# Apparent Contradiction in Helmholtz Free Energy

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?

• I wikipedia is one of your more reliable source of information you are already in trouble. I'd check out a thermodynamics textbook. Do you need a recommendation?
– hft
Apr 7, 2015 at 5:55
• If you have the answer in the textbook, I'd appreciate if you can post it here. I'm not sure if I can get a hand on those books easily.
– Neoh
Apr 7, 2015 at 8:13

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 = -S\text{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, $\text{d}A = -S\text{d}T -P \text{d}V$
• 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
• Even for a pure substance we can have spontaneous changes e.g. when supercooled water spontaneously solidifies. Apr 11, 2015 at 6:54
• Edited to be more precise: for a single pure substance at equilibrium, constraining T and V defines all the system properties and thus precludes any change. Sub-cooled water would not be at equilibrium. Apr 11, 2015 at 7:00