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Ideal gas law hot air balloon problem

  1. Jan 2, 2010 #1
    1. The problem statement, all variables and given/known data
    A hot-air balloon achieves its buoyant lift by heating the air inside the balloon, which makes it less dense than the air outside. Suppose the volume of a balloon is 1900 m3 and the required lift is 2500 N (rough estimate of the weight of the equipment and passenger). Calculate the temperature of the air inside the balloon which will produce the required lift. Assume that the outside air temperature is 0°C and that air is an ideal gas under these conditions.


    2. Relevant equations
    PV=nRT
    ??

    3. The attempt at a solution
    I really have no idea how to start this one......I know no one is allowed to give out answers or complete solutions but I really just need help starting the problem like a little hint to get me started. I'm not asking for the answer...Thanks in advance
     
  2. jcsd
  3. Jan 2, 2010 #2
    The 'lift' on the balloon is simply the upthrust acting on it. You can get the required mass of air in the balloon / density of air in the balloon, and the rest follows from the ideal gas law.
     
  4. Jan 2, 2010 #3
    Ok thanks a lot; I'll try that.
     
  5. Jan 2, 2010 #4
    I appreciate your help but I still don't know where I am going with this problem......
     
  6. Jan 2, 2010 #5
    I tried using the buoyancy force formula to find the density of the hot air and got 0.134kg/m3 but that didn't help
     
  7. Jan 2, 2010 #6

    diazona

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    Homework Helper

    Okay, well, that's a start. Do you know how to convert the ideal gas law into a form which includes density?

    If not, you have the volume of the balloon. You can use the density to find the mass of air in the balloon. How can you convert that mass into something you can plug into the ideal gas law?
     
  8. Jan 2, 2010 #7
    I found the number of moles of air but I am a bit confused as to what the pressure would be inside the balloon.
     
  9. Jan 3, 2010 #8

    diazona

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    Homework Helper

    Would it be greater than, less than, or the same as the pressure outside the balloon?
     
  10. Jan 3, 2010 #9
    Hello!

    I hope this helps:
    [tex] \Sigma F_y = F_b-F_g [/tex] Right?
    [tex] F = \rho_{air-outside} V_{Ballon} g - \rho_{air-ballon} V_{Ballon} g = g V_{Ballon} \left(\rho_{air-outside}-\rho_{air-ballon}\right) [/tex]

    This eq makes sense. If the density of the surrounding air is greater then the air in the ballon, then the ballon will lift (F is positive). If the density of air of the surroundings is less then the density of air in the ballon F is negative and you sink.

    Rearranging:
    [tex] \rho_{air-ballon}=\rho_{air-outside}-\frac{F}{V_{Ballon} g} [/tex]

    For density use [tex] \rho=\frac{P}{R' T} [/tex]
    You might have to assume that P is taken to be at sea level and use R' for dry air.
    Can you take it from there?
     
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