Entropy

[chilehed]

I'm somewhere between stages 2 and 3 of my experience with entropy and the 2nd Law of Thermo and could use a bit of a sense check.

 

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.

 

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.

[reor]

So, what exactly is it and how can you measure it?

[chilehed]

OUCH! Tough crowd...

 

 

You first. In my thought experiemnt, is W a constant? If not, why not?

[timo]

W is the number of microstates associated to a macrostate. In the example of your colored balls, W for the unmixed case is the number of ways you can arrange the balls to be seperated by color and W for the mixed case is the number of ways you can arrange the balls to be mixed.

[chilehed]

W is the number of microstates associated to a macrostate. In the example of your colored balls, W for the unmixed case is the number of ways you can arrange the balls to be seperated by color and W for the mixed case is the number of ways you can arrange the balls to be mixed.

Let me more fully explain what I'm thinking.

 

I understand that, for the colored balls, there is no difference in entropy between any arrangement of the colors, whether segregated or mixed (there's no heat content to the colors). Also, if you start with them segregated and then shake them into a mixed state, there is no generation of entropy due to the mixing of the colors. There is, of course, some generation of entropy due to the energy expended by your body, viscous drag through the air, and friction losses between the balls themselves, but none due to the mixing of the colors.

 

And in the same way, there is no entropy associated with the case of all of the balls being located in one quadrant of the box vs. them being all spread out (provided the box is of a size that allows the balls to not come in contact with each other).

 

If entropy was generated due to the arrangement of the colors, then you could never unmix the balls by shaking them, even if the number of balls was very small (say, four or six total). But one can take six balls of two colors, and on shaking them find them spontaneously segregated by color given enough trials. Therefore, the arrangement of the colors has nothing to do with either entropy or entropy generation.

 

Same thing if you have ten balls, or a thousand, or a mol. You just need more trials to get there.

 

Extend this to an ideal gas in a rigid, isolated chamber. At time=0 all of the molecules are in one half of the chamber. At a later time they are evenly distributed throughout. There were no intermolecular forces against which work was done during the expansion (it's an ideal gas), the chamber walls are rigid so that no work was done against either an external atmosphere or the box itself, and the system is isolated so that the temperature of each molecule was unchanged.

 

So dS= dQ/T gives a zero result. The process generates no entropy, and is thus reversible. And indeed, the slow compression and expansion of a gas is reversible: all of the work done on the gas is recoverable.

 

If that's the case, then, if the chamber holds two species of ideal gas on opposite sides of the chamber, no entropy is generated when they mix. Thus the mixing of ideal gases is reversible.

 

I've gotta have my head up my rear somewhere on this, right? Like I said, between stage 2 and stage 3: I think I know what entropy is, but I'm beginning to wonder again. At least I do (usually) get the right answers when doing cycle analysis.

[J.C.MacSwell]

Let me more fully explain what I'm thinking.

 

I understand that' date=' for the colored balls, there is no difference in entropy between any arrangement of the colors, whether segregated or mixed (there's no heat content to the colors). Also, if you start with them segregated and then shake them into a mixed state, there is no generation of entropy due to the mixing of the colors. There is, of course, some generation of entropy due to the energy expended by your body, viscous drag through the air, and friction losses between the balls themselves, but none due to the mixing of the colors.

 

And in the same way, there is no entropy associated with the case of all of the balls being located in one quadrant of the box vs. them being all spread out (provided the box is of a size that allows the balls to not come in contact with each other).

 

If entropy was generated due to the arrangement of the colors, then you could never unmix the balls by shaking them, even if the number of balls was very small (say, four or six total). But one can take six balls of two colors, and on shaking them find them spontaneously segregated by color given enough trials. Therefore, the arrangement of the colors has nothing to do with either entropy or entropy generation.

 

Same thing if you have ten balls, or a thousand, or a mol. You just need more trials to get there.

 

Extend this to an ideal gas in a rigid, isolated chamber. At time=0 all of the molecules are in one half of the chamber. At a later time they are evenly distributed throughout. There were no intermolecular forces against which work was done during the expansion (it's an ideal gas), the chamber walls are rigid so that no work was done against either an external atmosphere or the box itself, and the system is isolated so that the temperature of each molecule was unchanged.

 

So dS= dQ/T gives a zero result. The process generates no entropy, and is thus reversible. And indeed, the slow compression and expansion of a gas is reversible: all of the work done on the gas is recoverable.

 

If that's the case, then, if the chamber holds two species of ideal gas on opposite sides of the chamber, no entropy is generated when they mix. Thus the mixing of ideal gases is reversible.

 

I've gotta have my head up my rear somewhere on this, right? Like I said, between stage 2 and stage 3: I [i']think[/i] I know what entropy is, but I'm beginning to wonder again. At least I do (usually) get the right answers when doing cycle analysis.

 

Increased entropy is often associated with decreased order and often an order type of analogy is used to describe the second law. But it is just an analogy though a strong one. The second law is colourblind (notwithstanding the differing wavelengths associated with the colours in a real sense). Order in of itself is not the opposite of entropy.

[Anjruu]

Sorry for asking a dumb question, but what are these "stages" you are talking about?

[chilehed]

Sorry for asking a dumb question, but what are these "stages" you are talking about?

Somebody once said that there are three stages of understanding entropy:

Stage 1 - You haven’t the slightest idea what it is or what it means.

Stage 2 - You know what it is and what it means.

Stage 3 - You realize that you really don’t know what it is, but you get the right answers.

[chilehed]

Increased entropy is often associated with decreased order and often an order type of analogy is used to describe the second law. But it is just an analogy though a strong one. The second law is colourblind (notwithstanding the differing wavelengths associated with the colours in a real sense). Order in of itself is not the opposite of entropy.

Yeah, I know that the “entropy is a measure of disorder” statement is highly misleading, it’s only true given a very specific definition of disorder. It has nothing to do with how messy your room is.

 

At least part of my error is in forgetting that the position of the balls is only analogous to the microstates of the molecules, and while the former has nothing to do with heat energy the latter do.

 

And I know that unrestrained expansion is irreversible for a real gas, but I don’t get why it would be for an ideal gas. Nor do I understand whether or not the two-species thought experiment would be classified as a slow expansion.

 

I need to take a graduate-level course.

[timo]

Forget heat when you talk about entropy. dS=dQ/T is for chemicists. S=k ln(W) is the definition to go with when you want an understanding of the concept. There is simply more ways to arrange the balls such that they are disordered than there is to have them ordered. Therefore if any arrangement is likely probable, the chances to find the balls disordered are higher than to find them ordered. The ratio of the probabilities increase as the number of balls increases. Thermodynamics is defined in the limit of an infinite number of particles where the probability of finding the balls ordered goes to zero.

Important point once more: Thermodynamics is defined in the limit of infinite particles (10^26 = inifinite ), so your statements about "there still is a chance to find them ordered" are absolutely correct, but asymptotically irrelevant.

[chilehed]

Forget heat when you talk about entropy. dS=dQ/T is for chemicists. S=k ln(W) is the definition to go with when you want an understanding of the concept...

I'm beginning to see that, although since Boltzman's constant has units of J/K I don't think you mean "forget heat" literally.

 

Alright, I see that I'm actually at Stage 1 (or perhaps that means that I'm finally at Stage 3?).

 

Cripes, I have a headache....I'll be back later.