Surface tension and capillary rise

Why a surface pulls

A molecule deep inside a liquid is pulled equally in all directions by its neighbours. The pulls cancel. A molecule at the surface has neighbours below and beside it, but none above. So it is pulled inwards and sideways. All surface molecules are pulled into the liquid, and the surface tries to get as small as it can. For a given volume, the shape with the smallest surface is a sphere. This is why a drop or a bubble is round.

To make the surface larger, molecules must be brought from the inside to the surface, and this needs work. The surface of a liquid behaves as if it were a stretched elastic skin. Insects such as the pond skater walk on water, and a careful hand can float a steel needle on it.

Surface tension

The surface tension $\gamma$ is the force per unit length that acts along the surface, at right angles to a line drawn in it. It is also the energy needed to make a unit area of new surface. The two definitions give the same unit:

because $1\ \text{J} = 1\ \text{N m}$. The work to make a new area $\Delta A$ is $W = \gamma\,\Delta A$. In older books, $\gamma$ is in dyn/cm, and $1\ \text{dyn/cm} = 1\ \text{mN/m}$. Here are the values at about $20\ ^\circ\text{C}$:

Strong forces between molecules give a high surface tension. Water has hydrogen bonds. Mercury has metallic bonds. Hexane has only weak dispersion forces. Surface tension falls as the temperature rises, since the molecules move more and attract each other less. It becomes zero at the critical temperature. Soap and detergent molecules gather at the surface and lower the surface tension of water, which helps water to spread and wet fabrics.

Pressure inside a drop and a bubble

The inward pull makes the pressure inside a curved surface higher than outside. Let a drop of radius $r$ grow by a small amount $dr$. The work done against the extra pressure is $\Delta P \times 4\pi r^2\,dr$. The energy stored in the extra surface is $\gamma \times 8\pi r\,dr$. Setting them equal gives

for a drop, and for an air bubble in a liquid. A soap bubble has a thin film with two surfaces, an outer one and an inner one. Each adds $\dfrac{2\gamma}{r}$, so

A water drop of radius $0.50\ \text{mm}$ has an excess pressure of $\dfrac{2 \times 0.0728}{0.50 \times 10^{-3}} = 291\ \text{Pa}$. A soap bubble of radius $2.0\ \text{cm}$ and $\gamma = 25\ \text{mN/m}$ has $\dfrac{4 \times 0.025}{0.020} = 5.0\ \text{Pa}$. The smaller the drop, the greater the pressure inside.

Contact angle

Where a liquid meets a solid, there is a pull between the molecules of the liquid and those of the solid, adhesion, and a pull between the liquid molecules themselves, cohesion. The angle $\theta$ between the solid surface and the liquid surface, measured through the liquid, is the contact angle.

• If adhesion is stronger than cohesion, the liquid wets the solid. The angle is less than $90^\circ$. Water on clean glass has $\theta \approx 0^\circ$.
• If cohesion is stronger, the liquid does not wet the solid. The angle is more than $90^\circ$. Mercury on glass has $\theta \approx 140^\circ$.

Capillary rise

Put a narrow tube of radius $r$ in a liquid. The liquid surface meets the wall along a circle of length $2\pi r$. Along it the surface pulls the liquid upwards with a force $\gamma \cos\theta$ for each unit of length. The upward force is $2\pi r\gamma\cos\theta$. The weight of the column that is raised is $\rho\,\pi r^2 h g$. At balance, the two are equal:

We took the density of the vapour as nothing against that of the liquid. A positive $h$ is a rise. For $\theta > 90^\circ$ the cosine is negative, so $h$ is negative, and the liquid is pushed down below the outside level.

For water in a clean tube of radius $0.30\ \text{mm}$, with $\gamma = 0.0728\ \text{N/m}$, $\rho = 998\ \text{kg/m}^3$ and $\theta = 0^\circ$:

For mercury in the same tube, $\gamma = 0.486\ \text{N/m}$, $\rho = 13\,534\ \text{kg/m}^3$ and $\theta = 140^\circ$. The result is $h = -18.7\ \text{mm}$, a fall. The rise grows as the tube gets narrower, in proportion to $\dfrac{1}{r}$. This is why water climbs through the fine channels of soil and paper, and why a lamp wick draws up oil.

Start with $\gamma = 73\ \text{mN/m}$, $\theta = 0^\circ$ and $\rho = 1000\ \text{kg/m}^3$. Halve the radius, and the height doubles. Now set $\theta = 140^\circ$ and $\rho = 13\,600\ \text{kg/m}^3$ for mercury.

Measuring the surface tension

• Capillary rise. Measure $h$ in a tube of known radius, and use $\gamma = \dfrac{\rho g r h}{2\cos\theta}$.
• Drop count. A stalagmometer is a glass tube that lets a liquid fall in drops. A drop falls when its weight equals the pull of the surface, $2\pi r\gamma$, so the mass of a drop is proportional to $\gamma$. For the same volume of liquid, the mass of one drop is $\dfrac{\rho V}{n}$, where $n$ is the number of drops. So $\gamma \propto \dfrac{\rho}{n}$. If $n_1$ drops of liquid 1 and $n_2$ drops of liquid 2 have the same volume, then

A stalagmometer gives $42$ drops of water and $98$ drops of a liquid of density $789\ \text{kg/m}^3$. The surface tension of the liquid is $72.8 \times \dfrac{42}{98} \times \dfrac{789}{998} = 24.7\ \text{mN/m}$.

Practise with new numbers

Try it in Python

The program finds the rise or fall of water, ethanol and mercury in a tube of radius $0.30\ \text{mm}$.

Common mistakes

• Using the diameter in $h = \dfrac{2\gamma\cos\theta}{\rho g r}$. The formula has the radius.
• Mixing units. $\gamma$ in mN/m must be changed to N/m, and the radius from mm to m.
• Forgetting that $\cos\theta$ is negative for mercury. The liquid falls, and the answer is negative.
• Using $\dfrac{2\gamma}{r}$ for a soap bubble. A soap bubble has two surfaces, so use $\dfrac{4\gamma}{r}$.
• Saying that surface tension is a force on the whole surface. It is a force per unit length, or an energy per unit area.

Remember these

• $\gamma$ is a force per length (N/m) and an energy per area (J/m²). $1\ \text{dyn/cm} = 1\ \text{mN/m}$.
• Strong forces between the molecules mean a high surface tension. A higher temperature means a lower one.
• Excess pressure: $\dfrac{2\gamma}{r}$ in a drop or an air bubble, and $\dfrac{4\gamma}{r}$ in a soap bubble.
• Capillary rise: $h = \dfrac{2\gamma\cos\theta}{\rho g r}$. For $\theta > 90^\circ$, the liquid falls.
• Drop count: $\dfrac{\gamma_1}{\gamma_2} = \dfrac{\rho_1 n_2}{\rho_2 n_1}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Show that the unit N/m is the same as J/m².

2. Why does mercury fall in a glass tube, while water rises?

3. What is the excess pressure inside a water drop of radius $0.25\ \text{mm}$? Take $\gamma = 72.8\ \text{mN/m}$.

4. Which has the higher excess pressure, a soap bubble or an air bubble in water, if both have the same radius and the same surface tension?

5. Water rises $30\ \text{mm}$ in one tube. How high does it rise in a tube with half the radius?

References

• OpenStax, Chemistry 2e, Chapter 10, section 10.2 (surface tension, capillary action).
• The surface tensions and contact angles are standard values from the CRC Handbook of Chemistry and Physics.