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Derive an expression of excess pressure inside a liquid drop .
in Chapter2:Mechanical Properties of Fluids by Expert (11.6k points) | 332 views

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Consider a spherical drop of radius R and surface tension T

Let pi be the pressure inside the drop and p0 be the pressure outside it. As the drop is spherical in shape, the pressure, pi, inside the drop is greater than p0, the pressure outside.

Therefore, the excess pressure inside the drop is  pp0.

Let the radius of the drop increase from to + Δr, where Δis 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 (rr)2

∴ A2 = 4π (r2 + 2rΔ+ Δr2)

∴ A2 = 4π r2 + 8πrΔ+ 4πΔr2

As Δis very small, Δr2 can be neglected,

∴ A2 = 4πr2 + 8πrΔr

Thus, increase in the surface area of the drop is

dA A2 – A1 = 4πr2 + 8πrΔ- 4πr= 8πrΔ--- (1)

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

∴ dW TdA(8πrΔr) --- (2)

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

∴ dW FΔ--- (3)

The excess force is given by,

(Excess pressure) × (Surface area)

∴  = (pi – p0) 4πr2 --- (4)

Equating Eq. (2) and Eq. (3), we get,

T(8prΔr) = (pi – p0) 4πrΔr

∴ (pi – p0) = 2T/r --- (5)

This equation gives the excess pressure inside a drop. This is called Laplace’s law of a spherical membrane or Young Laplace Equation in spherical form

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