Another thought about entropy...

[chilehed]

Some of you may recall my thread about how the 2nd Law does not disprove evolution.

I've been meditating on the fact that there's no entropy difference between colored balls that are all mixed up vs. ones that are separated according to color. When I extended this to the case of an isolated, rigid chamber containing two segregated ideal gases, I concluded that there isn't a change in entropy when the gases mix.

Seems counterintuitive, which I know doesn't mean anything, but still...

Then I got to thinking about Boltzman's discovery of the relationship S = k*ln(W), which seems to confirm my conclusion because in this case W remains constant as the gases mix (and k is, of course, a constant itself).

Am I thinking clearly? Or missing something?

Thanks.

 

[RealityCheck]

No, you're not thinking quite correctly here.

Sets of colored balls do not order themselves into distinct sets, as they have no mechnism to do so. You, the observer, have to expend energy to order them. When they are mixed up, your order is ruined, and the balls cannot go backwards and reorder themselves. You can only have a 0 entroy increase if the process is completely reversible - which it is not.

Same with gases. If you keep two gases in separate chambers, and then allow them to mix, there is no mechanism for the gases to spontaneously reorganize themselves in the same pattern they were in prior to mixing. There is an infinitesimally small chance it could happen, since each gas molecule has kinetic energy and, theoretically, it could randomly happen that they all fall into the exact same distribution as before - gas 1 on one side, gas 2 on the other.

But the chances of it happening are so small they're effectively nil.

So entropy increases here.


Think of it this way - if you had to put energy into organizing them, entropy will increase once they break that organization.

 

[chilehed]

I'm afraid that you're mistaken on several points.

The arrangement of the colored balls has absolutely nothing to do with entropy, they can be (and are, given a sufficient number of trials) found to have re-separated after mixing, and there is absolutely no requirement for a "mechanism" for them to do so.

 

[RealityCheck]

I'm afraid that you're mistaken on several points.

The arrangement of the colored balls has absolutely nothing to do with entropy, they can be (and are, given a sufficient number of trials) found to have re-separated after mixing, and there is absolutely no requirement for a "mechanism" for them to do so.

Of course a mechanism is necessary, you have to supply the balls with some energy in order for them to move and rearrange themselves.

It's just like breaking a glass. Take an ordinary glass and drop it, it breaks. You can leave it there for however long you like, but it will not spontaneously rejoin to form a whole glass again (much less in the same shape as before). This is primarily because the original glass required a high input of energy and ordering to begin with, and the broken glass does not have this energy or a mechanism to order itself.

With the balls, you can jumble the balls around to simulate "breaking" the glass, but no matter how long you let the balls sit there, they won't reorder themselves.

Now it's true that if you keep shaking the box or whatever the container is, there is a small chance that at some random time, the arrangement of the balls will revert back to their original pattern. The probability of this, of course, decreases with an increasing number of balls in the box. But notice that in order to get the balls randomly back into their original pattern, entropy is increasing. Your entropy, because you are expending energy in order to re-order the balls. That is the basic principle of the 2nd law of TD - locally entropy can decrease, but any such decrease will be offset by an even larger increase elsewhere.

 

[chilehed]

Of course a mechanism is necessary, you have to supply the balls with some energy in order for them to move and rearrange themselves.

It's just like breaking a glass. Take an ordinary glass and drop it, it breaks. You can leave it there for however long you like, but it will not spontaneously rejoin to form a whole glass again (much less in the same shape as before). This is primarily because the original glass required a high input of energy and ordering to begin with, and the broken glass does not have this energy or a mechanism to order itself.

With the balls, you can jumble the balls around to simulate "breaking" the glass, but no matter how long you let the balls sit there, they won't reorder themselves.

Now it's true that if you keep shaking the box or whatever the container is, there is a small chance that at some random time, the arrangement of the balls will revert back to their original pattern. The probability of this, of course, decreases with an increasing number of balls in the box. But notice that in order to get the balls randomly back into their original pattern, entropy is increasing. Your entropy, because you are expending energy in order to re-order the balls. That is the basic principle of the 2nd law of TD - locally entropy can decrease, but any such decrease will be offset by an even larger increase elsewhere.

I took "mechanism" to mean that some means is necessary by which the normally random motion of the shaking balls is guided so that they fall into the segregated state. Such a claim would be false.

When you shake the balls, entropy is certainly generated due to the thermodynamic processes in your body, the viscous drag with the air, the friction between the balls and the box, and between the balls themselves. Those processes are somewhat irreversible.

But there is absolutely NO entropy generation due to the arragement of the colors. If there was, it would not be possible for the balls to ever be found in their original pattern. The fact that they can proves that the arrangement and re-arrangement of the colors themselves are reversible processes, which by definition means that they result in no generation of entropy.

Look at it this way. Imagine two perfectly identical boxes of perfectly formed balls of identical mass and physical properties, shaken at an identical constant temperature in an identical fashion with the result that each ball in each box travels over the exact same path as the corresponding ball in the other box. The only difference between the boxes is that one has balls of a single color and the other has balls of two colors that were segregated by color at the start of the experiment. In such a case, the entropy generation for each process would be identical. The arrangements of the colors has nothing to do with entropy.

The pattern of the colors has no heat content and no temperature, therefore is not the subject of thermodynamic law.

 

[us38]

I took "mechanism" to mean that some means is necessary by which the normally random motion of the shaking balls is guided so that they fall into the segregated state. Such a claim would be false.

When you shake the balls, entropy is certainly generated due to the thermodynamic processes in your body, the viscous drag with the air, the friction between the balls and the box, and between the balls themselves. Those processes are somewhat irreversible.

But there is absolutely NO entropy generation due to the arragement of the colors. If there was, it would not be possible for the balls to ever be found in their original pattern. The fact that they can proves that the arrangement and re-arrangement of the colors themselves are reversible processes, which by definition means that they result in no generation of entropy.

Look at it this way. Imagine two perfectly identical boxes of perfectly formed balls of identical mass and physical properties, shaken at an identical constant temperature in an identical fashion with the result that each ball in each box travels over the exact same path as the corresponding ball in the other box. The only difference between the boxes is that one has balls of a single color and the other has balls of two colors that were segregated by color at the start of the experiment. In such a case, the entropy generation for each process would be identical. The arrangements of the colors has nothing to do with entropy.

The pattern of the colors has no heat content and no temperature, therefore is not the subject of thermodynamic law.

There certainly is entropy generation when you mix the colors. If there wasn't, then they could go back to their original ordering (or any specific ordering, for that matter) without input of outside energy. But they can only sit there. Something else has to reorder them, and in so doing, raises the overall entropy of the universe.

 

[chilehed]

There certainly is entropy generation when you mix the colors. If there wasn't, then they could go back to their original ordering (or any specific ordering, for that matter) without input of outside energy. But they can only sit there. Something else has to reorder them, and in so doing, raises the overall entropy of the universe.

Nope, I'm afraid you're quite mistaken. The pattern of the colors has no heat content and no temperature, therefore is not the subject of thermodynamic law.

 

[us38]

Nope, I'm afraid you're quite mistaken. The pattern of the colors has no heat content and no temperature, therefore is not the subject of thermodynamic law.

You don't need heat content or temperature to measure entropy. If there was no icrease in entropy, then the balls could go back to their original arrangement without any outside intervention. They can't, so obviously there must be a change in entropy.

 

[Frumious Bandersnatch]

Some of you may recall my thread about how the 2nd Law does not disprove evolution.

I've been meditating on the fact that there's no entropy difference between colored balls that are all mixed up vs. ones that are separated according to color. When I extended this to the case of an isolated, rigid chamber containing two segregated ideal gases, I concluded that there isn't a change in entropy when the gases mix.

Seems counterintuitive, which I know doesn't mean anything, but still...

Then I got to thinking about Boltzman's discovery of the relationship S = k*ln(W), which seems to confirm my conclusion because in this case W remains constant as the gases mix (and k is, of course, a constant itself).

Am I thinking clearly? Or missing something?

Thanks.

There certainly is an entropy of mixing of ideal gases. It is given by DeltaS = -nR(Xa*ln(Xa) + Xbl*n(Xb)) Where R is the gas constant, n is the total number of moles and Xa and Xb are the mole fractions of gas a and b. One of the first lessons in a glass on statisitcal thermo or statistical mechanics will cover how to get this result from S =k*Ln(W), usually right after calculation of the entropy change associated with adiabatic expansion. The situation is different with colored balls.

I think a lot of confusion arises because calculations done with colored balls are used to illustrate the calculations. The problem is that W in Boltzmann's equation represents the number of accessable microstates (which depends on T through the Boltzmann distribution). These microstates even for a monatomic ideal gas involve position and momentum not just position. For a crystal they involve positition and vibrational energy levels.

For the colored balls sitting at rest the only accessible microstate related to the positions of the balls is the one they are in. They can't access any other microstate without the input of new energy into the system. Increasing T will increase the number of microstates for the molecules in the balls but not for the balls, at least until they start melting and then they aren't really balls any more.

This issue is addressed in some detail by Frank Lambert in an article he published in J Chemical Eduction that is online HERE.

F. B.

 

[Frumious Bandersnatch]

Sorry about the duplicate post. I got a database error message when I first tried to post it and didn't think it was posted.

F.B.

 

[chilehed]

You don't need heat content or temperature to measure entropy. If there was no icrease in entropy, then the balls could go back to their original arrangement without any outside intervention. They can't, so obviously there must be a change in entropy.

The balls can go back to their original arrangement.

Read my earlier thread, then let's talk:
http://www.christianforums.com/showthread.php?p=21640037#poststop

 

[chilehed]

There certainly is an entropy of mixing of ideal gases....

Hey, F.B, good to hear from you. Happy New Year.

I knew that my reasoning was wrong because of the rapid expansion thing, but I didn't know why. After I posted that I spent some time looking at statistical mechanics, which is way above my formal training but I was able to learn a little bit. Amazing stuff, I'd like to take some formal coursework but that'll have to wait. Thanks for the link, I'll copy it to my hard drive and read it as I'm able.

 

[us38]

The balls can go back to their original arrangement.

Of course they can, but not without something putting them there. Don't believe me? Get two colors of balls, and put them in a box, with one color on top of the other. Close the box and shake it up. Note that the balls are now mixed. Let them sit. I guarantee that they will not go back to their original arrangement unless you shake them up or put them their. The original shaking is an irreversible process, so the entropy change must be positive.

 

[chilehed]

Of course they can, but not without something putting them there. Don't believe me? Get two colors of balls, and put them in a box, with one color on top of the other. Close the box and shake it up. Note that the balls are now mixed. Let them sit. I guarantee that they will not go back to their original arrangement unless you shake them up or put them their. The original shaking is an irreversible process, so the entropy change must be positive.

My OP in the thread that I linked above dealt with that argument, and explained why the macroscopic arrangement of the balls has nothing to do with entropy.

Entropy is not a generic measure of disorder, and statements such as "entropy is a measure of disorder" are false unless the definition of "disorder" is severly restricted. It has nothing to do with the state of clutter of your room, patterns, "information" content, or the like.

Entropy is a property of matter, expressed in terms of heat and temperature. Patterns, such as the arrangement of the balls, have neither heat content nor temperature, therefore they have nothing to do with entropy.

 

[Frumious Bandersnatch]

Of course they can, but not without something putting them there. Don't believe me? Get two colors of balls, and put them in a box, with one color on top of the other. Close the box and shake it up. Note that the balls are now mixed. Let them sit. I guarantee that they will not go back to their original arrangement unless you shake them up or put them their. The original shaking is an irreversible process, so the entropy change must be positive.

The point is that they will not get mixed up unless you shake them. However, if they are packed tightly in the box they won't get mixed up will they?
But you are right the shaking up is an irreversible process and the entropy of the universe increases. But suppose you had 6 balls three red and three blue. Suppose in one case there were two reds and blue on one side and two blues and and red on the other. This is state 1. Now you shake the box up. After the shake they happen to end up with three reds on one side and three blues on the other. This is state 2. The balls are more "ordered." After the box returns to it original temperature does state 2 have more or less entropy than state 1? Of course the entropy of the universe has increased because of the energy you expended shaking the box but this doesn't really depend on the "state" the balls end up in.

The difference between these systems and gases is that boxes of balls are not ergodic. One of the assumptions that goes into calculating entropy from statistical mechanics is that all accessable microstates are equally probable. The problem with the box is that as I said, the only accessable state is the one it is in.

PS could you consider using a larger font?