Gibbs’ Paradox

Some notes outlining progress on a question concerning Gibbs’ Paradox. Not everything will be totally cogent / concise, as they are working notes.

Project supervised by A. Grosberg at NYU, Autumn 2013.


There are two paradoxes bearing Gibbs’ name, both arising in the context of entropy in statistical physics.

The first paradox concerns the (apparently spurious) entropy gain from processes which leave the thermodynamic entropy unchanged.

The second paradox is closely related, and concerns the mixing of particles. Since the distinction between identical and non-identical particles can be made arbitrarily small, it is mysterious that there exists a dichotomy between the two cases when dealing with entropy generation.

Aside: resolution according to Jaynes

In REF?, Jaynes says that entropy increase has to be treated more “subjectively”. Entropy production is not absolute: if we cannot distinguish the properties of two mixing gases, then there is no entropy increase and no work required to un-mix them. If we can distinguish the gases, then this is no longer true. To repeat, if the particles are experimentally indistinguishable for whatever reason, Gibbs’ paradox is resolved.1

At first this seemed absurd: a colour-blind person calculates zero entropy increase when a box of green balls mixes with a box of red balls, but clearly he is wrong. We could use the mixing to do work (see section \ref{sec:workout}): would the colour-blind person also be blind to winches lifting weights?

In Smith et al. (1992), Jaynes elaborates and his story (originally due to Gibbs) makes more sense. The problem is that when the gases go from being non-identical to identical, our definitions of “reversible” and “original state” change; i.e. we have double standards. In the non-identical case, we want to separate all molecules originally in \(V_1\) and put them back into \(V_1\), and the same for \(V_2\); but in the same-gases case we are happy to just reinsert the diaphragm without reference to the particles’ origins. But this is beside the point: remember that “reversible” applies to thermodynamic variables that we can measure, not to microscopic states. The entropy increase associated with the mixing process corresponds to the work required to recover the same family of microstates: in this case, the original separation of molecules into \(V_1\) and \(V_2\). The multiplicity \(W\) is the size of the aforementioned “family”; but this depends on what we’re actually measuring about the macrostate.
Example2: imagine there are two types of argon, but current technology cannot distinguish them, so mixing them gives \(\Delta S=0\). Then a new solvent, “whifnium”, is synthesised, and it is discovered that one type of argon is soluble in it, but not the other type. By putting this knowledge to use and constructing a setup involving whifnium, we can extract work from the mixing – an observable consequence which produces entropy, i.e. \(\Delta S>0\). So the work extractable depends on “human” information. Identical (even microscopically) physical processes can be assigned different entropy depending on what we’re interested in. But more knowledge lets us extract more work. Conclusion: entropy not a physical property of microstate (as energy is), but an anthropomorphic quantity.

To do:

  1. Make statement of paradoxes clearer and more informative.

  1. In the quantum realm, this indistinguishability may be true as a matter of principle, rather than being due to an insufficiently refined experimental capability.

  2. From Section 5 of Jaynes’ paper in Smith et al. (1992).


To develop an information-theoretic approach to Gibbs’ Paradox. Quantify “distinguishability” in terms of information, and understand the thermodynamic consequences.

How much “effort” do we have to put into measuring the properties of e.g. a particle, to distinguish it from a similar particle? How does this affect the net extractable work? Does considering this effort add to our understanding of the paradox?

Try to develop a model (à la Mandal et al. (2012)) where the information of a particle is explicitly manifest (e.g. a DNA molecule, or the particle is a binary string). Show how the information allows us to extract work (or not), and what energy cost is associated with using the information.

Work from mixing gases


What is the entropy increase associated with mixing dissimilar gases, and how do we extract usable work from this?

Once two distinct gases, \(A\) and \(B\), mix, there is an increase in entropy and it will take some work to separate them again. We can also extract work from their mixing (maximum if quasistatic).


  1. What is entropy change?

  2. Hence what is the maximum work extractable from the mixing process?

  3. How does it vary with dissimilarity of particles?

  4. How could one extract this work?


  1. For distinguishable particles, mixing two volume-\(\frac{V}{2}\) boxes of \(\frac{N}{2}\) ideal-gas particles each incurs an entropy increase of \(\Delta S = N\ln2\).

  2. The maximum work extractable from this process is \(W_{\rm max} = NT\ln2\). (This is also the minimum work required to recover the original state.)

  3. The reason I ask is this: perhaps it is the case that increasingly dissimilar particles have more different attributes which can be harnessed to extract more work. However, there is clearly a maximum amount of work that can be extracted from any mixing process – we cannot gain infinite energy! This leads me to conclude that \(NT\ln2\) is the maximum work out (for two containers of equal volume, with no initial pressure or temperature difference).

  4. \label{pnt:mixwork} Essentially the way to extract work from mixing particles A and B is to have it so that the diaphragm is permeable to particles A and not to B. Then the partial pressure of A will push it over to B’s side, reducing the total pressure in side A. Then we can do \(p\,dV\) work.

    • How much work? Say we have two containers, 1 and 2, which have initial volumes \(V_1^{\rm i}\) and \(V_2^{\rm i}\) respectively. Box 1 contains \(N^{\rm A}\) particles of type A and box 2 contains \(N^{\rm B}\) particles of type B. Everything is in contact with a thermal reservoir at fixed temperature, and we ensure all processes happen sufficiently slowly that equilibrium is maintined at that temperature. For now, assume that the particles obey the ideal gas equation of state.
      The protocol is as follows: the partition is permeable to particles of type A. They diffuse into box 2 until the partial pressures equalise. An impermeable but movable partition is then inserted, and the total pressures in the two boxes equalise by expanding the volume of box 2. This last stage is where work is done.
      After some calculation, I find that \(W=-\frac{V_1^{\rm i}}{V}N^{\rm A}T\ln\frac{N^{\rm A}}{N}\).

    • For the simpler case where the boxes are initially the same size with equal particle numbers, \(W=NT\ln2\) theoretically. But here, we get only \(W=\frac{1}{4}NT\ln2\) for the same conditions. What’s going on?
      Remember that at the end of this process, the gases are not totally mixed. Thus there is still some work left to extract. But does it add up when we take this into account? CHECK.

To do:

  • Extension beyond ideal gas? – Hard, and perhaps not much to learn from the exercise.