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Ideal Gas Law and balloons

  1. Feb 11, 2016 #1
    1. The problem statement, all variables and given/known data

    Two balloons with Helium gas are filled, first with 10 liters of He and second with 20 liters. Molecules of which balloon will be moving faster as compared to the other?


    2. Relevant equations


    3. The attempt at a solution

    Speed of molecules is directly proportional to temperature . Consider the fact that pressure inside balloon is same as ambient pressure i.e pressure is constant ( Atm. pressure) . Applying Ideal gas law PV = nRT ,temperature depends on 'n' and V . Volume is given but , 'n' is not given .

    Is the question incomplete or am I missing something ?

    Many Thanks
     
  2. jcsd
  3. Feb 11, 2016 #2

    Bystander

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    You "am" missing something.
     
  4. Feb 11, 2016 #3
    Are you suggesting that 'n' i.e number of moles can be calculated from the volume given ? Should n1 = 10/22.4 and n2 = 20/22.4 ?
     
  5. Feb 11, 2016 #4

    Bystander

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    Preference would be some unspecified constant, but otherwise, yes.
     
  6. Feb 11, 2016 #5
    But what I did (division by 22.4) was under the assumption that 1 mole of an ideal gas occupies 22.4 liters at "STP" i.e at 273K . Why should we assume that gas is filled at a constant temperature of 273K ?
     
  7. Feb 11, 2016 #6

    Bystander

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    Hence, the UN-specified constant.
     
  8. Feb 11, 2016 #7
    Sorry . I do not understand . Could you please explain ?
     
  9. Feb 11, 2016 #8

    Bystander

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    You multiply by a constant; you then divide by that same constant; what do you get? Rid of the constant.
     
  10. Feb 12, 2016 #9

    haruspex

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    There is not enough information in the problem statement. I assume the balloons are initially identical, but are we considering them immediately after filling or on return to ambient temperature?
    In the attempted solution it says the balloon contents are at atmospheric pressure, which seems unlikely. Is this a given or just part of the attempted solution?
     
  11. Feb 12, 2016 #10
    I interpret this as immediately after filling .

    I think the gas inside the rubber party balloons are always at a pressure just above the ambient pressure (approx. equal to ambient pressure ) . It is part of the attempted solution.
     
  12. Feb 12, 2016 #11

    haruspex

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    Well, that is rather a critical question, no? But if so, what would that say about the temperatures?

    Either the volumes are quite different or the pressures are. Or both. Without knowing the relaxed volumes of the balloons it is impossible to say how the two volumes and pressures will change. But as I hinted above, I don't think you need to investigate that beyond some basic assumptions.
     
  13. Feb 12, 2016 #12
    This is what we need to find :smile: . Translational speed of molecules is directly proportional to temperature . Higher the temperature higher the speed.

    Volume of one is double that of the other . Pressures are same . This is my thinking .
     
  14. Feb 12, 2016 #13

    haruspex

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    Sure, but you are only assuming the question relates to just after filling the balloons. That assumption affects the answer.
    How far do you think 'adiabatic' would apply here?
    It would be somewhat strange for a balloon to be filled to double the size without some increase in pressure. It need not be much to affect the answer.
     
  15. Feb 12, 2016 #14

    ehild

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    They are not party balloons I am sure. Can be meteorologic ones - of the same volume when filled, which means when they are taut. That also means about the same pressure, a bit higher than atmospheric. But then the temperature must be different
     
  16. Feb 12, 2016 #15
    In that case the 20L would have double the number of moles and half the temperature as compared to 10L one ??
     
  17. Feb 12, 2016 #16

    ehild

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    I think so.
     
  18. Feb 12, 2016 #17
    Fine . And molecules in 20L balloon would be moving at slower(half) average speed than the 10L balloon ??

    What if we change the question a bit and take two party balloons and we put in 1L and 2L of air ( i.e volume of air inside =volume of balloon) . Do you think we would consider temperatures to be same (ambient temperatures) ??

    Since Pressure and temperatures would be same in this case , the Volume would be directly proportional to the no of moles of air put in . The number of moles in 2L balloon would be double that of 1L balloon .But since temperatures are equal , in this case molecules in 2L balloon would be moving at same average speeds as that in 1L balloon

    Does that seem correct ??
     
  19. Feb 12, 2016 #18

    ehild

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    Yes, in case of ideal gas.
    But how can you fill one party balloon with twice as much gas so as the pressure is the same in both? The pressure should balance the tension of the rubber wall, which must be higher if the volume is greater.
     
  20. Feb 12, 2016 #19
    Ok. How would two meteorological balloons as we interpret in the OP have equal gas pressures i.e same as ambient pressure ? Are they closed or do they have an opening ?
     
  21. Feb 12, 2016 #20

    haruspex

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    How can it have half the temperature?!
    You put 10L in each balloon, so far they're the same. Now you put another 10L in one of them. Why should the temperature go down dramatically? Half the temperature (K) would be extremely cold.
     
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