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

Asked by seeni2005 | 19th Nov, 2021, 12:47: PM

Expert Answer:

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 , Pi =  Po + [ (4T) / r ]

Answered by Thiyagarajan K | 19th Nov, 2021, 09:01: PM