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Apparently, the air inside a soap bubble is under higher pressure than the surrounding air. This is for instance apparent in the sound bubbles make when they burst. Why is the pressure inside the bubble higher in the first place?

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If it wasn't, the bubble would collapse until it was. – immibis Jul 24 '14 at 4:47
Well, the air can't really get out of the bubble, so your answer is certainly not complete. – David Zwicker Jul 25 '14 at 1:50
exactly - because the air can't escape, the collapse would increase the pressure. – immibis Jul 25 '14 at 8:26
Well, but this doesn't really explain why it would contract in the first place. I guess your idea is similar to David's answer, but I think one has to be a little bit more precise to answer the question fully. – David Zwicker Jul 25 '14 at 13:24
If I thought it was a complete answer, I would have posted it as an answer... – immibis Jul 26 '14 at 3:27
up vote 37 down vote accepted

I drew an image to illustrate the forces at play.

bubble section under tension

For any curved surface of the bubble, the tension pulls parallel to the surface. These forces mostly cancel out, but create a net force inward. This compresses the gas inside the bubble, until the pressure inside is large enough to counteract both the outside pressure, as well as this additional force from the surface tension.

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I love this intuitive way of looking at it. It also ties in nicely with David Zwicker's mathematical answer. The larger the radius, the smaller the pressure difference, according to that final equation. In your diagram it's immediately obvious why, since the surface tension will become nearer and nearer to perpendicular to the pressure as the radius increases. – trichoplax Jul 22 '14 at 22:37
Pressure difference causes surface tension OR surface tension causes pressure difference? Also @DavidZwicker – user13107 Jul 23 '14 at 6:57
Quick question: what about the conservation of mass? Is the soap bubble "alright" with having a thicker depth after contraction? If not, is this modeled? Curious – Andres Salas Jul 23 '14 at 18:37
Just from observation, any excess fluid tends to run down a soap bubble due to gravity, so I would suppose the surface tension is approximately constant. I don't know of any detailed models for surface tension in a bubble. – David Jul 24 '14 at 15:31
@user13107 If you imagine a flat bubble film, this has surface tension without any pressure difference. I think it would be more accurate to say that the pressure difference causes the curvature of the bubble surface. – David Jul 24 '14 at 15:39

The increased pressure is caused by the surface tension between the soap and the surrounding air. This can be seen by a simple equilibrium energy argument. The total energy of the system reads $$ E = E_i + E_o + E_s \;, $$ where $E_i$ is the energy associated with the air inside the bubble, $E_s$ is the interfacial energy, and $E_o$ denotes the energy associated with the air outside of the bubble. Importantly, the contribution of the surface energy is given by $E_s = 2 \gamma A$, where $\gamma$ is the surface tension and $A$ is the surface area of the bubble. The factor of 2 emerges, since there are actually two interfaces (one facing the inside of the soap bubble and one facing the outside).

In equilibrium, the total energy will be minimal. We thus analyze the total differential of the energy. Here, the differentials of the partial energies of the air can be approximated by the ideal gas law, which yields $dE_i = -p_i dV_i$ and $dE_o = -p_o dV_o$. Next, we have to discuss the degrees of freedom of the system. Generally, the soap bubble wants to keep its spherical shape to minimize the surface area (and thus the surface energy $E_s$) at a given volume. This leaves us with a single parameter, the radius $R$ of the bubble, which can be varied in any process. The volume differentials then become $dV_1 = 4\pi R^2 dR$ and $dV_2 = -4\pi R^2 dR$. Additionally, the surface area changes by $dA = 8\pi R dR$. The differential of the surface energy thus reads $dE_s = 2\gamma 8\pi R dR$, provided that the surface tension stays constant.

Now we got everything and can express the differential of the total energy as $$ dE = -p_i 4 \pi R^2 dR + p_o 4\pi R^2 dR + 2\gamma 8\pi R dR \;. $$ Dividing by $4\pi R^2$ and noting that $dE/dR$ vanishes in equilibrium, we thus arrive at $$ p_i - p_o = \frac{4\gamma}{R} \;. $$ This expression shows that the pressure inside the bubble is large than outside. The pressure difference is two times the Laplace pressure $2\gamma/R$.

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Understood a few sentences...plausible. I like it. Question: can this be viewed intuitively from the perpective of the fact that in order for a bubble to form a higher-pressure must be applied to the air in front of the bubble; i.e. it must be blown into? Since this high-pressure air is the air that goes into the bubble, shouldn't this air be of this higher pressure for all time, rather like a balloon? I got this idea from someone else so I won't post it as an answer, but it may be right, so I will ask. – Andres Salas Jul 22 '14 at 19:08
It is true, that the ideas employed here are the same that would explain why the pressure inside a balloon is higher than outside. In order to inflate a balloon you want to blow extra air inside it. This is only possible when the pressure you create in your mouth is larger than the pressure in the interior of the balloon, since the air always flows from regions of large to small pressure in this example. Consequently, the fact that your have to blow quite a bit means that there is quite some pressure inside the balloon. – David Zwicker Jul 22 '14 at 19:21
Right, since an elastic balloon always needs a pressure differential. Solid math. Well done. Math always brings a chance of being wrong if it isn't founded on real life. I love your math here though its really well put together. Good work man. – Andres Salas Jul 22 '14 at 19:33
While it is possible to define the surface tension the way you did, it is not immediately obvious that it should make sense. You, implicitly, postulate that $\gamma$ equals the surface tension between soap and air. If I went and carried out experiments with bulk soap it is possible that I might get a different value for $\gamma$ as the thickness of the soap phase in your system is only ~100 nm, and might therefore be subject to interfacial phenomena. I would prefer to use $\gamma' = 2\gamma$, where $\gamma'$ just represents the tension in the film and is agnostic as to the number of interfaces – alarge Aug 8 '14 at 20:45
That is certainly true. Using the effective surface tension, the formula is true no matter how thin the film actually gets. – David Zwicker Aug 8 '14 at 21:27

It is like a balloon. The pressures of the inner and outer air tend to equilibrate, creating a force over the balloon surface from the higher pressure to the lower one, trying to make them equal (the force goes from inside to outside, when you inflate it, from outside to inside when you deflate it). That's why it changes its size, because the gas pressures tend to be the same. At the other hand, the elastic material of the balloon when it is inflated tries to return to its resting state (deflated), so it creates an inward force compressing the inner air. The balloon reaches its balance (it gets inflated at a certain level) when these two forces are equal. That means that the inner air will always be at a higher pressure than the outer air, being this difference related to the balloon material elasticity (the more the material resistance to inflate/deform is, the more the pressure difference will be). In a bubble this resistance to inflate is due to the surface tension (a contractive force that always tries to leave a liquid surface at its minimum; that's why bubbles and droplets are spherical). (I'm not a native english speaker, please be comprehensive with expression)

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The simple answer to "why is the pressure inside a soap bubble higher than outside," is that a higher pressure than the local atmosphere is required to make the bubble in the first place! This requirement comes from the need to counterbalance the surface tension force.

For stable conditions, $$F_i = F_o + F_s $$ Where $F_i$ is the force due to inside pressure, $F_o$ is the force due to outside pressure, and $F_s$ is the force due to surface tension.

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You certainly have the right idea, but I think that the way in which you present it is rather sloppy, since forces are vectorial quantities. Also pressure and surface tensions are not forces, but have different units. – David Zwicker Aug 8 '14 at 21:26

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