Entropy Change During Reversible Processes

Question

I'm confused about the Second Law of Thermodynamics. The Second Law of Thermodynamics prohibits a decrease in the entropy of a closed system and states that the entropy is unchanged during a reversible process.

Then why do we say that $\Delta S = \int_a^b{\frac{dQ}{T}}$ for a reversible process? Doesn't the second law simply state that $\Delta S = 0$?

(I'm a high school student teaching himself the principles of thermodynamics, but I am struggling with more challenging material due to my poor understanding of these basics)

Answer 1

To say the same thing David Zaslavsky said in slightly different words, the second law implies that entropy cannot be destroyed, but it doesn't prevent you from moving it around from place to place. When we write the equation $\Delta S = \int_a^b \frac{dQ}{T}$, we're assuming that this $dQ$ represents a flow of heat into or out of the system from somewhere else. Therefore $S$ (which, by convention, represents only the entropy of some particular system) can either increase or decrease. Since we're talking about a reversible process, the entropy of some other system must change by an equal and opposite amount, in order to keep the total constant. That is, $\Delta S + \Delta S_\text{surroundings} = 0$.

One other thing: in thermodynamics, "closed" and "isolated" mean different things. "Isolated" means neither heat nor matter can be exchanged with the environment, whereas "closed" means that matter cannot be exchanged, but heat can. In your question you say the second law "prohibits a decrease in the entropy of a closed system," but actually this only applies to isolated systems, not closed ones. When we apply the equations above, we're not talking about an isolated system, which is why its entropy is allowed to change. I mention this because you said you're teaching yourself, and in that case it will be important to make sure you don't get confused by subtleties of terminology.

Answer 2

What the second law really says (or part of it) is

The entropy of an isolated system during a reversible process is constant

However, this integral formula for $\Delta S$ also applies to processes in which a system exchanges energy with its environment, and thus is not isolated. That $\mathrm{d}Q$ is the amount of energy absorbed from the environment through heat exchange. If a system is isolated (and undergoing a reversible process), $\mathrm{d}Q = 0$ at every step of the way, and thus the second law is trivially satisfied.

Answer 3

Modern work by Jarzynski, Crooks, and others shows that a more accurate statement of the second law is $\langle e^{-\beta\Delta S}\rangle = 0$.

If we use Jensen's inequality, and expand this to first order, we find

$\begin{align} \langle e^{-\beta\Delta S}\rangle &\geq e^{-\langle \beta \Delta S \rangle}\\ 1 &= 1-\langle\beta\Delta S\rangle + O(\beta^2 \Delta S^2) \end{align}$

which shows $\langle \Delta S \rangle \geq 0$

In the thermodynamic limit (where $\Delta S$ is big), the classic version of the second law becomes true (we can drop the expectation brackets), but for small systems, you need to consider the modern form.

This comes from a more fundamental result that relates the probability of positive changes in entropy along a trajectory and negative entropy changes in the time reversed trajectory

$\frac{\mathbf{P}_+(+\Delta S)}{\mathbf{P}_-(-\Delta S)} = e^{+\beta\Delta S }$

Things can occur that generate negative entropy, and things can occur that generate positive entropy, but on average we expect them to have positive change in entropy.

Answer 4

You are getting it all wrong, Halliday's closed system concept holds for the second law. Even to a layman, it is glaring the entropy of an irreversible process never decreases