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