Name: Capillary Rise
Uploaded: 2019-07-19
Duration: 10 min 29 s
Description: Have you ever seen what happens to a celery stalk dipped in red-coloured water? We notice that after a day or two, the leaves turn reddish! Let's learn the r...

Question

Derive Laplace&rsquo;s law for a spherical membrane.

Accepted Answer

Let us consider a liquid drop that is spherical in shape with surface area $A_{1}$ . We know that due to 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 $A_{2 .}$

. Increase in area $= Δ A = A_{2} - A_{2} = 4 π {(r + Δ r)}^{2} - 4 π r^{2}$

$∴ Δ A = 4 π \times [{(r + Δ r)}^{2} - r^{2}]$

$∴ Δ A = 4 π \times 2 r Δ r$

$Δ A = 8 π r Δ r$

Work done in expanding the drop = gain in energy $Δ W = T Δ A = T 8 π r Δ r$ ------(1)

Here T is surface tension.

Excess pressure $= P_{i} - P_{0} $

Excess force= excess pressure×area

$Δ F = (P_{i} - P_{0}) 4 π r^{2}$

$W o r k d o n e = Δ W = Δ F \times Δ r = (P_{i} - P_{0}) 4 {πr}^{2}$ -----(2)

From 1 and 2 we get,

$T 8 π r Δ r = (P_{i} - P_{0}) 4 π r^{2} Δ r$

$\frac{2 T}{r} = P_{i} - P_{0}$ (Laplace's Law)

For a bubble, there are 2 surface areas, internal and external, the Laplace's Law gets changed as $\frac{4 T}{r} = P_{i} - P_{0}$ .

Derive Laplace’s law for a spherical membrane.

Important Questions on Surface Tension

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