Rub a cake of soap between wet hands, and a white lather grows. The lather is thousands of tiny bubbles, each wrapped in a thin film of soapy water. Stop rubbing, and no more lather appears. A garden sprayer is similar. Pump harder, and the jet breaks into finer droplets. Making a lather or a mist seems to take effort. Where does that effort go?
Work to make a surface
Go back to the U-frame of the last lesson. The film pulls the slider with \(2Tl\). Now draw the slider out slowly by a distance \(\Delta x\), with a force just equal to that pull. The work done is the force times the distance, \(W = 2Tl\,\Delta x\).
The film has grown. Each of its two surfaces has gained an area \(l\,\Delta x\), so the new surface is \(\Delta A = 2l\,\Delta x\). The work is therefore \(W = T\,\Delta A\). The work does not depend on the shape of the frame or the length of the slider. It depends only on how much new surface was made.
\[ F = 2Tl \]
Why. each of the two surfaces of the film pulls on the slider with \(Tl\).
\[ W = F\,\Delta x = 2Tl\,\Delta x \]
Why. the slider is drawn out slowly, with a force equal to the pull, through \(\Delta x\).
\[ \Delta A = 2\,l\,\Delta x \]
Why. each surface grows by the strip \(l\,\Delta x\), and there are two surfaces.
\[ W = T\,\Delta A \]
Read it. substitute \(\Delta A\) for \(2l\,\Delta x\).
The work is stored in the new surface as surface energy. Each square metre of surface holds energy \(T\). So surface tension has two faces. It is the force per unit length across a line in the surface, and it is the energy stored in each unit area of surface. The two units agree: \(1\,\text{N/m} = 1\,\text{N·m/m}^2 = 1\,\text{J/m}^2\).
The work needed to make unit area of new surface, slowly and at a steady temperature, equals the surface tension. So surface tension may be measured in N/m or in J/m².
The dimensions agree too. Force per length is \(\mathrm{MLT^{-2}}/\mathrm{L} = \mathrm{MT^{-2}}\). Energy per area is \(\mathrm{ML^2T^{-2}}/\mathrm{L^2} = \mathrm{MT^{-2}}\), the same. For the film of Fig 1.2, soap solution with a slider \(10\,\text{cm}\) long drawn out \(5\,\text{cm}\): \(\Delta A = 2 \times 0.10 \times 0.05 = 10^{-2}\,\text{m}^2\). The work is \(0.03 \times 10^{-2} = 3 \times 10^{-4}\,\text{J}\), the same as \(2Tl\,\Delta x\).
A surface stretched at a steady temperature also takes in a little heat from its surroundings. Stretched quickly, with no heat coming in, it would cool. So the total energy of a unit area of surface is somewhat larger than \(T\). The work \(T\,\Delta A\) is the part that must be supplied as work, and it is the part that matters in every problem of this chapter.
Which of these is a correct unit for surface tension?
Round mercury, flat puddles
A small drop of mercury on a table is almost a perfect ball. A large pool of it lies flat. The two energies explain it. The surface energy of a drop grows as its area, like \(r^2\). Its gravitational energy grows as its mass times the height of its centre, like \(r^3 \times r = r^4\). For a small drop the surface term wins, and the drop takes the least area, a sphere. For a large drop gravity wins, and the drop spreads to lower its centre.
The crossover comes where the two are comparable, near \(\sqrt{T/\rho g}\). For mercury this is about \(2\,\text{mm}\), and for water about \(3\,\text{mm}\). Drops smaller than this are round; larger ones are flattened.
Show that force per unit length and energy per unit area have the same dimensions. Then explain in words why they are the same quantity for a liquid surface.
Force per length has dimensions \(\text{MLT}^{-2}\) divided by L, which is \(\text{MT}^{-2}\). Energy per area has dimensions \(\text{ML}^2\text{T}^{-2}\) divided by \(\text{L}^2\), which is also \(\text{MT}^{-2}\). Physically, pulling a slider against the pull \(T\) per unit length makes new surface, and the work done, \(T\) times the new area, is stored in the surface. So the pull per metre and the energy per square metre are the same number.
Blowing a bubble
A soap bubble is a thin shell of soap solution with air inside and air outside. So it has two surfaces, an outer one and an inner one, both of radius \(r\) because the film is so thin. Its total surface area is \(2 \times 4\pi r^2 = 8\pi r^2\), and its surface energy is \(8\pi r^2 T\). A liquid drop has only its outer surface, so its surface energy is \(4\pi r^2 T\).
Take \(T = 0.03\,\text{N/m}\) for soap solution. Find the surface energy of a soap bubble of radius \(5\,\text{cm}\). Then find the work needed to blow it from a radius of \(5\,\text{cm}\) to \(10\,\text{cm}\), at a steady temperature.
A bubble has two surfaces, so its area is \(8\pi r^2\). The work to enlarge it is \(T\) times the increase in area.
\(E = 8\pi r^2 T\), and \(W = T \times 8\pi(r_2^2 - r_1^2)\).
\(E\) \(= 8\pi \times 0.05^2 \times 0.03\) \(= 6\pi \times 10^{-4}\) \(\approx 1.9 \times 10^{-3}\,\text{J}\). \(W\) \(= 0.03 \times 8\pi \times (0.01 - 0.0025)\) \(= 1.8\pi \times 10^{-3}\) \(\approx 5.7 \times 10^{-3}\,\text{J}\).
The bubble of \(5\,\text{cm}\) stores about \(1.9\,\text{mJ}\). Doubling its radius needs about \(5.7\,\text{mJ}\) more, three times as much, since the area grows four times.
Find the surface energy of a soap bubble of radius \(2\,\text{cm}\), with \(T = 0.03\,\text{N/m}\). Give the answer in millijoules.
Breaking a drop into many
Break a drop of radius \(R\) into \(n\) equal drops. The water is the same, so the volume is the same: \(n \times \tfrac43\pi r^3 = \tfrac43\pi R^3\). Each small drop has radius \(r = R/n^{1/3}\). Eight drops have half the radius; a thousand have a tenth.
The total area is not kept. It is \(n \times 4\pi r^2 = 4\pi R^2\,n^{1/3}\). So the area grows \(n^{1/3}\) times: twice for 8 drops, ten times for a thousand. Fig 2.1 shows it for a drop of radius \(1\,\text{mm}\). The extra area must be made, and that takes work \(T\,\Delta A\).
\[ n \cdot \tfrac43\pi r^3 = \tfrac43\pi R^3 \;\Rightarrow\; r = R\,n^{-1/3} \]
Why. the volume of water does not change.
\[ A_{\text{after}} = n \cdot 4\pi r^2 = 4\pi R^2\,n^{1/3} \]
Why. substitute \(r = R\,n^{-1/3}\), so \(n\,r^2 = R^2\,n^{1/3}\).
\[ \Delta A = 4\pi R^2\,(n^{1/3} - 1) \]
Why. subtract the area of the one drop we began with.
\[ W = T\,\Delta A = 4\pi R^2 T\,(n^{1/3} - 1) \]
Read it. each square metre of new surface needs work \(T\).
Take a water drop of radius \(2\,\text{mm}\) broken into 8 equal drops. Each has radius \(1\,\text{mm}\). The area grows from \(4\pi \times 2^2 = 16\pi\,\text{mm}^2\) to \(8 \times 4\pi \times 1^2 = 32\pi\,\text{mm}^2\), by \(16\pi\,\text{mm}^2\). The work is \(0.07 \times 16\pi \times 10^{-6} \approx 3.5 \times 10^{-6}\,\text{J}\). If nothing outside supplies it, the energy comes from the heat of the water itself, and the water cools very slightly.
This is why the sprayer of the opening needs more pumping for a finer mist. Every halving of the droplet size doubles the area of water surface to be made.
Drops that merge
Run the process backwards. When many small drops merge into one, surface disappears, and its energy is set free as heat. A thousand droplets of radius \(0.1\,\text{mm}\) make one drop of radius \(1\,\text{mm}\), since \(1000^{1/3} = 10\). The area lost is \(1000 \times 4\pi \times 0.1^2 - 4\pi \times 1^2\) \(= 40\pi - 4\pi\) \(= 36\pi\,\text{mm}^2\). The energy set free is \(0.07 \times 36\pi \times 10^{-6} \approx 7.9 \times 10^{-6}\,\text{J}\), and the drop warms very slightly.
How warm does the drop get? Its mass is \(1000 \times \tfrac43\pi (10^{-3})^3\) \(\approx 4.2 \times 10^{-6}\,\text{kg}\). Water needs about \(4200\,\text{J}\) to warm one kilogram by one degree, as the chapters on heat will show. So the rise is \(7.9 \times 10^{-6} / (4.2 \times 10^{-6} \times 4200)\) \(\approx 0.0005\,^\circ\text{C}\). Real, but far too small to feel.
Bubbles merge for the same reason. Two small bubbles have more surface than one bubble of the same total volume. When they touch and join, surface disappears, and the energy of the foam falls. That is why the tiny bubbles on a freshly poured fizzy drink gather into fewer, larger ones. Lesson 4 follows two bubbles as they join.
Twenty-seven water droplets, each of radius \(1\,\text{mm}\), merge into one drop. Take \(T = 0.07\,\text{N/m}\). How much energy is set free, in microjoules?
A water drop is broken into many tiny droplets inside an insulated box, with no work done from outside. What happens to the temperature of the water?
Benjamin Franklin had read that sailors calmed rough water by pouring oil on it. One windy day at a pond on Clapham Common, near London, he tried it. He poured no more than a teaspoon of oil on the water. It spread rapidly over perhaps half an acre, and the ripples there became as smooth as glass. Franklin reported the trial to the Royal Society in 1774. More than a century later, Lord Rayleigh used such films to estimate that they were only about one molecule thick.
- Counting one surface for a soap bubble.A bubble has two surfaces; its energy is \(8\pi r^2 T\).
- Taking the total area to stay the same when a drop is broken up.The volume stays the same; the area grows as \(n^{1/3}\).
- Taking the radius of each small drop as \(R/n\).Volume goes as \(r^3\), so \(r = R/n^{1/3}\).
- Thinking merging drops takes in energy.Merging loses surface, so it sets energy free as heat.