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Pv = nrt / ideal gasses

  1. Jan 30, 2008 #1
    ok this might should be an easy question, so here goes.

    Lets say you have air in a balloon, floating in a room. The pressure outside the balloon is 2 atm, and the pressure in the balloon is 2 atm the volume is 1 L and the temp is 300K (arbitrary values)

    Now lets decrease the pressure in the room to 1 atm. this should let the air in the balloon expand, increasing in volume to 2 Liters and the pressure to 1 atm. does anything happen to the temperature? im not sure. because the gas is expanding and decreasing in pressure, i would think that the temperature should drop. If so, by what factor? by 4? (divide by 2 for change in pressure and 2 for change in volume) and if not, why not
  2. jcsd
  3. Jan 30, 2008 #2
    ah nevermind i found my mistake... you divide by 2 for pressure then multiply by 2 for volume, so no change in temp

    so as a followup question, is there any circumstance that the decrease in pressure would lead to a decrease in temp instead of an increase in volume, or do they average, or what
    Last edited: Jan 30, 2008
  4. Jan 30, 2008 #3

    D H

    Staff: Mentor

    I was going to write a detailed response, but had second thoughts. This looks a bit too much like homework, and we don't do your homework for you. Instead, we help you do your homework.

    I will say one thing: The balloon will expand to 2 liters if and only if the balloon's temperature remains constant. This is a direct consequence of pV=nRT. What conditions are needed to keep the balloon's temperature constant? What happens if these conditions aren't met?
  5. Jan 30, 2008 #4
    hah, its not homework, im in college, and am done with chemistry. It is part of an experiment (outside of school) i am doing i would really like a detailed response if possible. anyways, I hope this doesnt sound conceited, but i dont get homework help, i like to turn in my own work.
    Last edited: Jan 30, 2008
  6. Jan 30, 2008 #5
    the reason this is a difficult problem, is because it is a real life application, and i cant hold the temperature constant
  7. Jan 30, 2008 #6
    i was thinking if the temperature was not held constant, that the decreased temp would cause a decrease in volume and it would settle out at a 50% increase in volume, 50% decrease in temp from a 100% decrease in pressure
  8. Jan 30, 2008 #7


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    Hi A(s),
    If you put a control volume around the balloon, you should note that the balloon expands while doing work (pV) against the atmosphere. If there's no heat transfer (ie: the expansion is adiabatic), the expansion of the balloon is isentropic. In real life there is some heat transfer, so you can either take that into account or neglect it. If the expansion is relatively fast, you can neglect it. If you want to take it into account, you can either apply the first law directly to the air in the balloon and determine the rate of heat transfer, or you could note that this is a polytropic expansion and take a WAG at the exponent.
  9. Jan 30, 2008 #8
    wow, i never even thought about the effect of heat transfer... that could be significant in my case...

    as far as polytropic processes go, i think that puts me on the right track, I know n does not equal zero, because a decrease in pressure is the trigger for the event, so pressure cannot be constant. i guess that leaves somewhere between 1 and 2, and infinity? so how can i calculate the exponent, guessing will do me no good here : )
  10. Jan 31, 2008 #9


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    The exponent is generally bounded by the minimum of 1 for an isothermal case, and a maximum of the ratio of specific heats for an isentropic case. Isothermal gives you complete heat transfer to keep the air at the same temperature and isentropic gives you adiabatic conditions meaning no heat transfer. So for a diatomic gas like air, the polytropic exponent can range from 1 to 1.4.
  11. Jan 31, 2008 #10
    ok, i think that helps, perhaps i should start a new thread, as this has deviated from the original question, but since in my scenario, the fluids are water, and air ( a bubble rising and thus expanding in water) that would change the exponent

    as the bubble rises in the water, the surrounding pressure would decrease, thus allowing it to expand. and also allowing transfer of heat to and from the water

    ps (please ignore drag on bubble, as well as integrity of a bubble)
  12. Jan 31, 2008 #11


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    Yes, I have to believe a bubble in water would result in much more heat transfer than a balloon in air. Depending on expansion rates, it may be closer to isothermal, but it's hard to determine without doing an analysis on the heat transfer.
  13. Feb 1, 2008 #12
    ok, thanks for the leads, dont put too much thought into it. i will get it figured out. there are a lot of factors to consider. i really appreciate the help, so thanks again
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