# Homework Help: Question, help please

1. Aug 19, 2011

### driventowin

1. The problem statement, all variables and given/known data

An air bubble has a diameter of 1cm at a depth of 10m below the sea surface.
Assuming that the atmosphere and water are at the same temperature and assuming STP.

A) Find the diameter of the bubble at a depth of 0.5m below the sea surface
b) find the boundary work done by the bubble as it expands
c) find the amount of energy the bubble absorbs in the process

2. Relevant equations

volume of a sphere = 4/3 x pi x r^3

r of air is 287j/kgk and y=1.4

density of sea water = 100kg/m^3

3. The attempt at a solution

for the first part of the question i got r to be 2.306m, anyone able to tell me if thats correct?

p x v = v
(98100) x (0.00052) = 4/3 x pi x r^3

r^3 = 12.26

r=2.306m, am i right so far?

2. Aug 19, 2011

### Staff: Mentor

Hi driventowin, welcome to Physics Forums.

You have a couple of problems with your solution. First, the problem gives you an initial diameter, not a radius, and they ask you to find a diameter, not a radius.

Begin by finding the pressures that bubble is supporting at both the 10m depth and at the surface (what equation is relevant?). Then see if you can think of an equation you know that relates a gas volume with pressure.

3. Aug 19, 2011

### driventowin

pressure at surface is atm, right? which is 101, so atm is present at both levels so i just canceled it out,

pressure at depth therefore is equal to - (1000)(9.81)(10) = 98100pa

am i right in that?

4. Aug 19, 2011

### Staff: Mentor

If you're dealing with ratios then you can't just cancel. That is,
$$\frac{a + b}{a} \neq \frac{b}{1}\;\;\; \text{'cancelling' a from numerator and denominator}$$
Leave the atmospheric pressure 'bias' in place.

5. Aug 19, 2011

### driventowin

ok then, pressure at surface is 101'000pa

at 10m depth is 199'100pa

have i got the pressures correct now? thanks for your help by the way, have been stuck in this for days but just cant seem to make headway

6. Aug 19, 2011

### Staff: Mentor

Your pressures look okay. Now, what general formula are you going to apply to find the volume of the bubble at its new pressure near the surface?

7. Aug 19, 2011

### driventowin

is the way i done it in the first post not correct? if not can you tell me the formula please as i have no idea what it is

8. Aug 19, 2011

### Staff: Mentor

No it's not correct, since pressure multiplied by volume is not equal to a volume (the units cannot match!).

Look up the Ideal Gas Law (and the version of it that is Boyle's Law).

9. Aug 19, 2011

### driventowin

ok, think i have got it, thanks for your patience to by the way, im making slow progress

p1v1=p2v2
(199100)(0.00052)=(101000)(v2)

v2=0.001

0.001=4/3 x pi x r^3

r^3=0.000239
r=0.062
d=0.124

have i finally hit the nail?

10. Aug 19, 2011

### Staff: Mentor

Looks much better!

11. Aug 19, 2011

### driventowin

for part two im absolutly clueless,(even worse than part1 if thats possible), can you show me how to do it or give some fairly major clues please

12. Aug 19, 2011

### Staff: Mentor

The boundary (the bubble surface) is expanding as the bubble rises. At all times during this expansion it is expanding against the pressure external to the bubble (which is changing with depth). So work is being done as the radius expands from its initial value to its final value against the force due to the pressure on the surface area of the bubble.

Things you know:
1. P*V = constant for this bubble (and you can calculate that constant).
2. Starting radius and ending radius for the bubble.
3. Formulas for surface area and volume of sphere.
4. Formula for work done: W = F*d

Given P*V = k, and since V is a function of r, you can arrive at an expression for P(r). The surface area of the bubble is also a function of r. The total force on the bubble at any instant is given by the pressure x the surface area: F(r) = P(r)*A(r).

How might you find the work done? (HINT: an integration is required)

13. Aug 19, 2011

### driventowin

sorry, have absolutly no idea, im using these questions as study for a fast approaching exam, is there anyway you could do the solution please as i havent any idea how to do it, the only way i think ill figure this one is to see it done and work out why, i had a vague understanding off the first part but not the slightest for this part, thanks again

14. Aug 19, 2011

### Staff: Mentor

Sorry, Physics Forums policy does not allow helpers to provide solutions to homework problems. We can only provide appropriate hints and guidance for you to arrive at your own solution.

If the question is part of your coursework then you should have the required background material and methods at hand, and the hints given so far should (I think) be sufficient to allow you to apply them. For this problem you need to know about the Ideal Gas Law, work and energy, force area and pressure, and how to integrate f.dr to find the work done. There should be sections in your texts pertaining to these topics.

Again, I'm sorry that I can't work though the problem step by step for you.

How about you start by writing an expression for the surface area of the bubble with respect to the bubble radius, and also an expression for the pressure for a given radius? (you know the volume for a given radius and also that P*V = constant).

15. Aug 19, 2011

### driventowin

ok, heres a long shot but:

v1=0.0005
v2=0.0010
r=287
t=(constant)

mass= vol x density = 0.0005 x 1000 = 5kg

5x287x (temp??) ln(0.001/0.0005) = wb

any tips for how to get temp or do i just cancel it as its constant?

also am i right in using v1 to find the mass??

or am i gone completly wrong

16. Aug 19, 2011

### Staff: Mentor

You shouldn't require a mass value in order to solve the problem; you can work with forces and distances in order to calculate the work (work = Integral{Force*dr) }. In order to determine the net force at any instant you need to have the surface area of the bubble and the pressure on that surface.

The starting radius and the ending radius give you the distance covered, and thus the limits of the integral. The radius at any instant gives you the surface area and volume at that instant. Given that P*V is constant, you thus also have the pressure at that radius. P*Area is the force at that radius.

17. Aug 19, 2011

### driventowin

so i should be using this formula to get the work done: -wb=r x t x ln(p1/p2)

am i correct in saying that? if so i think i might be getting this

18. Aug 19, 2011

### Staff: Mentor

What is wb? Work? If so, and if r is $\rho$ (the density) and t temperature, then the units don't match. Where does this equation come from?

Since you aren't given a particular temperature to work with, I'd have to say that this formula does not look promising here.

EDIT: Okay, I think I know where this is coming from. The work done will be the integral of the volume times the pressure (actually, the integral of v*dp):
$$W = -\int_{p_0}^{p_f} v \cdot dp$$
But
$$p \;v = nRT = E_o \;\;\;\;\; \text{{ a constant in this case }}$$
so that
$$v = \frac{E_o}{p}$$
$$W = -\int_{p_0}^{p_f} \frac{E_o}{p} \cdot dp$$
You're trying to work with the values of n, R, and T. You don't have to. You have the initial P*V and that is equal to nRT.

Last edited: Aug 19, 2011
19. Aug 19, 2011

### driventowin

r=gas constant wb=work done

can you give me the formula for work that i should be using, i think if i had that i can work the problem

20. Aug 19, 2011

### Staff: Mentor

See my edit to my previous post.