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Offline Supercryptid

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Entropy Thought Experiment
« on: 22/03/2008 19:39:26 »
Let's say that we have a box that is impermeable to all matter and energy. That is, nothing inside the box can interact with anything outside of the box and vice versa. In essence, the box represents a closed system. Therefore, the total amount of matter-energy inside of the box never changes. Inside of the box is a mixture of hydrogen and oxygen gas. The gas is ignited and all of the hydrogen and oxygen burns to form water vapor.

Since this is a closed system, all of the energy released by the reaction is still contained in the box and is not released. The energy required to break one water molecule down into hydrogen and oxygen is the same as the amount of energy released by the formation of one molecule of water from those two gases. Now for my question: is it possible, after some very long period of time, that the thermal energy in the box combined with the random molecular collisions between water molecules could eventually recreate the initial mixture of oxygen and hydrogen gas? After a trillion years? Or even after a googol years? What does this mean for entropy?

If it could eventually revert to its original state, what about more complex objects? If you burned a tree with oxygen in the box to form carbon dioxide, water and ashes, could the tree and oxygen eventually reform after a long period of time?


 

Offline DoctorBeaver

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« Reply #1 on: 22/03/2008 20:12:43 »
Doesn't fusion require more energy than fission? If so, then the energy released by separating water into hydrogen & oxygen would be insufficient to re-combine them.
 

Offline Supercryptid

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« Reply #2 on: 22/03/2008 20:32:53 »
1) Burning hydrogen with oxygen is a chemical reaction, not a nuclear one.

2) The idea that fusion generates more energy than fission isn't exactly relavent here. The fission of uranium and fusing of hydrogen are both exothermic reactions. They are not related to one another in the sense that they are opposites (fusing hydrogen doesn't produce uranium nor does the fissing of uranium produce hydrogen). The opposite of hydrogen fusing into helium would be the fissing of helium into hydrogen. The fusion of hydrogen is exothermic and the fissing of helium is endothermic.
 

Offline DoctorBeaver

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« Reply #3 on: 22/03/2008 20:48:04 »
Maybe fusion is the wrong word. I'm not a scientist and sometimes I say things wrong.

Maybe I should have asked if separating water into hydrogen and oxygen requires (not produces) less energy than re-combining them.
 

Offline Supercryptid

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Entropy Thought Experiment
« Reply #4 on: 22/03/2008 20:54:08 »
According to the conservation of energy, it must be the same. Otherwise, energy would be created or destroyed. Due to inefficiencies such as heat loss, it may require more energy to split the water using technology, but in a closed system, the matter-energy is still the same before and after a reaction.
 

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Entropy Thought Experiment
« Reply #5 on: 22/03/2008 21:34:22 »
I agree that the energy must balance.  In the real world, the leakages of energy from the system would mean that in reality to need to put more energy in than you got out, because some of the energy you obtained was leaked out of the system.  In this gedankenexperiment we are assuming no leakage.

Ofcourse, the fact that you will obtain as much energy out of the combustion as you will put into the disassociation does not mean you will not need more on a transient basis to trigger the disassociation, even if that extra is energy is later recovered.

I think the probability that all of the water will disassociate into hydrogen and oxygen is highly improbably small, but arguable probably non-zero; but the reality will be that some proportion of the water will continually be disassociating.  The issues that are going to matter are the ambient temperature and pressure, and how evenly the energy is distributed in the box, because you will need enough energy in any one locality in order to feed the disassociation process in that locality, but any energy in one region would have to mean less energy in another region, and so less likelihood that there will be sufficient energy elsewhere to drive disassociation.
« Last Edit: 22/03/2008 21:36:43 by another_someone »
 

Offline Make it Lady

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« Reply #6 on: 23/03/2008 00:08:26 »
I agree that the energy must balance.  In the real world, the leakages of energy from the system would mean that in reality to need to put more energy in than you got out, because some of the energy you obtained was leaked out of the system.  In this gedankenexperiment we are assuming no leakage.

Ofcourse, the fact that you will obtain as much energy out of the combustion as you will put into the disassociation does not mean you will not need more on a transient basis to trigger the disassociation, even if that extra is energy is later recovered.

I think the probability that all of the water will disassociate into hydrogen and oxygen is highly improbably small, but arguable probably non-zero; but the reality will be that some proportion of the water will continually be disassociating.  The issues that are going to matter are the ambient temperature and pressure, and how evenly the energy is distributed in the box, because you will need enough energy in any one locality in order to feed the disassociation process in that locality, but any energy in one region would have to mean less energy in another region, and so less likelihood that there will be sufficient energy elsewhere to drive disassociation.

This is the perfect answer. It is very unlikely that you will get all the water turning back into the original components due to the spread of energy throughout the box. You may get the odd water molecule breaking down but the whole lot is unlikely to get that intense energy needed to reverse the reaction completely.If there is a lot of energy present you will also get changes in both directions happening at once.
 

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Entropy Thought Experiment
« Reply #7 on: 23/03/2008 01:31:17 »
The energy released by the combination of Hydrogen and Oxygen molecules corresponds to a specific frequency - most of the photons released will interact with surrounding molecules by giving them Kinetic Energy and resulting in an increase in temperature. The spectrum of energy will be , essentially, black body radiation  corresponding to the final temperature after the combustion. The vast majority of the photons will have much lower energy than is required to cause dissociation of the H and O atoms (the specific frequency which was produced during combination) so the H2O molecules will stay as H2O. There is a small probability that some H2O molecules will dissociate, temporarily but they will re-combine soon.
So, effectively, one frequency has been spread out into a broad spectrum of frequencies; dis-order rules and the lower energy state will dominate.
 

another_someone

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Entropy Thought Experiment
« Reply #8 on: 23/03/2008 02:10:30 »
The vast majority of the photons will have much lower energy than is required to cause dissociation of the H and O atoms (the specific frequency which was produced during combination) so the H2O molecules will stay as H2O.

Probably so, but we have not been given enough information to say it is so.  We have been told that no energy can leek in or out of the box, but we have not been told how much energy was initially in the box (it must be a non-zero amount of energy, otherwise the hydrogen and oxygen would be in solid form and would be unable to mix).  Thus, without knowing the base energy level, we cannot say what the average energy in the final mix will be, only what the increment in energy would be (hence my caveat about the dependency on the ambient temperature and pressure).
 

Offline DoctorBeaver

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« Reply #9 on: 23/03/2008 08:31:44 »
I'll just shut up then  :(
 

Offline Make it Lady

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« Reply #10 on: 23/03/2008 10:27:49 »
I'll just shut up then  :(
Very wise! This was a physical chemistry question. I love a bit of subject interaction.
 

lyner

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Entropy Thought Experiment
« Reply #11 on: 23/03/2008 11:23:01 »
Quote
we have not been given enough information to say it is so.  We have been told that no energy can leek in or out of the box, but we have not been told how much energy was initially in the box
Yes, you are right; if the temperature were high enough it would all be ions and nothing would combine with anything. I sort of assumed STP to start with.
« Last Edit: 23/03/2008 14:17:19 by sophiecentaur »
 

Offline Soul Surfer

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Entropy Thought Experiment
« Reply #12 on: 23/03/2008 12:31:58 »
as you say the initial temperature of the hydrogen and oxygen gases in the box has not been specified.  let us assume that it is moderate say around room temperature and not very cold or very hot.  As the hydrogen and oxygen reacts to form water vapour the temperature will rise however if the temperature rises enough some of the water molecules will start to dissociate so eventually a stable state will be reached inside the chamber where the number of hydrogen and oxygen atoms forming water and liberating energy equals the number of hydrogen and oxygen atom dissociating and releasing energy.  That is the simplest model.

However the situation will be a bit more complex than this because hydrogen and oxygen generally form diatomic molecules the temperature will be so high that theres a fair chance that some of the atoms and molecules will be ionised and there will probably also be a few  HO molecules as well so the container will be filled with a mixture of  hydrogen atoms molecules and ions oxygen atoms molecules and ions HO molecules and ions H20 Molecules and ions and possibly even a few O3 (ozone) molecules ands ions all in a stable equilibrium

 
 

Offline Supercryptid

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Entropy Thought Experiment
« Reply #13 on: 23/03/2008 12:35:22 »
Assume the starting conditions are at STP.
 

Offline Soul Surfer

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« Reply #14 on: 23/03/2008 12:39:45 »
As I said  I just forgot to specify the gas pressure as well.
 

Offline Supercryptid

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« Reply #15 on: 23/03/2008 12:44:03 »
What if quantum tunneling is taken into account as well? Remember, I'm talking about extremely long periods of time. Let's say we wait for a time period long enough for quantum tunneling of particles to become a significant factor.

*Oh, let's also say that particles cannot tunnel out of the box.
« Last Edit: 23/03/2008 12:45:35 by Supercryptid »
 

Offline Soul Surfer

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« Reply #16 on: 23/03/2008 12:57:57 »
At STP hydrogen and oxygen to not spontaneously ignite but the reaction does not need much of a spark to start it so even if you did not start the reaction there would be the occsional cases where atoms reacted and the temperature would rise until the reaction wetnt rapidly to the stable configuration.  After that no change would take place  (subject to your total isolation conditions)
 

Offline Soul Surfer

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« Reply #17 on: 23/03/2008 13:06:25 »
You call it a thought experiment but it is possible to carry it out.  If you made the box out of a highly refractory noble metal like platinum and measured the peak temperature of the gas in the box immediately after the explosion and then raised the box to that temperature the final conditions of your experiment would be realised and the composition of the equilibrium gas mixture measured.
 

Offline JP

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« Reply #18 on: 25/03/2008 13:05:56 »
I think in any of the cases mentioned above, you'll always have a tiny probability of getting the box back to its initial state.  This is especially true since you're dealing with such tiny systems, small energies, and such long periods of time that the smallest probabilities become important.  This means you'll probably need to use quantum mechanics to get accurate predictions, and quantum mechanics allows weird things to happen: H20 molecules can spontaneously separate into hydrogen and oxygen even if they don't have enough energy to do so. 

Quote
What does this mean for entropy?
Entropy basically tells us what the most likely state for your box is: in this case, it is incredibly likely to stay as water, since it'll require a very unlikely series of events in order to revert back to hydrogen and oxygen.  If it takes 1 googol years in order to see the box change to hydrogen and oxygen, for the remaining 1 googol - 1 years you probably saw nearly all water in the box.  Entropy is still useful, since it tells you what you expect to see most of the time. 
 

Offline Bored chemist

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« Reply #19 on: 25/03/2008 19:38:08 »
Eventually the mixture will reach an equlibrium state where the energy released from the reaction will raise the temperature to a point that the reverse reaction (reformation of H2 and O2) is as fast as the forward reaction.
That temperature is easy enough to find. It's the adiabatic flame temperature for H2 and O2; about 3080K
http://en.wikipedia.org/wiki/Adiabatic_flame_temperature


In principle the whole lot might revert to the starting material but the odds are against it.
Calculating thr real probablity isn't easy so in the great tradition of such things I will calculate them for something simpler.
Imagine a box with a sliding partition down the middle with nitrogen on one side (left) and argon on the other. They are at NPT and just to make thigs easier we can pretend the N2 molecules have the same weight as the Ar (theyr'e pretty close and itf you aren't happy with that we can put the box in orbit so they are weightless).
Then you take the partition away and wait.
After a while the gases mix. How likely are they to "unmix"?
Well, their positions are decided by chance and the chance of any one atom being in, say , the left hand side is 50%. Consider each atom or molecule in turn.
There's a 50% chance that the first nitrogen is in the left hand side. There's a 50% chance that the second nitrogen is too so ther's a 1 in 4 chance they both are. look at a 3rd nitrogen and you have 1 in 8. Keep playing this game for all the molecules in the box. For the sort of size of box you might carry groceries in there will be about 10^23 nitrogen molecules. That means the likelihood of them all being on one side of the box is about 1 in 2^(10^23)
We are talking about pretty thin odds here. A google years doesn't come close.
In case you wondered, I hadn't forgotten about the argon. The likelihood of them all being on the right side of the box is pretty much the same number so the probablity of both things happening is roughly the square of that probablity
I think that's about 1 in 2^(2*(10^23))

OK that's a very rough aproximation to the original question but the odds are so remote that if I'm out by a hundred orders of magnitude, it barely matters. I'd need to know what the box was made of but tunneling might be important. The suposed half life for the decay of protons ( something like 10^35 years) would certainly be a factor.



« Last Edit: 25/03/2008 19:43:44 by Bored chemist »
 

Offline Supercryptid

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Entropy Thought Experiment
« Reply #20 on: 28/03/2008 01:35:13 »
Thanks for all the input, especially from Bored chemist.

I guess my main reason for wanting to know this is due to the implications of a possible hypothetical scenario as follows:

Let's say that we put a cat in this same box. In the box with the cat is regular air, with 21% oxygen, 78% nitrogen, and 1% argon at STP. Over time, the cat uses up all of the available oxygen and suffocates to death. It begins to rot and, you know, do whatever dead things do. Given the analogy with the original hydrogen-oxygen scenario, could it be possible that after eons and eons of random molecular collisions, quantum tunneling, and whatever else, the cat would eventually be reconstructed in whole and come back to life? I guess it's exponentially more likely that the cat's head alone would reform first while the body was still disorganized. It might even be that the cat's atoms would rearrange into a multitude of other life forms first.

As for proton decay, let's say that it does happen. Couldn't the proton potentially reform if the resulting decay products collided with one-another at the right place at the right time after bouncing around inside the box for untold periods of time? Would that not allow the reconstruction of the cat even if it takes longer than 10^35 years?

Remember, I don't want to know what is unlikely, extremely unlikely, or INCREDIBLY extremely unlikely. I want to know if it this possible in principle at all, since we can deal with potential infinite spans of time.

If you want, you can say that this "box" is actually a tiny universe in its own right. I guess that would keep things from "getting out" (unless quantum tunneling to other universes is possible). Any way, just assume that the box is impermeable to even quantum tunneling.
« Last Edit: 28/03/2008 01:38:00 by Supercryptid »
 

Offline Soul Surfer

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« Reply #21 on: 28/03/2008 11:07:24 »
It is not possible even in principle on the scale you envisage but it is possible on the scale of a planet provided you do not insist on the same atoms being in the same places.

The reason for it being not possible in the small box is that the assembly process takes time and if it started happening slowly decay processes would destroy it long before it could be assembled and there is no room to dump the extra entropy we would need to use to create the order.  We know it is possible on a planet sized lump because evolution has produced life including some cat sized functioning lumps and it is possible to create more cat sized functioning lumps by breeding them.  however this process requires an increase in entropy of other molecules that have been used in making the beasts.
 

Offline JP

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« Reply #22 on: 28/03/2008 18:32:26 »
I have to disagree.  It certainly would be possible in theory for the cat to reassemble.  It won't ever happen in an actual experiment because the probabilities are so small that it shouldn't happen in the time the universe has existed.

Entropy is useful because we will never actually see the cat reform, even though it's technically possible.  Entropy increases in closed systems because the probability of it doing otherwise is generally too small to ever happen on time scales we care about.  Therefore someone who tells you "based on entropy, the cat won't reform" is correct. 

Finally, I'd add that the reason that life can form on the earth is that the earth isn't a closed system on its own.  It's getting energy input from the sun, and so the entropy on earth can decrease.  (The earth-sun combination does have a net increase in entropy, however.)
 

Offline Onanist

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« Reply #23 on: 29/03/2008 03:29:39 »
also... wouldn't the energy in the box actually be more than that required to split up all those water molecules?

because the energy in the closed system would not only amount to all that released by the combustion reaction, but also the original energy applied to get over the reaction threshold?
 

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Entropy Thought Experiment
« Reply #23 on: 29/03/2008 03:29:39 »

 

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