Derive an expression for the excess …

Pnet = Pi - Pa Now, as the radius of the drop has increased, work has been done by the net internal pressure. So, this work done will be W = force x displacement or W = (pressure x area) x displacement and in this case W = (pressure x area) x increase in drop radius so, the work done will be W = (P net x 4?r 2 ) x dr (1) but, the increase potential energy will be twice dU = 16?r.dr x S (2) this is because in case of a bubble there are two free surfaces, the outside surface and the inside surface. The potential energy will thus change accordingly. so, as (1) equals (2), we have (Pnet x 4?r2) x dr = 16?r.dr x S or the net pressure inside a bubble will be Pnet = 4S / r [twice than the that inside the drop]