Question

Derive Laplace's law for spherical membrane.

Answer 1

Laplace's law for spherical membrane of a liquid drop

Let's consider a liquid drop of radius , r . due to spherical shape , the inside pressure is always greater than outside pressure. Let inside pressure is $P_i$ and outside pressure ,$P_0$, then excess pressure is $P_i-P_0$

Let radius of liquid drop increases from r to r + ∆r, where we assume ∆r is vary very small .hence inside pressure assumed to be constant.

now, initial surface area, $A_i=4\pi r^2$
final surface area , $A_f=4\pi(r+\Delta{r})^2$
$=4\pi r^2+4\pi \Delta{r}^2+8\pi r\Delta{r}$
we can assume 4π∆r² ≈ 0
so, $A_f=4\pi r^2+8\pi r\Delta{r}$

so, change in surface area, ∆A = 8πr∆r

now, workdone = T.∆A , T is surface tension
workdone = T.8πr∆r..........(1)

we also know, workdone = force × displacement
e.g., W = F × ∆r ......(2)
and force = excess pressure × surface area
F = $(P_i-P_0)\times4\pi r^2$.....(3)

from equations (1), (2) and (3),

$P_i-P_0=\frac{2T}{r}$
this expression is known as Laplace's law for spherical membrane of a liquid drop.