Definition and Resolution of the Gibbs paradox in thermodynamics

Question

The entropy of an ideal gas is known to be $$S(V,T)=S_{0}+nR\log \left|\frac{V}{V_0}\right| +C_v\log\left| \frac{T}{T_0}\right|.$$

Now let us have a cylinder of volume $V=V_1+V_2$ separated by an impermeable partition on whose either side there is an ideal gas of $n_1$ or $n_2$ moles. These gases are different but have the same temperature and pressure. On removing the partition the gases diffuse into each other. Since the inter-diffusion of ideal gases is like a Joule expansion into the full volume at constant temperature the resulting entropy increase is $$\Delta S_{12} = \Delta S_1+\Delta S_1 =n_1R\log \left|\frac{V_1+V_2}{V_1}\right|+n_2R\log \left|\frac{V_1+V_2}{V_2}\right| $$ Note that this $\Delta S_{12} >0$ always.

Now assume that the partition separates the same type of gas, since the two parts parts are already, by definition, in thermodynamic equilibrium, removing the partition causes no inter-diffusion. Consequently the entropy increase is zero, $\Delta S^* =0$.

What is the reason for this discrepancy and how it can be resolved in classical thermodynamics?

Answer 1

The back-and-forth in the comments is due to the fact that you are conflating the Gibbs paradox with the entropy of mixing - two distinct (albeit related) concepts. The former is specifically a quasi-paradox which arises in statistical mechanics, and so asking for its resolution in the context of thermodynamics does not make sense.

The answer to the question in the body of your post is that there simply is no discrepancy to resolve. Mixing two different gases is manifestly different from mixing two samples of the same gas; the former introduces entropy, and the latter does not. This is tied to the fact that the latter is reversible by simply re-introducing the partition, while the former is not.

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One then might ask, what do we mean by different? Or rather, how different must two gases be in order for this "entropy of mixing" to take effect? For instance, I might say that I have two boxes of N$_2$ at the same temperature and pressure, so when I remove the partition between them, the entropy of the system stays the same. But you might counter with the fact that nitrogen has several stable isotopes - perhaps one box has slightly more $^{15}$N than the other. In the limiting case, perhaps one box contains pure $^{15}$N and the other pure $^{14}$N. At what point are they sufficiently different to conclude that an entropy increase has occurred?

The answer to that is an often overlooked subtlety in thermodynamics - namely that entropy is, to some extent, subjective. Let's say that Alice and Bob are physicists who study completely identical systems, but their experimental capabilities are different. Alice has a brand new dual-comb spectrometer which can precisely measure the ratio of $^{15}$N to $^{14}$N in her systems, but Bob's lab has not yet invested in that instrument. As a result, when Alice does her calculations, she treats $^{14}$N and $^{15}$N as different atoms, while Bob treats them as the same because he cannot experimentally resolve the difference with the tools available to him.

If Alice receives a box of $^{14}$N and a box of $^{15}$N at the same temperature and pressure, then she would be able to tell the difference between them. If she allowed the boxes to mix, she would say that the total system has increased in entropy as per the standard formula. The results of her experiments would be compatible with the predictions she obtains from her thermodynamic calculations.

If Bob received the same boxes, he would not be able to make the same determination. To him, they are the same gas, and when he allows them to mix, he would say that no entropy increase has occurred. And as long as none of his experiments are sensitive to the difference between $^{14}$N and $^{15}$N, the predictions he makes using thermodynamics would match his observations, too.

This all comes down to the question of how you define a "system" in thermodynamics - and the operational answer is that two systems are different if and only if you have the experimental means to distinguish between the two. The fact that thermodynamics is somewhat agnostic to fundamental questions of e.g. identity is part of why it is so powerful and general a framework.

Answer 2

Replace the impermeable partition with two partitions: one permeable to gas A (but not gas B) and one permeable to gas B (but not gas A). Each partition tends to move through the gas it’s permeable to, to allow expansion of the gas it isn’t permeable to. Collect this pressure–volume work separately. Assuming the partition movement is quasistatic, the process is reversible and the total entropy constant. Note that the gases cool as they adiabatically expand.

Now dissipate the stored energy in the container (e.g., by running a current through a resistor). The entropy has now increased, and the gases are mixed, and the temperature has returned to its original value. The state is indistinguishable from that in which all partitions were removed and the gases allowed to mix irreversibly.

In the case of a single gas, no partition can be produced that’s both permeable and impermeable to the gas, so the above operation isn’t possible. The gas remains at its original entropy. This may clarify the discrepancy.

(An interesting aspect is what it means for two gases to be “different.” If I look only at transparency, for example, then nitrogen and oxygen look identical either separated or mixed. No tailored partition seems possible, and I calculate a constant entropy in all cases, due to my ignorance. Once I learn about atoms, I can in theory design a selective partition and perform the expansion described above. The same argument can be applied once I learn about isotopes of a single particular gas. If someday we learn of a new characteristic distinguishing two types of N-14, we could in theory develop a new type of partition, and we’d need to revise our entropy calculations. This has interesting implications for the objective/subjective nature of entropy. In a thermodynamics context alone, for the above thought experiment, “different” can mean only “separable by a selective partition.”)

Answer 3

You should clarify what you mean by "discrepancy" and what exactly you want to "resolve". Note that there are two distinct paradoxes which are both called "Gibbs Paradox" in the literature. One is about a false increase in entropy which, in statistical mechanics, is calculated when you remove the partition between two gases consisting of distinguishable particles. The other is about the discontinuous vanishing of $\Delta$S when the two gas types transition from similar to same. The first of these two paradoxes is resolved in my paper "Demonstration and resolution of the Gibbs paradox of the first kind" Eur. J. Phys. 35 (2014) 015023 (freely available at arXiv).

Answer 4

The simple answer is that there is no discrepancy. Your formula for $\Delta S_{12}$ only applies to the case of different gases. In that case, you can look at the problem as a double free expansion from the initial volumes $V_1$ and $V_2$ to the final volume $V_1 + V_2$.

If the gases are the same gas, the initial state is the gas in a volume $V_1 + V_2$, even if there is a separating wall, and the final state corresponds again to the same volume. No change of entropy.

Answer 5

Gibbs himself derived this paradox using his entropy of mixing, see page 166, eq. 297, section "Consideration relating to the Increase of Entropy due to the Mixture of Gases by Diffusion," in Gibbs: Collected Works, vol I. He writes

When we say that when two different gases mix by diffusion, as we have supposed, the energy of the whole remains constant, and the entropy receives a certain increase, we mean that the gases could be separated and brought to the same volume and temperature which they had at first by means of certain changes in external bodies, for example, by the passage of a certain amount of heat from a warmer to a colder body. But when we say that when two gas-masses of the same kind are mixed under similar circumstances there is no change of energy or entropy, we do not mean that the gases which have been mixed can be separated without change to external bodies. On the contrary, the separation of the gases is entirely impossible. We call the energy and entropy of the gas-masses when mixed the same as when they were unmixed, because we do not recognize any difference in the substance of the two masses.

In Gibbs' wholly classical view the heart of the problem lies at the (in)separability gases. This idea was followed by Denbigh who showed that:

It is only when mixing two or more pure substances along a reversible path that the entropy of the mixing can be made physically manifest. It is not, in this case, a mere mathematical artifact. This mixing requires a process of successive stages. In any finite number of stages, the external manifestation of the entropy change, as a definite and measurable quantity of heat, is a fully continuous function of the relevant variables. It is only at an infinite and unattainable limit that a non-uniform convergence occurs. And this occurs when considered in terms of the number of stages together with a ‘distinguishability parameter’ appropriate to the particular device which is used to achieve reversibility. These considerations, which are of technological interest to chemical engineers, resolve a paradox derived in chemical theory called Gibbs' Paradox.

Answer 6

Suppose, for the sake of argument, you're dealing with two distinguishable particles, and you just put one in each side of the box. You have one particle on the left ($L$), bouncing around, and another particle on the right ($R$), also bouncing around. Because there's only one configuration, we have

$$ S_\text{LR} = 0 $$

When you remove the membrane, suddenly we lose the information on which particle is in which box. Now we have four states: LL, LR, RL, and RR.

$$ S_\text{LR} = \ln(2^2) = 2 \ln (2) $$

so the entropy increases. (If you work out the entropy change for the $N$-particle system, it's $N\ln(2)$. I leave it to you to show this.)

You can ask yourself: is this actually a paradox? Or is this exactly what we expect? In my view, if you remove the membrane, you've increased the number of microstates accessible to the system. You've increased the phase space accessible to the system. So you've increased the entropy.

If you put the membrane back, you don't decrease the entropy again. If you view the entropy as "the information needed to determine the microstate of a system," well, you're just as ignorant about the microstate after you put in the membrane as you were before. So how could the entropy decrease again?