# Pressure of a soap bubble

## Homework Statement

Two spherical soap bubbles of radii a[1] and a[2] are made to coalesce. Show that when the temperature of the gas in the resulting soap bubble has returned to its initial values, r of the bubble is given by p*r^3+4*$$\gamma$$*r^2=p(a[1]^3+a[2]^3)+4$$\gamma$$(a[1]^2+a[2]^2) where p is the ambient pressure and $$\gamma$$ is the surface tension.

## The Attempt at a Solution

I cut the bubble in half and found that the surface tension along the circumference is 2$$\gamma$$ therefore the total force is 2$$\gamma$$*circumference (2$$\pi$$*r) we then have the pressure inside the bubble acting over the entire cross section which gives total force of p*$$\pi$$*r^2. adding the two together, my final equation is p*r^2+4*$$\gamma$$*r=p(a[1]^2+a[2]^2)+4$$\gamma$$(a[1]+a[2]). I have no idea how I did it but my entire equation has been divided by r somehow. Please help. Thanks.