Work done in expansion of a bubble

[Vriska]

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.

 

[haruspex]

2pi*r*L

Dimensionally wrong. What have you forgotten?

 

[Vriska]

Dimensionally wrong. What have you forgotten?

Dimensionally wrong. What have you forgotten?

L is force by length right?

 

[haruspex]

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.

 

[Vriska]

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?

 

[haruspex]

small path dA,

I deliberately wrote A, not dA.

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?

 

[Vriska]

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!

 

[haruspex]

dF = 2AL/r dr?

No, the force F is 2AL/r. The work done on A is F.dr.

2*4pi *r L dr

Yes.

 

[Vriska]

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?

 

[haruspex]

what's wrong with the integral I did?

At the end of post #5?

 

[Vriska]

At the end of post #5?

nevermind I got it, thanks for the help!