Name: Angle of Contact
Uploaded: 2019-07-19
Duration: 6 min 57 s
Description: What is the contact angle? Which factor does the contact angle depend on? Why does the contact angle for the same liquid on different surfaces differ? Let's...

Question

What do you mean by excess pressure? Derive an expression for excess pressure inside a soap bubble.

Accepted Answer

The pressure inside a spherical drop is greater than the pressure outside. The way in which the excess pressure P depends on the radius a of the drop, and the surface tension γ and density ρ of the liquid is amenable to dimensional analysis.

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

$P_{n e t} = P_{i} - P_{s}$

Now, as the radius of the drop has increased, work has been done by the net internal pressure.

So, this work done will be $W = f o r c e x d i s p l a c e m e n t$

or $W = (p r e s s u r e x a r e a) x d i s p l a c e m e n t$

And in this case $W = (p r e s s u r e x a r e a) x i n c r e a s e i n d r o p r a d i u s$

So, the work done will be $W = (P n e t x 4 π r^{2}) x d r . . . . .$ (1)

But, the increase potential energy will be twice $d U = 16 π r . d r x S . . . . . . . . . . . (2)$

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 $(P_{n e t} x 4 π r^{2}) x d r = 16 π r . d r x S$

Or the net pressure inside a bubble will be:

$P_{n e t} = \frac{4 S}{r}$ [twice than the that inside the drop]

What do you mean by excess pressure? Derive an expression for excess pressure inside a soap bubble.

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