step 6, change r from 0.5 mm to 0.25 mm and check that h doubles.
Key formulas and definitions
- Surface tension S = F / l (N/m)
- Surface energy per area = S (J/m²)
- Soap film on a wire: F = 2Sl
- Drop or air bubble in liquid: ΔP = 2S / r
- Soap bubble in air: ΔP = 4S / r
- Capillary rise: h = 2S cos θ / (r ρ g)
- Work to enlarge a soap bubble: W = S × 2 × ΔA
Worked examples
1. A wire of length 4 cm floats on water. What extra force (beyond its weight) is needed to pull it off the surface? (S = 0.072 N/m)
Step 1: The surface pulls along both sides of the wire, so total length = 2 × 0.04 m. Step 2: F = S × 2l = 0.072 × 0.08. Answer: 5.76 × 10⁻³ N.
2. Find the excess pressure inside a water drop of radius 1 mm. (S = 0.072 N/m)
Step 1: ΔP = 2S/r. Step 2: = 2 × 0.072 / 10⁻³. Answer: 144 Pa.
3. Find the excess pressure inside a soap bubble of radius 2 cm. (S = 0.03 N/m)
Step 1: Two surfaces, so ΔP = 4S/r. Step 2: = 4 × 0.03 / 0.02. Answer: 6 Pa.
4. Water rises in a glass tube of radius 0.5 mm. Find the rise. (S = 0.072 N/m, θ = 0°, ρ = 1000 kg/m³, g = 9.8 m/s²)
Step 1: h = 2S cosθ / (rρg). Step 2: numerator = 2 × 0.072 × 1 = 0.144. Step 3: denominator = 5 × 10⁻⁴ × 1000 × 9.8 = 4.9. Step 4: h = 0.144/4.9. Answer: h ≈ 0.029 m ≈ 2.9 cm.
5. A mercury column in a glass tube of radius 1 mm: find the depression. (S = 0.465 N/m, θ = 140°, cos 140° = −0.766, ρ = 13600 kg/m³, g = 9.8 m/s²)
Step 1: h = 2S cosθ / (rρg). Step 2: numerator = 2 × 0.465 × (−0.766) = −0.712. Step 3: denominator = 10⁻³ × 13600 × 9.8 = 133.3. Step 4: h = −0.712/133.3. Answer: h ≈ −5.3 × 10⁻³ m; the mercury falls about 5.3 mm below the outside level.
6. How much work is needed to blow a soap bubble of radius 3 cm? (S = 0.03 N/m)
Step 1: One sphere's area = 4πr² = 4 × 3.14 × 9 × 10⁻⁴ = 1.13 × 10⁻² m². Step 2: Two surfaces, so new area = 2.26 × 10⁻² m². Step 3: W = S × area = 0.03 × 2.26 × 10⁻². Answer: W ≈ 6.8 × 10⁻⁴ J.
7. 1000 tiny water drops, each of radius 0.1 mm, join to form one big drop. Find the energy released. (S = 0.072 N/m)
Step 1: Volume is conserved: 1000 × r³ = R³, so R = 10r = 1 mm. Step 2: Area before = 1000 × 4πr² = 1000 × 4π × 10⁻⁸ = 1.257 × 10⁻⁴ m². Step 3: Area after = 4πR² = 4π × 10⁻⁶ = 1.257 × 10⁻⁵ m². Step 4: Loss of area = 1.131 × 10⁻⁴ m². Step 5: Energy released = S × ΔA = 0.072 × 1.131 × 10⁻⁴. Answer: ≈ 8.1 × 10⁻⁶ J (it warms the drop slightly).
Common mistakes
- Using 2S/r for a soap bubble in air. A soap bubble has two surfaces, so ΔP = 4S/r.
- Forgetting cos θ in the capillary formula, or its sign. For mercury θ > 90°, cos θ < 0, so the level falls.
- Thinking surface tension and surface energy have different values. Numerically they are equal; N/m = J/m².
- Taking the diameter instead of the radius of the tube in h = 2S cosθ / (rρg).