Apparent Contradiction in Helmholtz Free Energy

Question

I am looking at some mathematics of HelmHoltz free energy. Naturally, wikipedia is one of the (hopefully) more reliable sources of information. In its derivation section,
the last equation states that

begin{equation}
    dA=-SdT-pdV
end{equation}

Then, in the later section, someone pointed out the apparent contradiction:

"...This result (the inequality $Delta A leq 0$) seems to contradict the equation $dA = -S dT - P dV$, as keeping T and V constant seems to imply $dA = 0$ and hence $A = text{ constant}$.
 In reality there is no contradiction. After the spontaneous change, the system, as described by thermodynamics, is a different system with a different free energy function than it was before the spontaneous change."

I am in no position to judge the accuracy of the wikipedia text, but I can understand the apparent contradiction that is pointed out.

However, I am not satisfied with the so-called explanation that immediately follows it. My question is, how can $A$ have "different" functions of state? If we are talking about the same system, isn't it that there should be just one function for $A$ ? If so, how do we resolve the apparent contradiction that was being pointed out?

user1476176 · Answer

The example they're considering is a spontaneous (irreversible) process which occurs within a rigid container in water bath - i.e., at constant $T$ and $V$.  Integrating the differential
begin{equation}
text{d}A = -Stext{d}T -P text{d}V 
end{equation}
along a path of constant $T$ and $V$, you get the result that $Delta A = 0$.  The apparent contradiction is that the process was described as spontaneous (irreversible), which implies that $Delta A < 0$.  $Delta A$ cannot simultaneously be zero and less than zero!

The resolution is the fact that the first equation only applies to a pure substance.  For a closed system consisting of a single pure substance at equilibrium, constraining $T$ and $V$ defines all the system properties.  The only "process" which can respect the const. $T$ and $V$ condition is the null process (doing nothing), and this process is reversible. The result that $Delta A = 0$ is not surprising, then; that's exactly what we expect for a reversible process.

To get an irreversible process which occurs at constant $T$ and $V$, you need to allow chemical reactions to occur. Since these reactions change the composition, the system is no longer a single pure substance, so the simple relation for $text{d}A$ above no longer holds.  Different substances have different $A$ values, and - with $T$ and $V$ held constant - the only reactions which will occur spontaneously are those for which the products have a lower $A$ than the reactants.

To summarize:

For a pure substance, $text{d}A = -Stext{d}T -P text{d}V$
For a system of any composition undergoing a process at constant $T$ and $V$:

$Delta A = 0$ if the process is reversible
$Delta A < 0$ if the process is irreversible
$Delta A > 0$ is impossible

Apparent Contradiction in Helmholtz Free Energy

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