The internal energy as a function of its natural variables is:

$$dU=-p dV+TdS$$

where $p$ is the system pressure and $dS$ includes only changes of the entropy due to heat transfer (the "reversible" heat).

This can be understood as a root finding process of $dU$ (corresponding to minimization process of $U$) under constant $V$ and $S$ with the Lagrange multipliers $-p$ and $T$:


From this, we can follow that under constant volume and entropy, there should be a minimum of the internal energy!

However, we could instead of using $V$ as a variable also switch to $p$ to yield a minimization function:

$\alpha=dU+\left.\frac{\partial U}{\partial p}\right|_{S}dp+\left.\frac{\partial U}{\partial S}\right|_{p}dS$

Following the same argumentation, $U$ also marks a minimum in a system of constant pressure and entropy. But the equilibrium state of such a system should actually be described by the enhalpy:


rather than the internal energy.

Where is my mistake?

It seems that we can always find a minimum of any thermodynamic potential in each 2D thermodynamic space, but the values of the 2 equilibrium variables at this point do not correspond to the physically correct values. Meaning that only the enthalpy minimum is located at the correct equilibrium pressure and entropy. But how can this be shown?

  • $\begingroup$ "$$dU=-p_\mathrm{ex} dV+TdS$$ "where $dS$ includes changes of the entropy due to irreversibility and heat transfer." It is only irreversible if $p_{ext}$ is not the equilibrium pressure of the system $\endgroup$
    – Bob D
    Mar 8 at 3:14
  • $\begingroup$ Please cite an actual peer reviewed reference for that first equation. $\endgroup$ Mar 8 at 3:21
  • $\begingroup$ Due to $U$ being a state function we should have $dU=\delta W_{rev}+\delta Q_{rev}=\delta W_{irr}+\delta Q_{irr}$ right? And for only volume work: $\delta W_{rev}=-pdV$ and $\delta W_{irr}=-p_{ex}dV$, difference is work lost to heat via irreversible processes. Also, for reversible processes $\delta Q_{rev}=-TdS$, including irreverability: $\delta Q_{irr}=\delta Q_{rev} + \Delta \delta Q$, where $\Delta \delta Q$ corresponds to irreverable heat that must compensate the irreversible work. $\endgroup$
    – Guiste
    Mar 8 at 3:35
  • $\begingroup$ In the book Modern Thermodynamics by Kondepudi they set $\delta Q_{irr}=TdS$, where $dS$ includes reversible and irreversible changes of entropy. $\endgroup$
    – Guiste
    Mar 8 at 3:38
  • $\begingroup$ In the equation for dU, dU=TdS-PdV, P is the thermodynamic equilibrium pressure of the gas at temperature T and specific volume V, and S is the thermodynamic equilibrium entropy at temperature T and specific volume V. Any other interpretation is invalid. The equation refers to the change in these parameters between two closely neighboring thermodynamic equilibrium states, and is a property of the material (the gas) and not the process which brought about the change. $\endgroup$ Mar 8 at 3:51

I think I got it but I would really appreciate other experts' opinions. Let's start with the regular form of $dU$:


where $d_eS$ indicates that entropy is only produced via heat transfer. We can rewrite this as:

$dU=-pdV+TdS-Td_i S$

where $Td_i S$ is the entropy produced by irreversible processes. Under constant $V,S$ we get:

$dU=-Td_i S\geq 0$

Since each irreversible processes increases the system's entropy balance. The internal energy is thus minimized during each process (This is the same interpretation as in Kondepudi, Modern Thermodynamics).

Now comes the new part, let's consider $U$ now to be given as a function of $p$ instead of $V$:

$dU=\left.\frac{\partial U}{\partial p}\right|_Sdp+\left.\frac{\partial U}{\partial S}\right|_pd_eS=\left.\frac{\partial U}{\partial p}\right|_Sdp+\left.\frac{\partial U}{\partial S}\right|_pdS-\left.\frac{\partial U}{\partial S}\right|_pd_iS$

This part is a bit strange, because somehow we are considering the entropy as a variable, but not the full entropy change in the differential, so not quite sure if it is correct like this, opinions please!

For constant $p,S$ this becomes:

$dU=-\left.\frac{\partial U}{\partial S}\right|_pd_iS=-p\left.\frac{\partial T}{\partial p}\right|_S d_i S-Td_i S$

with the help of variable switches and Maxwell's relations. So my understanding $\left.\frac{\partial T}{\partial p}\right|_S$ is not strictly positive, which means that a process can also go up in internal energy in a constant $p,S$ system which means that $U$ does not mark the equilibrium function.

Of course analogously one can show that for the enthalpy indeed again $dH=-Td_i S\geq 0$, which proves that this IS indeed the equilibrium thermodynamic potential.


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