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Assume two closed systems adjacent to each other together forming one adiabatic system. Both systems are assumed to have their volumes fixed and can therefore communicate with each other through heat transfer only. Now let study this simple example with three different formulation:

  1. The first formulation is what I had always thought to be obvious, when I thought I know Classic Thermodynamics! Assume $\delta Q$ is the heat flow from system $2$ into system $1$: \begin{align} &\Delta S=\Delta S_1+\Delta S_2\,,\qquad \Delta S_1\ge\int_1^2\frac{\delta Q}{T_1}\quad\text{&}\quad \Delta S_2\ge\int_1^2\frac{-\delta Q}{T_2}\\ &\Rightarrow\;\Delta S\ge\int_1^2\delta Q\,\Bigl(\frac{1}{T_1}-\frac{1}{T_2}\Bigr) \end{align} Now again the Clausius statement of the second law says if $\delta Q\ge0$ then $T_2\ge T_1$ and if $\delta Q\le0$ then $T_2\le T_1$. Therefore, eitherway we would have $$\Delta S\ge\int_1^2\delta Q\,\Bigl(\frac{1}{T_1}-\frac{1}{T_2}\Bigr)\ge0$$

  2. But no matter if the process is reversible or not, only assuming that the system has no work (volume is fixed, but also assume there is no friction or gravity work and etc.) we would obtain: \begin{align} &\delta Q-\delta W=dU=T dS - p dV\quad\Rightarrow\quad\delta Q=T dS\\ &\Rightarrow\quad \Delta S_1=\int_1^2\frac{\delta Q}{T_1}\;\;\text{&}\;\; \Delta S_2=\int_1^2\frac{-\delta Q}{T_2}\\ &\Rightarrow\quad \Delta S=\Delta S_1+\Delta S_2=\int_1^2\delta Q\,\Bigl(\frac{1}{T_1}-\frac{1}{T_2}\Bigr)\ge0 \end{align} The problem is that we have derived in the first line $\delta Q=T dS$ which is not correct in general and this suggests that even if the volume is fixed but yet another work should exist, but if the containers contain only solid material what work should be considered to resolve this paradoxical result?

  3. In the Emanuel's ``Advanced Classical Thermodynamics" it has been suggested to write the entropy change in a closed system as: $$dS=dS_{int}+\frac{\delta Q}{T_{surr}}$$ wherein, int and surr are the abbreviations for internal and the surrounding, respectively. Assuming the irreversibility caused by the flow of heat between a finite temperature difference $T_{surr}-T$ to be considered inside $dS_{int}$ then it concludes the second law of thermodynamics as: $$dS_{int}\ge0$$ Now returning back to our two-system problem we would have: \begin{align} &dS=dS_{int_{total}}+0\quad\text{&}\quad dS_1=dS_{int_1}+\frac{\delta Q}{T_2}\quad\text{&}\quad dS_2=dS_{int_2}-\frac{\delta Q}{T_1}\\ &dS=dS_1+dS_2\quad\Rightarrow\quad dS_{int_{total}}=dS_{int_1}+dS_{int_2}-\delta Q\,\Bigl(\frac{1}{T_1}-\frac{1}{T_2}\Bigr) \end{align} The last line requires $dS_{int_{total}}\le dS_{int_1}+dS_{int_2}$ which is not correct obviously and it should have been only "equal" instead of "lower or equal".

As is already clear, the second formulation of the problem should have a mistake but I just cannot find it, but the third formulation is rather more serious problem, to me either assuming the irreversibility of heat flow between finite temperature differences cannot be removed from it into $dS_{int}$ or if it is done then only one temperature should be used in the formulation, but then the problem rises that which one, $T_1$ or $T_2$? If it is arbitrary and I can write both $dS\ge\frac{\delta Q}{T_{surr}}$ and $dS\ge\frac{\delta Q}{T}$ then the first formulation above would run into problem!

Thanks

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  • $\begingroup$ I am sorry, I don't understand why you say that the second formulation is wrong. Can you specify your thoughts? $\endgroup$ – gatsu Jul 15 '14 at 8:04
  • $\begingroup$ @gatsu, it is wrong as a general formulation, but is valid for only reversible processes. My formulations were in differential form which were nonsense so I updated the answer by their integral form, now the difference between the first and the second formulation is more clear, the first inequality has turned into an equality which is not correct unless for reversible processes. My heat transfer process however is irreversible as $T_2-T_1$ is assumed to be finite. $\endgroup$ – topology Jul 15 '14 at 12:43
  • $\begingroup$ I am super sorry but I don't see the problem...I don't even understand why you specify that there should be no work to conclude that $\delta Q = TdS$. In my view, this last formula is an axiom of the second principle of thermodynamics that prescribes how to calculate entropy variations, nothing else. But as a consequence, there is no need to derive it since it is already given. $\endgroup$ – gatsu Jul 15 '14 at 13:06
  • $\begingroup$ @gatsu, no as far as I can say $\delta Q=T dS$ is not valid in general, but only if the heat transfer and the whole process is reversible. This equality is used to define entropy as a state function but in this definition a reversible process is used so that $S$ is path independent. In an irreversible process we will have $\Delta S > \int_1^2\frac{\delta Q}{T}$ so that the equality doesn't hold any longer. $\endgroup$ – topology Jul 15 '14 at 13:30
  • $\begingroup$ I guess it is a problem of semantic but my first point is that reversible or not I don't see why you specify that there should be no work to derive $\delta Q = TdS$. My second point is that, if you assume you can write $dS$ then you can write $dS=\delta Q/T$. This leads to the very well known (and misunderstood) formula that for any transformation from state $A$ to state $B$, $\Delta S_{AB} = \int_{\Gamma_{AB}} \frac{\delta Q(\Gamma)}{T(\Gamma)}$ where $\Gamma_{AB}$ denotes any reversible path going from $A$ to $B$. $\endgroup$ – gatsu Jul 15 '14 at 13:36
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It is best to always stick to the rigorous formulation of the second law, which comes in two parts. Quote from the book "Fundamentals of Statistical and Thermal Physics "by F. Reif:

1) In any process in which a thermally isolated system goes from one macrostate to another, the entropy tends to increase, i.e.,

$$\Delta S\geq 0$$

2) If the system is not isolated and undergoes a quasi-static infinitesimal process in which it absorbs heat $dQ$, then

$$dS = \frac{dQ}{T}$$

Combining 1) and 2) in the form of statements such as $dS\geq\frac{dQ}{T}$ are weaker than the second law as formulated above, as you need to make additional assumptions that would make this valid. It's best to never use such statements directly and always stick to the modern rigorous formulation of the second law that doesn't assume that temperature is definable during any non quasi-static change.

The paradox you consider is resolved when you consider that the system is not in thermal equilibrium. Note that you are in principle free to define the thermodynamic description of a given physical system. So, the physical system is described exactly by specifying its microstate, and the thermodynamic description is a course grained description of it where you keep only a few macroscopic parameters. Now, what then can happen is that a slightly less course grained description would have allowed equilibrium thermodynamics to be applicable to the system. This is the case here.

So, you have two choice. You if you describe the system as one isolated system, then temperature is not defined and you can only use the first part of the second law which doesn;t invoke a notion of temperature. However, if you take a sligly less course graiend look at the system, you see that you have two subsystems, the changes in entropy of the system can be described by the second part of the second law, provided the changes are quasi-static.

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  • $\begingroup$ sorry for being late, what if in the second statement the system is not isolated but also the process is not quasi-static and infinitesimal? Meanwhile, I don't see how it resolves the paradoxes, once you choose the whols system as one isolated system, so the first statement results in $\Delta S\ge0$ without defining a single temperature for the whole system, next we choose each system separately and they have temperature defined for them but your second statement doesn't necessarily hold for them. Now what? Am I missing a point? $\endgroup$ – topology Jul 22 '14 at 5:04
  • $\begingroup$ The second statement des hold if the subsystems are in thermal equibrium. So, we may consider quasistatic changes in these systems and add uo the chsnges in the entropies. $\endgroup$ – Count Iblis Jul 22 '14 at 16:33
  • $\begingroup$ Whenever temperature is invoked in a rigorous way, it must apply to a system in thermal equilibrium. So, in your case you can only use the second equation when the two subsystems undergo quasistatic change. This (implicitly) implies that both systems are in contact with a third system (the contact area through which heat is conducted) that is not in thermal equilibrium. If you assume that this contact area is small and/or that th change from well defined initial and final states of its thermodynamic variables are zero, then you can use the second expression for the subsystems. $\endgroup$ – Count Iblis Jul 22 '14 at 16:34
  • $\begingroup$ Dear Count Iblis, once I read your answer I was sure it cannot be the answer but then reading the book by Klotz I finally found out that your answer is indeed the very right answer to resolve the apparent paradoxes! Indeed, it mainly addresses the third formulation, the first two are usualy dealt with differently, though your explanation above have something to say in that regard as well. So thank you for answering the question and I accepted your answer. $\endgroup$ – topology Jul 23 '14 at 3:44

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