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Stuck on ideal gas law question

  1. Jul 10, 2008 #1
    1. The problem statement, all variables and given/known data
    On a warm day 90F, the air in aballon occupies a volume of .25m3 and exerts a pressure of 20 lb/in2. If the baloon is cooled to 30F in a refrigerator, the pressure drops to 14.2lb/in2. What is the volume of the balloon


    2. Relevant equations
    P1xV1/T1=P2xV2/T2-----((P1xV1)/T1)xT2/P2
    T1=305.2K
    T2=271.88K

    3. The attempt at a solution
    .31m3 is the answer i keep getting How is this possible when this volume is larger than the original.
     
  2. jcsd
  3. Jul 10, 2008 #2

    Tom Mattson

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    You've got two competing effects here. First, there's a temperature drop, which tends to make the balloon shrink. But you've also got a pressure drop, which tends to make it expand. It just so happens here that the effect of the pressure drop dominates the effect of the temperature drop, hence you have a larger balloon.
     
  4. Jul 10, 2008 #3
    But the pressure drop is inside the balloon.
     
  5. Jul 10, 2008 #4
    There is no way you are going to have a larger balloon when you cool it down in a freezer.
     
  6. Jul 10, 2008 #5

    HallsofIvy

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    What values did you use for T1 and T2? In the ideal gas law, termperatures must be given in degrees Kelvin.
     
  7. Jul 10, 2008 #6
    I listed them in the problem T1 305.2 T2-271.88
     
  8. Jul 10, 2008 #7

    Tom Mattson

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    Yes that's true. And how that pressure drop is achieved, I do not know as the problem doesn't say. All I know is that under the temperature and pressure drops given in the problem statement, you do in fact get a larger volume.

    It could be that the data given in the problem is unrealistic, but your calculation is nonetheless correct.
     
  9. Jul 10, 2008 #8

    dynamicsolo

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    I suspect this is yet another case of someone making up a problem without checking to see if the given information makes sense in terms of a natural process. I don't really believe that the pressure in the balloon would drop to that value by chilling it in the fridge that way. However, the answer of 0.314 m^3 is what you get from the ideal gas law using that information. (Yeah, right: I wanna see how they got a balloon with a diameter over 60 cm. into the refrigerator to start with... Maybe it's a "walk-in"...).

    The point is that, for the purposes of this problem [grinds teeth], the balloon would have to have expanded and cooled from its initial state in order to reach that final state. You did the problem right -- it's not your fault it makes little sense...
     
  10. Jul 10, 2008 #9

    Andrew Mason

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    Your answer is correct.

    The problem here is that you are imagining that this is occurring in the real atmosphere. It isn't.

    If the atmospheric pressure was held constant (i.e as in the real atmosphere), when the temperature decreases the internal pressure decreases and the balloon volume is reduced until the pressure inside balances the atmospheric pressure + pressure of the balloon material. This is what you are imagining is occurring. But it isn't. In this problem you end up with a lower pressure.

    In the problem, you have a balloon at 90F experiencing a reduction in the external pressure from 20 lb to 14.2 This will result in expansion of the gas and, therefore, cooling. In fact, the expansion due to pressure reduction will result in the gas cooling to even less than 30 F. So in the problem, heat actually flows into the balloon.

    AM
     
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