# Blowing up a balloon (thermodynamics and ideal gases)

1. Oct 8, 2012

### doctordiddy

1. The problem statement, all variables and given/known data
A balloon behaves such that the pressure is P = CV3
where C
= 100 kPa/m3
. The balloon is blown up with air from a starting volume of 1 m3
to a volume
of 3 m3
. Find the work done by the air.

2. Relevant equations
W=PdV

3. The attempt at a solution

What I tried is simply trying the above solution by multiplying the P which is given by P=CV^3 by the change in volume

so i got

W=(CV^3)(Vf-Vi)

I then realized I had no idea what would the V in this equation be, can someone tell me if I am even doing this question right and if I am how I can determine the V?

Thanks

2. Oct 8, 2012

### LawrenceC

Hint: This will be an integration problem because P is a function of V.

3. Oct 8, 2012

### doctordiddy

Do I use the formula W=∫VfV1PdV?

That's what I have been trying to do but i'm not sure how to do this with two different P because you can usually move the P out of the integral

edit: nvm i think i figured it out, should it looook something like this

PiVi = PfVf

so

Vf/Vi = Pi/Pf

and then replace Vf/Vi with Pi/Pf to get

W = nR*ln(Pi/Pf)

?

4. Oct 8, 2012

### LawrenceC

The pressure is a constant times volume. So the integral would be

W = integral(C * V * dV) from 1 m^3 to 3 m^3.

Check the units and you'll see that the units are work.

5. Oct 8, 2012

### doctordiddy

could I instead just find the initial and final pressures and then directly plug them into the modified version of the ideal gas formula like in my edit above?

Also i tried to do the integral you told me to do above, should I end up with 31| C*3V2?

Last edited: Oct 8, 2012
6. Oct 8, 2012

### LawrenceC

How do you determine n without knowing the temperature?

7. Oct 8, 2012

### doctordiddy

yeah I realizd that when i started doing the calculations

i tried to do the integral you told me to do above, should I end up with 31| C*3V2?

however when I plug everything in I end up with units in kPa instead of joules... unless the integral of V would change it's units?

8. Oct 8, 2012

### doctordiddy

also, my answer of 7800J seems unrealistically high so i think i might be doing something wrong.

9. Oct 8, 2012

### LawrenceC

You have

W = integral ( (kPa/Volume) * Volume * dVolume) = kPa * Volume after integration

kPa is Newtons/area

So you get Newtons-meters

10. Oct 8, 2012

### LawrenceC

Hope this has helped. I have to leave my computer now.

11. Oct 8, 2012

### doctordiddy

just one last question, can you give me an approximate value of Joules that would be needed in this question? I just need a reference to determine whether or not my answer is plausible and i have no idea how much 1 joule is.

Thanks.

12. Oct 8, 2012

### doctordiddy

ok nvm then, thanks alot for the help!