5
$\begingroup$

I know that this question has many other variations on this site, but I'm trying to see if I understood Helmholtz and Gibbs energy properly or not. The material I'm reading from is Physical chemistry by Thomas Engel and Philip Reid, third edition, and An introduction to Thermal Physics by Daniel Schroeder.

In Schroeder's book, Chapter $5$, the author defined Helmholtz energy, $A$, as $A= U-TS$, where,

$U$ is internal energy of the system and $S$ is the system's final entropy.

The author says that, It is the total energy needed to create the system minus the heat you can get free from the environment at temperature $T$. He further states that it is the available or "free" energy.

Then, for a system in constant pressure $(P)$ and temperature $(T)$ environment, he defines Gibbs energy, $G$, as $G= U-TS+PV$, where $PV$ is the atmospheric work term that's in enthalpy,

$H$ $=$$U+PV$.

Also, from Engel and Reid, chapter $6$, we have, for isothermal process,

$dA$ $\le$ $đw_{total}$ $...$$(1)$, including expansion and non-expansion work, where the equality is satisfied for reversible process. Equation (1) allows us a way to calculate maximum work that a system can do on the surroundings.

And in a similar manner $dG$ $\le$ $đw_{non-expansion}$ $....$$(2)$, where the equality is satisfied for reversible process.

It is stated that equation $(2)$ allows one to calculate maximum non-expansion work that can be produced.


Now, this is my understanding:

For reversible process, $A$ represents total available internal energy, and a part of $A$, which is $G$, is the available energy to do non-expansion work.For irreversible processes the inequalities (1) and (2) gives the lower bound for expansion and non-expansion work respectively. Is it the correct way of understanding or am I way off? Also, if $A$ does represent total available internal energy for reversible process then what does it represent for irreversible process? Thanks in advance

$\endgroup$
9
  • 1
    $\begingroup$ pV is an atmospheric work term only at isobaric conditions. $\endgroup$
    – Poutnik
    Jan 23, 2023 at 8:42
  • $\begingroup$ @Poutnik, I did write that the system is in constant temperature and pressure environment. $\endgroup$
    – Natasha J
    Jan 23, 2023 at 9:06
  • $\begingroup$ OK, but it is misleading formulation there, making impression G is defined for constant pressure (P) and temperature (T), what is not true. It is defined generally, having just special advantages at those conditions, like delta pV being the volume work term. The similar for A and constant T and V. $\endgroup$
    – Poutnik
    Jan 23, 2023 at 9:43
  • 1
    $\begingroup$ I will not probably pay enough attention today, so you may want to wait until US wakes up, there are some TD experts. // Yeat another interpretation of A and G is $\Delta S_\mathrm{tot} \ge -\Delta G_\mathrm{sys}/T$ for T,p const, resp. $\Delta S_\mathrm{tot} \ge -\Delta A_\mathrm{sys}/T$, for T,V const, where $\Delta S_\mathrm{tot}=\Delta S_\mathrm{sys} + \Delta S_\mathrm{surr}$ $\endgroup$
    – Poutnik
    Jan 23, 2023 at 11:45
  • 1
    $\begingroup$ Note that $-\Delta G$ represents the maximal amount of work the system may provide during the respective system state change. For isothermal change of ideal gas state, $\Delta G = W=nRT\ln{\frac{p_2}{p_1}}$ $\endgroup$
    – Poutnik
    Jan 25, 2023 at 12:42

0

Your Answer

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