Thermodynamic equilibrium state of constant $(p,S)$ system

Question

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 $T$:

$\alpha=dU+pdV-TdS$

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:

$\alpha=dH-Vdp-TdS$,

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?

Answer 1

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

$dU=-pdV+Td_eS$

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.

Answer 2

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

When we write $$dU=TdS-pdV$$ we imply that $U$ is a function of variables $S, V$: $$U=U(S,V).$$ Thus, minimizing $U$ for fixed $S, V$ is meaningless - it means minimizing value of the function at a specific point (aka: what is the minimum of $f(x)$ at $x=0$?)

The form of the first law of thermodynamics is actually a differential form (see this answer for a crash introduction to differentials): $$ dU(S,V)=\frac{\partial U}{\partial S}dS + \frac{\partial U}{\partial V}dV. $$ Minimizing/maximizing it indeed requires setting $dU=0$, which is equivalent to setting the partial derivatives to zero, so that the differential is zero regardless of the increments $dS, dV$: $$ \frac{\partial U}{\partial S}=T(S,V)=0,\frac{\partial U}{\partial V}=-p(S,V)=0 $$ Note that these are two equations, to find the position of the extremum of a function of two variables, $S, V$. (The conditions on the determinant of the of the second derivative than determine whether we deal with a maximum or the minimum.)

However, neither $T(S,V)$, nor $p(S,V)$ can be zero, so the function really doesn't have an extremum, but a rather a ridge, where to every value of $V$ there is corresponding value of $S$ minimizing the function. This line is easily found by setting: $$ TdS-pdV=0\Rightarrow \frac{dS}{dV}=\frac{p}{T}. $$ If we have the equation of state, e.g., if we deal with an ideal gas, $PV=nkT$, we then get $$ \frac{dS}{dV}=\frac{p}{T}=\frac{nk}{V}\Rightarrow S=S_0 + nk\log V. $$