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By definition,

dS=dqrev/T.

Based on my understanding, the subscript rev means that the dq is heat exchanged in a reversible pathway from a state i to another state f, regardless of the actual pathway by which the process proceed. I am always curious why ΔS is defined only for reversible process, it would be good if someone can explain it, nonetheless it is not our main concern here-let's just accept the definition first.

So, similarly,

dSsur=dqrev,sur/Tsur......(1)

From what I have read, V of surrounding can be regarded as constant, so

dqrev,sur=dUsur

and since dUsur is a state function,

dUsur=dqsur

and therefore dqrev,sur=dqsur implying

dSsur=dqsur/Tsur......(2)

(a) If V is constant, q=ΔU, so q is a state function at constant V?

(b) Let's stop at equation (1). Suppose we would not like to simplify it further, and since the equation (1) is an intermediate result it must be true, we have

dS=dqrev/T and

dSsur=dqrev,sur/Tsur

I suppose dS without any subscript is dS of system. By First Law,

dqrev= -dqrev,sur

and then if T=Tsur and the chemical reaction has not reached equilibrium (is this scenario possible?),

dS= -dSsur

By summing up all entropy changes for system and surrounding, we have ΔStotal=0 for non-equilibrium state? But as I know ΔStotal=0 if and only if surrounding is in equilibrium with system. What's my misconception here?

(c) How can dqsur,irrev=dqsur,rev? Does not make sense for me physically, without looking at the equation with dUsur

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  • $\begingroup$ (a) is true. (b) whenever you write $dS=dq/T$ you imply reversibility (equilibrium). For any process this is generally assumed for the surroundings (we assume it to have properties consistent with this), but not (generally) for the system. (c) Again, we generally assume that the surroundings can give or accept heat reversibly. $\endgroup$ – Night Writer Jan 22 at 11:40
  • $\begingroup$ (a) is not true in general. A battery loses internal energy when it is discharged even if it does not transfer any heat (i.e. perfect battery that does not warm up when discharged, or perfect insulation around the real battery). $\endgroup$ – Karsten Theis Jan 22 at 16:29
  • $\begingroup$ @KarstenTheis True, the assumption I made is that there is only PV work. $\endgroup$ – Night Writer Jan 22 at 18:57
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In thermodynamics, the surroundings are typically treated as an ideal constant temperature reservoir for which $\Delta S$ is always equal to $Q_{surr}/T_{surr}$, irrespective of whether the process that the combined system and surroundings experiences is reversible. As such, the surroundings is assumed to have an infinite heat capacity and thermal conductivity, and to not change significantly in temperature (its temperature is always $T_{surr}$), no matter how much heat it receives or rejects. So it always presents the temperature $T_{surr}$ to the system at its interface with the system. And, all the entropy generated in any process is confined to the system.

In evaluating the entropy change for system and surroundings, the system and the surroundings must first be separated from one another, and then each must be separately subjected to a reversible path. These separate reversible paths do not have to match up in any way whatsoever. For the surroundings, the final state is that which was obtained with the irreversible path when $-Q_{irrev}$ was transferred from the system. So the change in entropy for the surroundings is $-Q_{irrev}/T_{surr}$. And the change in entropy for the combination of system and surroundings is then given by: $$\Delta S=\Delta S_{sys}-\frac{Q_{irrev}}{T_{surr}}=\left[\int{\frac{dQ_{rev}}{T}}\right]-\frac{Q_{irrev}}{T_{surr}}$$

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  • $\begingroup$ For your last equation, you have simplified my equation (1) into equation (2). If I do not want to simplify equation (1) [this intermediate result must be true], and I plug the equation (1) in place of your delta S for surrounding, I will get both q are reversible, and by First Law the two must add up to zero and that gives delta S always=0. It was just like you put rev q for system although the process is irrev, can't I do that on delta S for surrounding? $\endgroup$ – The99sLearner Jan 23 at 5:03
  • $\begingroup$ No. It is impossible to devise a reversible process for the combination of system and surroundings that takes them both simultaneously between the same two thermodynamic end states as those obtained with an irreversible process. To get the entropy change between the end states, you must first separate the system from surroundings, and then subject each of them individually to a different reversible path. The reversible heats and works for these individual paths do not have to match. For more on this, see physicsforums.com/insights/grandpa-chets-entropy-recipe $\endgroup$ – Chet Miller Jan 23 at 11:40
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(a) It is correct to say that $q$ is equal to a state function under certain conditions. Your example of constant V making $q=\Delta U$ is one. Another specific case is at constant pressure, $q=\Delta H$. However, the fact that $q$ is equal in value to a state function in these specific cases does not mean that it is itself a state function. It still does not meet the definition of a state function. By enforcing the condition of constant V or constant P, you are selecting a path for the process, so it is no longer path independent.

(b) In order for a process to be truly reversible in the thermodynamic sense, it must be at equilibrium at all times. That is because $\Delta S_{universe}$ is only 0 at equilibrium. By this definition, though, no reversible process can happen at all, since we can never be away from equilibrium conditions. To keep the math easy and get around this issue, we introduce the idea of "microscopic reversibility", meaning that we move the system along a path via very very small deviations from equilibrium, and equilibrium is restored before we make the next increment change.

For example, if we are adding heat to a system, we create a nearly reversible process by starting with the system and surroundings at the same temp (at equilibrium) and then changing the temp of the surroundings by a very small amount and waiting for the system to come up to that temp before changing the temp again. This is not truly a reversible process, but as the increment approaches zero (and the time for the process approaches infinity), the limit is the reversible process.

Going back to your question, if the chemical reaction is not at equilibrium, then the process by which it gets to equilibrium is necessarily irreversible. Thus, the assumption that $q_{rev,sur}=-q_{rev,sys}$ is not valid. The surroundings are not in the same state that they would be for a reversible process that took the system from $i$ to $f$. Thus the two $q_{rev}$ values are not the same magnitude, which is question (c).

Finally, I don't think it's correct to say that $\Delta S$ is defined only for reversible processes. It's a state function, so we don't need to assume reversible to determine it if we have all the other information necessary. The only definition of $\Delta S$ that invokes a reversible process is $\Delta S = \frac{q_{rev}}{T}$, in which it's the heat transfer that must be from a reversible process, not the $\Delta S$. That $\Delta S$ would apply to an irreversible process if the final state is the same as for the reversible process with heat transfer equal to $q_{rev}$.

One last comment - you stated that V of surroundings can be assumed constant. That is not always true. We often consider cases where the surroundings are a local volume that is considered to be isolated from the universe for all practical purposes. For example, we could have a reaction occurring in a balloon inside a sealed container, and if the reaction resulted in gas production that increased the volume of the balloon, the volume of the interior of the box would correspondingly decrease. If we were interested in the energetics of the reaction, we would treat the interior of the box as the surroundings and the inside of the balloon as the system.

[Really last comment - a picky bit of nomenclature - since $q$ represents a transfer of heat, $dq$ is not the appropriate term in your equations. For example, $\Delta S =\frac{q}{T}$, not $\frac{dq}{T}$.]

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    $\begingroup$ $q = \Delta U$ is true if there is no work. Constant volume ensures that there is no pV work, but you also have to specify there is no non-pV work such as mechanical work, electrical work, magnetic work or gravitational work Wikipedia $\endgroup$ – Karsten Theis Jan 22 at 16:24
  • $\begingroup$ Good point. I was only considering PV work. $\endgroup$ – Andrew Jan 22 at 18:32
  • $\begingroup$ For (b), if the chemical reaction is at equilibrium, qrev,sur=−qrev,sys; if the chemical reaction is not at equilibrium, the equation does not hold, yet by definition we still have to put qrev,sys for dS of system, but for dS of surrounding we will put the actual q released/accepted by the system (surrounding=actual system) through the actual irreversible pathway, so the two will not add up to zero, is that what you meant? $\endgroup$ – The99sLearner Jan 23 at 4:50
  • $\begingroup$ Essentially yes. The actual heat transfer to the surroundings is in most cases going to be the same as what would be transferred in a reversible process to get to the final state of the surroundings, as you describe. This is based on the typical case of an irreversible process where energy gets transferred as heat (irreversible) rather than work (reversible), which is why the maximum possible magnitude of work is in the reversible process. $\endgroup$ – Andrew Jan 23 at 13:07

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