Mechanical Properties of Fluids · Lesson 7 of 8
Drops, Bubbles and Capillary Action
“Derive curved-interface pressure and capillary height, then distinguish local excess, gauge and absolute pressure.”
• Explain why a free small drop tends to become spherical at fixed volume. • Derive excess pressure for a liquid drop and distinguish a gas cavity from a soap bubble. • Combine hydrostatic pressure with interfacial pressure for a submerged bubble. • Derive capillary rise and depression using a consistent pressure convention and correct curvature radius.
Why small free drops tend toward spheres
Surface energy is S times exposed area when S is fixed. A freely adjustable drop of fixed volume therefore tends to reduce its area. Among closed shapes of the same volume, a sphere has the least surface area. This is why small free drops and bubbles are approximately spherical when gravity, airflow and other deforming forces are negligible. Surface tension does not require every falling raindrop or every large puddle to be spherical.
Compare a sphere of radius r with a cube of the same volume. The sphere volume is 4πr³/3 and its area is 4πr². The cube side is (4π/3)^(1/3)r, giving area 6(4π/3)^(2/3)r², about 1.24 times the spherical area. This comparison illustrates the advantage of a sphere over a cube; it is not a proof of the minimum-area theorem for every possible shape.
Deriving the pressure difference across a spherical drop
Consider a spherical liquid drop of radius r, with pressure P_in inside and P_out outside. Imagine increasing the radius very slowly by a tiny amount dr while keeping S constant. The surface area grows from 4πr² to 4π(r + dr)². Neglecting terms containing (dr)² leaves an area increase 8πr dr. The corresponding volume increase is 4πr² dr to first order. These small-change formulas are also the derivatives of sphere area and volume with respect to radius, multiplied by dr.
The net pressure difference does work (P_in − P_out)dV during expansion. At reversible mechanical balance this work equals the extra surface energy S dA. Equate the two, cancel the common 4πr dr, and solve for the pressure difference. A positive inside excess is required to balance the contracting tendency of the curved interface.
The formula says smaller radius requires larger excess pressure. It gives only the jump across the curved interface. It does not set the outside pressure and does not say inside absolute pressure equals 2S/r. For a bubble underwater, the outside water already has hydrostatic pressure before the interface jump is added.
Why a soap bubble has twice the excess pressure
A soap bubble in air contains a thin liquid shell. There is an inner gas–liquid interface and an outer liquid–gas interface. Expanding a thin bubble creates approximately twice the new area created for a single interface of the same radius. The surface-energy cost doubles, while the enclosed volume change is approximately unchanged. Thus the inside gas must exceed outside air by about 4S/r.
A modest pressure excess is needed to inflate a soap bubble. The required excess decreases as the bubble grows if S stays constant. Surface tension supplies the tendency to contract; a pressure difference can balance it. The formula alone does not promise a permanently stable bubble: drainage, evaporation and film rupture are separate physical processes. The important classification is the number of interfaces crossed, not merely whether an everyday description uses the word “bubble”.
Problem
A mercury drop has radius 3.00 mm and S = 0.465 N/m at the specified temperature. Outside pressure is 1.01 × 10⁵ Pa. Find excess and inside absolute pressure.
- 1.There is one interface, so ΔP = 2S/r = 2 × 0.465/0.00300 = 310 Pa.
- 2.Inside absolute pressure = 101000 + 310 = 101310 Pa = 1.01310 × 10⁵ Pa before rounding.
- 3.Rounded to the source’s three-significant-figure convention, the total is 1.01 × 10⁵ Pa; the separately reported excess remains 310 Pa. The numerical correction can be physically meaningful even if hidden by rounding the large total.
- 4.Use the 0.465 N/m specified here. Do not substitute a different mercury value from a general table without changing the problem conditions.
Problem
A soap solution has S = 0.0250 N/m. Find the excess in a soap bubble of radius 5.00 mm in air. Then find pressure in an air bubble of the same radius at depth 40.0 cm in that bulk solution, whose relative density is 1.20. Take P_atm = 1.01 × 10⁵ Pa and g = 9.8 m/s².
- 1.The soap bubble in air has two interfaces: ΔP = 4S/r = 4 × 0.0250/0.00500 = 20.0 Pa.
- 2.The submerged gas bubble has one interface, so its excess above the nearby solution is 2S/r = 10.0 Pa.
- 3.Solution density is 1200 kg/m³. At depth 0.400 m, outside liquid pressure is 101000 + 1200 × 9.8 × 0.400 = 105704 Pa.
- 4.Inside gas pressure is 105704 + 10.0 = 105714 Pa, or approximately 1.06 × 10⁵ Pa. Its gauge pressure is 4714 Pa, while its excess over the immediate surrounding liquid is only 10.0 Pa. These are different pressure differences.
Problem
A tube of internal diameter 2.00 mm ends 8.00 cm below a water surface. Find the pressure needed to form a hemispherical air bubble at its end. Use S = 0.0730 N/m, water density 1000 kg/m³, g = 9.80 m/s² and P_atm = 1.01 × 10⁵ Pa.
- 1.For a hemisphere spanning the circular tube end, bubble curvature radius equals the tube radius: r = 1.00 mm = 0.00100 m. There is one gas–water interface.
- 2.Outside water pressure at depth 0.0800 m is P_out = 101000 + 1000 × 9.80 × 0.0800 = 101784 Pa.
- 3.Interfacial excess is 2S/r = 2 × 0.0730/0.00100 = 146 Pa.
- 4.Required inside absolute pressure = 101784 + 146 = 101930 Pa ≈ 1.02 × 10⁵ Pa. The gauge pressure above atmosphere is 930 Pa.
- 5.The 146 Pa excess is measured relative to nearby water, not relative to atmosphere. This distinction is essential whenever a curved interface is submerged.
Capillary rise begins with a pressure jump at the meniscus
Put a narrow clean glass tube vertically in water. Water wets glass and forms a concave meniscus: the surface is higher at the walls than near the centre. The air-side pressure just above this meniscus is approximately atmospheric, while the water-side pressure just below it is smaller. The surrounding reservoir then pushes water up the tube until the extra height produces a matching hydrostatic pressure reduction. Capillary action is a consequence of surface tension, wetting and gravity together.
Use a for the tube’s internal radius and R for the spherical meniscus curvature radius. They are not generally equal. Geometry of the contact angle gives a = R cos θ for an acute wetting angle, or R = a/cos θ. Thus the air-to-water pressure jump is 2S/R = 2S cos θ/a. Keeping the two radii separate avoids replacing every capillary formula with a bubble formula using the wrong radius.
Let h be the meniscus height above the reservoir surface. At the reservoir-surface level, connected water inside and outside the tube has the same pressure P_atm. Move up height h inside the tube: hydrostatic balance gives P_liquid,top = P_atm − ρgh. The air above is also approximately P_atm, so its excess over water is ρgh. Equate this hydrostatic deficit to the curved-interface pressure jump.
The same expression follows from a force check. Surface tension acts around circumference 2πa. Its net upward component on the raised liquid is 2πaS cos θ. The weight of a cylindrical column is approximately ρgπa²h. Equating the two again gives the height relation. For complete wetting θ is close to zero, so cos θ is close to one. For θ = 90° there is no capillary height shift in this idealisation. For mercury on glass, θ is obtuse, cos θ is negative and the liquid level is depressed.
A smaller tube radius produces a greater height magnitude. A higher density produces a shorter column because more weight must be supported per unit height. Temperature, contamination and tube material can affect S or θ. Capillary rise is not an upward pressure force created from nothing: the lowered liquid pressure below the wetting meniscus and the interfacial force description are two consistent ways to describe the same equilibrium.
Problem
A clean tube of radius 0.05 cm is placed in water. Take θ ≈ 0, S = 0.073 N/m, ρ = 1000 kg/m³ and g = 9.8 m/s². Find the rise.
- 1.Convert a = 0.05 cm = 5.0 × 10⁻⁴ m. Complete wetting gives cos θ ≈ 1.
- 2.h = 2 × 0.073/(1000 × 9.8 × 5.0 × 10⁻⁴) = 0.0298 m = 2.98 cm.
- 3.Using diameter in place of radius would give half this rise. If the actual angle were 60°, the rise would also halve because cos 60° = 0.5.
- 4.The meniscus liquid pressure lies about ρgh = 292 Pa below atmospheric pressure. This negative gauge pressure is not a negative absolute pressure.
Quick Quiz
Quiz
Why is a small free drop approximately spherical when external deformation is negligible?
At fixed volume, minimising area minimises interfacial energy when S is fixed.
An air bubble in bulk water has how many liquid–gas interfaces?
Gas touches surrounding bulk liquid through one boundary; unlike a soap bubble, there is no thin shell with two faces.
A soap bubble in air has S = 0.030 N/m and radius 6.0 mm. Its ideal excess pressure is what?
Two interfaces give 4S/r = 4 × 0.030/0.0060 = 20 Pa.
For an otherwise unchanged wetting liquid, halving tube radius changes ideal capillary rise how?
h = 2S cos θ/(ρga) is inversely proportional to tube radius.
A wetting meniscus stands above the reservoir. The pressure in its liquid just below the meniscus is what?
Hydrostatic balance inside the raised column gives P_liquid,top = P_atm − ρgh.
Practice Problems
- Starting from small changes in sphere area and volume, derive ΔP = 2S/r for one interface. Explain why a thin soap bubble gives approximately twice that excess pressure.
- A liquid drop of radius 2.0 mm has S = 0.040 N/m. Find its excess pressure and inside absolute pressure when outside pressure is 1.00 × 10⁵ Pa. Repeat the excess-pressure calculation for a soap bubble with the same radius and S.
- Air is blown through a tube of radius 0.80 mm ending 0.12 m below a water surface. Find the pressure required for a hemispherical bubble using S = 0.072 N/m, ρ = 1000 kg/m³, g = 10 m/s² and P_atm = 1.00 × 10⁵ Pa. Report separately the excess over nearby water, gauge pressure and absolute pressure.
- A liquid of density 800 kg/m³ and surface tension 0.030 N/m wets a tube with θ = 60° and radius 0.25 mm. Find capillary rise using g = 10 m/s². State the corresponding pressure deficit at the meniscus.
- A nonwetting liquid has θ = 120°, density 13600 kg/m³ and S = 0.46 N/m in a tube of radius 0.50 mm. Find the signed capillary height using g = 9.8 m/s². Draw and explain the depressed meniscus and distinguish tube radius from curvature radius.
Key Takeaways
• Surface-energy minimisation favours a sphere when external deformation is negligible. • One spherical interface gives ΔP = 2S/r; a thin soap bubble has two and gives about 4S/r. • Add the local hydrostatic pressure before the interfacial excess when finding submerged-bubble absolute pressure. • Capillary height is 2S cos θ/(ρga); its sign distinguishes wetting rise from nonwetting depression.