#### 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 P_{o} and inside pressure be P_{i}, such that the excess pressure is P_{i} − P_{o}.

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 (A_{1}) = 4Πr^{2}

Final surface area (A_{2}) = 4Π(r + Δr)^{2}

^{ }= 4π(r^{2 }+ 2rΔr + Δr^{2})

= 4Πr^{2} + 8ΠrΔr + 4ΠΔr^{2}

As Δr is very small, Δr2 is neglected (i.e. 4πΔr^{2}≅0)

Increase in surface area (dA) =A_{2} - A_{1}= 4Πr^{2} + 8ΠrΔr - 4Πr^{2}

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=(P_{1} - P_{0})4πr^{2}

The increase in the radius of the bubble is Δr.

dW=dFΔr= (P_{1} - P_{0})4Πr^{2}*Δr ..........(Equ.2)

Comparing Equations 1 and 2, we get

(P_{1} - P_{0})4πr^{2}*Δr=T*8πrΔr

∴(P_{1} - P_{0})=2T/r

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