Answer 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 (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 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=(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 Laplace’s law of spherical membrane.  

Answer 2

1. 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.
2. As drop is spherical, P_i > P_o
   ∴ excess pressure inside drop = P_i − P_o 
3. Let the radius of drop increase from r to r + ∆r so that inside pressure remains constant.
4. Initial area of drop A₁ = 4πr² ,
   Final surface area of drop A₂ = 4π (r + ∆r)² 
   Increase in surface area,
   ∆A = A₂ − A₁ = 4π[(r + ∆r)² − r²]
   = 4π[r² + 2r∆r + ∆r² − r²]
   = 8πr∆r + 4π∆r²
5. As ∆r is very small, the term containing ∆r² can be neglected.
   ∴ dA = 8πr∆r
6. 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) × 4πr² 
   ∴ 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 = 4πr²)
   ∴ P_i − P_o = `(2T)/r` ….(2)

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