obtain an expression for the difference in pressure inside a soap bubble floating in water.

Asked by seeni2005 | 19th Nov, 2021, 12:51: PM

Expert Answer:

 
Suppose a spherical bubble of radius r, floating in water,  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. For example, an air bubble in a liquid, would have higher pressure inside it.
 
 
Hence we get , Pi =  Po + [ (2T) / r ]

Answered by Thiyagarajan K | 19th Nov, 2021, 08:51: PM