I have looked at other questions, videos and articles but I kept reading disorganization, messiness or energy distribution (probability) or a thermodynamical equilibrium and stuff. What do we mean by disorganization? And some say that it is a wrong definition. Also, there is this spontaneity and what is its relation to entropy? I am very confused and I would be very pleased if anyone explained it.

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    $\begingroup$ By disorganization we mean a huge number of possible states ($\Omega$), and entropy equals $k\ln\Omega$. $\endgroup$ – Ivan Neretin May 19 '18 at 13:23
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    $\begingroup$ To add a little to what @Ivan Neretin has written, words such as 'disorganisation, randomness' etc are not helpful and are best forgotten; from memory they arise from a mis-interpretation of a throwaway comment Boltzmann made. If a system at equilibrium has an energy $E$ to $E+\delta E$ and $\Omega$ accessible states then entropy, as first defined by Boltzmann, is uniquely defined by $S\equiv k_B\ln(\Omega)$. In a quantum system the number of states is completely determined. $\endgroup$ – porphyrin May 19 '18 at 14:18
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    $\begingroup$ One devotes the entirety of a course on thermodynamics to understand what entropy is and does. I'm afraid there is not space here to answer this question, although it is an important one. $\endgroup$ – Zhe May 19 '18 at 20:41
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    $\begingroup$ Possible duplicate of Entropy - "Wiggle"? $\endgroup$ – Mithoron May 19 '18 at 21:05

Entropy has two general precise definitions. That which is most useful here is from Boltzmann, S(E) = k ln W(E) where S is the entropy, E is the energy of an isolated system (which is the easiest system to consider), k is Boltzmann's constant, and W(E) is the number of states of the system that have energy E.

For example, if the system has exactly one state with energy E, then S = 0. If there are 2 states with energy E, then S = k ln 2. And so on.

Now "organization" has no precise definition, but is rather an descriptive label we might apply to categorize states of the system. Some of those states look "organized" to us, some look "disorganized." For example, consider a model of atoms adsorbed on a crystal surface. Let's model the surface as a square 3x3 array of sites and put 5 atoms on it. A priori, there are 9!/(9-5)!4! = 126 ways of putting 5 atoms on 9 sites. Now which of these states are "organized?" That's to some extent a matter of opinion, but a common definition is "even" or "symmetrical." So, for example, the state where there's an atom on every other site, in a checkerboard pattern, looks "organized" -- as if the sites formed a 2D crystal. There's only one such state.

So now let us supposed we observe this system, and we have some way of knowing, either from measurement or method of preparation, that it is in an "organized" (symmetric, checkerboard-like) or "unorganized" state. There is only one "organized" state, so if we know the system is in an organized state then its entropy is k ln 1 = 0. On the other hand, if we know the system is in an "unorganized" state there are 125 possible states it could be in, so its entropy is k ln 125, higher.

This is generally the case. States we are willing to call "organized" generally have severe restrictions on the degrees of freedom: their values are distributed symmetrically in some way, e.g. the atoms are distributed uniformly, or in a regular array, the velocities are all the same, and so forth. Because of these restrictions, it is almost always the case that there are fewer states of the system that satisfy the restrictions than that don't. So the entropy of the system when it's known to be "organized" is less than when it is known to be not.

In short, it might help if when you hear "organized" you mentally translate that to "has some symmetry restrictions on how the degrees of freedom (e.g. atoms) are arranged." Hopefully it is intuitive that there are far fewer ways of arranging atoms symmetrically than not, and therefore that the entropy of systems known to be symmetric is lower than the entropy of systems that are not.

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    $\begingroup$ Your discussion about 'organised' and 'symmetry' can be misleading. Entropy is only about counting the number of accessible states at a given energy. If the system has atoms on alternate sites only (as you describe) then this state has the lowest energy and no other sites can be reached. Entropy zero. If it is 'disorganised' as you call it then the energy must be higher and more states are occupied. This is a different situation as the energy is greater. As the energy is greater more states are occupied (including alternate sites) and so the entropy is greater. Both cases cannot co-exist. $\endgroup$ – porphyrin May 20 '18 at 7:04
  • $\begingroup$ I made the normal and physically realistic assumption that the adsorption energy is the same for every site. Hence, an "organized" state with 5 atoms on a certain set of 5 sites has the exact same energy (let us say -5e, with e some per-site adsorption energy) as a "disorganized" state with 5 atoms on any one set of a large collection of sets of 5 sites. $\endgroup$ – Christopher Grayce May 21 '18 at 8:07
  • $\begingroup$ Your energy is in effect the enthalpy, the free energy G is what is important in determining what the state of the system will be and this depends on the entropy. Your regular pattern will be at higher free energy than a disorganised one and so will not be at equilibrium. Only by lowering the temperature can the regular pattern state become the most stable. $\endgroup$ – porphyrin May 22 '18 at 8:58
  • $\begingroup$ Remember we're working in the microcanonical ensemble here, because I explicitly said the system is isolated. So working out the thermodynamic potentials is less useful than it usually is. Nevertheless, since there's no meaningful pV terms, you're right that H is E in this case, and for both kinds of states H is the same (since E is the same). S when the system is known to be "disorganized" is higher, as I said, so G for that situation is indeed lower, just what you'd expect from the general rule that G should be minimized for the macroscopic state of highest entropy for a fixed H. $\endgroup$ – Christopher Grayce May 23 '18 at 15:32

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