# Pressure difference inside a soap bubble when taking the limit $R \to 0$

The force exerted by a fluid on a surface normal to the fluid is:

$\pi R^2 \Delta p$ (1)

and the surface tension is:

$2\pi R \gamma$ (2)

and a sphere has two surfaces so, roughly:

$\Delta p = \frac{4 \gamma}{R}$

but if we take the limit:

$\lim_{R \to 0} \frac{4 \gamma}{R}$,

then obviously the pressure difference between the outside and inside surface of the bubble is $\infty$! How does that make sense? Is there any way I can get round this? Or have I done something wrong with my algebra?

• You have done something wrong with your algebra. – Chet Miller Feb 15 '17 at 17:36
• @ChesterMiller Yeah I assume so but I don't get what logical error I made! Everything makes sense to me :( – Lewis Feb 15 '17 at 17:38
• Why are you taking limit? You cannot have a soap bubble with zero radius. – Yashas Feb 15 '17 at 17:45
• @YashasSamaga Well it doesn't have to be zero but a radius of 0.000.....1 will produce a large value for pressure which doesn't make sense to me. – Lewis Feb 15 '17 at 17:52
• The limit of 1/x as x becomes infinite is zero, not infinity. – Chet Miller Feb 15 '17 at 18:53

Your equation is the Young-Laplace equation applied to a bubble in equilibrium.

Considering a non-equilibrium state in which $\frac{dR}{dt} < 0$ where $R$ is the radius of the bubble, the subsequent increase in pressure (as described by your equilibrium equation) will drive molecular diffusion of the gases inside the bubble to the outside of it (which can be thought of as acting to reduce the pressure in response to this increase). However, when enough gasses inside the bubble have diffused outward, the bubble collapses never really producing a bubble of near zero radius.