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Obtain an expression for the difference in pressure inside a saop bubble floating in air.
Asked by seeni2005 | 19 Nov, 2021, 12:47: PM
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 | 19 Nov, 2021, 09:01: PM
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