# Excess Pressure inside a soap bubble

The pressure inside a soap bubble and outside it are not identical due to the surface of tension of the soap bubble.

To calculate this pressure difference, let’s first consider an air bubble inside a liquid. If the pressure difference is ΔP , then the work done to increase the radius of bubble from r to ( r + Δr) is given by:

W = F Δr = (4 πr2 )ΔP . Δr

while the change in area,

ΔS = 4π (r + Δr)2 – 4πr2

= 8 πr Δr

From the definition of surface tension

$\displaystyle T = \frac{W}{\Delta S}$

$\displaystyle T = \frac{4\pi r^2 \Delta P \Delta r }{8\pi r \Delta r}$

$\displaystyle \Delta P = \frac{2 T}{r}$

For a soap bubble in air, there are two surfaces, and so,

$\displaystyle \Delta P = 2 (\frac{2 T}{r}) = \frac{4T}{r}$

Exercise : Two soap bubbles of different radii (r, R : r < R) are connected by means of a tube. What will happen to the larger bubble ?