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)


4 Answers 4


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.

  • 1
    $\begingroup$ I see. So is the 9th edition of Halliday and Resnick poorly phrasing the second law when it states that "If a process occurs in a closed system, the entropy of the system increases for irreversible processes and remains constant for reversible processes. It never decreases"? Should it really say "isolated system" as you pointed out? (This was taken from the chapter summary at the end of Chapter 20) $\endgroup$ Jan 14, 2013 at 12:44
  • 4
    $\begingroup$ Yes, it should say "isolated" instead of "closed". $\endgroup$
    – N. Virgo
    Jan 14, 2013 at 13:43

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.

  • 2
    $\begingroup$ If it is isolated and irreversible then what is dQ $\endgroup$
    – Koolman
    Dec 7, 2016 at 11:33

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.

  • $\begingroup$ I do not think this, although touching an interesting topic, answers the question, which asks about thermodynamic entropy and 2nd law, not about fluctuation theorems. $\endgroup$ Jul 31, 2015 at 19:03
  • $\begingroup$ I think it's important to point out that this: "Doesn't the second law simply state that $\Delta S = 0$?" is not actually true for a reversible process, only that its expectation is zero. $\endgroup$ Jul 31, 2015 at 19:09
  • $\begingroup$ It is true if $S$ is thermodynamic entropy. What you're getting at is a different concept used in statistical physics that is similar to it. $\endgroup$ Jul 31, 2015 at 19:12
  • $\begingroup$ In a real system, the second law (and $\Delta S = 0$ for reversible processes) is true only in the statistical sense. $\endgroup$ Jul 31, 2015 at 19:33

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

  • 1
    $\begingroup$ This answer (though it seems to be at least related to a comment on this answer by Nathaniel) could use with a bit more detail (and maybe a reduction in tone). $\endgroup$
    – Kyle Kanos
    Sep 29, 2015 at 13:11

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