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Calculating temp change releasing a compressed gas

  1. May 17, 2015 #1

    I understand when you release a compressed gas generally it comes out cooler than atmosphere.

    But here are my questions:

    When you fill the container, the gas warms up as the pressure builds. Generally it takes a long time to compress and I believe the temperature of the gas when it's released will vary by the rate of release. Is there anyway to calculate what the temp will be based on Total vol. of the container, pressure at the time of release and the rate of release?

    Also If you compress it and immediately release it, what is the temp change compared to if you compress it today and let the container cool to ambient tomorrow? Or if you put the container in the freezer overnight? How much cooler would the gas coming out be?

    (assume just general air for the gas.)

    Now what if the air being released it then feed into a vortex tube (aka Maxwell's Demon.) Will the vortex tube separate the released air into even colder air on the cold side then what comes out of the container? Ie, the air released would be concentrated into the warmest parts and coldest parts, so the resulting air would be even lower than what it comes out of the container? (Since what comes out of the container is an average of all the air molecules.)

    And if this true, then is there some way to calculate how much (% of the original containers vol.) going to the "hot" and "cold" sides?

    Sorry for such a long question, but it's a question I have been trying to understand now for a long time. Trying to quantify is as much as possible.

    Thank you.
  2. jcsd
  3. May 17, 2015 #2


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    Staff: Mentor

    As long as the pressure isn't ridiculously high, you can just ratio the pressures and temperatures using the ideal gas law. That answers most of your questions: not sure about the relevance of Maxwell's demon here.
  4. May 17, 2015 #3
    Pressure is 4500psi.

    Not sure if you consider that "ridiculously high" or not.
  5. May 18, 2015 #4
    Sorry boss, but I'm not getting it.

    I have 2 transformations.


    I can't solve this because I don't know V1.

    I only know the pressure I'm staring with 1ATM, Temp I'm starting with 100F, or 310K

    So T2=P2V2T1/P1V1

    But at this point I don't know T2 because I don't know V1.

    Then I have a 2nd transformation:


    T3 is what I'm looking for.

    So T3=P3V3T2/P2V2

    But using the above I found T2 is P2V2T1/P1V1 (as a function of V1)

    So T3=(P3V3)(P2V2T1)/P1V1)(P2V2) or P3V3P2V2T1/P1V1P2V2) or P3V3T1/P1V1 (P2's and V2's cancel) But V1=V3, so I'm left with P3T1/P1 but P3 and P1 are the same 1ATM so this shows T3=T1, Which can't be right because we all know the gas goes in ambient temp, compresses, heats up and then comes out colder, usually much colder. Right?

    So where did I go wrong?

    Also nothing in here takes into consideration TIME. The first transformation takes a long, long time. The 2nd takes just seconds. That has to have some effect on this.

    But it still does not answer the question: If I compress the gas, it heats up in the container to some higher temp than what I started with, to what I call T2. But then over time, T2 eventually ( I think) goes back to ambient or T1.(maybe I'm wrong, but that heat is going to escape that container no matter how well I insulate it.) So then we I start the 2nd transformation, T2 is not really T2, it could less (if I put it in the freezer) or more (if I warm it) II still need to know what the ultimate temp of the escaping gas will be. And how much cooling the container at the end of the first step will effect the final temp of escaping gas.
  6. May 18, 2015 #5
    Or is the problem my expectations? Do we only get a temp drop of it involves a phase change? Compressing "air" will not get a phase change. :-(
  7. May 18, 2015 #6


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    Staff: Mentor

    Yes, 4500 psi is "ridiculously high" and the ideal gas law won't work. What happens in this case will be highly dependent on the shape of the nozzle, but the potential exists for liquefying the air: that's how industrial gas plants do it.

    I was thinking more in terms of what you might find when you pump up and/or let the air out of your bike/car tires.
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