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I tried to write down my "own" fundamental rules of thermodynamics, which I can then use as solid ground to understand the rest of the topic. The main thing I thought of is separating entropy variation in two parts (see equation \eqref{eq3}) and define how to generalize this to a compound system. I'll try to state my train of thought clearly and concisely, could anyone please tell me where I'm partially or totally wrong?

For simplicity, I will use the following assumptions:

  • no external forces, so we can work only with internal energy $U$
  • closed system (if it is open, choose a bigger system)

0) Thermodynamical system

We define a system and its surroundings. The system has a state, which corresponds to a point in a state space, with two coordinates/state variables: entropy $S$ (defined below) and volume $V$.
For now, we only consider simple or homogeneous systems, which means we can define a unique temperature $T$ and pressure $p$ (defined below). We will also assume that every process on a simple system is quasistatic, which allows us to maintain homogeneity.

1) First law

Definition of energy

For every simple system, we have a state function $U(S,V)$ called "energy".
Since it is a state function, it has an exact differential, and defining $T$ and $-p$ as the conjugate variables of $S$ and $V$ respectively, we get: $$dU = TdS - pdV \tag{1}\label{eq1}$$

Variation of energy

Physics (observation) tells us that the energy of a simple system can vary in two different ways: receiving heat $Q$ or work $W$ from its surroundings. These are not state functions but process functions (depend on the path in state space), so they don't have an exact differential: $$dU = \delta Q + \delta W \tag{2}\label{eq2}$$
Note: In general, we can't associate terms, $\delta Q \neq TdS$ and $\delta W \neq -pdV$.

2) Second law

Definition of entropy

For every simple system, we have a state function $S(U,V)$ called "entropy". It also has an exact differential, which can be obtained from $dU$ above.

Variation of entropy

Physics (observation) tells us that we can separate entropy variations in two parts: $$dS = \delta S^p + \delta S^f, \tag{3}\label{eq3}$$ where $\delta S^p \ge 0$ is the produced or irreversible entropy and $\delta S^f = \frac{\delta Q}{T}$ is the flow or reversible entropy, due to heat exchanges with the simple system's surroundings.
Note: entropy is a state function, i.e. coming back to the same state always yields $\Delta S = 0$. Production and flow entropy are not, i.e. we can choose a closed but irreversible path s.t. $\Delta S^p \gt 0$, which will have to be compensated by entropy evacuation $\Delta S^f \lt 0$.

3) Combining 1st and 2nd law

Equating \eqref{eq1} and \eqref{eq2}, and inserting \eqref{eq3}, we get explicit relations for infinitesimal heat and work: $$\delta Q = T\delta S^f \qquad \delta W = T\delta S^p - pdV$$ which we can invert to also obtain explicit relations for produced and flow entropies.
Note: Here, we see that the work we can extract $-\delta W$ is always lowered by the produced entropy.
Note 2: We now see that, in the reversible case ONLY, we get the nice relationships $\delta Q = TdS$ and $\delta W = -pdV$
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4) Compound systems

Until now, we have only worked with simple systems. Since we know from observation that $U$, $S$ and $V$ are extensive quantites, we can expand to compound systems. Without loss of generality, consider only two systems, labeled 1 and 2. Then we define the total state variables by $U = U_1 + U_2$, etc., which generate a new state space of the total system. Differentials of these variables still exist, but they don't have a nice form anymore. More precisely, $dU \neq TdS -pdV$ and $\delta S^f \neq \frac{\delta Q}{T}$, since we can not define unique conjugate quantities $T$ and $p$ anymore.
However, we can verify that:

  1. the variation of $U$ still follows \eqref{eq2} if we define $\delta Q = \delta Q_1 + \delta Q_2$ and $\delta W = \delta W_1 + \delta W_2$ (the exchanges between subsystem 1 and 2 are opposite, so they cancel out, the only exchange which remains is with the surroundings).
  2. the variation of $S$ still follows \eqref{eq3}, but we can't simply add up the flow (neither production) entropies of the subsystems. This is because we would like to define the total flow entropy to represent heat flow with the surroundings: $\delta S^f = \frac{\delta Q_{1out}}{T_1} + \frac{\delta Q_{2out}}{T_2}$ and unfortunately the flow entropies of the subsystems will not cancel each other out. To put it in equations:

$$ \delta S_1^f + \delta S_2^f = \frac{\delta Q_{12} + \delta Q_{1out}}{T_1} + \frac{\delta Q_{21} + \delta Q_{2out}}{T_2} = \delta Q_{12}(\frac{1}{T_1} - \frac{1}{T_2}) + \delta S^f \\$$

So, defining the net entropy flow between the subsystems: $\delta S_{12}^f := \delta Q_{12}(\frac{1}{T_1} - \frac{1}{T_2})$, we get:

$$\begin{align} \delta S^f := \delta S_1^f + \delta S_2^f - \delta S_{12}^f \\ \delta S^p := \delta S_1^p + \delta S_2^p + \delta S_{12}^f \end{align}$$

which means that flow entropy can be turned into produced entropy if we consider the bigger system.

5) Questions

I applied "my formalism" to several systems and I seem to get correct results (at least for the ones I checked, e.g. thermal, pressure, chemical equilibrium between two subsystems, thermodynamic potentials and systems in contact with reservoirs, etc.).

My questions are:

  1. If all of this works, why is it never taught in this way? I have never explicitly seen the entropy divided in two parts (except maybe in the book I use: "Principles of Thermodynamics" by Ansermet, but not exactly in the same way).
  2. Also, from this formalism, we can deduce that entropy production is generated either by work/volume variation or by internal entropy flow in compound system, is this generally true?
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  • $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community Bot
    Sep 10, 2022 at 12:32
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    $\begingroup$ I think that you started from a "too special" form of the 1st principle. Try to have a look here basics.altervista.org/test/Physics/TD/td_principles.html $\endgroup$
    – basics
    Sep 10, 2022 at 12:38
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    $\begingroup$ unfortunately, thermodynamics is usually presented in a very bad way, imho. The formulas I gave you are general and then specialize if we are talking about a fluid $\delta \ell^{rev} = - p dv$, or a solid $\delta \ell = \sigma : \delta \varepsilon$. My $\delta U$ is what is usually called $\delta Q^{rev}$ that doesn't really means anything, coming from experience. I tried also to justify that internal energy doesn't include any contribution of kinetic energy, being the difference between the total energy (governed by 1st principle of TD) and kinetic energy (governed by KE thm from mechanics) $\endgroup$
    – basics
    Sep 10, 2022 at 16:54
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    $\begingroup$ Your "first law" has assumed that entropy is a state function, but that assumption implicitly assumes the second law. This point is often missed in simple treatments. If you want to build up the laws then the first law must mention energy but it may not make any mention of entropy. $\endgroup$ Sep 10, 2022 at 17:03
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    $\begingroup$ "Why is it never taught this way" is often a shaky premise and an unanswerable question. Allen et al.'s Kinetics of Materials distinguishes entropy transfer and entropy production, for example, and gives models for the entropy production rate. Regarding other textbooks and resources, I don't see how anybody can answer for the individual authors. $\endgroup$ Sep 11, 2022 at 20:39

4 Answers 4

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"Also, from this formalism, we can deduce that entropy production is generated either by work/volume variation or by internal entropy flow in compound system, is this generally true?"

No. Entropy projection is generated by transport process and chemical reactions proceeding irreversibly at finite rates. Such transport processes include (1) heat conduction at finite temperature gradients, (2) fluid deformations at finite velocity gradients, and (3) molecular diffusion at finite concentration gradients. For more details on this, see Transport Phenomena by Bird, Stewart, and Lightfoot, Problem 11.D.1 and Section 24.1, The Equation of Change For Entropy.

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Question 1 If all of this works, why is it never taught in this way?

Thermodynamics can be taught and understood in many different ways. If your way makes sense to you, use it. Personally I would not use it to teach thermodynamics for the following reasons.

  1. It begins by defining the state using entropy and volume, but does not define entropy.

  2. The choice of $(S,V)$ is not motivated. Why not $(S,P)$ or $(S,T)$? Or more naturally, $(T,P)$?

  3. Why is heat equal to $T dS$? This is not me asking –it's the student who reads your lecture notes.

  4. "Physics (observation) tells us that we can separate entropy variations in two parts [...]" This statement presumes that entropy has been given a definition by which we may analyze an experiment and conclude that $S$ can be separated into such parts. It seems to me that the above conclusion is based on the "standard" approach to thermodynamics, which infers the existence of $S$ and its properties from thought experiments about possible and impossible processes. In other words, it presumes knowledge outside this new formulation of thermodynamics.

Question 1 we can deduce that entropy production is generated either by work/volume variation or by internal entropy flow in compound system, is this generally true?

Yes, it is generally true:

  1. Irreversible expansion/compression generates entropy via what you call work/volume variation
  2. Bring two systems, each at its own $T$, to thermal contact. The flow of heat amounts to entropy generation via what you call entropy flow
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we can deduce that entropy production is generated either by work/volume variation or by internal entropy flow in compound system, is this generally true?

The following line of thinking could be useful in your framework:

Entropy is generated in the shifting of an extensive thermodynamic variable $Y$ (e.g., entropy, volume, mass, area, magnetization, polarization, volumetric strain, or charge) to transfer energy $E$ down a gradient in a conjugate intensive variable $X$ (temperature $T$, pressure, chemical potential, surface tension, magnetic field, electric field, stress, or voltage, respectively). The entropy production rate is

$$\dot\sigma=\vec J_E\cdot\nabla\frac{1}{T}-\sum_i\vec J_i\cdot\nabla\frac{X_i}{T},$$

where $\vec J$ indicates a flux and $i$ indexes the nonthermal conjugate pairs (source: Allen et al.'s Kinetics of Materials). Note that heat transfer is special because it both transfers and produces entropy.

Thus, a larger flux and larger gradient lead to faster entropy production. Now, it stands to reason that reducing either to zero would provide us with reversibility, but if we block the flux, then process evolution is precluded, and if we have no gradient, then there's no driving force to induce a flux in the first place. Thus, any actual macroscale process is irreversible.

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  • $\begingroup$ Doesn't flux require the existence of a gradient? Or alternatively, when can we expect a flux without a gradient or a gradient that does not induce a flux? $\endgroup$ Sep 11, 2022 at 22:49
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    $\begingroup$ Yes; that's why I wrote that a process doesn't proceed at a nonzero rate without a driving force. (I'll edit to make this clearer.) The standard example is the Carnot cycle, which lacks gradients and thus exhibits zero entropy generation, maximum potential efficiency, and (uh-oh) zero power output. $\endgroup$ Sep 11, 2022 at 23:54
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If all of this works, why is it never taught in this way? ...

The three most common paradigms to teach thermodynamics are laws, postulates, and statistical mechanics. They have withstood tests on their validity in various applications over and over and over again.

The difficulty that arises in teaching from any one of the above paradigms is that certain aspects may not be easily explained entirely just in that one frame. For example, in teaching only from the paradigm of the laws of thermodynamics, we might struggle to present a proposal that an increase in entropy leads to a disorder in molecular order in a material because we have no idea that molecules exist or how their configurations contribute to entropy. Only when we acknowledge the statistical mechanics perspective can we provide confidence in a molecular perspective.

You offer an that is closest to the postulates paradigm. Perhaps you could answer your own question by explaining where and how your approach is an improvement on it.

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