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Why Temperature increases by increasing pressure?

  1. Feb 7, 2010 #1
    Temperature of gas is due to kinetic energy of molecule & pressure is due to molecule collide to container wall with some kinetic energy.
    If I decrease volume of container it increase pressure & Temperature. For pressure it is clear that more no. of molecule is struck to container wall due to decrease in container area & so increase pressure.

    But how the Temperature increases?
    How kinetic energy of molecule increase by decreasing volume? Which phenomena is responsible for it at microscopic level?
     
  2. jcsd
  3. Feb 7, 2010 #2

    russ_watters

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    Take a billiards table where the balls are spread out on the table. Now sweep your arm across the table. What happens to the balls in the way? They move. You've just imparted kinetic energy to them.
     
  4. Feb 7, 2010 #3
    According to this, it is the mechanical action that puts the walls to move that delivers energy to the gas, is it?

    Therefore, if one is able to displace the wall a little bit so that during this process no collision happens there will be no energy delivered. I am correct thinking this way?

    Best Regards

    DaTario
     
  5. Feb 7, 2010 #4
    Take russ watters analogy a step further.Imagine the gas trapped in a cylinder by a piston.If the piston is stationary the molecules bounce off with the same KE.If the gas is compressed by moving the piston the molecules bounce off with increased KE and the gas warms up.The faster the compression the greater the temperature rise.
     
  6. Feb 7, 2010 #5

    Q_Goest

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    Sounds like a maxwell's demon type question.
     
  7. Feb 7, 2010 #6
    As stated by russ_watters, if you compress rapidly, the temperature will increase. This is a non-equilibrium process. However if the system thermally interacts with the environment [http://en.wikipedia.org/wiki/Canonical_ensemble" [Broken]], then once equilibration with the environment takes place, the average energy is independent of the volume :

    If the system thermally interacts with the environment, then after the system has reached thermal equilibrium, the average energy is given by (see http://en.wikipedia.org/wiki/Equipartition_theorem" [Broken]) :

    [tex]
    AverageEnergy = \frac{3}{2} NkT
    [/tex]

    where N is the number of particles. i.e. the mean energy, after equilibration, is independent of the volume.
     
    Last edited by a moderator: May 4, 2017
  8. Feb 7, 2010 #7

    russ_watters

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    It's not a question of speed. If you compress an insulated piston-cylinder arrangement at an arbitrarily slow speed, it will heat up.
     
  9. Feb 7, 2010 #8
    Yes you are correct. I was not trying to imply that speed was required for heating it up.

    Just to add, if the system is thermally interacting with the environment & is initially in equilibrium, then we can ensure that it remains arbitrarily close to equilibrium during compression by ensuring that the compression procedure is performed arbitrarily slowly. c.f. Reif (Fundamentals of statistical and thermal physics)

    "Such a process is said to be 'http://en.wikipedia.org/wiki/Quasistatic_process" [Broken]') that the system requires to attain equilibrium if it is suddenly disturbed. To be slow enough to be quasi-static implies that one proceeds slowly compared to the time [tex]\tau[/tex]"
     
    Last edited by a moderator: May 4, 2017
  10. Feb 7, 2010 #9
    In a diesel engine, I believe for air PV1.4 = constant and TV0.4 = constant during compression.. Look at the equations in

    http://www.ent.ohiou.edu/~thermo/Intro/Chapt.1_6/Chapter3c.html [Broken]

    Bob S
     
    Last edited by a moderator: May 4, 2017
  11. Feb 7, 2010 #10
    @Bob S : what you have stated is true for http://en.wikipedia.org/wiki/Adiabatic_process" [Broken] processes (i.e. processes where no heat is transfered).

    For a quasistatic process where the system is thermally interacting with the environment (i.e. heat transfer allowed & compression performed infinitely slowly), the temperature of the system does not change.
     
    Last edited by a moderator: May 4, 2017
  12. Feb 7, 2010 #11

    rcgldr

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    For the opposite case, in the case of a "free expansion" of a gas, such as removing a panel separating a chamber with gas from chamber with no gas (vacuum), no work is done and there is very little change in temperature.
     
  13. Feb 7, 2010 #12

    morrobay

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    Just considering the temperature increase in terms of heat, it is a result of the conservation of energy. A gas absorbs heat in expansion and must release same in compression.
    Consider the heating by compression and cooling by expansion in a refrigerator
     
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