Why is the pressure inside a bubble greater than the pressure outside?

[aaaa202]

As you may know the pressure inside a bubble of is greater than the pressure outside due to the surface tension. Now in my book the difference is derived by minimizing the Helmholtz energy such that:

dF = - p_inside * dV1 - p_outside * dV2 + σdS, where S is the bubble surface area.

This can then be used to find the correct equation for the difference in pressure. But intuitively I don't understand why. To find the pressure one must balance the different forces and is the above equation doing that? Also why is the last term positive? I.e. the negative signs come from the fact that p_inside is directed in the opposite direction as dV1 and similarly for the second term. When the surface expands does it then do positive work on the system?

 

[BvU]

You are asking some good questions! This one ended up in the homework part of PF, where they want you to use a template. Never mind.

If p_inside>p_outside and all the d's are pointing in the same direction, $\sigma$ is positive, which is nice.

Compare it with blowing up a balloon: the elasticity of the material resists expansion; you have to do work to overcome that. And yes, when expanding, the surface has to push away the surrounding air, which requires work (p_outside * delta V). Inside pressure has to do more work: stretching the surface and pushing away the outside air.

In bubbles the attracting forces between the molecules (van der Waals forces) tend to contract the surface.
In drops you have the same thing: they tend to form spheres because that minimizes surface area.