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Before i get to the questions I have; I am a high school student. We are not asked to understand thermodynamics nor real gases. I'm simply trying to understand the topic without diving too much into the calculations/derivations.

  • 1): Why does sudden expansion of (most) gases cause a cooling effect?

The horrible paint

  • 1.1): Looking at the horrible paint drawing of two chambers and a tube connecting them (V1=V2),(P1 > P2). Why is (T2 < T1) after the expansion?
  • 2): Why do helium and hydrogen cause a positive temperature difference while all other gases cause a negative one?

My best attempt to explain these problems was the following: Looking at the Lennard-Jones potential graph, https://www.researchgate.net/figure/Plot-of-the-Lennard-Jones-potential-as-stated-in-eq-11-Using-the-parameters_fig1_281032446 If atoms are super close; there is a repulsion (positive potential energy) and when they are at medium range they attract each other (negative potential energy).

Gases flow through c1 to c2 because of the pressure difference (or because of a piston) and as they go through the tube this pressure difference is converted to negative potential energy for gas molecules. Therefore when leaving the tube, they use their own kinetic energy to get out of this bond (it is not a bond, but I don't think I have any other way of phrasing it); which cools the right chamber. Making T2 < T1

For hydrogen and helium, as they have only one shell, they can get nearer to each other which creates a positive potential energy. This potential energy then turns into kinetic energy when entering the second chamber. Making T2 > T1

I feel like I'm quite off from what's actually going on, so any help is appreciated.

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  • $\begingroup$ Just want to congratulate you for seeking to understand physics conceptually, it is really important to do this, and not get blind by big calculations. $\endgroup$
    – Socrates
    Commented Jul 11, 2022 at 17:56
  • $\begingroup$ Welcome to Physics SE! You are doing very, very well for high-school-level studies. I'm sure you'll be interested that this exact point is the topic of McClure's "The Joule-Thomson coefficient—a molecular interpretation," American Journal of Physics (1971), which considers only a square potential. Note that your dichotomy between H & He and the other gases applies only around room temperature; every gas has its own inversion temperature. $\endgroup$
    – Chemomechanics
    Commented Jul 11, 2022 at 20:28
  • $\begingroup$ See also this discussion. $\endgroup$
    – Chemomechanics
    Commented Jul 11, 2022 at 20:32
  • $\begingroup$ @Chemomechanics I tried to avoid Joule–Thomson coefficient, but i guess i can't run forever. $\endgroup$
    – Quante
    Commented Jul 12, 2022 at 8:21

1 Answer 1

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In the JT flow situation, expansion of the gas is not the only thing that is happening. There is also viscous flow of gas passing through the small diameter tube, and associated dissipation of mechanical energy to internal energy. The gas expansion tends to cool the gas because it is doing work on surrounding parcels of gas. The viscous dissipation tends to increase the temperature of the gas. Whichever of these two effects wins out determines if net cooling or net heating occurs. In the case of an ideal gas, the two effects exactly cancel, but in the case of hydrogen and helium, the viscous heating wins out.

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  • $\begingroup$ There is also viscous flow of gas passing through the small diameter tube, and associated dissipation of mechanical energy to internal energy. Hello, can you explain what you mean by this section? I can't seem to follow. $\endgroup$
    – Quante
    Commented Jul 12, 2022 at 8:38
  • $\begingroup$ Are you familiar with the concept of viscous "heating?" Have you had any experience with fluid mechanics? $\endgroup$
    – Chet Miller
    Commented Jul 12, 2022 at 9:41
  • $\begingroup$ Looked it up. My question is when these fluids are doing work on the adjacent layers (which is transformed into heat); isn't this work done by using the internal energy of the said molecules? Are they not using their kinetic energy? $\endgroup$
    – Quante
    Commented Jul 12, 2022 at 12:01
  • $\begingroup$ No, it is not at the expense of internal energy. It is a result of viscous work being done locally (non-PV work) within the fluid. This dissipation work produces an increase in internal energy/temperature. $\endgroup$
    – Chet Miller
    Commented Jul 12, 2022 at 12:29

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