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In the second stage of the Carnot cycle, a gas is thermally insulated and allowed to expand and do work on the piston.

I understand the reason people give is that because entropy is $\,dS = \,dQ/T$ then a reversible adiabatic expansion doesn't change the entropy because there is no heat energy flowing in or out of the system.

But quantum mechanically (particle in a box), we know that increasing the volume causes an increase in the number of available energy levels i.e. microstates, and since entropy is essentially a measure of the number of microstates, how can this process not cause the entropy to increase?

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The available volume goes up, but the momenta of the particles go down. Since the number of available states is proportional to phase space volume, which includes both momentum and position factors, the total number stays constant. You can do an explicit calculation to show this.

However, it's tricky to see why, classically, these two effects must cancel exactly. It's much more clear in quantum mechanics, where the reasoning goes like this:

  • For a sufficiently slow process, the adiabatic theorem applies, so that initial microstates are mapped one-to-one to final microstates.
  • The entropy is proportional to classical phase space volume, which is proportional to the number of quantum microstates.

Thus the entropy doesn't change.

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  • $\begingroup$ Thanks for your excellent answer. Do you have any good references that explain the above facts you mentioned in good detail? $\endgroup$ – user1654183 Feb 2 '16 at 13:38
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The entropy of a gas does not simply depend on the number of ways to arrange the particles that make up the gas but also on the number of ways of distributing the available energy between those particles. In an adiabatic expansion there is no heat transfer but he gas does do work on its surroundings. This reduces the internal, and so reduces then number of ways that energy be distributed between the particles of the gas.

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  • $\begingroup$ So you're saying that in fact the entropy reduces? Or are you saying that the entropy increase due to volume exactly cancels the entropy decrease due to doing work on the surroundings, and the entropy change is therefore zero? I have never seen an expression that shows that entropy depends on the work done, can you show me where I can find this please? $\endgroup$ – user1654183 May 23 '15 at 13:14
  • $\begingroup$ No I'm saying that for a reversible adiabatic expansion the increase in the number of ways particles can be arranged in space and the decrease in the number of ways the energy of those particles can be distributed compensate for each other. $\endgroup$ – By Symmetry May 23 '15 at 13:16
  • $\begingroup$ To say "entropy depends of work done" doesn't really make sense because work done is not a function of state. Entropy does depend on the energy available to particles in a gas and that energy can be changed either by heating the gas or by doing work. $\endgroup$ – By Symmetry May 23 '15 at 13:22
  • $\begingroup$ That's somewhat cleared it up, but do you know a proof for your second comment? I find it somewhat implausible that they exactly cancel out. $\endgroup$ – user1654183 May 23 '15 at 14:21
  • $\begingroup$ Most of the time they don't exactly cancel, and you change in entropy. Its simply that a reversible adiabatic change is the the very special case where they do. The reason it works in this case is because of the Adiabatic principle $\endgroup$ – By Symmetry May 23 '15 at 14:45
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Basics

Suppose we study a fixed volume:

  1. Reversible means that no entropy is produced inside the volume.

  2. Adiabatic means that no entropy is transported in or out of the volume.

  3. Isentropic means that entropy inside the volume stays constant.

If you have 1 and 2 you also have 3. But it is also possible to have an isentropic change of state where entropy is produced inside the volume. (when exactly the produced entropy leaves the volume)

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Part 2

we know that increasing the volume causes an increase in the number of available energy levels i.e. microstates, ..., how can this process not cause the entropy to increase?

For the change in internal energy of a gas we have the eqn:

$dU=C_v \, dT=T \, ds - p \, dV$

So entropy stays constant if temperature decreases as volume increases in a certain matter:

$\frac{dT}{dV}=-p/C_v$

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