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With thermodynamics, systems are studied using macroscopic variables (pressure, temperature, volume, etc.) which do not need a mechanical explanation, which is what statistical mechanics deals with. So I wonder if it is possible to do thermodynamics without atomic theory (as positivists like Mach would have done. To follow this program it would be necessary first of all to do chemistry without atomic theory. So is it possible to do chemistry entirely with equivalents? I tried. The concept of equivalent is complex, but a definition can be given for each type of reaction. Despite this, however, I was not able to completely eliminate the mole. for example the mole (or the number of molecules) already takes place in the equation of state of ideal gases. In addition, the first equation of state for real gases is that of Van der Waals. It was obtained with molecular hypotheses. However, it is true that the equation of state for real gases can be obtained with a series development of the gas compression factor. But only statistical mechanics tells me if this series is convergent. In addition I am not sure if we can eliminate the concept of atom, molecule and subatomic particles (electrons):

  1. in chemical thermodynamic systems that deals with the equilibrium of reactions,
  2. in thermodynamic that deal with electric and magnetic systems (however, it is true that in this case electromagnetism uses continuous charge distributions)
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    $\begingroup$ Thermodynamics and statistical physics are two different ways of looking at the same thing. The former indeed does not require atomic theory, see, e.g., [here](physics.stackexchange.com/questions/714220/… and here $\endgroup$
    – Roger V.
    Commented Jun 24, 2022 at 6:27
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    $\begingroup$ @RogerVadim, why not post your comment as an answer? -NN $\endgroup$
    – niels nielsen
    Commented Jun 24, 2022 at 7:26
  • $\begingroup$ Well, e.g. temperature dependence of molar heat capacity of multiatomic gases (progressive involving of quantized rotational and later vibrational modes) cannot be addressed without molecular quantum approach. Or generally, any macroscopic behavior which is result of behaviour or properties of atoms or molecules. $\endgroup$
    – Poutnik
    Commented Jun 24, 2022 at 7:28
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    $\begingroup$ @Poutnik it cannot be explained without using microscopic approach, but it can be measured and used for practical purposes, as did the Gibb's contemporaries throughout the industrial revolution in XIXth century (while the development of statistical mechanics took place in the beginning of the XXth). $\endgroup$
    – Roger V.
    Commented Jun 24, 2022 at 7:31
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    $\begingroup$ @Poutnik indeed, and people built palaces and sailing ships long before Newton. It ultimately boils down to the difference between thermodynamics and statistical mechanics, a phenomenological and a microscopic theory - I won't pursue this discussion, since I explained it in the answers linked. $\endgroup$
    – Roger V.
    Commented Jun 24, 2022 at 7:37

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Thermodynamics and statistical physics are two different ways of looking at the same thing. The former indeed does not require atomic theory, see, e.g., here and here.

Also relevant: Timeline of thermodynamics and statistical mechanics

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  • $\begingroup$ It does not require it, but there will be left mysteries for more advanced courses. BTW, I remember having in (deep) past an undergraduate course Statistical thermodynamics. $\endgroup$
    – Poutnik
    Commented Jun 24, 2022 at 7:40
  • $\begingroup$ @Poutnik well, even Landau&Livshits do not discriminate the two in their text, which leads to lots of confusion. Yet, many texts do it in parallel, and the two certainly existed independently for some time. I recommend Prigogine's book as a modern example of doing thermodynamcis without statistical physics. $\endgroup$
    – Roger V.
    Commented Jun 24, 2022 at 7:44
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    $\begingroup$ The name trascription is Lifshitz (Russian: Евге́ний Миха́йлович Ли́фшиц ). Thanks for all your responses. $\endgroup$
    – Poutnik
    Commented Jun 24, 2022 at 7:49
  • $\begingroup$ @Poutnik yes, you are right. $\endgroup$
    – Roger V.
    Commented Jun 24, 2022 at 7:50

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