Why can't entropy decrease for an adiabatic and closed system?

In equations (10) and (11) of this article by Curzon on the adiabatic piston, Curzon uses that a system which is thermally insulated cannot have its entropy decrease (irrespective of anything else conceivably "doable" to the system). Why is this so, or am I misinterpreting?

The crux of the derivation is that we want a characterization of the (elusive) final equilibrium state. In (10) and (11), Curzon is therefore writing (since this is what is being used in equation (5)) that any virtual, quasistatic process away from the final equilibrium state (to other equilibrium states on the constrained manifold describing the composite system) must obey $$dS_i \geq 0$$. But why isn't it flatly $$dS_i =0$$? If we are supposing that the virtual process is quasistatic (since the entire point is to distinguish the final equilibrium state from all the other equilibrium states on the constrained manifold), then $$TdS_i = dU_i - PdV_i = dQ - dW - PdV_i = dQ = 0 \implies dS_i = 0$$. So why the $$\geq$$? I ask because I feel like I must surely be missing something.

• The title asks a different question from the question in the text, which is "why is $dS_i\geq 0$ instead of $dS_i=0$." Commented May 16, 2023 at 10:46

The only way to decrease the entropy of a system is by transferring heat to its surroundings, which is precluded if the system is adiabatic.

On the other hand entropy can be generated (increase) within the system by means of irreversible adiabatic processes, or there can be no change in entropy for reversible adiabatic (isentropic) processes.

Hope this helps.

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– Buzz
Commented May 15, 2023 at 22:38

Curzon is therefore writing (since this is what is being used in equation (5)) that any virtual, quasistatic process away from the final equilibrium state (to other equilibrium states on the constrained manifold describing the composite system) must obey $$dS_i\geq 0$$.

Curzon is not referring to quasistatic process when writing $$dS_i\geq 0$$. The relevant passage reads:

Consideration will now be given to the case where the barrier is perfect thermal insulator, all other properties of he barrier remaining the same as before. In this circumstance the barrier will only transmit mechanical energy; that is, $$dU_1$$ and $$dV_1$$will be independent. An equation to represent this interdependence may be obtained from the principle of increase of entropy, according to which the entropy of a thermally isolated system can only remain the same or increase. Subsystems 1 and 2 are each thermally isolated because by assumption the piston is made of a perfect thermal insulator. Therefore, from the principle of increase of entropy $$dS_1\geq 0,\tag{10}$$ and $$dS_2\geq 0.\tag{11}$$ Equations (5), (10), and (11) can only be consistent if $$dS_1 = dS_2 = 0.\tag{12}$$

The problem discussed here by Curzon is the equilibrium state of an adiabatic system partitioned into two parts via a fixed adiabatic partition. Initially each compartment contains a gas at different temperature and pressure. The partition is then made movable and the question is to determine the pressure at equilibrium by application of the second law. The argument then is that during this process each compartment is a closed adiabatic system, therefore the second law applies individually to each compartment. At the final state each compartment must be in equilibrium, i.e., $$dS_i=0$$ with respect to variations about the equilibrium state. The maximization condition does not imply that $$dS_i=0$$ along the entire path, only at the state of equilibrium.

The purpose of the paper is to refute prior statements that the condition for mechanical equilibrium is $$P_1/T_1=P_2/T_2$$. Curzon argues that the correct condition is $$P_1=P_2$$ regardless of whether $$T_1=T_2$$ or not. This of course is obvious from a mechanical standpoint but Curzon's point is that the result is demanded by the second law.

• I don't agree that the second law applies directly to each compartment -- after all, they are not, individually, isolated systems (they can exchange volumes). Therein lies the difficulty for me and the reason that (I think) the adiabatic hypothesis is crucial.
– EE18
Commented May 16, 2023 at 13:12
• Adiabatic exchange of volume can only increase entropy. Why? At best we can conduct the process reversibly, in which case $\Delta S=0$. Anything less than reversible will result in $\Delta S\geq 0$. Still, if this is your question, that's not what I understand it to be based on your last sentence of the originalpost. Commented May 16, 2023 at 13:24
• I apologize for the confusion, but yes I suppose in the end I'm wondering why $\Delta S \geq 0$ for this adiabatically closed (but not closed to volume exchange) subsystem?
– EE18
Commented May 16, 2023 at 21:07
• No apology necessary – I need to know what the question is. Do you accept the Clausius inequality, $\Delta S \geq dQ/T$? If so, for adiabatic process $dS\geq 0$. This applies regardless of whether the system is isolated or it exchanges volume. Commented May 16, 2023 at 22:01
• The point is that reversible mechanics is isentropic and irreversible mechanics generates entropy. Only heat can increase or decrease entropy. Commented May 16, 2023 at 22:07

It is something silly.

He is invoking 2nd Law. The form of it that does not assume that it is reversible, and hence why $$\mathrm dS_\text{each}\geqslant0$$

You nearly sent me on a wild goose chase, because $$\mathrm dS_i$$ tended to mean $$\mathrm dS_\text{irrev}$$

• $\mathrm dS_\text{each}\geqslant0$ supposes that the system is isolated. Here, the subsystems can interact with each other. Thus the answer must have something to do with the fact that the two subsystems cannot exchange heat.
– EE18
Commented May 15, 2023 at 4:55
• If they cannot exchange heat, they cannot exchange entropy directly, only volumes. So that really comes down to $\mathrm dS_\text{irrev}\geqslant0$, which is the statement of the 2nd law. I thought of all of these before writing the answer, dude Commented May 15, 2023 at 4:58