surface tension capillarity and Laplace pressure

Molecules at the surface of a liquid have fewer neighbours than in the bulk, so their potential energy is higher, and the liquid tends to reduce the area of its free surface. This is described quantitatively by the surface tension $\sigma$: the force acting along a contour is $F = \sigma L$, and the surface energy is $E = \sigma S$.

A curved surface creates a pressure jump — the Laplace pressure: for a general surface $\Delta p = \sigma\left(\dfrac{1}{R_1} + \dfrac{1}{R_2}\right)$, for a spherical drop $\Delta p = \dfrac{2\sigma}{R}$, and for a soap bubble with two surfaces $\Delta p = \dfrac{4\sigma}{R}$.

In a capillary of radius $r$ this pressure jump is balanced by the hydrostatic pressure of a liquid column of height $h$, which gives Jurin's law: $h = \dfrac{2\sigma\cos\theta}{\rho g r}$, where $\theta$ is the contact angle. For $\theta < 90^\circ$ the liquid rises, for $\theta > 90^\circ$ it is depressed; the smaller the capillary radius, the greater the rise height.

A wire frame with a movable bar of length $L = 5{.}0$ cm is covered by a soap film. The surface tension of the soap solution is $\sigma = 0{.}025\ \mathrm{N/m}$. Find the surface-tension force acting on the bar.

Water rises in a glass capillary of radius $r = 0{.}25$ mm. Assuming complete wetting ($\theta = 0^\circ$) and taking $\sigma = 0{.}072\ \mathrm{N/m}$, $\rho = 1000\ \mathrm{kg/m^3}$, $g = 9{.}81\ \mathrm{m/s^2}$, find the rise height $h$.

A glass capillary of radius $r = 0{.}50$ mm is placed in mercury. Given $\sigma = 0{.}465\ \mathrm{N/m}$, $\rho = 13600\ \mathrm{kg/m^3}$, $\theta = 140^\circ$, $g = 9{.}81\ \mathrm{m/s^2}$, determine whether the mercury rises or is depressed, by how much, and explain the sign of the result using the Laplace pressure.

Self-study activity. Build a capillary-rise model in a spreadsheet. Columns: $r$ (from $0{.}05$ to $1{.}0$ mm), $\sigma$, $\theta$, $\rho$, $g = 9{.}81\ \mathrm{m/s^2}$, and Jurin's formula for $h$. Plot $h$ versus $1/r$ and verify that the graph is linear; obtain $\dfrac{2\sigma\cos\theta}{\rho g}$ from the slope. Repeat the calculation for mercury ($\theta = 140^\circ$) and explain the change of sign. As a safe addition, observe water rising in a strip of paper towel dipped in water and measure the height with a ruler. Do not use boiling or hot liquids; perform any experiment with mercury only in a simulation.

To show the link between surface curvature and Laplace pressure, and to explain how its balance with hydrostatic pressure produces capillary rise or depression.

Diagram: a vertical capillary of radius $r$ in a container of liquid; it shows the meniscus, the contact angle $\theta$, the height $h$ from the container level to the bottom of the meniscus, and arrows for the pressures $p_{\mathrm{atm}}$ and $p_{\mathrm{atm}} - \Delta p$. Beside it, a graph of $h$ versus $1/r$: a straight line with slope $\dfrac{2\sigma\cos\theta}{\rho g}$ that goes into the negative region for $\theta > 90^\circ$.

A $200\,\text{mL}$ spill of water on a counter. A kitchen paper towel is built from cellulose fibres with pores of radius about $r = 3.0\times10^{-5}\,\text{m}$. Water wets cellulose well, contact angle $\theta \approx 20^\circ$, with $\gamma = 0.072\,\text{N/m}$ and $\rho = 998\,\text{kg/m}^3$. The column climbs to

$$h = \frac{2\gamma\cos\theta}{\rho g r} = \frac{2(0.072)(0.94)}{(998)(9.81)(3.0\times10^{-5})} \approx 0.46\,\text{m},$$

i.e. about $46\,\text{cm}$ — taller than the sheet itself, so the towel saturates through its thickness instead of only getting damp on top.

This is capillarity in a narrow pore. Surface tension acts along the meniscus perimeter with $F = 2\pi r\,\gamma\cos\theta$, lifting the column against its weight $\rho g h\,\pi r^2$. Setting those two expressions equal gives Jurin's law. The $\propto 1/r$ dependence explains why tighter pores wick water higher: in a towel with $r = 3\,\mu\text{m}$ the same formula would give $h \approx 4.6\,\text{m}$.

The speaker and microphone openings of a smartphone are sealed by a mesh with pores of radius $r = 1.5\times10^{-5}\,\text{m}$ and a hydrophobic coating. For water $\gamma = 0.072\,\text{N/m}$, so a pressure jump appears at each pore mouth:

$$\Delta p = \frac{2\gamma}{r} = \frac{2(0.072)}{1.5\times10^{-5}} \approx 9.6\times10^{3}\,\text{Pa}.$$

That pressure matches a water column of $h = \dfrac{\Delta p}{\rho g} = \dfrac{9600}{(1000)(9.81)} \approx 0.98\,\text{m}$. A raindrop or a splash cannot produce that jump, so water stays out of the case.

This is Laplace pressure at a curved meniscus: the pressure difference between water and air is $\Delta p = 2\gamma/r$ for a spherical interface. The formula also gives the limit of the protection: drop the phone deeper than about $1\,\text{m}$ and the hydrostatic pressure $\rho g h$ exceeds the Laplace barrier, so the mesh starts to leak. The hydrophobic coating with $\theta > 90^\circ$ raises the breakthrough threshold further, because the meniscus then bulges towards the water side.

A cyclist rides through rain at $8\,\text{m/s}$; on a water-repellent jacket the drops bead up into spheres of radius $r = 0.5\,\text{mm} = 5.0\times10^{-4}\,\text{m}$. Inside such a bead

$$\Delta p = \frac{2\gamma}{r} = \frac{2(0.072)}{5.0\times10^{-4}} \approx 2.9\times10^{2}\,\text{Pa},$$

less than $0.3\,\%$ of the atmospheric pressure $1.01\times10^{5}\,\text{Pa}$. Shrink the same drop to a fog droplet of $r = 1\,\mu\text{m} = 1.0\times10^{-6}\,\text{m}$, however, and $\Delta p = 2(0.072)/(1.0\times10^{-6}) \approx 1.4\times10^{5}\,\text{Pa}$ — more than one atmosphere.

This is the law $\Delta p = 2\gamma/r$: the smaller the radius of curvature, the harder surface tension squeezes the droplet contents. At $r = 50\,\text{nm}$ the pressure inside reaches $2.9\times10^{6}\,\text{Pa}$ — about $29\,\text{atm}$. That pressure rise also raises the equilibrium vapour pressure over small droplets (the Kelvin effect), which is why fog droplets do not nucleate in clean air without supersaturation or dust condensation nuclei.