derive an expression excess pressure inside a bubble in a liquid
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 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]
which is the required expression
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