Surface Tension Of Raindrops

Classical Mechanics Level 3

Small raindrops are actually perfectly spherical. The pressure difference between the inside and outside of a spherical raindrop of volume $4.2 \times 10^{-9}~\mbox{m}^3$ is $280~\mbox{Pa}$ . What is the surface tension in N/m of the water on the outside of the raindrop?

Image credit: Samantha Krieger

The answer is 0.14.

6 solutions

Anish Puthuraya
Jan 19, 2014

Consider a Hemisphere of raindrop. It is clear that there are 3 forces that act on the hemisphere, The outer pressure, inner Pressure, and Surface Tension.

Outer Pressure = $P_o\times\pi r^2$
Inner Pressure = $\left(P_o+P\right)\times\pi r^2$ , where P is the Excess Pressure ,
Surface Tension = $T\left(2\pi r\right)$

NOTE: The Surface Tension acts Perpendicular to the flat surface of the Hemisphere and is in the same direction as the Outer Pressure.

Thus, for Equilibrium of the drop,
Inner Pressure $=$ Outer Pressure $+$ Surface Tension

Therefore,
$\left(P_o+P\right)\times\pi r^2 = P_o\times\pi r^2 + T\left(2\pi r\right)$

Thus, $P = \frac{2T}{r}$

Using the given values, we get,
$T = \boxed{0.14}$

Good solution Anish. Can you change the nomenclature in your solution? It is not "outer pressure", but should be something like: "net downward force on the upper hemisphere caused by the external pressure". Similarly, "inner pressure" maybe replaced by something like "upward force on the upper hemisphere by the internal pressure". And "Surface tension" with "downward force due to surface tension on the upper hemisphere". Don't want to be pedantic, but I guess that would help someone looking up the solution.

Muralidhar Kamidi - 7 years, 4 months ago

I guess that makes sense. But, I prefer to write this way, because, it shows everything in a very short way, and I guess that the reader will have more time to digest the solution, rather than the long sentences... But thanks, anyway.

Anish Puthuraya - 7 years, 4 months ago

ok so I understand that with outer/inner pressure you mean a force, but why do you multiply by pi{r}^2?? Shouldn't be 4pi{r}^2, that is the area of the sphere?? I don't get that

Pietro Pelliconi - 7 years, 3 months ago

I mentioned that I considered a hemispherical part of the sphere.

Anish Puthuraya - 7 years, 3 months ago

Sorry, I didn't see it. Got it, thanks!

Pietro Pelliconi - 7 years, 3 months ago

Shashwat Mishra
Jan 21, 2014

Pressure difference = $\Delta P= \Large \frac{2T}{R}$

$R= \sqrt[3]{\frac{3V}{4\pi}}= 1\times10^{-3}m$

$T= \frac {\Delta PR}{2}=\boxed{0.140 Nm^{-1}}$

Jason Fernandes
Feb 25, 2014

By Laplace's law:
P=2T/R
we can find R from volume, we get R=0.001
thus T = RP/2 = 0.001*280/2 = 0.14 N/m

Abbas Bagwala
Jan 29, 2014

The formula for difference of pressure betwixt a concave and a convex side of a spherical liquid surface is,

P1-P2 = 2T/r

We have P1-P2 = 280 Pa, and the radius can easily be found out from the formula V = 4/3 * pi * r^3

Thus, by mere substitution, we get, S.T = 140 E-3 = 0.140

Timothy Zhou
Jan 22, 2014

Since the raindrop does not collapse or expand, the surface tension and pressure must balance. So taking measurements along the center of the sphere, 2 pi r y = P pi r^2 where r is radius, y is tension, and P is pressure. Since we know the volume, we can calculate the radius as V = 4/3 pi r^3. Solving for y we find the tension is 0.140 N/m

Muralidhar Kamidi
Jan 20, 2014

Because it is sphere, radius can be calculated from the volume. Then it is just a matter of using 2 T/r = difference in pressure = 280 Pa. Hence, T = 280*r/2 = 140 r = 0.140 N/m

Surface Tension Of Raindrops

Image credit: Samantha Krieger

The answer is 0.14.

6 solutions

0 pending reports