Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Atomic-S on 16/05/2017 00:02:39
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Is there any way to prepare a system, place a system in a box that isolates it from its environment, wait a while, open the box, and find that the entropy of the contents is less than when it was put in -- if we allow the entropy outside the box to increase by at least the amount lost on the inside of the box? (The process inside the box would violate the Second Law of Thermodynamics if considered by itself, but if the entropy outside increases by a greater amount than that lost inside, it appears that no violation occurs.) For example, do something to a container of gas that sets in motion the generation of a lot of heat outside the box, then put the gas in the box, close the box, open it a while later and find the gas has liquified (i.e., now is at a lower entropy) but the Second Law is not violated because the heat generated by the initial process more than offsets the cold generated inside.
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Another example: If you have two bottles of different gasses, and connect them by a narrow hose, and put them in the isolation box, then after a time they will tend to mix together and become an indistinguishable mixture when you take them out again. What we would want to do is put them into the box initially mixed, and then on taking them out later find that they had unmixed themselves into their bottles. That, of course, is contrary to the Second Law, but it is here proposed that if one somehow processes the mixed gasses before insertion into the isolation box, that they could in theory be set on the road to self-unmixing, and that because the process for initializing the gasses would generate much external heat, the Second Law would not be violated. It is essential, of course, that the initialization process not itself unmix the gases . The gases at the time of insertion into the box must still be mixed in order for the experiment to be valid.
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There is no reason that entropy within a designated region cannot decrease, as long as the system is not "closed." (by closed I mean that no matter, energy, or information can enter or exit the region). But if the region (or box, as you call it) is totally isolated (closed), then it is extremely unlikely (to the point of impossible) for the entropy to decrease (for any macroscopic system).
I can think of several cases in which only matter is contained, but heat is allowed to flow in or out of the region, that would allow for condensation or un-mixing to occur within the container (but preventing the flow of heat prevents the change of state). For instance, a steel bottle filled with steam, will lose heat to the surrounding environment, eventually allowing condensation back to water within the bottle.
...it is here proposed that if one somehow processes the mixed gasses before insertion into the isolation box, that they could in theory be set on the road to self-unmixing, and that because the process for initializing the gasses would generate much external heat, the Second Law would not be violated. It is essential, of course, that the initialization process not itself unmix the gases . The gases at the time of insertion into the box must still be mixed in order for the experiment to be valid.
A few points related to the above quote:
- If the gases are already predisposed to un-mix, how are they prevented from doing so until the time has come?
- The second law of thermodynamics holds across any (every) time interval. This means that you cannot expect to see entropy decrease today because it got higher than expected yesterday.
I recently came across this video, which I think describes entropy quite nicely (it's not perfect, but I like the probabilistic approach), and might help you think through these questions:
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If the two systems are not linked in such a way that they can transfer entropy between each other, then each system is essentially closed. We would therefore not expect the total entropy of either system to decrease. However, if you wait a really long time, entropy can spontaneously decrease even in a closed system (Poincare recurrence theorem).
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However, if you wait a really long time, entropy can spontaneously decrease even in a closed system (Poincare recurrence theorem).
Yes, emphasis on really long time. I think for most macroscopic systems, this is as long (or longer) than the expected wait until the heat death of the universe.
I will point out, that for very small and simple systems (atoms and molecules), spontaneous de-entropizing is relatively fast and common (both of which allow it to be observable), but with any sufficiently large system (still much, much smaller than single cells) there is no way that this process can happen on an observable time scale.
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Is there any way to prepare a system, place a system in a box that isolates it from its environment, wait a while, open the box, and find that the entropy of the contents is less than when it was put in -- if we allow the entropy outside the box to increase by at least the amount lost on the inside of the box? (The process inside the box would violate the Second Law of Thermodynamics if considered by itself, but if the entropy outside increases by a greater amount than that lost inside, it appears that no violation occurs.) For example, do something to a container of gas that sets in motion the generation of a lot of heat outside the box, then put the gas in the box, close the box, open it a while later and find the gas has liquified (i.e., now is at a lower entropy) but the Second Law is not violated because the heat generated by the initial process more than offsets the cold generated inside.
One situation where this can occur is with a super cooled liquid. If we placed a supercooled liquid into a closed container, you could induce a liquid to solid transition allowing the entropy to decrease. This can also happen by starting with a supersaturated solution, then inducing it to precipitate out its dissolved contents. These are not stable steady states to begin with, and are subject to spontaneous change into higher order.
In chemistry, entropy is part of a larger equation; G=H-TS. G is the free energy, H is enthalpy, S is entropy and T is temperature. In a closed system, G will equal 0 and therefore the equation simplifies to H=TS. Enthalpy H is a measure of internal energy. The supercooled liquid has too much internal energy at that temperature. This keeps it a liquid even when it should not be.
As the system tries to lowered its enhanced enthalpy; negative enthalpy, we also get a negative entropy, so G=0 for the closed system.
Entropy is a state variable, meaning for a given state of matter, the entropy is always a specific value. Entropy is not random, but is a specific characteristic of a state. Our supercooled versus the frozen water reflect two different states of matter, with supercooled state wanting to change into the solid state. If we did this experiment a hundred times, the entropy will always be the same in each state.
In that respect, entropy is almost like the facade of a state of matter; its look. Enthalpy is more what is on the inside.
Information entropy is different in that you use language; spoken or computer, with language not a state variable, like matter. This type of entropy is more random. This difference tends to confuse people. With languages there is no universal correlation for word sounds and meanings. This is arbitrary and assigned by humans. States of matter are always consistent and defined by specific parameter and natural laws.
The reason the entropy of the universe increases is the free energy of matter is always trying to lower. This makes heat available for entropy to increase. The second law is an extension of energy conservation as matter changes states into lower free energy states; -TS implies entropy increase has a negative impact on free energy.
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One situation where this can occur is with a super cooled liquid. If we placed a supercooled liquid into a closed container, you could induce a liquid to solid transition allowing the entropy to decrease. This can also happen by starting with a supersaturated solution, then inducing it to precipitate out its dissolved contents. These are not stable steady states to begin with, and are subject to spontaneous change into higher order.
This is an interesting answer, and might actually work. I will have to think more about this. :-)
The first gut objection I have is that both processes (freezing of supercooled liquids, or precipitation of a supersaturated solution) are exothermic. Actually, this is the underlying principle behind several types of instant heat packs. ( ). If the heat is not allowed to escape from the system, I wonder if the process would stop part way through, resulting in a mixture of solid and liquid, but overall at a higher temperature (so higher entropy). If, at any point during the phase change, the entropy decreases steadily, then I would say that this answer is correct. But if the increase in temperature of the surrounding liquid cancels out the increased order of crystallization at every step of the way (as I suspect it would), then this example doesn't fit the criteria.
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Any attempt to arrange the atoms with a predisposition to un-mix will be thwarted by Heisenberg. The best you can hope for is a Poincare event.