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Work done by inflated balloon

  1. Aug 29, 2014 #1
    1. The problem statement, all variables and given/known data

    An initially deflated and flat balloon is connected by a valve to a storage tank containing helium gas at 1 MPa at ambient temperature of 20°C. The valve is opened and the balloon is inflated at constant pressure of 100 kPa (atmospheric pressure) until it becomes spherical at D1 = 1 m. If the balloon is larger than this, the balloon material is stretched giving a pressure inside as:

    P = P0 + C (1-D1/D) D1/D

    The balloon is slowly inflated to a final diameter of 4 m, at which point the pressure inside is 400 kPa. The temperature remains constant at 20°C. Determine the work done during the overall process.

    2. Relevant equations

    W = ∫PdV
    PV = nRT

    3. The attempt at a solution

    I solved for the volumes of the ballon at D1 = 1 m and at D1 = 4 m to use for the definite integral. Then I the formula for P in terms of V using the ideal gas law:

    P = nRT/V

    At this point, I realize that I don't know the number of moles in the balloon and that I haven't used the expression given in the problem for P. I know how to evaluate the calculus, but I'm terrible at setting up the integrals. Anyone have a hint for me? Thanks!
     
  2. jcsd
  3. Aug 29, 2014 #2

    Simon Bridge

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    Since the number of moles changes, the state of the gas in the balloon is given by four numbers, (P,V,T,n) related by the ideal gas equation. You have an additional equation to tell you a final P in terms of an initial one. You also know how to get the volumes from the diameters.

    The balloon moves between two states where you know the something about the state variables.

    As usual, list what you know.
    Start with the initial state:
    P0=
    V0=
    T0=
    n0=

    put a question mark next to anything you don't know, then look for an equation that relates an unknown to the other stuff you do know.
     
  4. Aug 30, 2014 #3
    Temperature is constant at 20°C.

    P0 = ?
    V0 = .5236 M3
    n0 = 21.49 mol

    Pf = 400kPa
    Vf = 33.51 m3
    nf = 5502 mol

    My inclination is that I should back-solve for P0 using the given expression, but I don't know what C is. Heat capacity is my only thought, but neither pressure or volume is constant. Am I even close?
     
  5. Aug 30, 2014 #4

    Simon Bridge

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    What's wrong with using the ideal gas equation to get P0?
    Hang on - aren't you told the initial pressure is 100kPa??
    How did you get the number of moles?

    What you need is the equation for P as a function of volume.
    This means you need a value for C - you can determine that from the information you have.
    (Check your idea about C by doing a dimensional analysis - what units should it have?)
     
    Last edited: Aug 30, 2014
  6. Aug 30, 2014 #5
    Not real sure how to interpret the problem there. I know the initial pressure in the tank is 1MPa and that the pressure added to the balloon is 100kPa, but I assume that P0 is the pressure when the balloon becomes spherical. Am I misinterpreting?

    EDIT: Just realized that I calculated n0 using 100kPa as P0.

    That said, it can't be right because filling the balloon at that pressure doesn't mean that the balloon is spherical at 100kPa.
     
  7. Aug 30, 2014 #6
    From the given expression, C should be in Pascals.
     
  8. Aug 30, 2014 #7

    Simon Bridge

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    The problem statement says the baloon is spherical at 100kPa - which it calls "atmospherc pressure".
    The clue is that the balloon if filled at that pressure until it stops getting bigger, and it has not stretched yet.

    This means you have a complete description of the initial and final states of the air in the balloon.
    From that you can compute C.
    ... i.e. C should have units of pressure.
    What are the units of "heat capacity"?

    The work done on the gas is the area under the P-V graph.
    This is where the P function comes in.
     
  9. Aug 30, 2014 #8
    Heat capacity has units of J/g*K.

    I plugged 100 kPa in for P0, and 400 kPa for P and solved for C yielding a value of 1600 kPa.

    So then I plugged those values into the given expression and simplified yielding:

    P = 100 kPa + 1600 kPa/D - 1600 kPa/D

    So I guess I need this expression in terms of volume to solve for work?
     
  10. Aug 30, 2014 #9

    Simon Bridge

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    It is important to think about what you write down.
    Does this equation make sense? Don't the last two terms cancel out?
     
  11. Aug 30, 2014 #10
    The last term should have been 1600/D2.

    Sorry.
     
  12. Aug 30, 2014 #11

    Simon Bridge

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    OK:
    ... does this make sense?
    If the diameter were D=6m, would you expect the pressure to make that happen to be bigger or smaller than the pressure for D=4m?
    Now what does the equation tell you?
     
  13. Aug 30, 2014 #12
    You're right. Doesn't make much sense since the pressure would get smaller as diameter increases.

    Back to the drawing board.
     
  14. Aug 30, 2014 #13
    Kind of stuck. Not sure where to go from here.
     
  15. Aug 30, 2014 #14

    Simon Bridge

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    Well, in the problem statement, initially D1=100kPa and finally D1=400kPa ... what does that suggest to you?
     
  16. Aug 31, 2014 #15
    I realize that pressure and volume are changing. I assume that I need an expression for pressure in terms of volume. Since the given expression for pressure is in terms of diameter, I solved the equation for volume of a sphere for diameter, but, plugging that into the given expression for pressure yields an expression that results in a far more complex integration which suggests to me that something isn't right. (Most of the solutions to the problems in this book are conceptual challenges with simpler computations.)
     
    Last edited: Aug 31, 2014
  17. Aug 31, 2014 #16

    Simon Bridge

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    There are two diameters in your equation - which is the diameter that the problem statement says is changing?
     
  18. Aug 31, 2014 #17
    Not sure but tend to think it's D1.
     
  19. Aug 31, 2014 #18

    Simon Bridge

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    So what happens if you redo your derivation using D1 as the variable?
     
  20. Sep 2, 2014 #19
    When I integrate the expression for pressure in terms of volume, I get an answer that is close to the correct answer but still off by a few thousand joules.

    I think my expression is wrong. I took the equation for volume of a sphere and solved it for diameter. Then I took that solution and plugged it into the given expression for D1. Is that not right?
     
  21. Sep 3, 2014 #20

    Simon Bridge

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    Well that's what I would have done - after getting the correct value for C.
     
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