Blowing up a balloon (thermodynamics and ideal gases)

In summary, the problem involves a balloon with pressure P = CV^3, where C = 100 kPa/m^3. The balloon's volume is increased from 1 m^3 to 3 m^3, and the task is to find the work done by the air. The solution involves using the equation W = ∫PdV and integrating from 1 m^3 to 3 m^3. The resulting work is measured in Newtons-meters.
  • #1
doctordiddy
54
0
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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  • #2
Hint: This will be an integration problem because P is a function of V.
 
  • #3
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)

?
 
  • #4
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
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:
  • #6
How do you determine n without knowing the temperature?
 
  • #7
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?
 
  • #8
also, my answer of 7800J seems unrealistically high so i think i might be doing something wrong.
 
  • #9
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!
 

Related to Blowing up a balloon (thermodynamics and ideal gases)

1. How does blowing up a balloon relate to thermodynamics?

Blowing up a balloon involves the transfer of energy from the person's lungs to the air inside the balloon. This process follows the principles of thermodynamics, specifically the conservation of energy and the laws of gas behavior.

2. What is the ideal gas law and how does it apply to blowing up a balloon?

The ideal gas law, PV=nRT, relates the pressure (P), volume (V), number of moles (n), temperature (T), and gas constant (R) of an ideal gas. When blowing up a balloon, the gas inside is heated and expands, causing an increase in volume and pressure. This follows the ideal gas law as the temperature and volume of the gas increase.

3. What happens to the temperature of the air inside a balloon when it is blown up?

When air is blown into a balloon, the pressure and volume of the gas increase, causing the temperature to also increase. This is because the molecules in the gas are moving faster and therefore have a higher kinetic energy, resulting in a higher temperature.

4. How does the elasticity of the balloon material affect the process of blowing it up?

The elasticity of the balloon material determines how much the balloon can expand before it bursts. If the material is very elastic, the balloon can be blown up to a larger size before it bursts. However, if the material is less elastic, the balloon may burst at a smaller size.

5. Is blowing up a balloon considered an endothermic or exothermic process?

Blowing up a balloon is considered an endothermic process because it requires the input of energy (from the person's lungs) to inflate the balloon. This energy is used to increase the temperature and volume of the gas inside the balloon.

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