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Derive Laplace’S Law for Spherical Membrane of Bubble Due to Surface Tension. - Physics


Derive Laplace’s law for spherical membrane of bubble due to surface tension.

Derive Laplace’s law for a spherical membrane.

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Solution 1

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(r+ 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.


∴dW=T*8πrΔr         .......(Equ.1)

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.

Solution 2

Let us consider a liquid drop which is spherical in shape with surface area 

We know that due surface tension, liquids try to expose the minimum surface area to the air. Hence they have a tendency to contract.
Due to this contracting force, the inside pressure is greater than the outside pressure.
Let us assume that due to this excess pressure from inside, the drop is expanding and the bigger surface area becomes 

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