Derive an expression for excess pressure inside a drop of liquid.

#### Solution

Consider a liquid drop of radius R and surface tension T.

Due to surface tension, the molecules on the surface film experience the net force in the inward direction normal to the surface.

Therefore there is more pressure inside than outside.

Let p_{1} be the pressure inside the liquid drop and p_{o} be the pressure outside the drop.

Therefore excess of pressure inside the liquid drop is,

p = p_{1}– p_{0}

Due to excess pressure inside the liquid drop the free surface of the drop will experience the net force in outward direction due to which the drop will expand.

Let the free surface displace by dR under isothermal conditions.

Therefore excess of pressure does the work in displacing the surface and that work will be stored in the form of potential energy.

The work done by an excess of pressure in displacing the surface is,

dW = Force x displacement

= (Excess of pressure x surface area) x displacement of the surface

= p × 4 πR^{2} × dR ....(1)

Increase in the potential energy is,

dU = surface tension x increase in area of the free surface

= T[4π(R + dR)^{2} - 4πR^{2}]

= T[4π (2RdR)] ....(2)

From (1) and (2)

p × 4 πR^{2} × dR = T[4π (2RdR)]

`=> "p" = "2T"/"R"`

The above expression gives us the pressure inside a liquid drop.