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Entropy typically is an extensive thermodynamic variable. Thus, if I combine two subsystems 1 and 2, the total entropy $S_{total} = S_1 + S_2$. This follows directly from the Boltzmann-entropy when we assume that the two subsystems are independent. In that case the partition sum of the full system is the product of the partition sum of the two subsystems: $$ \Omega_{total} = \Omega_1 \cdot \Omega_2 $$ And thus $$ S_{total} = k_B\ln\Omega_{total} = k_B\ln\Omega_1 + k_B\ln\Omega_2 = S_1 + S_2 $$ However, the assumption of independent subsystems might be not hold for all systems. For example, a phase transition is typically associated with a diverging correlation length, making independent subsystems impossible. The extreme case would be the perfect crystal, where the crystal orientation in one subsystem defines the orientation in any other subsystem. However, the perfect crystal has zero entropy, thus making it a bad example.

Do systems exist (as theoretical or artificial as they might be) for which the entropy is an intensive variable?

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  • $\begingroup$ Even for a phase transition, it is still possible to determine the entropy per unit mass (or per mole), which is an intensive property. It depends on the temperature (or pressure) and the mass fraction of liquid. So it is a weighted average (weighted in terms mass fraction) of the entropy per unit mass of the liquid and the entropy per unit mass of the vapor. $\endgroup$ – Chet Miller Aug 31 '17 at 12:07
  • $\begingroup$ The fact that entropy is extensive is actually a definition from which one can prove the validity of all the other laws of thermodynamics. $\endgroup$ – gented Aug 31 '17 at 12:21
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    $\begingroup$ I do not know about intensive but collisionless gases like many plasmas appear to be well modeled by distribution functions (i.e., kappa distributions) derived from a non-extensive statistical mechanics. $\endgroup$ – honeste_vivere Aug 31 '17 at 14:49
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The thermodynamics of small systems has entropy being intensive

http://aip.scitation.org/doi/pdf/10.1063/1.1732447

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    $\begingroup$ Hi! Would you mind elaborating a bit more and making your answer as self-contained as possible? (at lease the readers do not need to surf any other websites.) $\endgroup$ – Shing Oct 6 '18 at 18:05
  • $\begingroup$ From what I can tell, the example in the linked paper of an intensive entropy is for an "ensemble" of a single particle, which seems like cheating ;). $\endgroup$ – Rococo Oct 6 '18 at 18:35
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I didn't study non-equilibrium thermodynamics so some of my views might be too simplistic.

The classical definition by Clausius explicitly states that entropy should be an extensive quantity. Also entropy is only defined in equilibrium state.

If you take one container with oxygen and one with hydrogen their total entropy will be the sum of the entropies. If you mix the containers together the total entropy will increase because this is a non-equilibrium process. The entropy can further increase if the mixture ignites... After the system reaches equilibrium entropy is extensive again.

The total entropy grows in non-equilibrium processes. Equilibrium process is a limit of slow quasistatc processes. If the limit doesn't exist then the process will always be non-equilibrium.

It seems that entropy must be always extensive. The only way how it can be intensive is that it is intensive and extensive at the same time. Intensive is not oposite of extensive. There are quantities that are neither extensive nor intensive (but as far as I know none of them is a state variable). Intensive means equal in all subsystems. Hence following equation must be satisfied for the system where entropy is both extensive and intensive: $$S_{tot}=S_1+S_2=S_1=S_2$$ The equation can only be satisfied if all the quantities are zero.

Entropy is intensive in systems with zero entropy. I dare say that only in those systems.

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  • $\begingroup$ The entropy of an ideal gas, as given by the Sakur-Tetrode equation (en.wikipedia.org/wiki/Sackur–Tetrode_equation), is only extensive in the thermodynamic limit. This is often the case. $\endgroup$ – Rococo Oct 6 '18 at 18:39

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