PRESSURE INSIDE A BUBBLE :
Consider the figure below
we know that as the drop/bubble remains in an equilibrium condition work done = increase in potential energy If we consider the atmospheric pressure to be 'Ps' and the internal pressure to be 'Pi'
then the net pressure would be written as
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 4Ïr 2 ) x dr ........(1)
but, the increase potential energy will be twice dU = 16Ïr.dr x S ...........(2) t
his 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 = 16Ïr.dr x S
or the net pressure inside a bubble will be:
Pnet = 4S / r [twice than the that inside the drop] which is the required expression