1
$\begingroup$

Textbook Thermal Physics (Danil V. Schroder), pages 150-151. I have a problem with the following extract:

“Helmholtz free energy ($F = U – TS$) is the total energy needed to create the system, minus the heat you can get for free from an environment at temperature $T$. This heat is given by $TΔS = TS$, where $S$ is the system’s (final) entropy; …”.

Let’s think that the system is an ideal gas, which will be created in a closed cylinder that is in thermal contact with the environment at temperature $T$.

  1. So, $TΔS = TS$. Then the initial entropy of the system is zero, because the system doesn’t exist yet, right?

  2. Then we start creating the system (its particles) at nearly zero temperature (still zero entropy ?). Appearing in the environment at temperature $T$, the system will absorb heat from the environment. This will increase thermal energy of the system so that its temperature rises from 0 to T. If it’s correct, we can write $TΔS = U_{\rm th}$. Consequently, $F$ – is work needed to make non-thermal internal energy of the system (chemical and nuclear energy). Then, we can write $$F = U - U_{\rm th}$$

(Interesting thing, in textbook Principles & Practice of Physics by Eric Mazur these energies and something else are grouped to so called source energy. So, if all the above is correct, I guess one can write $F = U_{\rm source}$.)

  1. Why is this heat equal to $TΔS$? The process is isothermal for the environment but not for the system. So, we can use the simplest expression for entropy change for the environment only: $$ΔS_{\rm env} = -\frac{Q}{T}$$ but we cannot use it for the system. Here $ΔS_{\rm env}$ is not equal to $-ΔS_{\rm sys}$ because the process doesn’t seem to be reversible.

Where I am wrong? Or how exactly the system should be created so that $F$ is just work to create this system and absorbed heat is equal to $TS_{\rm sys}$?

$\endgroup$
2
  • $\begingroup$ It's my understanding that the equation for the Helmholtz free energy is $F=U-TS$ not $F=U-T\Delta S$. See hyperphysics.phy-astr.gsu.edu/hbase/thermo/thepot.html#c1 $\endgroup$
    – Bob D
    Feb 9, 2022 at 13:49
  • $\begingroup$ You are wasting your time by trying to ascribe a physical interpretation to F. Just accept the definition and moves on to how F is used. $\endgroup$ Feb 9, 2022 at 14:00

1 Answer 1

1
$\begingroup$

Let's operate by analogy from the enthalpy $H\equiv U+PV$, where the additional term (the mechanical work $PV$) may be easier to conceptualize.

We often say that the enthalpy includes a system's internal energy $U$ plus the work we had to do to move the atmosphere out of the way. Atmospheric pressure $P$, system volume $V$: OK. Now, the system volume doesn't explode into existence; rather, it must expand infinitesimally slowly, with the pressure $P$ always balanced. (If this weren't the case, then there'd be some arbitrary entropy generation term that would depend on the details of the expansion; this would be undesirable.) We might imagine a perfectly balanced and frictionless mechanism mediating this reversible expansion.

Based on this reasoning, we can now consider the terms in the Helmholtz energy $F\equiv U-TS$, which is sometimes described as the system's internal energy $U$ minus free heating we get from the environment. By analogy to the $PV$ analysis, we can now say that suddenly introducing the system at 0 K and allowing rapid (and irreversible) heat transfer is not the correct way to interpret the $TS$ term, as you've deduced above. Instead, we need to reversibly transfer the system entropy $S$ while maintaining balance with the surrounding temperature $T$. Here, we might imagine a perfect-efficiency (e.g., a Carnot) heat engine extracting $TS$ from the environment and delivering it to the system.

The common conceptual element in both cases is the reversible mechanism that provides the $PV$ or $TS$ transfer to or from the environment, respectively.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.