Question

Derive an expression for the excess pressure inside a liquid drop.

Answer 1

Consider a spherical drop as shown in the figure. Let p_i be the pressure inside the drop and p₀ be the pressure out side it. As the drop is spherical in shape, the pressure, p_i, inside the drop is greater than p₀, the pressure outside. Therefore, the excess pressure inside the drop is p_i - p₀.

Let the radius of the drop increase from r to r + ∆r, where ∆r is very small, so that the pressure inside the drop remains almost constant.

Let the initial surface area of the drop be A₁ = 4πr² , and the final surface area of the drop be A₂ = 4π (r+∆r)².

∴ A₂ = 4π(r² + 2r∆r + ∆r²)

∴ A₂ = 4πr² + 8πr∆r + 4π∆r² (As ∆r is very small, ∆r² can be neglected)

∴ A₂ = 4πr² + 8πr∆r

Thus, increase in the surface area of the drop is

dA = A₂ -  A₁ = 8πr∆r

Work done in increasing the surface area by dA is stored as excess surface energy.

∴ dW = TdA = T(8πr∆r)  .....(1)

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

∴ dW = F∆r  .....(2)

The excess force is given by,

(Excess pressure) × (Surface area)

∴ F = (p_i – p₀) 4πr² .....(3)

Equating Eq. (1), and Eq. (2), we get,

T(8πr∆r) = F∆r

∴ T(8πr∆r) = (p_i - p₀)4πr²∆r  ....(using Eq. (3))

∴ (p_i - p₀) = \(\frac{2T}r\)

This equation gives the excess pressure inside a drop of liquid.

Answer 2

Suppose that ABC is a half drop of a liquid, whose radius is r and surface tension is T. On the molecules of liquid towards the part, the cohesive force F₁ would be downwards.

Thus, the pressure inside the drop is \(\frac { 2T }{ r } \) more than the pressure outside the drop.