Consider a system(S) + reservoir(R) that surrounds the system(S). The system(S) is not in thermodynamic equilibrium but the combined system (system + reservoir or say SR) is isolated i.e. the walls of the reservoir(R) are made impermeable for all forms of matter, energy, force exchange with the outside world. The system (S) is allowed to interact with the reservoir(R) by exchanging particles, energy etc.. If we use first and 2nd law of thermodynamics to show that the Gibbs free energy of the system(S) always decreases (and becomes minimum when it reaches equilibrium), there is an assumption that the internal energy and volume of the combined system (SR) do not change with time. I have the following questions:

  1. I do not understand why the volume of the combined system (SR) remains constant with time ? Is this coming from conservation of mass ? I do understand that the boundaries of the combined system (SR) can not change as such a change requires external work on the boundaries of the combined system (SR) which is prohibited by the definition of isolated system.
  2. Assume that there is an internal process that leads to discontinuities (for ex. voids, holes, cracks etc.) within the system(S)/reservoir(R) and therefore changes the volume (= original volume of combined system - volume of discontinuity) of the combined system. In such a case, the volume of SR is not constant. However, this situation leads to creation of additional internal boundaries. Is this even a possible scenario or is it contradicting some of the assumptions ? Creation of internal boundaries appears to contradict some of the assumptions.

It would be great if some one can throw light on these questions perhaps with the help of equations with clear assumptions.

  • $\begingroup$ Your description "system + reservoir" is too vague please elaborate. $\endgroup$
    – trula
    Apr 10 at 21:06
  • $\begingroup$ The system is open and it is allowed to exchange particles, energy with the reservoir and its volume is allowed to change. Can you please ask a more specific question if my answer is not sufficient ? Thanks $\endgroup$
    – Shravan
    Apr 10 at 21:11
  • $\begingroup$ What do you mean by the "combined" system? Exactly what does the system consist of? You stated to trula that the system is "open". A system cannot be open and isolated at the same time. Your post lacks sufficient detail to understand your question no less than answer it. $\endgroup$
    – Bob D
    Apr 10 at 21:17
  • $\begingroup$ Suppose that the combined system is bounded by a rigid insulated room. $\endgroup$ Apr 10 at 21:20
  • 1
    $\begingroup$ If your system’s volume isn’t constant, don’t make that assumption. But then your system needs to be surrounded by a vacuum, or work done on or by the surroundings violates the premise of isolation. Does this get at what you’re asking about? $\endgroup$ Apr 10 at 23:08

1 Answer 1


If you use the Gibbs free energy to determine the equilibrium state of the system (a subsystem of SR), it implies that you are considering a process at constant temperature and pressure.

  1. You mentioned:

... I do understand that the boundaries of the combined system (SR) can not change as such a change requires external work ...

That is incorrect. Because if you consider some free particles placed in a vacuum, they can expand freely without any external work being done. The reason why the volume of SR remains constant is that you are considering a process at constant pressure. Suppose all particles in SR are ideal gases, then you can understand why the volume of SR must remain the same, or else the pressure would change.

  1. I don't know what you mean. But you cannot break a space, I guess.

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