Answer 1

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 (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 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 the Laplace’s law of spherical membrane.

Answer 2

Let us consider a liquid drop which is spherical in shape with surface area A₁​

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 A₂​. 

Increase in area = `triangle "A" = "A"_2 - "A"_1 = 4 pi ("r" + triangle "r")^2 - 4pi"r"^2`

$`therefore\; triangle\; "A"\; =\; 4pi\; xx\; [("r"\; +\; triangle\; "r")^2\; -\; "r"^2]`$

`therefore triangle  "A" = 4 pi xx 2"r" triangle "r"`  ...(∵ Δ r is very small Δr² is still smaller and hence ignored and considered as zero) 

$\Delta \; A\; =\; 8\; \pi \; r\; \Delta \; r$

$Work\; done\; in\; expanding\; the\; drop\; =\; gain\; in\; energy\; \Delta \; W\; =\; T\; \Delta \; A\; =\; T\; 8\; \pi \; r\; \Delta \; r\; ....(i)$

$Here\; T\; is\; surface\; tension.$

Excess pressure = P_i - P₀ 

$Excess\; force\; =\; excess\; pressure\; times\; \times \; area$

Δ F = (P_i - P₀) 4 πr² 

Work done = ΔF × Δr = (P_i - P₀) 4 πr² Δ r  .......(ii)

From 1 and 2 we get , T 8 π r Δ r = (P_i - P₀)  4 πr² Δ r 

$`therefore\; "2T"/"r"\; =\; "P"\_"i"\; -\; "P"\_0`\; ......\; Lapalce\text{'}s\; Law$

$For\; a\; bubble,\; there\; are\; 2\; surface\; areas,\; internal\; and\; external,\; the\; Laplace\text{'}s\; Law\; gets\; changed\; as\; `(4"T")/"r"\; =\; "P"\_"i"\; -\; "P"\_0`$