Derive Laplace’s law for spherical membrane of bubble due to surface tension.
Derive Laplace’s law for a spherical membrane.
Consider a spherical liquid drop and let the outside pressure be Po and inside pressure be Pi, such that the excess pressure is Pi − Po.
Let the radius of the drop increase from r Δto r, where Δr is very small, so that the pressure inside the drop remains almost constant.
Initial surface area (A1) = 4Πr2
Final surface area (A2) = 4Π(r + Δr)2
= 4π(r2 + 2rΔr + Δr2)
= 4Πr2 + 8ΠrΔr + 4ΠΔr2
As Δr is very small, Δr2 is neglected (i.e. 4πΔr2≅0)
Increase in surface area (dA) =A2 - A1= 4Πr2 + 8ΠrΔr - 4Πr2
Increase in surface area (dA) =8ΠΔr
Work done to increase the surface area 8ΠrΔr is extra energy.
This work done is equal to the product of the force and the distance Δr.
dF=(P1 - P0)4πr2
The increase in the radius of the bubble is Δr.
dW=dFΔr= (P1 - P0)4Πr2*Δr ..........(Equ.2)
Comparing Equations 1 and 2, we get
(P1 - P0)4πr2*Δr=T*8πrΔr
∴`(P_1 - P_0) = (2T)/R`
This is called the Laplace’s law of spherical membrane.
Let us consider a liquid drop which is spherical in shape with surface area
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