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Consider a system that can be defined by $U$, $V$, ${N}_{i}$ $\left(i=1,2,\ldots k\right)$ and that this system changes from state A to state B via a quasi-static reaction path $C$.

Here, $U$ represents internal energy stored in the system, $V$ represents the volume of the system, and ${N}_{i}$ $\left(i=1,2,\ldots k\right)$ is the amount of $i$-th particles in the system.

At this time, comparing the following two, the term of the chemical potential seems to remain.

  • the $\delta Q$, obtained by integrating along the reaction path (See (1-9), below) and,
  • the first law of thermodynamics (See (1-11) below)

In the reactions where the number of particles does not change, the "chemical potential term" will be zero because ${dN}_i$ $\left(i=1,2,\ldots k\right)$ are zero. However, for example, if chemical reactions occur in the system, the number of particles ${N}_{i}$, $\left(i=1,2,\ldots k\right)\ $ will be change.

【My Questions】

  1. Where did the term of chemical potential disappear in the first law of thermodynamics?(See (1-11))†
  2. Perhaps if we replace the definition of work that takes place during this process from (1-6) to following (1-6)'. And the definition of internal energy will be change into “the sum of kinetic energy and potential energy of particles in the system”※.

$${{W}^{\prime}}_{C}\ = \int_{C} -PdV + {\sum}_{i=1}^{r}{{\mu}_{i}{dN}_{i}}\tag {1-6'}$$

†.In some comments, (1-11) is hold only in closed / isolated systems. (The definitions are as described in 【Note2】.) But, how about the reaction, like followings?

  • CO + O2 →CO2 (coefficients are omitted.) or
  • H2O(Vapor)→H2O (liquid)

In these reactions, seems not necessary for particles to go/come-from outside. Quasi-static paths may exist at least theoretically. Perhaps that theoretical pathway is very tricky like "Down to absolute zero, react in absolute zero, then increase temperature to final state.” but, we can use the path to calculate the relative entropy between the Initials and final state. Am I right?. (I add this sentence at 2019/11/27(JST) and modified at 2019/11/28(JST).)

※In my elementally textbooks (written in Japanese), the definition of U is "the sum of kinetic energy of particles in the system." They ignore the potential energy. (I add this sentence at 2019/11/27(JST).)

Is there any inconvenience if you rewrite the definition like this?

【Details】
Let $\delta Q$ be the thermal energy that flows into the system during a quasi-static micro-process where $U$, $V$, and ${N}_{i}$ $\left(i=1,2,\ldots k\right)$ change slightly. Then,

$$\delta Q\ =T\ dS\ \tag {1-1}$$

Here, T represents the temperature of this system. On the other hand, let $S[U,\ V,\ {N}_{i}]$ be the entropy of this system, then,

$$dS\ =\ \frac{1}{T}dU+\frac{P}{T}dV-{\sum}_{i=1}^{k}{\frac{\mu_i}{\ T_i}dN}\tag {1-2}$$

Here, Multiply $T$ for both sides of above equation, we can get $$TdS\ =\ dU\ +\ PdV\ -\ \sum_{i=1}^{r}{\mu_idN_i} \tag {1-3} $$

From equations (1-1) and (1-3), we get $$\delta\ Q\ = \ dU+ PdV - \ \sum_{i=1}^{r}{{\mu}_{i}{dN}_{i}}\tag {1-4}$$

The $\delta Q\ $ is incomplete derivative therefore, integration of this will depends on the reaction path even though the reaction path is quasi-static but, we can do line integration along a quasi-static reaction path, $C$. We denote ${Q}_{C}$ as the result of this line integral (See (1-5)) then, the ${Q}_{C}$ will represent the thermal energy that flows into the system during a quasi-static process $C$.

$${Q}_{C} = \int_{C}{\delta Q}\tag {1-5}$$

Similarly, the work, ${W}_{C}$ that the system receives in this process is: $${W}_{C} = {\int}_{C} - PdV\tag {1-6}$$

The dU is a complete derivative therefore, the change in internal energy before and after this reaction, $$\Delta{U}_{A,B}={U}_{B}-{U}_{A}\tag {1-7}$$

satisfies $$\Delta{U}_{A,B}\ =\ {\int}_{C}{\ \ dU}\tag {1-8}$$

Therefore, when equation (1-4) is integrated along reaction path $C$, $${Q}_{C}\ =\ \Delta{U}_{A,B}\ -\ \ W_C\ -\ \sum_{i=1}^{r}\int_{C}{\ \ \mu_idN_i}\ \tag {1-9}$$

therefore, $$\ \ \Delta{U}_{A,B}\ ={Q}_{C}\ +\ \ {W}_{C}\ +\ {\sum}_{i=1}^{r}\int_{C}{\ \ {\mu}_{i}{dN}_{i}}\tag {1-10} $$

On the other hand, the first law of thermodynamics is; $$\Delta{U}_{A,B}\ = {Q}_{C}\ +\ \ {W}_{C}\tag {1-11}$$

【Notes】 (Added on 2019/11/28 JST)

Note 1:Definitions of quasi-static process, reversible process, and reversible reaction:

Some respondents mention the feasibility of quasi-static process that change ${N}_{i}$ in a closed system. I assume that a chemical reaction or physical change has occurred. However, as of 2019/11/28 (JST) it is under discussion whether these are appropriate example in which the number of particles changes in a closed system.

In order to unify terms in our discussions, the definitions of quasi-static process and reversible process are described according to my text.

The definition of reversible reaction is not written in my text, but it came out in my comment, so I quote the definition from Wikipedia.

Here, as a definition of a quasi-static process, my text reads as follows.

Def1-1. Definition of a quasi-static process: "A thermodynamic concept that refers to the process by which a system slowly changes from one state to another while maintaining a thermodynamic equilibrium."(Originally in Japanese and I translated into English)

For reference, the definition of the reversible process is as follows.

Def1-2. Definition of the reversible process: "Even if a substance changes from one state to another, it is possible to return to the original state again and leave no change in the external world during this time."(Originally in Japanese and I translated into English)

These positions seems the same as the Wikipedia article on Reversible process. This article describes the relationship between quasi-static and reversible processes as follows:

"In some cases, it is important to distinguish between reversible and quasi static processes. Reversible processes are always quasi static, but the converse is not always true.[1]"

Here, the reference [1] of above-mentioned quotation is: Sears, F.W. and Salinger, G.L. (1986), Thermodynamics, Kinetic Theory, and Statistical Thermodynamics, 3rd edition (Addison-Wesley.)

Note that, there seems to be some confusion of terms; There seems to be some literature that uses reversible processes and quasi-static processes synonymously.

Another confusing term is the term reversible reaction. According to the Wikipedia, the definition is as follows.

Def1-3. Definition of reversible reaction: "A reversible reaction is a reaction where the reactants form products, which react together to give the reactants back."

■Note2. Definition of isolated system and closed system:

In my textbook the definition of isolated systems is as follows.

Def2-1. Definition of isolated systems: "A system in which neither material nor energy is exchanged with the outside world." (Originally written in Japanese and I translated into English.)

A system that exchanges energy but does not exchange substances seems to be called a "closed system" to distinguish it in my text book.

According to my textbook, the definition of closed system is as follows.

Def2-2. Definition of closed systems: " A system that exchanges energy with the outside world but does not exchange materials." (Originally written in Japanese and I translated into English.)

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    $\begingroup$ Please provide an example of how you would carry out a reaction in a closed system (no mass entering or leaving) reversibly (quasi-static reaction path). $\endgroup$ – Chet Miller Nov 26 '19 at 12:48
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    $\begingroup$ Yes the chemical potential term is part of the reversible work done on the system. What definition of internal energy were you using if not "the sum of kinetic energy and potential energy of particles in the system"? $\endgroup$ – By Symmetry Nov 26 '19 at 15:32
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    $\begingroup$ What about the chemical reaction. If the process is supposed to be quasi-static, how do you make the reaction go at a quasi-static rate? Regarding a closed system, let's first see if you can make you point for this kind of system before extending your arguments to the more complicated case of an open system. $\endgroup$ – Chet Miller Nov 27 '19 at 12:21
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    $\begingroup$ It's not. The increase in the amount of liquid is equal to the decrease in the amount of gas. So there is no change in the number of particles. And, the reversible reaction you are referring to is a different context from a reversible process. $\endgroup$ – Chet Miller Nov 28 '19 at 13:36
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    $\begingroup$ Liquid water and water vapor at equilibrium have the same chemical potential, for whatever relevance this has. $\endgroup$ – Chet Miller Nov 28 '19 at 14:36
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Let's imagine a system X consisting of 2 mol of H atoms and and 1 mol of O atoms held at constant $T$ and $P$. We can think of two states:

  1. In state "0," all of the H is in the form of H$_2$ and all of the O is in the form of O$_2$
  2. In state "eq," only H$_2$, O$_2$, and H$_2$O are present, and the three are in chemical equilibrium (which will tend towards mostly H$_2$O)

If we heated the system enough, the transition from state "0" to state "eq" would proceed very rapidly by irreversible chemical reactions (combustion), and would therefore not be described by equations 1-1, 1-4, 1-9, or 1-10. To make these equations apply, we would need to make the transition from state "0" to state "eq" reversible. To make the chemical reactions reversible, we would need to ensure that the chemical energy released was converted into work rather than thermal energy. We could do this using an ideal fuel cell (which converts chemical energy into electrical work).

With this change, the system would do work on its surroundings in two ways: boundary work due to expansion/contraction and electrical work from the fuel cell. For this particular reversible process, we could then identify the last two terms in Equation 1-10 as $$ \Delta{U}_{A,B} ={Q}_{C} + \underbrace{{W}_{C}}_{\substack{\text{Inward}\\\text{Boundary}\\\text{Work}}} + \underbrace{{\sum}_{i=1}^{r}\int_{C}{\ \ {\mu}_{i}{dN}_{i}}}_{\substack{\text{Electrical work done by}\\\text{surroundings on fuel cell}\\\text{to maintain reversible}\\\text{chemical reactions}}} \tag{1-10 applied} $$ where we expect the final term to be negative for process (0 $\rightarrow$ eq).

As alluded to in Aaron Stevens' answer, given that you define $W_c$ as the boundary work, Equation 1-11 is not the correct form of the First Law for a closed system. The correct form is instead $$ \Delta U_{A,B} = Q_C + \underbrace{\sum_i W_i}_{\substack{\text{Sum of}\\\textit{all}\text{ works}}} \tag{1-11 corrected} $$ With this correction, equations 1-10 and 1-11 become consistent: the last term of 1-11 encompasses the last two terms of 1-10.

Note that, if the chemical reactions are not carried out reversibly (and chemical energy in the fuel is converted directly into thermal energy in the combustion products, rather than being carried away as work), then the there is no electrical work output. The First Law remains consistent, however, because all of the energy that would have been carried away as work is instead stored as thermal internal energy. The energy lost from the system goes down, but the energy retained goes up by the same amount. Mathematically, the final term in (1-10) is moved to the other side and then grouped back as part of $\Delta U$.

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  • $\begingroup$ Thank you for your answer. Somehow my understanding is being made. Somehow understanding is being made. But let me confirm some definitions. ◆First, the "boundary work" you said is same as the "pΔV work" of [this site]( nuclear-power.net/nuclear-engineering/thermodynamics/…)? (Therefore, Neither VΔp Work nor chemical potential “work” is included ?in the "boundary work"). And can I understand that all of them are included in your "all work" of (1-11 corrected)? $\endgroup$ – Blue Various Nov 28 '19 at 13:39
  • $\begingroup$ >Equation 1-11 is not the correct form of the First Law for a closed system ◆Second, is the "closed system" you said on above the one described in my Def2-2 ? If so, Qc and ${W}_{i}$  of (1-11 corrected) inevitably become 0 because work and heat cannot be exchanged with the outside? What kind of closed system do you envisage specifically? $\endgroup$ – Blue Various Nov 28 '19 at 13:47
  • $\begingroup$ ◆Third,if following prerequisites is true, then, we can use (1-1), to (1-10)? And if following prerequisites is not true, then,in such case, at least Qc and Wc can't be written in the form of my (1-5) or (1-6), but does (1-10 applied) itself hold? ●Prerequisites: The chemical reactions are not carried out reversibly (in the sence of my Def 1-2) and chemical energy in the fuel is not converted directly into thermal energy. $\endgroup$ – Blue Various Nov 28 '19 at 14:13
  • $\begingroup$ ◆Forth, what is the "Inward" of Inward Boundary Work"? $\endgroup$ – Blue Various Nov 28 '19 at 14:58
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It is always important to know the assumptions behind an equation. Rarely does an equation hold for all possible scenarios.

The form of the first law you give only considers work done by/on the system and heat that enters/leaves the system. The first law is just a statement of energy conservation, so in general all such energy changes must be taken into account. In your case you have performed correct analysis, as repeated here.

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  • $\begingroup$ Thank you for your answer. >The form of the first law you give only holds for isolated systems without transfer of matter.: Does this "first law" you mentioned is equation (1-11)?  What is "isolated system" you mean? In my textbook the definition of isolated systems is "a system in which neither material nor energy is exchanged with the outside world." If so, both δQ and δW would have to be zero. $\endgroup$ – Blue Various Nov 27 '19 at 9:36
  • $\begingroup$ A system that exchanges energy but does not exchange substances seems to be called a "closed system" to distinguish it. According to my textbook, the definition of closed system is " A system that exchanges energy with the outside world but does not exchange materials." But, more importantly, ${N}_{i}$ can change in both isolated and closed systems in the sense of above. $\endgroup$ – Blue Various Nov 27 '19 at 9:37
  • $\begingroup$ How about the reaction, "CO + O2 →CO2 (coefficients are omitted.)" or "H2O(Vapor)→H2O (liquid)" . In these reactions, it is not necessary for particles to go/come from outside. Quasi-static paths shall exist at least theoretically. We can use the path to calculate the relative entropy between the Initials and final state. $\endgroup$ – Blue Various Nov 27 '19 at 9:38

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