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Consider a thermodynamical classical isolated system, made by a small subsystem and a way large reservoir. The two could exchange heat. Usually in such situation we say that the system is closed or is a $(N,V,T)$ system.

What perplexes me is that for the $(N,V,T)$ system, $N$ and $V$ are constant even if the system+reservoir are not yet at equlibrium. But that's not true for the temperature $T$.

I know that the larger reservoir imposes its temperature to the system. But this needs a step more.

Please, where I am wrong?

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  • $\begingroup$ Deviation from equilibrium could arise from the density being not equal to that of the reservoir. In this case, you will see that the system approaches equilibrium through diffusion of particles into or out of the system, and through the pressure causing the subsystem to change its volume. $\endgroup$
    – symanzik138
    Commented Nov 3, 2016 at 17:12
  • $\begingroup$ If I understand what you're saying, neither $V$ nor $N$ could be considered contstant ab initio. But, in fact, in the classical abstraction, the system has perfectly rigid boundaries - so these quantities are in fact constant, before the equilibrium is approached. $\endgroup$
    – Lo Scrondo
    Commented Nov 3, 2016 at 17:46
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    $\begingroup$ That is a choice, suitable for modeling certain systems, for example a test-tube with reagents in a heated bath. If you were modeling say a cell that passively exchanges mass with its surroundings, you would model it with a grand canonical ensemble (ensemble with variable particle number), where density may not be in equilibrium with the surroundings, whereas the temperature is. $\endgroup$
    – symanzik138
    Commented Nov 4, 2016 at 11:39
  • $\begingroup$ I understand what you're saying...in fact, I think it's a matter of definitions. A closed system is a system able to exchange heat but not matter with the surroudings. A closed system in contact with a heat bath is a $NVT$ system...not alway such difference is clearly cited in textbooks. $\endgroup$
    – Lo Scrondo
    Commented Nov 4, 2016 at 14:54

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It is a matter of boundary conditions: you are considering a subsystem with walls which are diathermal (they allow heat exchange) but impermeable (they don't allow matter exchange) and rigid (they don't allow changes in volume).

Let $S$ be the large system and $s$ be the small subsystem. With this choice of boundary conditions, the generic initial state of $s$ will be $(N_s^i, V_s^i, T_s^i)$ and that of $S$ will be $(N_S, V_S, T_S)$.

After thermal equilibrium is reached, the state of $s$ will be $(N_s^f=N_s^i, V_s^f=V_s^i , T_s^f = T_S \neq T_s^i)$ while the state of $S$ will be unchanged (we are here assuming that $S$ is so large compared to $s$ that its temperature $T_S$ remains unchanged during the reaching of thermal equilibrium).

You could also consider different boundary conditions: in this case you could have $N_s^i \neq N_s^f$ and/or $V_s^i \neq V_s^f$.

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  • $\begingroup$ Thank you @valerio92, it's obviously all clear to me. In fact, as I pointed out in the last comment to my question, I was believing that "closed system" and "NVT system" were "synonyms" - so that the definition of closed system included the presence of a heath bath. I hope that I'm right now in considering the difference among the two concepts. $\endgroup$
    – Lo Scrondo
    Commented Nov 5, 2016 at 16:11
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    $\begingroup$ Well, yes. Actually, the only condition for a system to be closed is that it must not be able to exchange matter with the surroundings, i.e. $N=const.$ An $NVT$ system is a closed system with rigid walls in contact with an heath bath. $\endgroup$
    – valerio
    Commented Nov 5, 2016 at 22:39
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The thermodynamic parameters $N, V, T$ are all mentioned at the equilibrium states only. So, when your system is in contact with a heat bath, there causes an exchange of energy between the system. According to the definition of a heat reservoir, its temperature is not affected by any slight exchange of heat. Hence the system will eventually come in thermal equilibrium at a temperature $T$ of the heat bath.

The system is closed in this sense means that the combined system of system+reservoir is isolated from the rest of the universe.

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  • $\begingroup$ Thank you @Unnikrishnan, it's pretty clear to me. In fact, as I pointed out in the last comment to my question, I was believing that "closed system" and "NVT system" were "synonyms" - so that the definition of closed system included the presence of a heath bath. I hope that I'm right now in considering the difference among the two concepts. $\endgroup$
    – Lo Scrondo
    Commented Nov 5, 2016 at 16:11

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