# Mixing of two bubbles

When a small bubble comes in contact with a larger bubble, what happens to their size?
(a) smaller bubble becomes smaller & eventually vanishes while larger one becomes larger
(b) both bubbles increase in size
(c) both bubbles decrease in size
(d) larger bubble becomes smaller while smaller one becomes larger

In our textbook the answer is mentioned as (a) while our teacher told us it is (b).

Who is correct and what is its explanation??

## Answers and Replies

Related Classical Physics News on Phys.org
That is a good question. I wonder if it depends on contact dynamics and soap film composition. I don't know.
I've seen soap bubbles collide together and stick to each other without any apparent change in their sizes(then again it was a brief observation versus actual experiment)
And I've seen one or both soap bubbles "pop" upon collision with each other, but I never paid close enough attention to the end results.

Mapes
Homework Helper
Gold Member
(a) is correct. The pressure inside a smaller bubble is larger because of the energy penalty of having more surface area (relative to volume) than a larger bubble (see http://en.wikipedia.org/wiki/Laplace_pressure" [Broken]). When the two bubbles meet, there is thus a driving force for material to move from the smaller bubble to the larger bubble.

The same mechanism drives http://en.wikipedia.org/wiki/Ostwald_ripening" [Broken] of precipitates in solids.

Last edited by a moderator:
but our teacher explains that as the molecules move from smaller bubble to larger bubble due to pressure difference, the pressure inside the smaller bubble decreases.
Hence as PV = constant, its volume will increase.

I did this question in a science olympiad type exam once.
But it had an equation to look at, which I am so far unable to dig back up (I'll have another look for the paper itself tomorrow).
Anyway, I remember the answer I worked out to be counter-intuitive, i.e. that the small bubble shrinks and the large one grows. Until I find that equation I can't do any better than Mapes on this one.

Sorry I couldn't help more.

Why is PV=C

Mapes
Homework Helper
Gold Member
Hence as PV = constant, its volume will increase.
PV = constant assumes a constant amount of gas, among other things (PV = nRT is the ideal gas law, where n is the number of moles of gas). This obviously doesn't hold if the gas is leaving the bubble. It sounds like you have a pretty poor teacher.

I did this question in a science olympiad type exam once.
But it had an equation to look at, which I am so far unable to dig back up (I'll have another look for the paper itself tomorrow).
Anyway, I remember the answer I worked out to be counter-intuitive, i.e. that the small bubble shrinks and the large one grows. Until I find that equation I can't do any better than Mapes on this one.
The final equation is in the link I gave. $\Delta P=\gamma \kappa$, where $P$ is the pressure, $\gamma$ is the surface energy, and $\kappa=\frac{1}{R_1}+\frac{1}{R_2}$ is the curvature, which simplifies to $\frac{2}{R}$ for a sphere.

PV = constant assumes a constant amount of gas, among other things (PV = nRT is the ideal gas law, where n is the number of moles of gas). This obviously doesn't hold if the gas is leaving the bubble. It sounds like you have a pretty poor teacher.

The final equation is in the link I gave. $\Delta P=\gamma \kappa$, where $P$ is the pressure, $\gamma$ is the surface energy, and $\kappa=\frac{1}{R_1}+\frac{1}{R_2}$ is the curvature, which simplifies to $\frac{2}{R}$ for a sphere.
Yep that's the equation.