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

Consider a liquid drop of radius R arid surface tension T. Due to surface tension the molecules on the surface film experience the net force in inward direction normal to the surface.

Therefore there is more pressure inside than outside. Let p_i 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 = p₁– p₀

Due to excess of 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 excess of pressure in displacing the surface is, 

dW  = Force x displacement
       = (Excess of pressure x surface area) x displacement of surface
       
       
Increase in the  potential energy is,
dU = surface tension x increase in area of the free surface