Question

Derive Laplace's law for spherical membrane of bubble due to surface tension.

Answer 1

Consider a spherical liquid drop and let the outside pressure be $P_0$ and inside pressure be $P_i$, such that the excess pressure is $P_i-P_o$.
Let the radius of the drop increase from $\Delta r$, where $\Delta r$ is very small, so that the pressure, inside the drop remains almost constant.
Initial surface area $\left(A_1\right)=4\pi r^2$
Final surface area $\left(A_2\right)=4\pi r^2$
$=4\pi (r+\Delta r{)}^2$
$=4\pi (r^2+2r\Delta r+\Delta r^2)$
$=4\pi r^2+8\pi r\Delta r+4\pi \Delta r^2$
As $\Delta r$ is very small $\Delta r^2$ is neglected (i.e. $4\pi \Delta r^2=0$)
Increase in surface area $\left(dA\right)=A_2=-A_1=4\pi r^2+8\pi r\Delta r-4\pi r^2$
In crease in surface area $\left(dA\right)=8\pi r\Delta r$
Work done to increase the surface area $8\pi r\Delta r$ is extra energy.
$\therefore dW=TdA$
$\therefore dW=T\times 8\pi r\Delta r$ .....(1)
This work done is equal to the product of the force and the distance $\Delta r$
$dF=(P_i-P_o)4\pi r^2\times r^2$
The increase in the radius of the bubble is $\Delta r$.
$dW=dF\Delta r=(P_i-P_o)4\pi r^2\times \Delta r^2$ ...(2)
Comparing equation (1) and (2), we get
$(P_i-P_o)4\pi r^2\times \Delta r=T\times 8\pi r\Delta r$
$\therefore (P_i-P_o)=2T/r$
This is called the Laplace's law of spherical membrane.