0
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ 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? $\endgroup$
    – hft
    Apr 7, 2015 at 5:55
  • $\begingroup$ 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. $\endgroup$
    – Neoh
    Apr 7, 2015 at 8:13

1 Answer 1

1
$\begingroup$

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
$\endgroup$
2
  • $\begingroup$ Even for a pure substance we can have spontaneous changes e.g. when supercooled water spontaneously solidifies. $\endgroup$
    – Nanite
    Apr 11, 2015 at 6:54
  • $\begingroup$ 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. $\endgroup$ Apr 11, 2015 at 7:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.