Work done in expansion of a bubble

Click For Summary
SUMMARY

The discussion revolves around calculating the work done in the expansion of a bubble, specifically addressing the forces involved and the integration process. The participants clarify that the force exerted by the atmosphere on the bubble is represented as 2L/r, where L is the force per unit length. The correct approach involves integrating the force over the surface area of the bubble, leading to the expression for work done as 4πL(R2² - R1²). Missteps in dimensional analysis and the integration process are highlighted, emphasizing the importance of considering the correct geometry and forces acting on the bubble.

PREREQUISITES
  • Understanding of basic physics concepts such as force, work, and pressure.
  • Familiarity with calculus, particularly integration techniques.
  • Knowledge of surface tension and its role in fluid mechanics.
  • Concept of pressure differentials in spherical geometries.
NEXT STEPS
  • Study the principles of fluid mechanics, focusing on surface tension and pressure differentials.
  • Learn about the integration of forces in spherical coordinates.
  • Explore the concept of work done in expanding systems, particularly in thermodynamics.
  • Investigate applications of surface energy in physical systems.
USEFUL FOR

Students and professionals in physics, particularly those studying fluid dynamics, thermodynamics, and applied mathematics. This discussion is beneficial for anyone looking to deepen their understanding of work done in expanding systems and the forces involved in bubble dynamics.

Vriska
Messages
138
Reaction score
2

Homework Statement


as title

Homework Equations



force exerted by atmosphere on bubble = 2(on account of dual layer) * 2pi*r*L(surface tension)
work is the integral of force *dx

The Attempt at a Solution


so we're looking at int ( 4pi*r*L *dr) but since not it 1 particle is traveling against atmospheric force rather a hole bunch in a bubble dr = 2pi*r*dr. (broke the dimensions :(, what else should i do?)
so now we have :
integral (8 pi^2*r^2*L*dr)

This is ofc the wrong answer. I want to do it this way instead of using the "surface energy". If i just used dr I'd be implying only one particle is moving against the atmosphere which is wrong.
 
Physics news on Phys.org
Vriska said:
2pi*r*L
Dimensionally wrong. What have you forgotten?
 
haruspex said:
Dimensionally wrong. What have you forgotten?
haruspex said:
Dimensionally wrong. What have you forgotten?

L is force by length right?
 
Vriska said:
L is force by length right?
Sorry, I was thinking of surface energy, which is per unit area.

I did not follow the rest of your argument.
You found the force between two hemispheres. But you cannot integrate that wrt r to find work. That force is not parallel to dr. Think about the force on a small patch A of the bubble, and the work done in pushing A out by dr.
 
haruspex said:
Sorry, I was thinking of surface energy, which is per unit area.

I did not follow the rest of your argument.
You found the force between two hemispheres. But you cannot integrate that wrt r to find work. That force is not parallel to dr. Think about the force on a small patch A of the bubble, and the work done in pushing A out by dr.

Yeah the original one was wrong, I actually wanted to multiply it by 4pi*r^2 to get the total particles that do the same work but the force is not parallel!

okay now I have a small path dA, the force is difference in pressure times that. delta p :

= 2pi*r*L/pi*r^2 = 2L/r.

So my force on that dA = 2L/r *dA.

but dA = 8*pi*r dr (because dA/dr = 8pi*4)

so dF= 16piL dr.

Soo work done could be

int r * df =int 16piL*r dr now to R1 to R2 - 8piL(R2^2- R1^2)

This is wrong, right? like by a factor of 2?
 
Vriska said:
small path dA,
I deliberately wrote A, not dA.
Vriska said:
dA = 8*pi*r dr
No.
You have a force 2LA/r acting on A. What work is done on that as the bubble expands by dr?
What work is done on the whole surface as it expands by dr?
 
haruspex said:
I deliberately wrote A, not dA.

No.
You have a force 2LA/r acting on A. What work is done on that as the bubble expands by dr?
What work is done on the whole surface as it expands by dr?
?

but isn't A =4pi*r^2? dA = 8pi*r dr right?

okay if we have a force dF = 2AL/r dr? so for the entire thing that is 2*4pi *r L dr . 4 piL (R1^2 - R2^2). now I get it why surface energy is a thing, thanks!
 
Vriska said:
dF = 2AL/r dr?
No, the force F is 2AL/r. The work done on A is F.dr.
Vriska said:
2*4pi *r L dr
Yes.
 
haruspex said:
No, the force F is 2AL/r. The work done on A is F.dr.

Yes.

Hm, yep that was a slip, my bad.

Can you tell me what's wrong with the integral I did?
 
  • #10
Vriska said:
what's wrong with the integral I did?
At the end of post #5?
 
  • #11
haruspex said:
At the end of post #5?

nevermind I got it, thanks for the help!
 

Similar threads

Work done to insert a point charge 𝑞 at the center of a conducting sphere
  • · Replies 12 ·
Replies
12
Views
1K
How can you solve soap bubble oscillations using an energy approach?
  • · Replies 18 ·
Replies
18
Views
4K
Fluid - Work done to blow a soap bubble
  • · Replies 1 ·
Replies
1
Views
2K
Why Is Work Positive When Done Against the Electric Field?
  • · Replies 22 ·
Replies
22
Views
4K
Work done pushing a spring from the side
  • · Replies 9 ·
Replies
9
Views
2K
Calculating Work Done by Expanding Gas in Spherical Bubble
  • · Replies 6 ·
Replies
6
Views
5K
Pressure inside a soap bubble just under surface
  • · Replies 3 ·
Replies
3
Views
3K
Integral for work done leading to potential energy sign confusion
  • · Replies 5 ·
Replies
5
Views
2K
Work done by non-conservative force while particle moves in circle
  • · Replies 25 ·
Replies
25
Views
2K
Minimum velocity to throw a object to another planet
  • · Replies 34 ·
2
Replies
34
Views
4K