Help with A), B) & C) on Air Bubble Diameter & Expansion Energy

[driventowin]

Homework Statement

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

Homework 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

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 that's 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?

 

[gneill]

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.

 

[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?

 

[gneill]

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?

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

 

[driventowin]

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

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 can't seem to make headway

 

[gneill]

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?

 

[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

 

[gneill]

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

You had written:

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

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).

 

[driventowin]

ok, think i have got it, thanks for your patience to by the way, I am 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?

 

[gneill]

Looks much better!

 

[driventowin]

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

 

[gneill]

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)

 

[driventowin]

sorry, have absolutly no idea, I am using these questions as study for a fast approaching exam, is there anyway you could do the solution please as i haven't 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

 

[gneill]

sorry, have absolutly no idea, I am using these questions as study for a fast approaching exam, is there anyway you could do the solution please as i haven't 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

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).

 

[driventowin]

ok, here's a long shot but:

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

mass= vol x density = 0.0005 x 1000 = 5kg5x287x (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 completely wrong

 

[gneill]

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.

 

[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

 

[gneill]

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

What is wb? Work? If so, and if r is [itex]\rho[/itex] (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):
[tex] W = -\int_{p_0}^{p_f} v \cdot dp[/tex]
But
[tex] p \;v = nRT = E_o \;\;\;\;\; \text{{ a constant in this case }} [/tex]
so that
[tex] v = \frac{E_o}{p} [/tex]
[tex] W = -\int_{p_0}^{p_f} \frac{E_o}{p} \cdot dp[/tex]
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.

 

[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

 

[gneill]

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

See my edit to my previous post.

 

[SteamKing]

Aaargh matey! What kind of seawater are ye using? The density of proper seawater is not 1000 kg / m^3. It's close to 1025 kg / m^3. There's salt in it, ye know!

 

[driventowin]

ok, new day, new energyok, so -wb = m x r x t x ln(v2/v1) mrt = pvso p x v x ln(v2/v1)

so (199100)(0.0005) ln (0.0010/0.0005)

wb= -69??

apart from the negative answer is this the direction you were pointing me in?

 

[gneill]

ok, new day, new energy


ok, so -wb = m x r x t x ln(v2/v1) mrt = pv


so p x v x ln(v2/v1)

so (199100)(0.0005) ln (0.0010/0.0005)

wb= -69??

apart from the negative answer is this the direction you were pointing me in?

Yes. I was hoping to have you arrive at the required equation by deriving it from the problem scenario rather than using a 'canned' equation that you might or might not understand the origins of. If you understand how to derive the equation you won't have to memorize it for exam situations (and hope that the problem given precisely matches what you've memorized).

Hopefully you understand how to derive the equation via integrating p*v either as p*dv or v*dp.

Regarding the result that you found, you are losing accuracy in your calculations because significant figures are being truncated when you use decimal notation with a set number of decimal places and the values are small. It would be better to use scientific notation and keep a consistent number of significant figures for all your values.

 

[driventowin]

so I've got part 2 pretty figured out now aswel, part 3 next...

is energy absorbed equal to the work done? that seems to easy?

 

[gneill]

so I've got part 2 pretty figured out now aswel, part 3 next...

is energy absorbed equal to the work done? that seems to easy?

It is easy, but it is also an important concept. Pressure x Volume yields a value with units of energy; the energy content of the bubble (due to gas pressure) at 10m depth is thus P_oV_o. As the bubble rises (presumably slowly) and expands, in order for the expanding gas to remain at the same temperature it has to absorb energy from its surroundings to keep the gas warmed. Otherwise the gas in the bubble would grow colder as it expanded, reducing the pressure and amount of expansion and you'd have another set of equations to deal with the simultaneous changes in all three variables: volume, pressure, and temperature. nRT would not be constant for the rising bubble.

So, the work done by the expanding gas in enlarging the bubble volume against the pressure of the surrounding water has to be precisely balanced by energy absorbed in order to keep the temperature constant (and thus nRT constant, and P*V constant).

 

[driventowin]

so wd = -69, so energy absorbed is -69j?? or have i misunderstood ur last post?

 

[gneill]

so wd = -69, so energy absorbed is -69j?? or have i misunderstood ur last post?

You didn't take my suggestion and keep more significant figures. The work done should be slightly larger (about 0.071J, assuming water density is taken to be 1.0 /cm³). And yes, the energy absorbed will be the same.

The sign of the energy calculated depends upon your point of view: is the bubble doing work on the medium surrounding it, or the medium doing work against the bubble expansion. The difference is in the assumed direction of the force acting (medium pushing in, bubble pushing out). Since you're interested in the work done by the bubble, take the value as positive (bubble pushing out in same direction as the change in radius).