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Let us consider a system exchanging work via pressure force only. It is assumed to be at mechanical equilibrium with the environment.

We can write down the differential of the thermodynamic potential as (I don't go into the derivation it will be my starting point):

$$d\Phi=dU+PdV-TdS \leq 0$$

I start with an example I understand: constraints are constant temperature and volume. I thus have:

$$d \Phi=dU-TdS $$

An appropriate potential under those constraints is thus $\Phi=F=U-TS$, it leads to the good differential. The free energy must thus be minimal at equilibrium.

Now, I am looking for transformation at constant energy and volume. I thus have:

$$ d \Phi = -TdS. $$

On the other hand I know that I should find a maximal entropy (just be cause I know that maximal entropy is what happens for $U,V$ constants). How can I find it here ?

The equilibrium will be reached for $\Phi$ being minimal. But what is $\Phi$ ? As the temperature is not assumed constant, I don't see how we could say that $\Phi=-T.S$. Indeed I would have $d \Phi=-TdS - SdT$ which is not the good differential. Thus I don't see how I could find the entropy being maximal.

How to prove from the general construction of thermoynamic potential that the entropy is the good candidate for constraints $U,V$?


I think I have the answer but I would like to check.

Basically for $(U,V)$ constants, I must have: $d \Phi = - T dS < 0$. It is not possible to directly find the potential $\Phi$ here.

However this equation implies that $TdS >0 \Leftrightarrow dS>0$ before the equilibrium. And at equilibrium it becomes an equality thus $dS=0$.

It shows that entropy must be maximal at equilibrium.

So for the entropy the construction is slightly different but we can show it this way.

Would you agree ?

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We can write down the differential of the thermodynamic potential as (I don't go into the derivation it will be my starting point):

$$d\Phi=dU+PdV-TdS \leq 0$$

The thermodynamic potential Gibbs Free Energy is defined as

$$G=U+PV-TS$$.

For constant temperature and pressure the differential change is

$$dG=dU+PdV-TdS$$

Which is the left side of your first equation.

I start with an example I understand: constraints are constant temperature and volume. I thus have:

$$d \Phi=dU-TdS $$

To get your first equation you already imposed the constant temperature constraint. By adding the constant volume constraint your second equation becomes the differential change in the Helmholtz Free Energy thermodynamic potential for constant volume and temperature, or

$$dF=dU-TdS$$

An appropriate potential under those constraints is thus $\Phi=F=U-TS$, it leads to the good differential. The free energy must thus be minimal at equilibrium.

As I already indicated $F$ is the Helmholtz Free Energy.

Now, I am looking for transformation at constant energy and volume. I thus have:

$$ d \Phi = -TdS $$

The equilibrium will be reached for $\Phi$ being minimal. But what is $\Phi$ ?

I don't know what the term is since, to the best of my knowledge, $\Phi$ in this equation does not represent any of the four recognized thermodynamic potentials, Internal energy, $U$, Enthalpy, $H$, Gibbs Free Energy, $G$, and Helmholtz Free Energy. Refer to the following http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thepot.html#c1.

What's more, each of the four thermodynamic potentials are system properties. In your last expression the only term you have on the right is heat, $TdS$. Heat is not a property of a system. Therefore I can't see how your last $\Phi$ is a property of a system. But I must admit I'm not very conversant in the subject of thermodynamic potentials so maybe there is some application for your last expression that I am not aware of.

Hope this helps.

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  • $\begingroup$ Thank you for your answer. I edited, I indeed meant $(T,V)$ constant as constraint (and not $(U,V)$. The potential is thus indeed the Free energy. About entropy, I can reformulate it a little bit. How do we show that entropy must be maximal under $(U,V)$ constants. I can find all the appropriate potential for all the other constraints $(T,V)$, $(E,V)$, $(T,P)$ but not for the entropy which confuses me. $\endgroup$
    – StarBucK
    Mar 25, 2020 at 19:51
  • $\begingroup$ I edited, I hope it makes things more clear (and I hope I didn't add any more typo). $\endgroup$
    – StarBucK
    Mar 25, 2020 at 19:55
  • $\begingroup$ I guess you mean $\Phi=TS$ instead of $\Phi=TdS$. I never wrote $\Phi=T*dS$. I agree that $T*dS$ is the heat for a reversible isothermal process. Now if you meant $\Phi=TS$ is not a thermodynamic property I am not sure to see why. $\endgroup$
    – StarBucK
    Mar 25, 2020 at 20:46
  • $\begingroup$ @StarBucK Sorry, I meant $dΦ=dQ$. I'll re-write my previous comment. $\endgroup$
    – Bob D
    Mar 25, 2020 at 20:55
  • $\begingroup$ @StarBucK First of all, do you understand and agree that $dΦ$ in your equation dΦ=𝑇𝑑𝑆 is not a thermodynamic property and is, in fact, heat dΦ=𝑑𝑄? An example would be heat transfer for a reversible isothermal process. So your equation should be 𝑑𝑄=𝑇𝑑𝑆. $\endgroup$
    – Bob D
    Mar 25, 2020 at 20:56

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