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Ideal Gas Law and a balloon

  1. Mar 23, 2004 #1
    Hey guy,
    I didnt think i would ever have to deal with gas laws after i got out of AP chem, but it has showed up again in physics. Tell me what to do.

    Suppose the volume of a balloon decreases so that the temperature of the balloon decreases from 280K to 240K and its pressure drops from 1.6x10^4 Pa to 1.7x10^4 Pa. What is the new volume of the gas.

    My question is what was the initial? How do i work this prob.? Thanks for any help.
  2. jcsd
  3. Mar 23, 2004 #2
    Let V1 be the unknown initial volume. Let V2 be the unknown final volume. At this point, I am not sure exactly what equation you are using. I assume the following:

    PV = nRT, where n = number of moles, R = universal gas constant, P = pressure, and V = volume.

    It does not seem to me that you need to know the initial volume. Just solve for V = nRT/P and put in the final values for pressure and temperature. Or am I missing something or you left something out?
  4. Mar 24, 2004 #3
    I need to know the New Volume of the gas. We have to use PV=NKbT where Kb is 1.38x10^-23. I was trying to use P1V1/T1=P2V2/T2. I just dont know where to start.
  5. Mar 24, 2004 #4


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    The problem is that he doesn't know what "n" is. He has 2 variables and only 1 equation.

    To solve the problem, I'm guessing you would need a formula for n or V that does not have the other in it. Then you could solve 2 variables with 2 equations.
  6. Mar 24, 2004 #5
    The ideal gas law says the volume of an ideal gas is proportional to its temperature and inversely proportional to its pressure. So when you know the initial volume V1 and the fraction by which the temperature increases T2/T1 you can calculate the final voulume: V2=V1(T2/T1). Considering a simultanuous change of pressure from p1 to p2 the final volume can be calculated to be V2=V1(T2/T1)(p1/p2).
  7. Mar 24, 2004 #6


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    He said in his first post that he doesn't know what V1 is.
  8. Mar 24, 2004 #7

    Doc Al

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    da_willem's answer is the best you can do with the given information.
  9. Mar 24, 2004 #8
    P1V1 = nKbT1
    P2V2 = nKbT2

    These are 2 equations with 3 unknowns, no? Need another independent equation. I suspect it involves n for air.

    Or, maybe all you need is the RATIO of final volume to initial volume:

    V2/V1 = (P1/P2)(T2/T1)
  10. Mar 24, 2004 #9
    Hey guys... uh, i really didnt think the prob. would be soo difficult. Let me tell you my reasoning behind the way i worked it, and you tell me if you think it is ok.

    #1. The pressure inside the balloon is 1 m3. I just assumed that.
    #2. I plugged that into the P1V1/T1=P2V2/T2 equation, which gives me that V2=.807 m3 i believe, i dont have my work in front of me.

    Does that sound good?
  11. Mar 24, 2004 #10
    I ment that the volume is 1 m3 not the pressure
  12. Mar 25, 2004 #11


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    PV=nRT, the n remains the same.

    so V=RT/P.

    In each case, before and after, we can find V since we know both T and P. So


  13. Mar 25, 2004 #12

    Doc Al

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    No. While the n remains the same, since you don't know what it is you can't use it to find V. (And you can't just drop it from the equation.) The best you can do is find the final volume in terms of the initial volume.

    Unless, of course, you are willing to just assume a value for n or for the initial volume. (Like Virmeretrix did when he arbitrarily set the initial volume equal to 1 cubic meter--why?)
  14. Mar 25, 2004 #13
    I don't know what Viry's textbook says, but "one gram-mole of an ideal gas occupies a volume of 22,400 cm3 or 22.4 liters at "standard conditions" or at 'normal temperature and pressure' (NTP), that is, at a temperature of 0degreesC = 273degreesK ((sic)) and a pressure of 1atm = . . ."

    The first example in my textbook for the ideal gas law involved oxygen, and treated it as tho it were an ideal gas.

    I believe Viry now has everything he needs to solve the problem.
  15. Mar 31, 2004 #14
    Ok got the answer.

    GeneralChemTutor is right.
    The n, which in this case is the molecules in the balloon, remains constant sense the balloon is a closed system. Once again, we were not to use R, the universial gas constant, but rather Kb, know as the bezionne or something constant.
    Sense the question did not state weather or not the system was at STP, you have to assume its not.
    So, [P1V1=N1Kb1T2]=[P2V2=N2Kb2T2]. Sense the balloon is a closed system the n(number of moles/atoms/molecules) no matter what the numerical value, devides out. So in each case, V1=KbT1/P1 which would give you the initial volume to be 2.4x10^-25 m3. and V2=KbT2/P2 gives you 1.9x10^-25 m3.
    Pretty small balloon huh?
  16. Apr 1, 2004 #15
    Uh, oh, does the -trix ending mean Viry is a female? If so, I apologize.

    No, you still need to know n or N. The answers you got is for N = 1. One atom, I think? That would be a small balloon, all right.
  17. Apr 5, 2004 #16
    Actually, I'm a male. The name is a combination of two latin words. Vir- meaning man, and meretrix, meaning W*ore. So, put the two together. And you say im still wrong. Then what do you suggest?
  18. Apr 5, 2004 #17
    It's clear by now that either you copied the question wrong, or forgot to include vital information. Whether you use R or Kb, you still need to know n or N, which you don't. As has already been said, the most you can find is the ratio between the old volume and the new volume, or the new volume as a function n or N.
  19. Apr 5, 2004 #18
    That is incorrent. Ignoring for a second that P1V1 does not equal P2V2 or N2Kb2T2, you cannot say that because:
    2x = 2 = y = 2
    You can divide only two parts of the equation by 2, to get:
    2x = 1 = y = 1
    If you want to divide by 2, you must divide all parts:
    x = 1 = y/2 = 1
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