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A Free expansion of a real gas vs Joule-Thomson effect

  1. Dec 7, 2018 at 4:42 PM #1
    I am writing a document on the Joule-Thomson effect. But in my research for it I have come across something that I am not completely clear on. In my document I explained the free expansion of an ideal gas and am trying to transition/segue into the Joule-Thomson effect. However now I am having trouble differentiating between the two (free expansion and J-T effect).

    I understand that there is no change of temperature in the free expansion of an ideal gas because the internal energy does not change and the internal energy of an ideal gas is solely a function of temperature, therefore temperature doesn't change

    But what about the free expansion of a real gas? The existence of inter molecular forces in a real gas means that the temperature will change. Can someone explain to me how this is different from the Joule-Thomson effect? Or are they essentially the same and the free expansion of a real gas will inevitably produce the Joule-Thomson effect?

    Any clarification is much appreciated!
     
  2. jcsd
  3. Dec 7, 2018 at 8:20 PM #2
    The starting equations to use for the two-chamber JT effect and the porous plug JT effect are as follows:
    $$dU=C_VdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$
    and $$dH=C_PdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$
    Are you familiar with these equations?
     
    Last edited: Dec 7, 2018 at 8:36 PM
  4. Dec 7, 2018 at 8:40 PM #3
    I am not unfortunately. I assume these are the ones that can be derived from closed and open forms (respectively) of the first law of thermodynamics? I think I can see how I can work with the dH equation to get the formula for dT/dP which is defined as the J-T coefficient.

    edit: just checked that the form for the J-T coefficient got from your second formula is correct.
    I was able to get the same J-T formula from dH = (dH/dp)dp + (dH/dT)dT. Is this formula and the one you showed me equivalent?
     
    Last edited: Dec 7, 2018 at 8:48 PM
  5. Dec 7, 2018 at 8:55 PM #4
    Actually, these equations were derived from $$dU=TdS-PdV$$and$$dH=TdS+VdP$$. The closed system version of the first law of thermodynamics is used to show that, for the two-chamber system, $$\Delta U=0$$ The open system version of the first law of thermodynamics issued to show that, for the porous plug continuous flow system, $$\Delta H=0$$where, in this case H is the enthalpy permit mass of gas passing through the plug.

    Now, for free expansion. In the two-chamber situation, if the volume of the 2nd (vacuum) chamber is made very large, you approach free expansion. I don't quite know how to relate the porous plug system to free expansion.
     
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