1
$\begingroup$

I was reading the Wikipedia page on Helmoltz Free Energy, http://en.wikipedia.org/wiki/Helmholtz_free_energy, and run across a point I cannot unravel. The discussion goes as follows, I reproduce it herein for your convenience:

"<...> the internal energy increase, $\Delta U$, the entropy increase $\Delta S$, and the total amount of work that can be extracted, performed by the system, $W$, are well-defined quantities. Conservation of energy implies: $\Delta U_{\text{bath}} + \Delta U + W = 0$, The volume of the system is kept constant. This means that the volume of the heat bath does not change either and we can conclude that the heat bath does not perform any work. This implies that the amount of heat that flows into the heat bath is given by: $Q_{\text{bath}} = \Delta U_{\text{bath}} =-\left(\Delta U + W\right) $, The heat bath remains in thermal equilibrium at temperature T no matter what the system does. Therefore the entropy change of the heat bath is: $\Delta S_{\text{bath}} = \frac{Q_{\text{bath}}}{T}=-\frac{\Delta U + W}{T} $, The total entropy change is thus given by: $\Delta S_{\text{bath}} +\Delta S= -\frac{\Delta U -T\Delta S+ W}{T} $, Since the system is in thermal equilibrium with the heat bath in the initial and the final states, T is also the temperature of the system in these states. The fact that the system's temperature does not change allows us to express the numerator as the free energy change of the system: $\Delta S_{\text{bath}} +\Delta S=-\frac{\Delta A+ W}{T} $, Since the total change in entropy must always be larger or equal to zero, we obtain the inequality: $W\leq -\Delta A$, If no work is extracted from the system then $\Delta A\leq 0$, We see that for a system kept at constant temperature and volume, the total free energy during a spontaneous change can only decrease, that the total amount of work that can be extracted is limited by the free energy decrease, and that increasing the free energy requires work to be done on the system.<...>"

Two doubts arise:

1) How can the system perform work on the environment without changing volume? The author of the entry expands on this immediately after the quoted section, but his explanation involving a change of free energy is arcane to me

2) The entropy increase of the bath is computed by knowing the temperature $T$ is constant and by computing the heat flow as the difference of performed work and decrease in energy. Could one not compute the change in entropy for the system in the same way/ Why $\Delta S$ differs, in magnitude, from $\Delta S_{bath}$? Is additionally entropy generated in the system? And how could this nt be accounted by the relationship $\delta S = \delta Q / T$? So many thanks

$\endgroup$
  • $\begingroup$ I appreciate that δS=δQ/T applies only to reversible processes, and it is to be changed to an inequality, but then how can one conceptually justify that the thermal bath acts reversibly, on an irreversible system? $\endgroup$ – Smerdjakov Apr 17 '14 at 19:58
2
$\begingroup$
  1. I believe the author is thinking here of things like electrochemical work that don't involve a change of volume (but rather, in this case, moving a charge through a potential difference). In this case a full thermodynamic description involves the chemical potential $\mu$, as a later equation shows. (I agree that the description of “different free energies $A$” isn’t terribly enlightening!)

  2. The nice thing about entropy is that it is a state function, so although $\mathrm{d}S = \delta Q/T$ is only applicable to a reversible change, you can integrate it over a reversible path to find the change in entropy for any path between the same endpoints. In this example, $\Delta S$ and $\Delta S_\text{bath}$ would indeed have equal magnitude in the case of a reversible process. The difference is that, since the bath’s temperature is constant by definition, we can imagine a reversible path between its initial and final states and integrate to give $$ \Delta S_\text{bath} = \int \frac{\delta Q_\text{bath}}{T} = \frac{1}{T}\int\delta Q_\text{bath} = \frac{Q_\text{bath}}{T} $$ as in the Wikipedia page. (This answers your question in the comment: we’re not proposing that the bath “acts reversibly, on an irreversible” process, but rather imagining a reversible path between the bath’s initial and final states, simply for use as a calculation tool.)

    For the system’s entropy change, on the other hand, we can’t propose a reversible path between its initial and final states without knowing the details of the spontaneous change it undergoes, which is why we leave that as $\Delta S$. Your question about entropy being generated doesn’t quite make sense, because total entropy is not a conserved quantity; indeed by the Second Law it is not surprising that it should increase in a spontaneous process.

$\endgroup$
  • $\begingroup$ Just a remaining doubt: yes the entropy is a state function and as such its differential can be integrated exactly, over a reversible path. Now, why would the constant temperature of the bath imply a reversible path exists? Even the system is at constant temperature. $\endgroup$ – Smerdjakov Apr 17 '14 at 20:52
  • $\begingroup$ We can introduce any amount of heat into the bath by lowering its temperature by an infinitesimal amount; this is reversible because raising the temperature by an infinitesimal amount will make the heat flow back the other way. The system is also at constant temperature, but may have other internal changes (like a chemical reaction going on), and without taking these into account we can’t calculate $\Delta S$ in the same way, because we can’t guarantee that the equivalent reversible path will end up in the right spot. $\endgroup$ – Ant Apr 17 '14 at 22:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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