Question

Derive Laplace's law for a spherical membrane.

Answer 1

Let us consider a liquid drop which is spherical in shape with surface area $A_1$

We know that due 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 $=\Delta A=A_2-A_1=4\pi (r+\Delta r{)}^2-4\pi r^2$

$\therefore \Delta A=4\pi \times \left[\right(r+\Delta r{)}^2-r^2]$

$\therefore \Delta A=4\pi \times 2r\Delta r$ $\because \Delta r$ is very small, $\Delta r^2$ is still smaller and hence ignored and considered as zero.

$\Delta A=8\pi r\Delta r$

Work done in expanding the drop $=$ gain in energy $\Delta W=T\Delta A=T8\pi r\Delta r$ -------------------------1

Here T is surface tension.

Excess pressure$=P_i-P_o$

Excess force$=$ excess pressure$\times$area

$\Delta F=(P_i-P_o)4\pi r^2$

Work done$=\Delta W=\Delta F\times \Delta r=(P_i-P_o)4\pi r^2\Delta r$-------------------2

from 1 and 2 we get, $T8\pi r\Delta r=(P_i-P_o)4\pi r^2\Delta r$

$\therefore \frac{2T}{r}=P_i-P_o$ -----------------------$Laplac{e}^{\prime }s$ $Law$

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