# Paradox with free expansion of ideal gas - where is the mistake?

Suppose we have an isolated chamber of volume $$V$$ with a wall in the middle of the chamber and an ideal gas on one side of the wall. In a quasi-static process we expand the gas to the entire volume $$V$$.

My professor writes in his lecture notes, that since the energy of an ideal gas doesn't depend on it's volume we have $$dU=0 \rightarrow dT=0$$ during the process. Also there's no change in particle number. So from the fundamental thermodynamic equation

$$dU=TdS+pdV$$ we get $$dS=\frac{p}{T}dV=\frac{nR}{V}dV \rightarrow \Delta S=nR\log(2)$$

Now I will present a different line of reasoning: Since the system is isolated there is no heat flow in the system $$\delta Q=0$$ Using $$dS=\frac{\delta Q}{T}=0 \rightarrow \Delta S=0 \quad (*)$$ I think this paradox comes from using $$(*)$$ in a non-reversible process. But it's surprising to me that the fundamental thermodynamic relation yields the right answer, since it's basically derived by using $$(*)$$ and the first law.

$$dU=\delta Q+\delta W\stackrel{(*)}{=}TdS-pdV$$

Can someone shed light on this?

• Just to be clear, there is a vacuum on the other side of the wall? Commented Dec 4, 2019 at 20:10
• And, if so, what is the agent that allows the gas to expand quasi-statically against a vacuum? Commented Dec 4, 2019 at 20:11
• Yes there is a vacuum on the other side. Think of there being multiple walls. Each one with very small distance to the other one. We start pulling out wall after wall. Since the walls are very close to each other in each step we are close to equilibrium.
– user224659
Commented Dec 4, 2019 at 20:17
• It's a purely academic example, used later on to model the mixture of two ideal gases. I'd give reference to the book, but it's in German and doesn't contain any additional information.
– user224659
Commented Dec 4, 2019 at 20:19
• The process as you describe it does not take place quasi-statically no matter how much you wish is was. Commented Dec 4, 2019 at 20:53

The free expansion is an irreversible process so that there are no equilibrium states joining the initial and the final one. For this reason the notation $$dU$$ and similar ones is inappropriate: no infinitesimal changes exist here.

During the free expansion the work of (and on) the gas is evidently $$0$$ and the net heat received by the system is $$0$$ since the system is isolated. From the first principle you find $$\Delta U=0$$ (notice $$\Delta$$ not $$d$$). Using the expression of $$U$$ of an ideal gas, you immediately conclude the the final temperature coincides with the initial one (no intermediate temperatures can be defined during the expansion).

The variation of entropy can be computed out of the known formula for an ideal gas as a state function of temperature and volume, again using only the initial and the final equilibrium states, finding your first result (your derivation is hower wrong if, as it seems, you integrate the infinitesimal variations along the actual transformation). Your last argument is untenable as you are improperly using the def of entropy as you also finally declare.

• What do you think of the method the OP has devised to make the expansion quasistatic (reversible)? Commented Dec 4, 2019 at 20:51
• I think that it is an oxymoron. Quasi static means that the ideal transformation goes through equilibrium states only, here instead there is always a non-equilibrium state between the two (infinitesimally close to each other) equilibrium states considered at each step. Commented Dec 4, 2019 at 21:29
• Great. My feeling too. Expanding into a vacuum, even a small one, is not quasistatic. Commented Dec 4, 2019 at 21:41
• However this is a quite subtle issue where one realises how the intuitive physical ideas are not always equivalent to the mathematical development of the theory. To compute the entropy (I mean to mathematically compute it) for instance of an ideal gas, one uses a smooth curve in the space of the states: each point is an equilibrium state there. Do these transformations exist in real world? In my opinion they do not exist. However trying to approximate these ideal transformations, physics textbooks sometimes end up with ideas like OP's one. Commented Dec 4, 2019 at 21:45
• "Do these transformations exist in real world?" I think most of us would agree no. Disequilibrium is essential for processes to occur in the real world. But they are assumed to exist in the case of the ideal reversible process. We speak of infinitesimally small temperature differences for a reversible heat transfer, and infinitesimally small pressure differences for reversible work. What I am interested in knowing is your opinion on whether or not the OP example is equivalent to these latter two examples. I sense you may not be sure. Commented Dec 4, 2019 at 22:33

You are right in that * is not valid for irreversible processes, however there is no paradox here. The reason is that entropy is a function of state (it does not depend on how the system reaches a given state, only on the state itself), so you can calculate the change in entropy of an irreversible process by using a reversible process that has the same initial and final states.

• What you say is true if it is a free expansion. You could use a reversible isothermal process. But the OP says it’s a quasistatic process. What do you think of the method the OP has devised to make the expansion quasistatic (reversible)? Commented Dec 4, 2019 at 20:55
• @BobD it is a great question, I really need to think the answer. If you know it please let me know
– user65081
Commented Dec 4, 2019 at 21:56
• @BobD I read the explanations from the other answers and I tend to agree with them, it is not quasistatic in the thermodynamical sense because te intermediate steps are not in equilibrium, however small. But I might be wrong
– user65081
Commented Dec 4, 2019 at 22:30
• See my discussion with Valter Moretti above. I feel the OP's example of a quasi-static process is somehow not equivalent to the examples used for reversible heat and work, but I can't quite put my finger on it. I'm going to run this by Chet Miller, whom I consider a thermodynamics mentor of mine. Commented Dec 4, 2019 at 22:47
• @BobD, yes, I read it, and it seems pretty reasonable. If you get a response from Chet Miller please post it somewhere as a comment in one of the answers, I would love to know his opinion. I am still learning what I thought I knew well, lol
– user65081
Commented Dec 4, 2019 at 23:02

In a quasi-static process we expand the gas to the entire volume V.

In response to my question as to how you would expand the gas quasi-statically, you responded:

Think of there being multiple walls. Each one with very small distance to the other one. We start pulling out wall after wall. Since the walls are very close to each other in each step we are close to equilibrium.

I would submit to you that your scheme would not result in a quasi-static (reversible) process. The reason is for a process to be quasi-static, thermal and mechanical disequilibrium needs to be minimized, i.e., approach zero. The procedure you describe doesn't, in my view, achieve this.

If I wish to compress a gas quasi-statically, the pressure of the surroundings must be infinitesimally greater than the pressure of the gas. Likewise, if I wish to expand a gas quasi-statically, the pressure of the gas must be infinitesimally greater than the pressure of the surroundings. In the scheme you propose you are making the displacements infinitely small, but no matter how small you make the volume of the evacuated space between the removable walls, it does not reduce the pressure differential between the gas and the vacuum when a wall is removed.

Bottom line: When you remove the wall no matter how small the volume, the pressure differential remains finite, not infinitesimally small. Therefore, in my view, the process is not quasi-static.

Hope this helps.

I think that the reason is: In the 1st law, P in PdV is the pressure in the gas, but not the external pressure which is zero It was proven in my paper:I. A. Stepanov, Determination of the isobaric heat capacity of gases heated by compression using the Clément-Desormes method, Journal of Chemical, Biological and Physical Sciences. Section C. 10(3), (2020), 108  116,. Free online. https://DOI.org10.24214/jcbsc.C.10.1.10816

Indeed the entropy of the outside (the environment) has not changed because the walls are adiabatic. This is what your second computation (*) showed, as you have referred to heat exchanged with the environment.

However, the entropy is increasing? Why? In a (very metaphorical, not physical!) way, there is heat transfer (although a subtle one [and one that I would not call heat in a very proper sense]) between the two halves of the container as you remove the separation. Basically, as there is the expansion, you might imagine there being a flow of moving molecules (i.e. kinetic energy) and thus a flow of heat (i.e. kinetic energy) between the two halves. So there is entropy production inside the container, despite no heat coming from the outside.

That "hidden" heat is hard to define for an irreversible process. That is why we refer to that of a reversible one.

edit: as pointed out in comments, I do not intend to say that there actually is heat flow, but rather to give a "picture" of what is physically happening. The proper way to "see" where the entropy is coming from would be to refer to other definitions of entropy (the $$S=klogW$$ with $$W$$ number of microstates being the best one) or to compute it using a reversible path (isothermal+adiabatic). However that does not really explain why the entropy increases without any heat flowing in.

• What in the evacuated half is receiving the “flow of heat between the two sides”? Commented Dec 4, 2019 at 20:47
• I admit (am editing now) it was an un-physical answer, it was meant to be a "figure of speech" to understand the discrepancy between the two definitions of entropy. Commented Dec 4, 2019 at 20:49
• however, in a way, the container on the left goes from "some temperature" to the final temperature. We just have difficulties defining things in a non quasi-static process. Commented Dec 4, 2019 at 21:02