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This question comes from a doubt resulting from some physics forums. Similar to and related to this question, I have a doubt on Joule expansion raised by reading too much bad faith arguments.

I'm not going to discuss why a rocket works but I'm going to ask a similar question. Suppose you have a big box with two separate compartments inside. The first (say 1) filled with air and the second one (say 2) is just vacuum. The box is rigid, fixed in place and cannot exchange heat with the exterior. A door suddenly opens and air from compartment 1 can pass to compartment 2 as a "free" "adiabatical" expansion.

There is of course no work done nor heat, $\mathrm{d}U=0$. My question now is related to the momentum of the gas. Before the door opens, the gas has an almost fixed center of mass (in compartment 1) that is still, then when the door opens the center of mass accelerates and moves with a certain speed. Later the COM of the gas just stops again (now it is between 1 and 2).

How is the motion of the COM possible without no change in energy? Is $\mathrm{d}U=0$ only if we compare the initial and final stages?

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  • $\begingroup$ Yes. You are exactly right. The initial and final states are both thermodynamic equilibrium states, and, in the final state, there is no "organized" kinetic energy associated with movement of the COM. $\endgroup$
    – Chet Miller
    Commented Dec 19, 2018 at 13:01
  • $\begingroup$ @ChesterMiller so in between initial and final states the com is just badly defined? $\endgroup$
    – Mauricio
    Commented Dec 19, 2018 at 15:38
  • $\begingroup$ I wouldn't say it is "badly defined." It is well-defined mathematically, but we just don't know where it is situated at any one time, and we don't know its velocity. Presumably, if we solved the partial differential equations of fluid mechanics (i.e., the Navier Stokes equations) and energy, we could establish its location as a function of time until it stops moving in the final state. But, in the kinds of thermodynamics calculations we are doing, that is not our objective. $\endgroup$
    – Chet Miller
    Commented Dec 19, 2018 at 15:50
  • $\begingroup$ @ChesterMiller the COM is not still for a few number of particles but for a large number it would be certainly in the geometrical center of the box (assume cube for simplicity). $\endgroup$
    – Mauricio
    Commented Dec 20, 2018 at 16:16
  • $\begingroup$ Not during the transient part of the process. It can't simply jump for the center of the left container to the center of the combined container in zero time. $\endgroup$
    – Chet Miller
    Commented Dec 20, 2018 at 16:19

1 Answer 1

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$U$ is the internal energy (https://en.wikipedia.org/wiki/Internal_energy), not the total energy, which would be $U_{total}=U+U_{CM}$. U does not change at any point, but $U_{CM}$ does. It does because you are applying an external force to keep the box fixed.

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  • $\begingroup$ why would U_CM change? Does it increase/decrease? $\endgroup$
    – Mauricio
    Commented Dec 20, 2018 at 16:57
  • $\begingroup$ In the final state, the kinetic energy associated with CM movement is zero, since the center of mass is no longer moving. $\endgroup$
    – Chet Miller
    Commented Dec 20, 2018 at 17:00
  • $\begingroup$ Let us say the gas expands to the right. Then the CM starts moving in the same direction because the pressure of the gas on the box is on tthe left only, so the box tries to move to move left but a force stops it from moving so the CM moves to the right. Eventually the gas reaches the other end of the box and the presure is on both ends of the box, but larger on the right end, so the force to keep the box from moving is now to the left which stops the CM from moving $\endgroup$
    – user65081
    Commented Dec 20, 2018 at 17:18
  • $\begingroup$ The bottom line is that, in the final thermodynamic equilibrium state, the CM is no longer moving and the KE associated with movement of the CM is zero. $\endgroup$
    – Chet Miller
    Commented Dec 20, 2018 at 18:26
  • $\begingroup$ @Wolphramjonny the Newtonian vision (using forces) is seemingly trivial, I do not understand it from an energy perspective $\endgroup$
    – Mauricio
    Commented Dec 21, 2018 at 9:52

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