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

Questions

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

Derive Laplace’s law for a spherical membrane.

Solution

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=TdA

∴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

∴(P1 - P0)=2T/r

This is called the Laplace’s law of spherical membrane.

Is there an error in this question or solution?

APPEARS IN

2017-2018 (March)
Question 3.1 | 3 marks
2015-2016 (March)
Question 4.3 | 3 marks

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Solution for question: Derive Laplace’S Law for Spherical Membrane of Bubble Due to Surface Tension. concept: Surface Tension. For the courses HSC Science (Computer Science), HSC Science (Electronics), HSC Science (General)
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