We all know the classical Joule free expansion experiment for an ideal gas: we trap an ideal gas inside half of a cylinder. Then, we open the door and the gas expands to the whole cylinder. After equilibrium has been reached, its volume doubles, temperature remains the same, pressure halves etc.

Another effect of this free expansion is that the entropy will increase. In this case, $ΔS = Nk\ln2$ where $N$ is the number of particles in the gas and $k$ is Boltzmann's constant.

We know from the second law of thermodynamics that this expansion is irreversible. In other words, if no work is done in this system, the gas will never on its own return to be confined in half the cylinder again. This, however, is a probabilistic statement. We know that entropy will not decrease in an isolated system because the probability of this happening is unbelievably, extremely low.

But... how really unlikely is it? In a more quantitative way?

Is there any way we can estimate the average time needed to wait until the molecules all return to that half even if they stay there only for an infinitesimal amount of time? I suppose the answer, in seconds, would be a number several orders of magnitude larger than the estimated age of the universe. Perhaps it will be larger than 10 to the Graham's number seconds or something like this. But this is just my intuition. Is there a way to solve this and find an actual estimation?


1 Answer 1


From the statistical physics point of view, the entropy of a physical system is just the Shannon entropy of the distribution of microstates, given some set of thermodynamic observables such as volume, temperature, etc. Given some gas at equilibrium, the state where all the molecules are concentrated on one side of the cylinder is just another microstate, and there is nothing special about it. Reaching that microstate doesn't reduce the entropy at all, because entropy counts the number of microstates it could be in, not the particular microstate it is in. A lot of sources seem to miscommunicate this point, unfortunately.

With that said, the entropy of a system can spontaneously decrease. That is, the distribution over microstates can evolve to a distribution with less entropy. But this can only happen for non-equilibrium systems. Because, by definition, the microstate distribution of an equilibrium system is unchanging over time. So it can't happen for a cylinder filled with thermally equilibrated gas. It can, however, happen for a system that we've just driven out of equilibrium - say, by pushing or releasing a piston.

Notice the subtlety here - in an away-from-equilibrium system, the system will tend to increase entropy over time as it approaches equilibrium, as per the 2nd law. But the path towards maximum entropy isn't monotonic - there are dips and bumps along the way.

With that framing in mind, if we have a non-equilibrium system, there is definitely a way to compute the probability of its entropy decreasing. The probability is given by the fluctuation theorem. Which has a very simple statement. Let the probability that the entropy will increase by some rate $A$ at some time be $P(A)$. Then:

$$ \frac{P(A)}{P(-A)} = \exp(At) $$

Where $t$ is temperature.

Now finally on to your actual question. Because the state where the gas is concentrated doesn't decrease entropy, the fluctuation theorem doesn't apply. Instead, we can analyze it using the specifics of the system in question i.e. an ideal gas. For an ideal gas, the molecules/particles move independently and don't interact. Any particular gas molecule has at a 1/2 probability, at any point in time, of being on one half of the cylinder. Thus thus probability that all particles are in the same specific half of the cylinder is simply the product of all these probabilities:

$$ p = 2^{-N} $$

Where $N$ is the number of gas molecules. If you want the probability that they are all either in the left half or the right half, you would multiply that number by 2. Or you could even consider the probability that they are arranged in other ways. But the point is that it's on the order of magnitude of $2^{-N}$, which for macroscopic quantites of gas is a very small number.

  • $\begingroup$ Thank you! But I believe that if the gas particles went all to the same side, equilibrium would be lost for a sma amount of time, no? $\endgroup$ Oct 21, 2018 at 23:24
  • $\begingroup$ Equilibrium is defined as a static distribution over microstates. When the particles all go to one side (let's call it state S), it's just the system sampling randomly from the space of all possible microstates, of which state S is one. The system will soon sample from the other microstates. It may be easiest to visualize this when you only have a few particles, say 2. In such a situation you actually find that the equilibrium system is in state S frequently (25% of the time!). But that doesn't change the fact that it's in equilibrium. $\endgroup$
    – Al Nejati
    Oct 21, 2018 at 23:48

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