Obtain an expression for the difference in pressure inside a saop bubble floating in air.

Suppose a spherical bubble of radius r, floating in  air,  is in equilibrium as shown in figure.

If its radius increase by Δr, then  extra surface energy is

 

 

[4π(r + Δr) 2- 4πr2] T = 8πr Δr  T ...........................(1)

 

 

Where T is surface tension. If the bubble is in equilibrium this energy cost is balanced by the energy gain due to

expansion under the pressure difference (Pi – Po) between the inside of the bubble and the  outside.

 

The work done is
 

 

W = (Pi – Po) 4πr2Δr   ...........................(2)

 

By equating the surface energy given by eqn.(1)  and workdone as in eqn.(2) , we get

 

(Pi – Po) = (2 T ) / r

 
In general, for a liquid-gas interface, the convex side has a higher pressure than the
concave side. In the case of air bubble floating in air , we have two liquid-air interfaces , one inside the bubble and 

another in outside of bubble . 
 

(Pi – Po) = (4 T ) / r

Hence we get , P_i =  P_o + [ (4T) / r ]