Blowing up a balloon (thermodynamics and ideal gases)

AI Thread Summary
The discussion revolves around calculating the work done by air when blowing up a balloon from a volume of 1 m³ to 3 m³, given that pressure is defined as P = CV³. The user initially attempts to apply the work formula W = PdV but struggles with integrating due to pressure being a function of volume. They later consider using the modified ideal gas formula and seek clarification on determining the number of moles without temperature information. The conversation concludes with the user expressing uncertainty about their calculated work value and requesting a reference for an approximate joule measurement to validate their results. The discussion emphasizes the importance of integration in thermodynamic calculations involving variable pressure.
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1. Homework Statement
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. Homework 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
 
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Hint: This will be an integration problem because P is a function of V.
 
LawrenceC said:
Hint: This will be an integration problem because P is a function of V.

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)

?
 
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.
 
LawrenceC said:
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.

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:
How do you determine n without knowing the temperature?
 
LawrenceC said:
How do you determine n without knowing the temperature?

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

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

kPa is Newtons/area

So you get Newtons-meters
 
  • #10
Hope this has helped. I have to leave my computer now.
 
  • #11
LawrenceC said:
You have

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

kPa is Newtons/area

So you get Newtons-meters

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

ok nvm then, thanks a lot for the help!
 

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