Hello dear students we are familiar about the force due to surface tension that acts on the liquid surface. It is depends upon the density of liquid and also the surface in contact. When a liquid is placed on flat and smooth surface, it tries to occupy the spherical shape due to surface tension. Sir Laplace’s studied the formation of drops and bubbles in his law called as Laplace’s law of spherical membrane.
In this article we are going to discuss the formation of drops and bubble and the equation for pressure outside and inside free surface of liquid drop/bubble.
- Streamline flow
- Turbulent flow
The flow of liquid in which all the layers of liquid are flows with constant velocities and the velocity of that individual layers is constant at all the points also different layers of liquid appear like the parallel lines, known as stream and the flow is called as streamline flow or laminar flow.
Let’s discuss the concept of Drops and Bubbles.……………!
Due to the surface tension free liquid drops & bubbles are spherical, if effect of gravity & air resistance are negligible.
As it is spherical in shape, the inside pressure will be greater than outside (Pi>P0) & excess pressure is (Pi-P0)
Let radius of liquid increases from r to (r + ∆r)
Initial surface area, A1 = 4πr2
Final surface area, A2 = 4π(r+∆r)2
= 4π (r2+2r∆r+∆r2)
∴ A2 = (4πr2+8πr∆r + 4π∆r)
As ∆r is very small, ∆r2 is neglected.
∴Increase in surface area = dA = A2 – A1
= (4πr2+8πr∆r-4πr2)
∴ dA =(8πr∆r) ………(1)
Now, work done in increasing surface area,
∴ dW = T.dA= T.(8πr∆r)………(2)
By definition, we know that pressure is force acting per unit area,
Excess pressure, (Pi – P0) = Excess force / Area
∴ Excess force = Excess pressure (Pi – P0) × area
∴ dF = (Pi -P0) × 4πr2 ……….(3)
And, Work done,
∴ dW = dF.∆r
∴ dW = (Pi – P0) × 4πr2.∆r ……….(4)
On Comparing, eqn (2) & (4)
∴ (Pi – P0) × 4πr2.∆r = T× (8πr∆r)
∴ (Pi – P0) = 2T/r ……….(5)
Equation (5) is called as Laplace’s Law of spherical membrane for drops.
For, soap bubble, there are two sides.
Change in surface area A = 2 (8πr∆r) = 16 πr∆r
∴ dW = 16 πr∆r.T
Work done, (Pi – P0) x 4πr2.∆r= T.(16πr∆r)
(Pi – P0) = 4T/r ……….(6)
Equation (6) is called as Laplace’s Law of spherical membrane for bubbles.
Go through the following numerical to understand this in details….!
Ex: 1) Find the diameter of drop of oil having excess pressure of 2.1 × 102 Pa and the surface tension of 0.065 N/m.
Solution:
Here,
Pi-Po= 2.1 × 102 Pa , T=0.065 N/m
Then by Laplace’s law,