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# derive an expression for excess pressure inside a bubble in a liquid

Asked by Shabnakp 12th January 2013, 9:36 PM

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

Answered by Expert 12th January 2013, 9:58 PM
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