Entropy increases when two substances mix with each other.

For example, the entropy of mixing of two different gases are given by $$\Delta S= 2Nk\ln\frac{V_f}{V_i}\;.$$

But, the entropy doesn't increase when the two gases mixing are same.

This is pointed by Daniel V Schroeder:

It's important to note that this result applies only if the two gases are different like helium and argon. If you start with the same gas on both sides, the entropy doesn't increase at all when you remove the partition.

Now, why is this so? Both the gases, though the same, increase their individual entropies when they expand, don't they?

So, why did Schroeder say there is no entropy of mixing? What actually happens such that there is no change in entropy when the gases are the same?

  • $\begingroup$ Is this some classical limit? I think the multiplicity would increase regardless of whether or not the two gases are composed of identical particles, upon which entropy depends (at least in statistical mechanics). What if one of the gases is hotter than the other? Does that mean they are different? Schroeder's statement seems odd, though that does not mean it is incorrect. $\endgroup$ Commented Nov 14, 2015 at 21:17
  • $\begingroup$ @honeste_vivere It is safe to assume that we are dealing with gasses at the same temperature: if they differed in temperature, the entropy would increase upon mixing, until equilibrium is reached. $\endgroup$
    – Andrea
    Commented Nov 15, 2015 at 1:14
  • $\begingroup$ @honeste_vivere multiplicity does not increase: when the partition is up, you have to arrange 2N identical particles in 2V of volume, when the partition is down, you have to arrange 2N identical particles in 2V of volume. $\endgroup$
    – Andrea
    Commented Nov 15, 2015 at 1:16
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    $\begingroup$ I think it's worth noting that entropy DOES increase when you lift the barrier; it's just that this increase is pretty neigligable. There are more ways to arrange 2N things in a 2V box then there are ways to arrange N things in a V box twice. $\endgroup$ Commented Nov 15, 2015 at 1:46
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    $\begingroup$ @AndreaDiBiagio - I guess I misunderstood the question. I thought the OP was implying that one added an arbitrary amount of the same gas to the original amount. I did not realize they were implying it was a separated container and the partition was simply being introduced/removed. $\endgroup$ Commented Nov 15, 2015 at 3:38

6 Answers 6


When two identical gases mix, the state is generally indistinguisable from the previous state. If one molecule from the left of the partition changes places with a molecule from the right of the partition, does the mixture actually look any different? If the left and right molecules are identical, you would never know which ones started where.

So entropy doesn't change. It doesn't change because you can't tell the two states apart.

Contrast this with the mixing between two different gases. When a partition is removed and an argon and several helium molecules switch places in space due to collisions, now you can actually see a difference. The state is identifiable and different from the previous state. This means the entropy has changed.

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    $\begingroup$ This is a pretty good answer, but still it bugs me. One should explain why two particles interchanging with each other doesn't count as a new state. This is not obvious in classical physics. It would be really great if someone could explain why, even in a purely classical theory, one must essentially regard particles as indistinguishable for thermodynamics to work out right. $\endgroup$
    – DanielSank
    Commented Nov 15, 2015 at 3:01
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    $\begingroup$ @DanielSank I think of it this way -- if I have 10 ping pong balls in a bag and you look inside it, you see 10 ping pong balls that are identical. If I take the bag away for awhile and then give it back to you, you can look inside and see 10 ping pong balls. Did I change them? Maybe, but you can't tell because the balls are indistinguishable. Maybe I changed one, or all ten, or none. It's the same case with indistinguishable molecules in a box. By virtue of being absolutely identical, there is no way to know that anything changed. I realize that's just an analogy... $\endgroup$
    – tpg2114
    Commented Nov 15, 2015 at 3:04
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    $\begingroup$ In quantum mechanics the indistinguishability is completely obvious, as explained in this other Physics.SE post. I think in the classical case your ping-pong ball analogy works pretty well. If the various atoms/molecules in the gas are all of the same species then I suppose there really isn't a difference between two microstates with two particles swapped. I can't say why, but this still bugs me a little. $\endgroup$
    – DanielSank
    Commented Nov 15, 2015 at 3:07
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    $\begingroup$ @DanielSank The part that always bugged me is that I felt like I could sit and "watch" the box evolve and then I could manually track each molecule, or a subset of them, as they moved. And then I would be able to distinguish the changes. But, when I thought about it, the virtue of explicitly tracking a molecule or subset of molecules makes those molecules unique and therefore distinguishable, violating the underlying assumption. But that also raises another concern -- does/should entropy depend on an observer. In the end, entropy is kind of weird stuff for me sometimes. $\endgroup$
    – tpg2114
    Commented Nov 15, 2015 at 3:11
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    $\begingroup$ @tpg2114. Alternatively, (warning: direct download link) here is a paper by E.T. Jaynes exactly on the entropy of mixing. His point is that the entropy doesn't increase in the case of identical gases mixing, but if at some point we discover that the identical gases actually weren't identical, then all of a sudden, there is an increase in entropy, and this has real consequences: namely, we can now extract more work, and it is a direct consequence of the fact that we have gained new knowledge about the character of the substance. $\endgroup$
    – march
    Commented Nov 16, 2015 at 18:33

Suppose the separated volumes of identical gas are a low-entropy state and the mixed volume is high-entropy. Imagine the reverse process from mixing. You have a single tank full of helium. You insert a partition, so now you have two half-tanks of helium. This can be done reversibly, but it takes you from the high-entropy to the low-entropy state. The entropy decreased without any increase of entropy elsewhere in the universe, so you have violated the second law.

The same argument doesn't work for different species. If you have mixed helium and argon and insert a partition, you'll simply have two volumes of mixed helium and argon, so you haven't gone back to the low-entropy state you started in.

Another way to say this is that if mixing identical gases gives an entropy increase of $\Delta S$, then you ought to be able to get work out of the process. Specifically, you should be able to convert $T\Delta S$ Joules of heat from the environment into an equal amount of work. But what physical mechanism would let you do that?

If the gases are different, you can use a semi-permeable membrane. Set up the gases with helium on the left and argon on the right. Make the membrane permeable only to helium. Make the membrane free to translate. It will start moving to the left as helium moves through it to the right to mix with the argon. You can apply an external force to the membrane and it will still move as long as the force is not too great. In this way you extract work. But if you have helium on both sides, the setup fails; the membrane won't know which way to go.

If the molecules are all distinguishable, then these arguments fail. That is, if you can say "molecules 1, 3, 5, 7 etc are all on the left and molecules 2, 4, 6, etc are all on the right", then allowing them to mix does cause an entropy increase because you lose information about the molecules' position. You can still insert a partition into the gas, but you won't be able to do it in such a way that the molecules are separated the way they were before. They'll just be randomly mixed. So ultimately, whether the entropy increases or not is about whether you're losing information. Having the molecules be different species simply gives us a way to tell them apart, and this distinguishability is what makes the mixing irreversible.

Both the gases, though are same, when expand, make their entropy increase, isn't it?

Yes, if you had two individual volumes of helium and you let both of them double in volume, that would be an entropy increase. But that's not what's happening. When gas expands, entropy increases because you know less about the positions of the molecules. Imagine two separated liters of helium. Choose a molecule. How accurately do you know its position? It could be anywhere in the left liter or anywhere in the right liter. You have an uncertainty of two liters. After the mixing, you still have an uncertainty of two liters. Nothing has changed.

With the two different species, argon on the right and helium on the left, entropy increases because each helium molecule's position uncertainty goes from one liter to two, so it's a different scenario.

  • $\begingroup$ you should be able to convert TΔS Joules of heat from the environment into an equal amount of work- sir, isn't entropy an unavailable energy? How could you use it to do work? $\endgroup$
    – user36790
    Commented Nov 15, 2015 at 2:20
  • $\begingroup$ Entropy is not "unavailable energy". Entropy and energy don't have the same units. The statement in my answer says that if you are increasing the entropy by a certain amount, you should be able to decrease the entropy of the environment by just slightly less than that and still have the process be spontaneous. $\endgroup$ Commented Nov 15, 2015 at 2:25
  • $\begingroup$ I'm not intending to tell entropy is energy; what energy has been used to increase entropy is lost for all purposes. $\endgroup$
    – user36790
    Commented Nov 15, 2015 at 2:34
  • $\begingroup$ I'm afraid your comments do not make much sense to me. The fundamental principle is that spontaneous processes have positive entropy change for the universe. That is the principle behind this answer. $\endgroup$ Commented Nov 15, 2015 at 2:35
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    $\begingroup$ @user36790 is correct. The lost work is $T\Delta S$, and the amount of available work would be $\Delta H - T\Delta S$ where $\Delta H$ is the enthalpy of mixing. $\endgroup$
    – user98538
    Commented Nov 15, 2015 at 4:30

There is something that is lurking in the background here, which is an important statement of physics: all helium atoms are indistinguishable. You cannot tell two helium atoms apart from each other, they are so identical, they are basically the same helium atom, twice.

Now let's use that to understand what is going on.

Two experiments
Imagine you have a rectangular box, with a partition in the middle separating it in two sides. In the first experiment you have argon on one side an helium on the other side. Remove the partition, allowing the gases to mix, and wait. Now close the partition again and look in one of the two sides. This is clearly a very different situation, from the one you started with.

Now do a second experiment, but this time have helium on both sides to begin with. Open the partition, wait, close again. Look in one of the sides. What have you got? The side is full of helium: that's exactly what you started with! Nothing has changed, you still have a box full of helium.

But doesn't the helium expand?
It is tempting to think that when you removed the partition in the second experiment, the "helium on the right" started to expand and diffuse on the left side, and the "helium on the left" was expanding into the right side. But the "helium on the right" and "helium on the left" are both just "helium"! They are identical and they are the same, they cannot "mix". It's a bit of a mind bending concept, but is real.

Bonus: another proof
If you rather not think about indistinguishability, there is a way of convincing yourself that entropy cannot increase in experiment two: by contradiction.

If entropy were to be increased by removing the partition of the box containing only helium, then you could put the partition back, and reopen it again, increasing the entropy of the gas. Do it again and again. Where is all this entropy coming from? No work is being done, no heat is transferred, and you still have the same volume of helium at the same temperature. Entropy is a state variable, therefore it must be the same.

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    $\begingroup$ I just added your 'bonus proof' independently as a comment on the main question, then deleted it when I saw this - so +1 from me. $\endgroup$
    – Baldrick
    Commented Nov 15, 2015 at 20:18

Now, why does this happen so? Both the gases, though are same, when expand, make their entropy increase, isn't it?

Yes, the measure of set of accessible states increases and the "ln W" entropy increases as well. This kind of entropy however has the inconvenient property that it does not simply add when equal systems are brought together; but the entropy of sum is higher than the sum of entropies before.

That is contrary to convention in thermodynamics, where thermodynamic entropy of system made of single chemical species is defined to be additive regardless of whether the two parts of it are mixed or not.

The main point of statistical physics is to explain the formula of thermodynamics and so it is customary to alter the definition of statistical entropy to make the resulting statistical entropy behave the same as the thermodynamic entropy.


This is an old question, but reading through the answers I think one important consideration is missing in the answers discussing the statistical mechanical entropy. The fact is that even including the factor of $ \frac{1}{N!} $ to account for indistinguishability in the statistical mechanical (Boltzmann) definition of entropy for an isolated system is not a strictly additive entropy. Consider two identical ideal gas particles, each in a separate container of volume $ V $. The total entropy of such a composite system is $$ S_{1+2} = 2\ln\frac{V}{\Lambda^3} $$ If we now consider the "mixed" system of two particles in a volume $ 2V $, we have that the entropy is $$ S_{12} = 2\ln\frac{V}{\Lambda^3} + \ln 2 $$ As the number of particles increases, the second term in $ S_{12} $ becomes negligible relative to the total entropy. Because thermodynamics is a theory of macroscopic systems, N is always very large. In the general case of two containers of volume $ V $ with $ N $ particles each, the entropy is $$ S_{1+2} = 2\ln\frac{V^N}{\Lambda^{3N}N!} $$ Using Stirling's approximation this can be simplified to $$ S_{1+2} = 2N\left( \ln \frac{V}{N\Lambda^3} \right) $$ For the "mixed" system, $$ S_{12} = \ln\frac{(2V)^{2N}}{\Lambda^{6N}(2N)!} $$ With Stirling's approximation, $ S_{12} $ becomes $$ S_{12} = 2N\left( \ln \frac{V}{N\Lambda^3} \right) $$ $ S_{1+2} $ and $ S_{12} $ are now equal, but only because we were able to apply Stirling's approximation assuming $ N $ is large.

In terms of the micro-states, this makes sense. The newly accesible micro-states are extremely rare (all gas particles found in half the volume is very unlikely for a system of many particles). So the additivity of the Boltzmann entropy only holds in the limit of a large number of particles.

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    $\begingroup$ As a note $\Lambda$ is not defined in your answer. Nevertheless, I think this hits the point better than the others, the large N limit is key. $\endgroup$
    – KF Gauss
    Commented Aug 11, 2022 at 16:36
  • $\begingroup$ Apologies, you are right about $\Lambda$. $\Lambda$ is the deBroglie thermal wavelength. It doesn't really have any bearing on the answer, but as a physical chemist I included it out of habit. $\endgroup$
    – scmartin
    Commented Aug 28, 2022 at 23:01
  • $\begingroup$ The derivation above by @scmartin proves that the entropy of mixing of two classical identical gases is zero in the thermodynamic limit. Quantum-mechanically, because of tunnelling, the partition is not sufficient to distinguish a gas particle in one container from a particle in the other. Hence, the states before and after the removal of the partition (assumed to contribute negligibly to the system) are the same state. Either way, the entropy of mixing is zero. $\endgroup$ Commented Oct 5, 2023 at 14:30

A short, alternative, and qualitative explanation:

Consider the entropy of a system as not "the disorder" of a system, but the wiggle room in a system. If two gases of the same size get opened to each other, do they both experience more wiggle room? I'll ask this a different way: can individual molecules wiggle less, just as much, or more than in the two separate partitions?

I submit that these molecules have just as many states they can be in (or they can wiggle just as much) together as apart.


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