Two spherical soap bubbles of radii a and a 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*[tex]\gamma[/tex]*r^2=p(a^3+a^3)+4[tex]\gamma[/tex](a^2+a^2) where p is the ambient pressure and [tex]\gamma[/tex] 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[tex]\gamma[/tex] therefore the total force is 2[tex]\gamma[/tex]*circumference (2[tex]\pi[/tex]*r) we then have the pressure inside the bubble acting over the entire cross section which gives total force of p*[tex]\pi[/tex]*r^2. adding the two together, my final equation is p*r^2+4*[tex]\gamma[/tex]*r=p(a^2+a^2)+4[tex]\gamma[/tex](a+a). I have no idea how I did it but my entire equation has been divided by r somehow. Please help. Thanks.