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Question about the ideal gas law

  1. Apr 11, 2006 #1

    I have a question about the ideal gas law.

    I have been under the impression that if volume goes down then pressure and temperature go up. But, if you look at the equation T=PV/nR, it seems that if volume, say, doubles, pressure will be halved and vice-versa. While that makes sense, it also leaves the temperature constant. Doesn’t that conflict with real life observations?

    Thank you,

  2. jcsd
  3. Apr 11, 2006 #2


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    Gold Member

    It depends on the specific situation.

    For example, if you have a cylinder of gas that is at a fixed volume then obviously V can't change. In that case when you double pressure you will double temperature because PV=nRT.
    Last edited: Apr 11, 2006
  4. Apr 11, 2006 #3


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    Pressure, density (n/V), and temperature are 3 quantities related by the gas law. If you change one of them, you don't know how the other two change; you need more information. So for example, if you slowly compress a gas that's in contact with a heat reservoir, then T=constant and so PV = constant. OTOH, if you do the compression while not allowing heat to escape, it's an adiabatic process, and temperature will increase. It turns out that in that case, [itex]PV^\gamma[/itex] = constant, where gamma is a constant > 1 related to the specific heats.
  5. Apr 11, 2006 #4
    Follow-up question

    Thank you for your replies.

    Basically, I am looking for the right way to calculate how much the temperature of a body of gas will increase in a cylinder if it’s compressed so much by a piston.

    Could you point me in the right direction?

  6. Apr 12, 2006 #5


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    Well if you consider the compression to be adiabatic (which it is likely to be in a piston) you can use the following equation;

    [tex]\frac{T_{f}}{T_{i}} = \left( \frac{V_{i}}{V_{f}} \right)^{\gamma -1} = \left( \frac{P_{f}}{P_{i}} \right)^{\frac{\gamma -1}{\gamma}}[/tex]

  7. Apr 12, 2006 #6


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    Staff: Mentor

    It depends on the situation. If the piston is in equliibrium with a constant-temperature reservoir, and heat can flow rapidly enough between the piston and the reservoir, then the temperature of the gas inside the piston remains constant. This is more likely to be possible if you compress or expand the gas slowly.

    On the other hand, if you insulate the piston so as to isolate it thermally from its surroundings, or if you compress/expand the gas rapidly enough that there's not enough time for a significant amount of heat to flow in or out, then you have an adiabatic process described by Hootenanny's equation.
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