Example: The Laplace Bubble Law


Understanding the situation

In our discussion of surface tension, we have shown that the surface of a liquid pulls on itself. This also works for a taut membrane. Applying the idea of surface tension to a curved membrane gives and interesting result; one that has significant biological implications. In order to get the idea of how this works, consider a simple model: a thin spherical membrane holding in a fluid (gas or liquid) under pressure.

Presenting a sample problem

Suppose a spherical membrane contains a fluid under pressure. (Imagine a balloon filled with air.) If the surface tension in the membrane is $\gamma$, how the the pressure differential across the membrane relate to the size of the bubble? Ignore gravity.

Solving this problem

So here's our picture: We have a spherical bubble containing a fluid. The pressure outside is $p_{outside}$ and the pressure inside is $p_{inside}$. How big is the bubble? If the inside pressure is greater than the outside it is trying to make the bubble bigger. This is balanced by the surface tension in the membrane holding it in. Let's see what the balance tells us.

The bubble is shown in the figure on the left. Now consider the top half of the bubble. What are the forces on it? The inside pressure is greater than the outside, so there is an upward force on the top half of the bubble as shown by the pink arrows. If the arrows were all in the same direction the total pressure force would be $F = Δp A = (p_{inside} - p_{outside}) (2πR^2)$ where $R$ is the radius of the bubble. (Since the surface area of a whole sphere is $4πR^2$ and we only have half the sphere.) But the forces don't point in the same direction. The horizontal components of all the little vectors cancel. Getting this correct requires doing a vector integral*, which we won't do here, but the result is that half of the forces cancel leaving the result of the pressure forces being 

$$F_{up} = \pi R^2 \Delta p$$

The downward force that prevents the top half of the bubble from being blown up by the pressure difference is the surface tension force from the bottom half of the membrane holding it down. By our surface tension rule, the downward force of surface tension will be gamma times the length, or

$$F_{down} = \gamma 2 \pi R$$

Setting the up and down forces equal gives $πR^2 Δp = γ 2πR$ or 

$$Δp = 2γ/R$$

This is what we call the Laplace Bubble Law.

Physical implications

The Bubble Law seems quite unintuitive. It says that if the surface tension remains constant, that a smaller bubble will have a larger pressure inside. Although it's not quite true that as a balloon gets bigger the surface tension doesn't change, it's not too bad an approximation. Take a look at the youtube video at the left. In it is shown two balloons of different sizes connected by an open tube. The open tube has a stopcock that allows the tube to be closed and opened. When the stopcock is opened, the air will flow in the tube from high pressure to low. Which way do you think it will flow? Will the smaller one get bigger to make the balloons more even? Or will the bigger one get even larger? Why? Decide before you watch!

Law of Laplace

Medical implications

The Laplace Bubble Law has medical implications. The heart is like a bubble of muscle creating tension on the fluid inside it (blood). An enlarged heart (bigger $R$) will require more tension (bigger $\gamma$) to create the same pressure differential. As a result, an enlarged heart will have to work harder (more muscular tension) to push the blood around the body.

*Here is a thought experiment showing why the net pressure adds up to $πR^2 Δp$ on the top half of the sphere. Imagine a hemisphere made of iron, not flexible like the bubble. The hemisphere is closed, but hollow. That is, take a hollow sphere, cut it in half, take half, and add a circular sheet of iron to the exposed equator to seal it again. The round part of the hemisphere feels some new force from pressure which seems complicated to calculate. The flat, circular part of the hemisphere, though, feels a force from pressure that is $πR^2 Δp$ because that's simply the area x pressure for that circle. The net force on the entire hemisphere from pressure must be zero because we know that if we sit it on a table, it won't just go flying away. So the net force on the spherical part must also be $πR^2 Δp$, just in the opposite direction as the force on the flat part.

Joe Redish 10/25/16

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Last Modified: February 24, 2019