Is entropy change always zero after a quasi-static evolution?

Question

If I am thinking in terms of a idealized, perfect carnot cycle I know that in sum

$$\Delta S_{\mathrm{total}} = 0.\tag{1}$$

But that does not mean that there is no entropy generated during the individual steps

$$\Delta S = \frac{Q}{T}.\tag{2}$$

It is just that in total the entropy terms 'cancel out' such that equation (1) holds.

Yet when I read about other processes sometimes they seem to imply that 'quasi-static' always means $\Delta S =0$.

For example as I was reading about the Jarzynski equality he states at the very beginning of the paper:

When the parameters are changed infinitely slowly along some path γ from an initial point A to a final point B in parameter space, then the total work W performed on the system is equal to the Helmholtz free energy difference ∆F between the initial and final configurations.

$$W = \Delta F \tag{3}$$

Now this means that we have a system which moves (slowly) from point A to point B in the parameter space. The only way I can think of to obtain eq. (3) is to use the definition of the Helmholtz free energy

$$F = E - TS\tag{4}$$

re-write it in terms of

$$\Delta F = \Delta E - T\Delta S\tag{5}$$

and iff $\Delta S =0$ then I obtain $\Delta F = \Delta E$ where additionally $\Delta E = W$ because we are in a thermally isolated system yielding the solution of eq. (3).

So my question: Does the fact that we do a quasi-static, slow evolution automatically imply that $\Delta S =0$.

Because literally in the next paragraph Jarzyski says that if the evolution is not quasi static we obtain

$$W \geq \Delta F\tag{6}$$

which (assuming my investigation was correct) would mean that here $\Delta S \geq0.$

Answer 1

The system is not thermally isolated like you think it is. What we actually have is a system always in contact with a heat bath. This means there is heat exchange, but for the slow process we are always in thermal equilibrium with the heat bath. The process is then an isothermal process, not an adiabatic one.

Therefore $\Delta E=W+Q=W+T\Delta S$ so that the change in free energy ends up being $\Delta F=W$. Notice this does not assume $\Delta S=0$. Quasi-static does not mean no change in entropy.

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Just to help even further, technically your change in the free energy is fully given by $$\Delta F=\Delta E-T\Delta S-S\Delta T$$ but since the temperature does not change for the slow process, we get the relationship you give. But for the fast process $\Delta T\neq0$ , which is why you get $\Delta F>0$ (and this analysis technically should be done with differentials)

Answer 2

The equality $\Delta S=0$ is always true for a cyclic process (entropy is a state function).

You have to make a difference between exchanged entropy ${{S}_{e}}=Q/T$ (isothermal process) and created entropy ${{S}_{c}}$which is always positive. For a reversible process, it is the created entropy which is $0$.

With the free energy, you have (isothermal process) :

$\Delta F=\Delta E-T\Delta S=Q+W-T\Delta S$ and $\Delta S={{S}_{e}}+{{S}_{c}}=\frac{Q}{T}+{{S}_{c}}$ or $Q=T\Delta S-T{{S}_{c}}$

Substituting in the first principle you get :

$\Delta F=\Delta E-T\Delta S=Q+W-T\Delta S=W-T{{S}_{c}}$

Conclusion : $T{{S}_{c}}=W-\Delta F\ge 0$

Hope it can help. I don't know "Jarzynski equality" at all and sorry for my poor english.