1
$\begingroup$

The definition of a reversible thermodynamic process requires in any instant the mechanical equilibrium (equal pressures) and thermal equilibrium (equal temperatures) of the system in a quasi-static processr.

But there are cases of processes in which one of the two kinds of equilibrium cannot be reached.

Can these processes be considered "reversible" anyway?

I'll make two examples

  1. Quasi-static process in a completely adiabatic tank with two different gases at different temperatures: mechanical equilibrium always present, but thermal equilibrium (between the two gases) not necessarily reached. enter image description here
  1. Isochoric quasi-static process of a gas in rigid and diatermic tank: thermal equilibrium always present, but mechanical equilibrium (between the gas and the environment) not necessarily reached.

enter image description here

Improve this question
$\endgroup$
5
  • $\begingroup$ The internal wall in the first example is diathermal, right? What process is occurring in the second example? Since the tank is rigid and with diathermal walls, I don't see any evolution. $\endgroup$
    – Diracology
    Commented Jun 20, 2016 at 16:13
  • $\begingroup$ @Diracology No it is adiabatic too. I'm talking about a generic process, for istance some work or heat is supplied to the systems and the point is, as far as I can see, that thermal (in the first) and mechanical (in the second) equilibrium is not reached $\endgroup$
    – Sørën
    Commented Jun 20, 2016 at 16:16
  • $\begingroup$ But if the walls are all adiabatic, there is no heat exchanged by the systems. $\endgroup$
    – Diracology
    Commented Jun 20, 2016 at 16:19
  • $\begingroup$ @Diracology Yes sorry I meant work (and no heat) is exchanged with system $1.$ for istance moving the internal wall, and heat (no work) is exchanged with $2.$ for istance with an electric resistance dissipating power inside the box $\endgroup$
    – Sørën
    Commented Jun 20, 2016 at 17:00
  • 1
    $\begingroup$ I think you are thinking of the "adiabatic piston" problem. It has an enormous literature, just search google.com/?gws_rd=ssl#q=adiabatic+piston+problem and enjoy! $\endgroup$
    – hyportnex
    Commented Jun 20, 2016 at 17:45

2 Answers 2

Reset to default
2
$\begingroup$

A reversible process is characterized by a continuous sequence of thermodynamic equilibrium states for whatever system you are considering. So, for your system to experience a reversible process, its pressure and temperature must differ only slightly from that of its surroundings throughout the entire process. And there can be no spatial temperature or pressure variations within the system during the process (unless these different parts of the system are isolated from one another both thermally and mechanically throughout the process).

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks for this clear and complete answer! There is a thing that I did not understand completely: is the thermodynamical equilibrium required to be inside the system or between the system and the environment or both of them? As far as I know the system must be in equilibrium and it must be in equilibrium with the environment too, is that possibly correct? $\endgroup$
    – Sørën
    Commented Jun 24, 2016 at 21:12
  • 1
    $\begingroup$ If the system experiences a reversible process, the system itself has to pass through a continuous sequence of thermodynamic equilibrium states. But the surroundings does not have to. For example, if you manually apply a gradual adiabatic compression of a gas by hand, the system (comprised of the gas) experiences a reversible process, but your body, which has all kinds of thermodynamic irreversibilities associated with it, does not experience an irreversible process. $\endgroup$
    – Chet Miller
    Commented Jun 24, 2016 at 23:17
1
$\begingroup$

It depends on what you consider to be the system. If the system is the entire container, then there are no thermodynamic operations, quasistatic or not, on the system by the external environment. And as you said, the system is not in thermal equilibrium.

If you talk about a thermodynamic operations you need to define a system and an environment, in this case one of the subparts will be the system and the other the environment. In you example the quasistatic process is reversible and the system (the gas of your choice) is in thermodynamical equilibrium along the process.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Ok probably I missed this point, but in $1.$ if I choose as system the whole box (both the parts and so both gases), and as environment the external space (say at temperature $T_0$), then the middle wall is moved exchanging work in a quasi static way, the temperatures of the two gases at any equilibrium state would be different with each other and will differ from $T_0$, that is $T_1\neq T_2 \neq T_0$ in any intermediate state of the (reversible?) process. Would this scenario be a good one for my doubt? $\endgroup$
    – Sørën
    Commented Jun 20, 2016 at 18:07
  • $\begingroup$ yes, agreed that the temperatures will differ, the entire system will not reach thermodynamic equilibrium because the temperature is not uniform, and will stay so even until the wall eventually becomes non adiabatic (which is a good approximation for short processes but not if you wait long enough in a real physical system. $\endgroup$
    – user65081
    Commented Jun 20, 2016 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.