The equation $dS = \frac{\delta Q}{T}$ is only defined for a reversible path. Given a irreversible path we typically calculate the entropy by choosing a reversible path from the same initial to final state. All examples I have seen involves some sort of reservoir that loses entropy such that $\Delta S_{system} =\Delta S_{original}+\Delta S_{res}= 0 $ for the total system in the reversible case, where by `original' I mean the system that was originally in consideration that I had to attached a reservoir to in order to find a reversible path.

My question is this:

(1) In the scheme of attaching a reservoir to our system, how would we generally find a reversible path to calculate the change in entropy for a system?

(2) Depending on the answer to the above: Specifically, if I mix a bunch of liquids together (clearly irreversible) what is the reversible path I could use to find the change in entropy for the irreversible case, and does it involve a reservoir?

  • $\begingroup$ The first equation you written is defined for any (sub)system, not path. You can then find the total change in entropy by summing all dS for the subsystem $\endgroup$
    – unsym
    Dec 17, 2013 at 19:21
  • $\begingroup$ hwlau - I am not sure what you mean by "...(sub)system, not path..." - Since $\delta Q$ is a path dependent quantity, don't I have to, in practice, specify what path I am taking in order to use the equation $dS = \frac{\delta Q}{T}$? $\endgroup$
    – DJBunk
    Dec 18, 2013 at 14:33

1 Answer 1


I will start by answering the second question. Let's consider the case of two species of liquid in a box, with a partition separating them. The irreversible process you describe is to remove the partition. A reversible process would be to have the partition actually composed of two separate independently movable partitions. One of these does not interact with the first species, but is impenetrable to the second species, while the other does not interact with the second species, but is impenetrable to the first. Then if the first species is initially on the left, what you do is move the patition that blocks the first species all the way to the right, and you move the other partition all the way to the left, and you end up with both species mixed. Now after this process, you will not be in the same final state as if you just removed both partitions instantly, because doing things the second way, the liquid has done work on the partitions. Thus the system must be connected to a thermal bath to maintain the temperature. Notice that your situation here is just two independent free exapansions (one for each species). Each of these free expansions is "made reversible" the same way as a single free expansion would be, independently from one another.

I am not so confident about answering the first question, but I would say by looking at my answer to the first part, it looks like you start with a isolated system and change one property of the system suddenly. This is the irreversible process.

To get the reversible process, you need to do two steps.

The first step is to attach a large reservoir to you system, which only interacts via the quantity which was suddenly changed. So if you suddenly changed volume the reservoir would be a piston providing a pressure (as in the case of a free expansion or the above mixing example). If you changed particle number suddenly, the reservoir would a particle reservoir with some chemical potential. Then you change the property (pressure, chemical potential, ...) of this reservoir until your system as the correct equilibrium extensive parameter (volume, particle number, ...). Now detach this reservoir. This ends the first step.

Now for the second step. Since the entropy remains constant but your extensive property of they system has changed, some other extensive property must have change. This property must be the energy, since no other property was allowed to be exchanged with the environment. To get the energy back to its original value, you attached a thermal reservoir with the same temperature of your system currently (at the beginning of this second step), and you slowly change the temperature of the system back to its original value at the beginning of the first step (note this is what we did in the example above). Now your system is in the same final state as it was at the end of the reversible process, so the changes in entropy are the same, and if you know the change in entropy for the reversible process, you know the change in entropy for the irreversible process. I am pretty sure that what I have outlined is not completely general and there are more cases to consider. Some one will probably point that out in the comments. Another thing to note is that for simple systems like an ideal gas, you could calculate the entropy of either state explicitly and just subtract the two to get the change in entropy.


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