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Curiosity question. What happens to entropy in the following situation?

A gas fills an entire container and is in equilibrium. Suddenly all particles are removed from half the container. As such, there are now 1/2 the original number of particles, but all occupy one side of the container.

What happens to entropy? Does it rise or fall?

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  • $\begingroup$ It depends (partly) on the kind of particles - ideal gas vs. real molecules. The latter has to deal with binding forces; the former does not. But anyway, take a look at the equation which calculates entropy for the initial state, and decide what (if anything) changes when you cut the number of particles in half. $\endgroup$ – Carl Witthoft May 17 at 14:36
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Let the container be insulated and rigid, and let the initial entropy of the 2N particles be 2$S_0$. If a rigid partition is first inserted to separate the two sets of particles without letting the remaining N particles expand to the full container, the final entropy of the container contents will be $S_0$ (since entropy is an extensive property), and the change in entropy of the container contents will be $\Delta S=-S_0$. If we then remove the partition and allow the container contents to expand to double its initial volume in the container, its internal energy will not change, and, if it is an ideal gas, its temperature will not change, and its entropy will increase by $Nk\ln{2}$, so that the final entropy of the container contents will be $S_0+Nk\ln{2}$, and the overall change in entropy of the container contents will have changed by $\Delta S=-S_0+Nk\ln{2}$. Without adding the partition first, but still have initially removing half the contents and then allowing equilibration, the entropy change of the contents would have been the same.

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  • $\begingroup$ Just for completeness, you should define $k$ here. $\endgroup$ – Carl Witthoft May 17 at 18:46
  • $\begingroup$ Yes. k is the Boltzmann constant. $\endgroup$ – Chet Miller May 17 at 18:53
  • $\begingroup$ Using $S_0$ as the initial entropy instead, we get $S = \frac{S_0}{2} + \frac{Nkln2}{2}$. If the entropy per particle was $S^{'}_0$ before, the entropy per particle afterwards is $S^{'}_1 = S^{'}_0 + kln2$, which is interesting. $\endgroup$ – Yakk May 18 at 0:01
  • $\begingroup$ Thank you for the very clear answer. My conceptual issue with this problem is with $\Delta S = -S_0$. But perhaps my issue is that I am unclear about what system I am referring to. $\endgroup$ – Lux Aeterna May 18 at 0:07
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The question is ambiguous, since talking about entropy we have to say entropy of what we discuss. In most cases "what" is omitted, since the answer is obvious, but not here. More specifically, assuming that initially we had $2N$ molecules:

  1. One can discuss the entropy of the $2N$ molecules after half of them was removed. Obviously, it will depend on the process of removal, which is non-quasistatic. We can confidently say that this entropy has increased.
  2. One could also discuss the entropy of the remaining $N$ particles. The entropy of this new system might well be less than that of the original one, but it is not the same system. Obviously, its entropy will further increase as the molecules spread over the container.

Let me also note that the discussion above and the question assumed that we are dealing with an isolated container. One could also formulate a more realistic problem of a subsystem being able to exchange particles with the reservoir, in which case the entropy difference between $2N$ and $N$ particles can be calculated using the grand canonical ensemble.

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  • $\begingroup$ Thank you. I will need to think about the system I refer to when calculating the entropy. $\endgroup$ – Lux Aeterna May 18 at 0:08
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With the thermodynamic definition of entropy:

$$dS = \frac{dQ}{T},$$

where $Q$ is heat and $T$ the temperature, the answer of what is the entropy of half the volume of your container will be , half of the entropy of the full one.

Now saying

Suddenly all particles are removed from half the container.

the state stops being "in equilibrium" and has to be studied dynamically, as it will be not an isothermal process to treat entropy as $Q/T$.

Depending on the mechanism of "removed" the entropy can increase, over the nominal half, given its definition in terms of "Entropy as a Measure of the Multiplicity of a System" as shown in the link. counting the microsystems of the changing gas conditions.

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