Derive Laplace's law for a spherical membrane.

Explanation

Consider a spherical liquid and let the outside pressure be `P_(0)` and inside presure be `P_(i)`, such that the excess pressure is `P_(i)-P_(0).`

Let, the radius of the drop increase from r to `r=Delta r`, where `Delta r` is very small, so that the pressure inside the drop remains almost constant.
Initial surface area `(A_(1))=4pi r^(2)`
Final surface area `(A_(2))=4pi (r+Delta r)^(2)`
`=4pi (r^(2)+ 2r Delta r +Delta r^(2))`
`=4pi r^(2)+8 pi r Delta r +4 pi Delta r^(2)`
As `Delta r` is very small, `Delta r^(2)` is neglected (i.e., `4 pi Delta r^(2) =0`)
Increase in surface area (dA)
`=A_(2)-A_(1)=4pi r^(2)+8pi r Delta r-4pi r^(2)`
Increase in surface area (dA)
`=8 pi r Delta r`
Work done to increase the surface area `8pi r Delta r` is extra surface energy.
`therefore dW=TdA `
`therefore dW=Txx 8 pi r Delta r " " ` ...(i)
This work done is equal to the product of the force and the distace `Delta r. `
Excess `" " ` Force = Excess Pressure `xx` Area
`dF =(P_(1)-P_(o))4pi r^(2)`
The increase in the radius of the bubble is `Delta r`.
`dW=dF Delta r=(P_(i)-P_(o))4pi r^(2) xx Delta r " " `...(ii)
Comparing equations (i) and (ii), we get
`(P_(i)-P_(o)) 4pi r^(2)xx Delta r =T xx 8 pi r Delta r`
`therefore (P_(i)-P_(o)) =2T//r`
This is called the Laplace's law of spherical membrane.