Question

Derive the expression for excess pressure inside a liquid drop

Answer 1

Consider a liquid drop of radius 'r' . The molecules on the surface of the drop experience a resultant force acting inwards due to surface tension. Therefore, pressure inside the drop will be greater then pressure outside the drop. It provides a force acting outwards perpendicular to the surface, to balance the resultant force due to surface tension. Let the drop be divided into two equal halves.
Considering the equilibrium of the upper hemisphere of the drop, the upward force due to excess pressure will be P π r² , where P is the pressure.
If 'T' is the surface tension then force acting downward along the circumference of the circle due to surface tension is T2πr .
At equilibrium,
 P π r² =  T 2 π r
P = 2T/r
which is the required pressure.
Hopefully this answer helps.
Thanks.

Answer 2

Let the pressure outside the liquid drop be Po and

the inside pressure be Pi

The excess pressure inside the drop = (Pi - Po)

Let T be the surface tension of the liquid

Let the radius of the drop increase from r to (r + Δr)due to excess pressure

The work done by the excess pressure is given by

dW = Force x Displacement

= (Excess pressure x Area) x (increase in radius)

=[(Pi - Po) × 4πr²] × Δr .......(1)

Let the initial surface area of the drop be A1=4πr²

The final surface area of the drop is A2=4π(r+Δr)²

A2 = 4π(r² + 2rΔr + Δr²)

= 4πr² + 8πrΔr +4πΔr²

As Δr is very small, Δr² can be neglected, i.e. 4πΔr² ≡ 0

Therefore, A2 = 4πr² + 8πrΔr

Therefore, increase in the surface area of the drop dA = A2 - A1

= 4πr² + 8πrΔr - 4πr²

= 8πrΔr

Work done to increase the surface area by 8πrΔr is the extra surface energy.

dW = T x dA

where T is the surface energy

dW = T x 8πrΔr .........(2)

Comparing (1) & (2), we get

[(Pi - Po) × 4πr²] × Δr = T × 8πrΔr

Pi - Po = 2T/r

The above equation gives the excess pressure inside a liquid drop.