Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Joule Thompson effect

  1. Sep 10, 2017 #1
    I'm trying to understand the Joule-Thomson effect in a more qualitative way. Here's my attempt at an explanation: Real gases experience intermolecular forces. If we expand a gas whose attractive interactions dominate, it'll cool down. This is due to the potential energy increasing and the kinetic energy decreasing. If we do the same with a gas whose repulsive interactions dominate, it'll heat up. This is due to the kinetic energy increasing and potential energy decreasing. This is where I'm kinda stuck. Why exactly does the kinetic energy increase? I know potential energy is a function of separation (I'm assuming these are just coloumbic attractions), but what about the kinetic energy? Does expansion cause the molecules to speed up?
     
  2. jcsd
  3. Sep 10, 2017 #2
    Mmm... I see how it's confusing. It's hard to explain, but I will try my best to just add some pointers. If you figure out what kinetic energy is first and what it's purpose is, then you can figure out how it can increase. Maybe an unbalanced force (like Newton's First Law Of Motion)? Or maybe something that can trigger it (like how if you shake a bottle of soda and it explodes)?
     
  4. Sep 10, 2017 #3
    Well, I know kinetic energy is the energy of motion. It can be changed by changing the velocity or mass. In this case, it'd be the velocity (i.e. we're accelerating the molecules). So expanding the gas somehow causes the molecules to accelerate. When the repulsive interactions dominate, that is. After we expand the gas, the pressure is lower and there are fewer molecules present but the somehow have more kinetic energy. Also, we're assuming the internal energy is constant. I think the repulsions are due to the size of the molecule. Still stuck...
     
  5. Sep 10, 2017 #4
    Are you trying to understand JT with respect to a porous plug or with respect to removal of a barrier in a rigid closed container?
     
  6. Sep 10, 2017 #5
    I didn't even consider the latter condition :nb). A gas actually increasing in temperature upon expanding in a vacuum?! I was just thinking about a porous plug.
     
  7. Sep 10, 2017 #6
    For an ideal gas, the cooling resulting from gas expansion is exactly cancelled by viscous heat generation within the pores.
     
  8. Sep 11, 2017 #7

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    If there is repulsion then that must happen because the energy has to go somewhere. But is that a likely situation? I only know of attraction between gas molecules. Do you know of examples where there is repulsion in a gas?
    I could think of a situation in which a solid is released from compressive forces could decrease its internal potential energy and thus it could heat up a surrounding gas.
     
  9. Sep 11, 2017 #8

    Lets consider the case of repulsive interactions.

    Two molecules approach each other and slow down while doing that. A collision with a container wall may occur when molecule is moving slowly, in which case thermal energy goes from the wall to the molecule. The wall gets the energy back at some later time. The energy was stored in the gas for a while, the gas has some extra heat capacity when that process is going on.

    Now, if the heat capacity of an object goes down, then the temperature of said object goes up.
     
  10. Sep 11, 2017 #9
    Gases such as hydrogen and helium will experience a warming effect upon expansion under conditions near room temperature. The sign of the Joule-Thomson coefficient depends on whether one works at temperatures above or below the Joule-Thomson inversion temperature at a given pressure.
     
    Last edited: Sep 11, 2017
  11. Sep 11, 2017 #10
    @Chemmjr18 Let's look at this from a more fundamental point of view.

    1. Do you accept the first law of thermodynamics?
    2. Are you familiar with the open system (control volume) version of the first law of thermodynamics that tells us that the change in enthalpy of a gas passing through a porous plug is zero? Do you accept this result?

    I'll continue after you answer these two questions.
     
  12. Sep 11, 2017 #11
    Does it have anything to do with those intermolecular "bonds" breaking and reforming?
     
  13. Sep 11, 2017 #12
    1. Yes, I do
    2. No, at least, I don't think so.
     
  14. Sep 12, 2017 #13

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Wouldn't that imply a change of state?
     
  15. Sep 12, 2017 #14
    Well, I can tell you that the key starting point in understanding the Joule Thompson effect in continuous flow of a real gas through an insulated porous plug is the application of the open system version of the first law of thermodynamics. This tells us that the enthalpy change between the inlet and outlet of the plug (per mole of gas) is zero: $$\Delta H=H(T_2,P_2)-H(T_1,P_1)=0$$This relationship can be used as the basis for accurately determining ##T_2##, given the values of ##T_1##, ##P_1##, and ##P_2<P_1##. But, before I show how this can be done and demonstrate the relationship between JT and the compressibility factor z of the real gas, I need you to read up on the derivation of the open system version of the first law of thermodynamics. This derivation can be found in all decent thermodynamics textbooks. Please let me know when you have assimilated this information.

    Chet
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Joule Thompson effect
  1. Joule Thomson Effect (Replies: 3)

  2. Joule Thomson Effect (Replies: 4)

Loading...