Mechanical Properties of Fluids · Lesson 6 of 8
Surface Energy, Surface Tension and Wetting
“Use surface energy to derive tension, explain measurements and understand contact angle and wetting.”
• Explain why creating liquid surface requires energy and why surface tension is an interfacial property. • Derive the two-surface film relation F = 2Sl from work and area change. • Interpret surface-tension measurements and distinguish force, energy and area. • Use contact angle and the interfacial-force balance to explain wetting and nonwetting.
Everyday clues at a liquid surface
Water forms beads on a waxy leaf but spreads over clean glass. A cotton wick draws in oil. Paintbrush hairs can remain separate while submerged and bunch together when lifted through the water surface. Small drops round up. These effects draw attention to an interface: a boundary between a liquid and air, another liquid or a solid. Surface tension is associated with the interface, not merely with the amount of liquid present.
Liquids can form a free surface because they keep approximately fixed volume without filling all available space. A gas by itself spreads through its container and has no comparable free liquid surface. A liquid–gas interface nevertheless involves both phases. Wetting in cloth and plant tissues involves surface effects, but capillary action alone is not a complete explanation of water transport to the tops of tall trees; biological transport can involve other mechanisms.
The molecular origin of surface energy
A molecule well inside a liquid has neighbouring liquid molecules in all directions. Their attractions bind it into the liquid, giving a negative potential energy relative to widely separated molecules. Removing molecules far from one another requires energy: vaporising one mole of water needs energy of order tens of kilojoules. The average attractive environment near a free surface differs because there are fewer liquid neighbours on the gas side.
A molecule near the surface therefore has higher, or less negative, interaction energy than a comparable bulk molecule. Bringing additional molecules into the surface region costs energy. The liquid tends to reduce its exposed area when its volume and external constraints permit. A useful rough molecular picture relates the energy cost of a surface molecule to a fraction of its vaporisation energy, sometimes pictured as about one half because roughly half its liquid neighbours are absent. This is an intuition, not a precise universal factor or a way to calculate surface tension accurately.
The actual equilibrium interface is not a perfectly sharp mathematical skin. Molecular density falls from its bulk value to its much lower gas value over a few molecular sizes. Short-range repulsive and attractive contributions balance in equilibrium. One should not imagine surface molecules accelerating inward forever because an attractive-force sketch omits other effects. At macroscopic scales we represent this thin interfacial region by a surface with an associated energy and tension.
Deriving the surface force using a movable film
Make a soap film in a U-shaped frame closed by a light movable bar of length l. A soap film has two liquid–air faces: its front and back. Pull the bar outward by a small distance d at constant temperature and approximately fixed film volume. Each face gains area ld, so the total surface-area increase is 2ld. Let S be the surface energy per unit area of one interface, treated as constant over this small change.
If a force F balances the film’s pull during a slow displacement, external work is Fd. In the ideal mechanical account that work is stored as additional surface energy S(2ld). Equating the two and cancelling d gives F = 2Sl. The factor two counts the two interfaces. It is not a doubling of the bar’s length and must not be attached indiscriminately to every liquid boundary.
At a boundary, surface tension acts in the tangent plane of the interface and perpendicular to the boundary line. Draw an imaginary line across a flat film: the two parts pull on one another across that line. Their forces are equal and opposite when both parts are included. On a bar at the film’s edge, the film’s pull is toward the remaining film. Do not confuse this in-plane tension with the normal pressure force exerted by a bulk liquid.
S is an interfacial property determined by the substances on both sides, their condition and temperature. For fixed materials and temperature it is approximately independent of the total area; the total surface energy changes with area. Doubling the area doubles S times area, not S itself. A larger soap film with the same slider length and the same two faces can therefore pull with the same force even though its total surface energy differs.
Problem
A film supports a total downward load of 1.5 × 10⁻² N on a slider 30 cm long. Find its surface tension. Then compare differently shaped frames with the same 40 cm slider if one frame supports 4.5 × 10⁻² N at the same temperature.
- 1.There are two film surfaces. With l = 0.30 m, S = F/(2l) = 0.015/(2 × 0.30) = 0.025 N/m. The stated load includes the slider’s own weight.
- 2.For frames with the same liquid, temperature and two-surface contact length, F = 2Sl is unchanged by the fixed frame’s outline.
- 3.Thus all three 40 cm-slider frames support 4.5 × 10⁻² N in the ideal arrangement, whether rectangular, rounded or roof-shaped above. Their areas and total energies may differ, but the slider’s local boundary pull is the same.
- 4.If the contact length, number of interfaces or contact geometry changes, repeat the force accounting rather than relying on the frame-shape comparison.
Measuring surface tension and comparing materials
A thin vertical glass plate can hang from one arm of a balance just above a liquid. Raise the liquid until it contacts the lower horizontal plate edge. Wetting adds a downward surface-tension force. The extra balancing weight needed to bring the edge just clear provides a measurement. For a very thin plate of edge length l, complete wetting and the simplified two-face geometry, the wetted contact length is approximately 2l and the vertical pull is 2Sl.
The subscript la identifies the liquid–air interface. A liquid–solid interface and a solid–air interface generally have different interfacial energies. Replacing the air or changing the solid can alter wetting even if the liquid is unchanged. The film and balance measurements express this same energy cost as a force per length.
| Liquid | Temperature (°C) | S (N/m) | Vaporisation energy (kJ/mol) |
|---|---|---|---|
| Helium | −270 | 0.000239 | 0.115 |
| Oxygen | −183 | 0.0132 | 7.1 |
| Ethanol | 20 | 0.0227 | 40.6 |
| Water | 20 | 0.0727 | 44.16 |
| Mercury | 20 | 0.4355 | 63.2 |
The values illustrate differences among interfaces. Surface tension usually decreases as liquid temperature increases. The table also lists vaporisation energies to connect binding with surface-energy scale; those molar energies are not numerically equal to S because the units and microscopic geometrical information differ. Some chapter exercises specify a different numerical mercury surface tension; use the value supplied in each question rather than silently replacing it with a table value.
Contact angle is measured through the liquid
Where a liquid surface meets a solid, the interface is usually curved near the contact line. Draw the tangent to the liquid surface at that line and measure the angle between that tangent and the solid through the liquid. This is the angle of contact θ. An acute angle corresponds to relatively good wetting; an obtuse angle corresponds to poor wetting in the usual droplet picture. The angle concerns a particular liquid–solid–gas combination and its surface condition.
To understand the angle, name the three interfacial tensions: S_la for liquid–air, S_sa for solid–air and S_sl for solid–liquid. At equilibrium on a smooth, ideal solid, balance the horizontal components at the contact line. The liquid–air contribution has component S_la cos θ. Combining it with the solid–liquid contribution gives the solid–air contribution. The solid also supplies a normal reaction, so one should not demand that these three horizontal terms alone describe every vertical force.
If S_sa is greater than S_sl, the cosine is positive and θ is acute for a finite equilibrium angle. If S_sa is less than S_sl, the cosine is negative and θ is obtuse. If replacing solid–air area by solid–liquid plus liquid–air area lowers the total energy sufficiently, the liquid can spread completely rather than remain as a finite-angle drop. Qualitatively, strong liquid–solid attraction favours wetting, whereas relatively stronger cohesion within the liquid favours beading.
Water commonly wets clean glass; mercury on glass commonly forms an obtuse angle and beads. Water on a waxy leaf also forms droplets. Surface contamination and roughness can change these observations, so “water always wets every solid” and “mercury cannot wet any material” are too broad. Wetting is a property of the interface combination, not a permanent label attached to a liquid alone.
Soaps, detergents and dyeing agents can improve wetting by changing the interfacial energies and reducing the contact angle. This helps the liquid penetrate fibres or contact a surface. Waterproofing treatments aim to make the water contact angle large. A liquid’s surface tension and its contact angle are related through interfacial balance but are not the same quantity; lowering one isolated parameter does not replace checking all three interfaces.
When wet paintbrush hairs emerge from water, the liquid bridges between them can reduce their exposed interfacial area by drawing the hairs together. While fully submerged, those particular liquid–air bridges are absent. Oil wicking into cotton similarly benefits from favourable wetting and narrow spaces; the pressure mechanism for capillary rise will be derived next. These examples connect the molecular energy picture to visible shape changes.
Problem
For an ideal solid–liquid–air combination, take S_la = 0.040 N/m, S_sa = 0.060 N/m and S_sl = 0.040 N/m. Find θ and predict the qualitative wetting behaviour.
- 1.Young’s relation gives cos θ = (0.060 − 0.040)/0.040 = 0.50.
- 2.Thus θ = 60°, measured through the liquid. It is acute, so the liquid wets the ideal solid more readily than a combination giving an obtuse angle.
- 3.The calculation depends on all three interfacial values. Comparing the two numbers S_sl and S_la alone would give no correct general criterion.
Quick Quiz
Quiz
Surface tension has which pair of equivalent units?
Energy per area equals force per length because 1 J = 1 N·m.
A soap-film slider of length l feels total pull 2Sl because of what?
The thin film has front and back interfaces, each contributing Sl.
A film’s area doubles at the same temperature with unchanged materials. Which statement is appropriate?
Surface tension is a local material/interface property; total surface energy depends on total area.
An equilibrium contact angle is measured where?
The through-liquid convention fixes whether the angle is acute or obtuse.
Which condition gives an obtuse finite contact angle in Young’s ideal relation?
cos θ = (S_sa − S_sl)/S_la is negative when S_sa < S_sl, for positive S_la.
Practice Problems
- A two-sided soap film has a slider 0.20 m long. It is extended outward by 3.0 mm at constant S = 0.030 N/m. Find total extra interfacial area, required force and work. Explain each factor of two.
- A thin, completely wetted plate with horizontal lower-edge length 5.0 cm requires an extra balancing mass of 0.50 g to detach under the simplified two-face model. Estimate surface tension using g = 9.8 m/s². State the neglected effects.
- Two frames hold films of the same liquid at the same temperature. One is rectangular and one triangular, but both meet a straight horizontal slider along 25 cm with two faces. If one supports 0.020 N, find the load supported by the other and explain why their different areas are irrelevant to this comparison.
- For an ideal smooth surface, S_la = 0.050 N/m and S_sa = 0.030 N/m. Calculate the contact angle if S_sl = 0.055 N/m. Explain whether it indicates good or poor wetting and how you chose the angle convention.
- Explain the molecular and energy reasons for (a) the tendency to minimise exposed liquid area, (b) water beading on wax but spreading on clean glass, and (c) paintbrush hairs coming together after emergence from water. Distinguish surface tension from pressure and from viscosity.
Key Takeaways
• Extra interfacial area costs energy because surface molecules have a different environment from bulk molecules. • Surface tension acts tangentially, has units N/m = J/m² and depends on both materials and temperature. • A two-sided film gives F = 2Sl; force depends on contact length rather than total film area. • Measure contact angle through the liquid and use all three interfacial tensions to interpret wetting.