Chemistry · Lesson
Liquids, viscosity, surface tension and vapour pressure
Explain flow resistance, the energetic cost of extra surface and the exchange between liquid and vapour. Keep temperature and equilibrium conditions attached to every property.
Back to the Chemistry mapBy the end, you can
- Distinguish viscosity, surface tension and vapour pressure.
- Explain how evaporation and condensation can balance while both continue.
- Connect boiling to external pressure rather than one universal temperature.
Viscosity describes resistance to flow
Tilt a container of water and one of a more viscous liquid: the second may flow much more slowly under comparable conditions. Viscosity quantifies resistance to relative motion within a fluid. It depends on the substance, molecular arrangement, interactions and temperature; it is not the same property as density.
For ordinary liquids, viscosity generally decreases on heating. A value needs its temperature, composition and measurement setting. Mixtures such as honey vary with composition, so their everyday names do not specify one fixed viscosity. Gases need their own temperature dependence rather than this liquid generalization.
Creating more surface costs energy
Molecules at an interface have a different environment from molecules in the bulk. Surface tension describes the energetic cost per added area, or equivalently a force per length. Small droplets tend toward shapes with small area, while gravity, contact surfaces and wetting affect the actual shape.
A membrane or skin analogy can help describe resistance to increasing surface area, but the liquid surface is not a separate solid sheet. At a liquid–vapour interface, surface tension depends on temperature and can change with dissolved substances or surface contamination. Read the conditions alongside the value.
| Property | Question it answers | Units or setting | Example or distinction |
|---|---|---|---|
| Dynamic viscosity | How strongly does the fluid resist relative flow? | Pa s; often mPa s for liquids | Water at 25 °C: approximately 0.890 mPa s |
| Surface tension | What energy is needed per added interface area? | J m⁻², equivalently N m⁻¹ | Water at 25 °C: approximately 71.99 mN m⁻¹ |
| Equilibrium vapour pressure | What vapour partial pressure coexists with the liquid at fixed T? | Pressure; both phases present at equilibrium | Different from total pressure when air is also present |
Definitions and one source-tabulated example for water at 25 °C. The vapour-pressure row describes the equilibrium setting without inventing a numerical value.
A closed container can keep exchanging molecules
Below its boiling point, a liquid can still evaporate: some molecules leave its surface. Vapour molecules can also return through condensation. In a closed container at fixed temperature with both liquid and vapour present, the two rates can become equal. This is dynamic equilibrium: exchange continues even though the average amounts stop changing.
The vapour’s equilibrium partial pressure is then the vapour pressure of the liquid at that temperature. For a pure liquid, it is not automatically the total pressure in a container that also holds air. An open container or a system still approaching equilibrium need not have that equilibrium vapour partial pressure.
Boiling depends on the surroundings
Boiling becomes possible when the liquid’s equilibrium vapour pressure matches the external pressure. Increasing temperature generally increases vapour pressure. Lower external pressure can therefore allow boiling at a lower temperature. Ordinary water’s familiar boiling point near 100 °C is approximate and tied to pressure near one atmosphere.
Evaporation below boiling and boiling through the liquid are related changes of phase, with different observable settings. A claim about boiling temperature needs the substance and pressure. Likewise, surface tension and viscosity numbers at 20 °C and 25 °C should not be compared as though temperature were irrelevant.
Worked examples
Express surface tension as energy per area
Convert the tabulated water value 71.99 mN m⁻¹ at 25 °C into J m⁻².
- Convert milli: 71.99 mN m⁻¹ = 0.07199 N m⁻¹.
- Use 1 J = 1 N m, so J m⁻² = N m⁻¹.
- The unit conversion changes the expression, not the physical property.
0.07199 J m⁻² at the stated temperature. This is a tabulated approximation for water, rather than an exact constant for every water sample.
Try it yourself
Are viscosity and density the same property?
Show the answer and reasoning
No.
Density relates mass to volume; viscosity describes resistance to flow.
At liquid–vapour dynamic equilibrium, do evaporation and condensation both stop?
Show the answer and reasoning
No.
Their average rates are equal, so continuing exchange leaves the average amounts unchanged.
With air also present, must water’s vapour pressure equal the container’s total pressure?
Show the answer and reasoning
No.
Water vapour contributes a partial pressure; the other gases contribute as well.
For the same liquid, why can lower external pressure reduce boiling temperature?
Show the answer and reasoning
The required vapour pressure is lower.
The liquid reaches that matching vapour pressure at a lower temperature.
What to carry forward
- Viscosity, surface tension and vapour pressure describe different properties.
- State temperature and composition with property values.
- Dynamic equilibrium keeps both directions active.
- Boiling temperature depends on external pressure.
References
Teaching resources and reference material supporting this lesson’s explanations and examples.