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Ideal Gase Rigid Container Experiment

  1. Nov 19, 2012 #1
    Let's assume that we have a non insulated glass container where there is any gas inside.

    If the beginning pressure is equal to the atmospheric pressure and then the gas is pressurized at ,say, 3 Bar . Then what kind of process would occur if we let a small amount of gas escape over a very short period of time ?


    Please let me know if my thinking is correct :

    When the container is opened then clearly work is done ( i.e if we put a light object on the opening,it would move by a certain distance).Therefore the process could be anything but isochoric,right ?

    Pressure is also not constant because it becomes smaller by an amount,say, dP.

    So the process could either be adiabatic or isothermal,correct ?

    dU + dW = dQ so the work done is equal to dP*dV , but we said that the volume is constant so this is where i become confused . And don't forget that the container is not thermally insulated.
    ~

    Thanks in advance for your help
     
  2. jcsd
  3. Nov 19, 2012 #2
    I think this system as you have described it is neither isochoric, adiabatic, nor isothermal.

    There is certainly work done, the gas expands, but I think you're pushing the limits of the definition of a global thermodynamic equilibrium. Without further constraints or assumptions (eg. the container is insulated, and the gas expands into a vacuum, the gas is ideal) I think your question is so general the answer probably depends on Navier-Stokes equations rather than thermodynamics. What is the density of the gas that it expands into? What are the temperatures? What is the Reylonds number? Surely all these things affect what happens.

    I might be wrong but IMO there is no "process" that describes what you've asked which allows the problem to be reduced to a set of equations in global parameters (that are uniform over space).
     
  4. Nov 20, 2012 #3
    Thank you very much for your answer, although i have to say it wasn't what i expected, which is not your fault . It just seems to complicated.

    Would anything change if we said that the container is thermally insulated ?
     
  5. Nov 20, 2012 #4

    haruspex

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    No, it's dPV = PdV + VdP.
     
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