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

Derive Laplace’s law for a spherical membrane.

#### Solution 1

Consider a spherical liquid drop and let the outside pressure be Po 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 to Δr, 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 by 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 Laplace’s law of spherical membrane.

#### Solution 2

- The free surface of drops or bubbles is spherical in shape.

Let,

P_{i}= inside pressure of a drop or air bubble

P_{o}= outside pressure of the bubble

r = radius of drop or bubble. - As drop is spherical, P
_{i}> P_{o }

∴ excess pressure inside drop = P_{i }− P_{o } - Let the radius of drop increase from r to r + ∆r so that inside pressure remains constant.
- Initial area of drop A
_{1}= 4πr^{2},

Final surface area of drop A_{2}= 4π (r + ∆r)^{2 }

Increase in surface area,

∆A = A_{2}− A_{1}= 4π[(r + ∆r)^{2}− r^{2}]

= 4π[r^{2}+ 2r∆r + ∆r^{2}− r^{2}]

= 8πr∆r + 4π∆r^{2} - As ∆r is very small, the term containing ∆r
^{2}can be neglected.

∴ dA = 8πr∆r - Work is done by a force of surface tension,

dW = TdA = (8πr∆r)T ….(1)

This work done is also equal to the product of the force F which causes an increase in the area of the bubble and the displacement Δr which is the increase in the radius of the bubble.

∴ dW = FΔr

The excess force is given by,

(Excess pressure) × (Surface area)

∴ F = (P_{i}– P_{o}) × 4πr^{2 }

∴ dF = (P_{i}– P_{o})A

dW = F∆r = (P_{i}− P_{o}) A∆r

From equation (1),

(P_{i}− P_{o}) A∆r = (8πr∆r) T

∴ P_{i}− P_{o}= `(8pirDeltarT)/(4pir^2Deltar)` ........(∵ A = 4πr^{2})

∴ P_{i}− P_{o}= `(2T)/r` ….(2)

**Equation (2) represents Laplace’s law of spherical membrane. **