Air pressure/temperature relationship

  1. When you compress air it heats, and when you decompress it it cools, but how much?
    Working from's_law I find the equation
    so I derived for example
    if the air is 10°C (283.16°K) and the pressure is 14PSI so...
    14/283.16 = 0.04944
    increase the pressure by 90 PSI
    104/2103.5 = 0.04944
    maintian pressure, but decrease temperature to almost ambient (16.84°C)
    104/290 = 0.3586
    decrease pressure to ambient
    14/39.04 = 0.3586
    i.e. I compress a tank of air to 90psi above ambient air pressure of 14psi at 10 degrees celcius, I let it cool to almost ambient (just for easier math) and when I shoot the air out it decompresses and cools to 39 Kelvin. It doesn't sound right.
    Then I realised that equation required mass and volume to remain constant. gives me a more complete equation
    (pressure x volume)/temperature = constant
    but that would mean when I increassed the pressure by compacting the air I would, for example, decrease the volume by ten times thus increasing the pressure ten times and the temperature would remain constant.

    Please, what am I missing???
  2. jcsd
  3. Andrew Mason

    Andrew Mason 6,921
    Science Advisor
    Homework Helper

    During an adiabatic compression, work is done on the gas. As you point out, this increases the temperature (internal energy) of the gas above the temperature of the surroundings. When you then let it cool by letting it remain in thermal contact with the surroundings, you are letting that internal energy decrease to its original value (ie. removing all the internal energy you added by doing work on it). Because its temperature goes down, the pressure must go down. You appear to be assuming that the pressure remains unchanged.

    Because the gas is contained in a smaller volume the pressure is still much greater than the original pressure (Pf/Pi = Vi/Vf if T is the same). To find the compressed volume, you have to use the adiabatic condition: [itex]PV^\gamma = \text{Constant}[/itex] where [itex]\gamma = C_p/C_v[/itex].

    The gas then expands and does work on its surroundings. Since no heat flow occurs, this work is being done at the expense of its internal energy. So its internal energy must decrease (ie temperature goes down) by an amount equal to the work it does. (This is essentially the way a refrigerator works). Again, you have to apply the adiabatic condition to find the change in volume (the final volume is smaller than the original because it stops expanding when the pressure reaches the ambient pressure). From that final volume, you can determine the final temperature.

    Last edited: Dec 19, 2010
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