At a mela, a child dips a plastic ring into a cup of soap solution and blows. At first nothing seems to happen: the film only bulges, and the child must blow firmly. Then, quite suddenly, a bubble swells and floats away, and gentle blowing is enough to make it grow. A tiny bubble resists being blown; a big one does not.
A surface that squeezes
A stretched surface that is curved pulls towards the inside of its curve. Think of a balloon: its stretched rubber squeezes the air inside, so the air inside is at a higher pressure than the air outside. The surface of a drop does the same. So the pressure inside a drop, \(P_{\text{in}}\), is higher than the pressure outside, \(P_{\text{out}}\). The difference \(\Delta P = P_{\text{in}} - P_{\text{out}}\) is called the excess pressure.
The excess pressure from energy
Let the radius of a drop grow by a small \(dr\), slowly, at a steady temperature. The excess pressure pushes the surface out over an area \(4\pi r^2\) through a distance \(dr\). It does work \(\Delta P \cdot 4\pi r^2\,dr\). That work goes into the new surface. The area grows from \(4\pi r^2\) to \(4\pi (r + dr)^2\), by \(8\pi r\,dr\), which needs \(T \cdot 8\pi r\,dr\).
\[ dW = \Delta P \cdot 4\pi r^2\,dr \]
Why. the net push of the pressures on the surface is \(\Delta P \times 4\pi r^2\), and the surface moves out by \(dr\).
\[ dA = 4\pi (r + dr)^2 - 4\pi r^2 \approx 8\pi r\,dr \]
Why. expand the square and drop the tiny \((dr)^2\) term.
\[ \Delta P \cdot 4\pi r^2\,dr = T \cdot 8\pi r\,dr \]
Why. the work done equals \(T\) times the new area.
\[ \Delta P = \dfrac{2T}{r} \]
Read it. cancel \(4\pi r\,dr\) from both sides.
The radius sits in the denominator. So the smaller the drop, the larger the step in pressure. A water drop of radius \(1\,\text{mm}\) has \(\Delta P = 2 \times 0.07/10^{-3} = 140\,\text{Pa}\). At \(0.1\,\text{mm}\) the step is ten times more, \(1400\,\text{Pa}\). A droplet of mist with a radius of \(1\,\mu\text{m}\) has \(1.4 \times 10^5\,\text{Pa}\), more than a whole atmosphere. Fig 3.1 shows the step growing as a drop shrinks.
The graph in Fig 3.1 is a hyperbola: halve the radius, and the excess pressure doubles. It stays gentle for large drops, then climbs steeply as the drop becomes tiny. This one curve explains most of this lesson and the next.
The same result from forces
Cut the drop in two with a plane through its centre, and look at the right half alone, as in Fig 3.2. Two things act on it across the cut. The left half pushes on its flat face. The excess of that push over the push of the outside air is \(\Delta P\) times the area of the face, \(\Delta P\,\pi r^2\). The surface of the left half pulls back on the rim of the right half, a circle of length \(2\pi r\), with a total pull \(2\pi r T\). The half drop is at rest, so the two balance: \(\Delta P\,\pi r^2 = 2\pi r T\), and again \(\Delta P = 2T/r\).
Find the excess pressure inside a water drop of radius \(0.5\,\text{mm}\). Take \(T = 0.07\,\text{N/m}\).
Two water drops have radii \(1\,\text{mm}\) and \(2\,\text{mm}\). Which has the higher pressure inside?
An air bubble inside water
A small air bubble inside water has one surface too, the boundary between the air and the water. Its surface is curved the same way, towards the air inside. So the air in the bubble is at a higher pressure than the water just outside, again by \(2T/r\). An air bubble of radius \(1\,\text{mm}\) in water has an excess pressure of \(140\,\text{Pa}\), the same as a water drop of that size in air.
Two surfaces, twice the step
A soap bubble is a thin shell of soap solution with air inside and outside. Crossing it from outside to inside, you cross two surfaces: first into the film, then out of the film into the air inside. Each surface is curved the same way, with the same radius \(r\), because the film is so thin. Each adds a step of \(2T/r\). So the pressure inside a soap bubble exceeds the pressure outside by \(4T/r\).
Soap solution has a small surface tension, \(0.03\,\text{N/m}\), and a bubble is large. A soap bubble of radius \(1\,\text{cm}\) has \(\Delta P = 4 \times 0.03/0.01 = 12\,\text{Pa}\). That is barely a ten-thousandth of an atmosphere. Compare a water drop and a soap bubble of the same radius: \(2 \times 0.07 = 0.14\) against \(4 \times 0.03 = 0.12\). The drop has the larger step, in the ratio \(7 : 6\).
A water drop (\(T = 0.07\,\text{N/m}\)) and a soap bubble (\(T = 0.03\,\text{N/m}\)) have the same radius. Find the ratio of the excess pressure in the drop to that in the bubble.
The first puff at the mela
Now the child with the ring. The film across the ring starts flat. A flat surface has no curvature, so it needs no excess pressure. As the child blows, the film bulges and its radius of curvature shrinks. The smallest radius it can have is the radius of the ring itself, when the film is a hemisphere. There the excess pressure is largest. Blow past that point, and the bubble's radius grows, so the pressure it needs falls. For a ring of radius \(1\,\text{cm}\), the hardest moment needs \(12\,\text{Pa}\); after that, less and less.
The air inside a soap bubble (\(T = 0.03\,\text{N/m}\)) is \(24\,\text{Pa}\) above the air outside. Find the radius of the bubble, in centimetres.
A bubble deep in water
An air bubble sits at a depth \(h\) below the surface of water in an open tank. The water just outside the bubble is at \(P_0 + \rho g h\), as the chapter on pressure in fluids showed. Here \(P_0\) is the pressure of the air above the surface. The air in the bubble is higher still, by the step across its one surface, \(2T/r\).
An air bubble of radius \(0.1\,\text{mm}\) is \(10\,\text{cm}\) below the surface of water. Atmospheric pressure is \(10^5\,\text{Pa}\), \(g = 10\,\text{m/s}^2\) and \(T = 0.07\,\text{N/m}\). Find the pressure of the air inside the bubble.
Two steps add to the atmosphere: the water above the bubble, and the curved surface of the bubble.
\(P_{\text{in}} = P_0 + \rho g h + \dfrac{2T}{r}\).
\(\rho g h = 1000 \times 10 \times 0.10 = 1000\,\text{Pa}\), and \(2T/r = 2 \times 0.07/10^{-4} = 1400\,\text{Pa}\). So \(P_{\text{in}}\) \(= 100\,000 + 1000 + 1400\) \(= 102\,400\,\text{Pa}\).
The air in the bubble is at \(1.024 \times 10^5\,\text{Pa}\). For so small a bubble, the surface adds more than the \(10\,\text{cm}\) of water above it.
An air bubble of radius \(0.2\,\text{mm}\) is \(50\,\text{cm}\) below the surface of a pond. Take \(P_0 = 10^5\,\text{Pa}\), \(g = 10\,\text{m/s}^2\) and \(T = 0.07\,\text{N/m}\). Find the pressure of the air in the bubble, in pascals.
Use \(2T/r\) for an air bubble inside a liquid, not \(4T/r\). It has only one surface, the boundary between its air and the liquid. The \(4T/r\) belongs to a soap bubble, a film with air on both sides.
The air we breathe ends up in about 500 million tiny sacs in the lungs. Each is around \(0.1\,\text{mm}\) in radius and lined with a thin layer of liquid. With a surface tension like that of water, each sac would need an excess pressure of about \(1400\,\text{Pa}\) to stay open. The small ones would also empty into the large ones. The lining contains a natural soap-like substance that lowers its surface tension about ten times. Babies born very early often lack it, and doctors give it to them to help them breathe.
In 1805 the English physician Thomas Young showed in words that a curved liquid surface must have a higher pressure on its concave side. A year later the French mathematician Pierre-Simon Laplace, famous for his work on the motions of the planets, gave the result its full mathematical form. For a surface curved differently in two directions, the step is \(T(1/R_1 + 1/R_2)\), and for a sphere this is \(2T/r\). The law is still called the Young-Laplace equation.
- Using \(4T/r\) for a drop or an air bubble in a liquid.Count the surfaces. A drop or an air bubble has one surface, \(2T/r\); a soap film bubble has two, \(4T/r\).
- Thinking a bigger drop has more pressure inside.\(\Delta P = 2T/r\): the smaller the drop, the larger the step.
- Leaving out \(\rho g h\) or \(P_0\) for a bubble under water.Start from the atmosphere, add the water above, then add the surface step.
- Using the diameter in place of the radius.The formulas use the radius \(r\).