Talk:Clausius-Clapeyron relation

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Note to all contributors:

Quite a few people have "improved" this page by saying that increased pressure actually is how ice skating works, or otherwise modified the example given and removed mention of sumo wrestlers. However, the page as it stands (6th April 2006) is correct--that is, ice will not melt from the pressure of the ice skates. I chose the example I did because it is the solution to a thermodynamics assignment problem I had in my final year as an undergraduate. So before you change it, you might want to send an e-mail to Prof. Don Melrose of the University of Sydney, director of the Research Centre for Theoretical Astrophysics, and tell him that he can't do thermodynamics.

--ckerr (the article's original author)


The thermodynamics are not in question. I suggest that you are discussing an over simplified model of a skate on ice. Consider the nature of the contact between ice skate blade and the ice in reality. Firstly ice skates, particularly hockey skates, have a curve along their length (to facilitate turning) so the area of contact is much smaller than the full 'length' would allow. Secondly blades tend to be squared off in section and you skate on the corners of the square, further reducing the area of contact. Finally we must consider the surface of the ice, not a smooth flat surface, but a maze of ridges and bumps. The area of contact is thus much smaller than the total blade area, consequently the pressure is much higher than has been discussed here. Contacts often generate much higher pressures than simple considerations predict (for example plastic deformations in ordinary steel ball race bearings are such that lubricants can be forced to undergo local glass transitions i.e. solidify, at the point of contact in normal operation. This requires pressures several orders of magnitude higher than predicted by a simple picture of lubricated sphere on plane) --DaveGwyddel 14:23, 10 May 2006 (UTC)

I don't dispute that many types of ice skates are curved, nor that one often skates on the edges of the skate. However, since ice skating is possible (and indeed, not noticeably different) when neither of these conditions hold, I think it's clear that they're not a crucial component of the explanation. As for the fact that the ice is uneven, that's certainly true, and I hadn't considered that. However, I wonder exactly how many of those "ridges and bumps" survive the pressure of the ice skate, and I find it hard to imagine that ice skating works because of small quantities of water that appear at the tips of tiny bumps on the ice. Anyway, we could certainly turn the article into a comprehensive analysis on the physics of ice skating, but I don't think that's the point. The Clausius-Clapeyron relation itself is an approximation (it assumes the slope is constant, which is never quite true), so I don't see any harm in using a non-rigorous example. Personally, I hate it when I'm looking for information on a physics topic and I find a mathematically flawless, indisputably correct article which is also entirely incomprehensible and useless for solving a practical problem.
Ckerr 16:00, 18 May 2006 (UTC)


You have to account for the heat differential, produced by friction, between the blades and the ice. This would obviously reduce delta P.

That example is absolutely terrible. It does not demonstrate what you claim. First off, you claim that 27 MPa would be the pressure generated by a sumo wrestler wearing a stilleto heel. What does a sumo wrestler and a stilleto heel have anything to do with an ice skate. The example of a stilleto heel is USELESS UNLESS YOU COMPARE THE HEEL'S CONTACT AREA WITH AN ESTIMATE OF THE CONTACT AREA OF THE SKATE. For example, what if the skate had a contact area of, say, 10 mm^2??? Then your sumo wrester would create a dP of almost 150 MPa standing on one ice skate, which is well over 27 MPa. Another discussion author recognizes that contact pressures are not simply uniform, and he is right, but I'm saying even if you're giving us rough, first order approximations, your example does not merit the conclusion you come to. Think about the problem yourself. Remember that homework problems are sometimes wrong. I don't care who your professor was. Like the another discusser said, it's not the thermodynamics that's in question; rather it's your statement of the example problem, flawed logic, and lack of explanation.

-Sean

Look, if the pressure of ice skates was sufficient to melt the ice, then guess what? You couldn't stand on the ice, you'd just fall through! (This can actually happen; if you hang a heavy weight from a wire suspended over an ice rod, the wire will cut through the rod, which in turn refreezes as the wire passes through it.) So it appears that, as a matter of fact, the theoretical explanation agrees with experiment. Could you come up with a more complicated theoretical model that also agrees with experiment? Yes, but what's the point? Why not use the simplest model that works?
As for "homework problems are sometimes wrong", that's true. But don't you think that many (if not most) of the people who read this article read it because they have a homework problem on the Clausius Clapeyron relation? For that audience, having a worked-through problem will be very helpful, regardless of anything else.
Ckerr 16:00, 18 May 2006 (UTC)

Exactly. But the point is not how ice skating works, but rather how it doesn't work. Personally, I don't believe that the coefficient of friction between ice and steel is all that high--in other words, I doubt the ice does melt.

From the article

PLEASE FIX!

(note: though this statement might be correct (I don't know if it is), it has neglected the fact that an ice skate functions on two knife edges. These are likely to be razor sharp so much smaller than 0.5 cm2 in area. I concede that an example was a good idea seeing it shows how the equation is used, so should not be removed. Could someone please correct this example so it has some grounds in reality or find a new one to replace it.)

Note from the original author: the example is paraphrased from H.B. Callen's popular text "Thermodynamics and an Introduction to Thermostatistics." Considering that ice skates are, say, 20cm long (an underestimate), this implies a width of 0.5/(20*2)=0.125 mm. You could easily cut yourself on that, and I've never had to take that much care handling ice skates.

The original example from Callen stated that the ice skates were 25 cm long and 2 mm wide (area = 10 cm2); I believe this to be a gross exaggeration, and the numbers I used are, I believe, much more reasonable.

_________________________________________________________________________________________________________________________________________________________________

"Further research in materials revealed the true nature of skating. Because the atomic structure of ice is a crystalline structure, it turns out that having this structure abruptly stop when it reaches the top of the ice is not the most entropically favorable form. Instead, there is always a thin film of liquid water ranging in thickness from only a few molecules to thousands of molecules on top of the ice. This allows a smoother transition from the structured ice to the completely random structure of the air molecules. The thickness of this liquid layer depends almost entirely on the temperature of the surface of the ice (higher temperatures give a thicker layer), and the liquid layer disappears around -20°C (-4°F). At temperatures below -20°C, ice skating becomes impossible because friction drastically increases and it feels like skating on glass. Experiments show that ice has a minimum of kinetic friction at -7°C, and many indoor skating rinks set their system to a similar temperature." From the Wikipedia Article, "Ice Skating"

Contents

[edit] complete article

this looks more complete article with graphs

[1]

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[edit] Lose the ice skating example!

That example may be correct, but it's a pretty poor illustration of the concept. You don't even say where you got all of the variables.

The variables don't matter (they are, however, reasonable). It's an example of how to use the equation in the sort of problem that a student is likely to encounter. While obviously I haven't sampled a statistically significant number of thermodynamics courses, both of the ones I've done (at different universities) have used problems similar to that, and hence I think it would be beneficial to a great many users. Far too many physics and maths articles, in my opinion, present a technically correct but utterly arid explanation which is of practically no use to understanding the concepts involved. Students who actually have to use the Clausius-Clapeyron relation in an assignment will be grateful for the example, and other users won't care. It certainly doesn't detract from the article from a practical point of view. Ckerr 20:35, 8 July 2006 (UTC)

[edit] Dubious

The ice-skating example doesn't agree with what ice ice skating says about ice's sliperyness.—Ben FrantzDale 17:19, 6 July 2006 (UTC)

Yes, it does, actually. It says: "the precise mechanism by which the low-friction is generated is not fully understood, though a number of plausible theories abound usually involving explanations of air-ice boundary layer water and/or friction generated through the skate bottom." Nothing about pressure. As a result I'm reverting the edit. Look, I'm not opposed to the idea of having a disputed tag if it can actually be shown to be in dispute. But since you appear to be mistaken (this article is consistent with the ice skating article), and since all other people who have criticised the ice skating example have failed to answer my replies, I don't believe--yet--that it's warranted. Ckerr 20:35, 8 July 2006 (UTC)

[edit] Requested move / Clapeyron equation

I think this article actually discusses the Clapeyron equation. I think the Clausius-Clapeyron equation is an approximation (based on the Clapeyron equation) for liquid-vapor (and perhaps solid-vapor) equilibrium assuming various things (e.g. ideal gas vapor and negligible liquid (or solid?) volume). See, for example, [2], [3], and [4].--GregRM 00:02, 14 July 2006 (UTC)

It appears there are several different conventions in use. I have never heard of it referred to as the "Clapeyron equation" until now, and I checked the original sources I used when writing this article, and they do refer to the equation given (dP/dT=L/TdV) as the "Clausius-Clapeyron equation". Perhaps a vote could determine which convention is more widespread, but I think whichever is chosen, both terms should redirect to the same page. Ckerr 03:45, 14 July 2006 (UTC)
The textbook Introduction to Chemical Engineering Thermodynamics (6th ed.) by Smith, Van Ness, and Abbott makes a distinction between the exact equation (which it calls the Clapeyron equation; it is equivalent to the one discussed in this article) and the approximate equation (which it calls the Clausius/Clapeyron equation; it seems to be equivalent to the one discussed in Clausius-Clapeyron equation (meteorology)). I just looked at my physical chemistry text (Physical Chemistry, 5th ed. by Ira N. Levine), and it claims that according to most texts for physical chemistry, the approximate equation is referred to as the Clausius-Clapeyron equation. It goes on to say, however, that the exact equation is called the Clausius-Clapeyron equation in most texts for engineering thermodynamics and physics. (On the previous page, discussing the exact equation, the book said the exact version was called the Clapeyron equation, but said that it was also known as the Clausius-Clapeyron equation.) Thus it seems that the term "Clapeyron equation" refers to the exact equation, but the exact equation is not always called the Clapeyron equation (it may be called the "Clausius-Clapeyron equation"). Likewise, it seems that when it is named, the approximate equation is always called the Clausius-Clapeyron equation, but the term "Clausius-Clapeyron equation" does not necessarily refer to the approximate form (it may refer to the exact form). I prefer the scheme of calling the exact equation the Clapeyron equation and the approximate equation the Clausius-Clapeyron equation, since it provides an easy distinction between the exact and approximate equations. The best solution may be to explain the different usages in the article, and perhaps use disambiguation header or redirects (as you suggested).--GregRM 23:09, 20 July 2006 (UTC)
  • Actually, none of the above explanations are quite correct, as both equations can legitimately be called the Clausius-Clapeyron equation. Now it's true that Clausius derived the approximate form (with T-squared in the denominator), but he also put Clapeyron's equation into the form given here, which is quite different from Clapeyron's original equation (which did not explictly include temperature at all). Also, Clausius derived it on correct principles, whereas Clapeyron did not. (But neither used the commonly-taught derivation given here. Entropy and the Second Law were still unknown at the time). So it's actually more correct (historically speaking) to call the one given here "Clausius-Clapeyron" rather than just "Clapeyron". As regards usage, as far as I can see, both are always referred to as "Clausius-Clapeyron" unless the text handles both and must distinguish them (i.e. in thermodynamics textbooks), in which case the exact equation is usually labelled "Clausius". So it's easy to jump to the erroneous conclusion that Clausius' contribution was merely the final approximation. I thought so myself, but it turns out that this is quite incorrect, and there's no undue credit being given. (I'll provide some refs if someone wants it) --BluePlatypus 20:01, 11 October 2006 (UTC)

[edit] The Ice Skating example, redux

I have removed the ice skating example, not for any perceived computational or physical flaws, but because it is non-encyclopedic, quasi-editorial, and verges on original research. Ξxtreme Unction 16:12, 13 October 2006 (UTC)

Sigh. I don't dispute that the example is not the sort of thing one would find in a paper encyclopedia, though it is not original research (the reference is H.B. Callen, "Thermodynamics and an Introduction to Thermostatistics") and the definition of "editorial" is pretty squishy. I think Wikipedia is stricken by a misguided attempt to make it just like a paper encyclopedia--only worse, because the people editing it are often not experts. I think the advantage Wikipedia has over paper encyclopedias is precisely that it can include things like examples that will be useful to a great many people. That said, consensus (or at least Jimmy Wales) seems to be against me, and I'm not going to try to push my agenda by editing single articles. I think it's a shame, though, that so much useful information is excluded from Wikipedia because of the historical vagaries of paper-based encyclopedias. Ckerr 02:05, 14 October 2006 (UTC)
The problem is not that it's something we wouldn't find in a paper encyclopedia. The problem is that (A) it's not really an example of how the equation is used in situ; (B) it draws a conclusion ("Clearly this is not how ice skating works.") that is not germane to the article; and (C) it draws a conclusion that appears (based on my reading of this talk page) to be one thermodynamics professor's interpretation of the mechanics and thermodynamics of ice skating, which generally runs afoul of Wikipedia:Reliable sources. (I concede that this last issue may be a misinterpretation on my part, but even if so, that still leaves us with the other problems.)
All the best,
Ξxtreme Unction
03:12, 14 October 2006 (UTC)
With regard to your points: (A) I'm not quite sure what you mean by this; it is exactly how the equation may be used. It is of course not the only way, but it is one of the most elementary applications. (B) This line is easily removed or modified; I'm happy with that. (C) Many common thermodynamics texts (I already mentioned Callen; Kittel is another, I recall) have nearly identical examples.
I agree that the example focused too much on ice skating rather than on the application of the equation. I have modified the article accordingly; tell me what you think. Ckerr 09:04, 19 October 2006 (UTC)
Extreme Unction--I'm not sure if "non-trivial" quite captures the concept. To put it another way, the pressure required to melt ice at -7 C (the temperature which ice rinks are typically at, according to ice skating) would require balancing a car on a thimble. (I might change the example to this.) If you think "tremendous" is too informal of a word, that's fine; but the concept is that melting ice by applying pressure is a very difficult thing to do, not just a non-trivial thing to do. Ckerr 08:28, 22 October 2006 (UTC)
It's a tone issue more than anything. There's a certain "OMG!" factor to words like "tremendous" or to ending a sentence with an exclamation point that I don't feel is encyclopedic in tone. It gives the impression that you're making a point about ice skating, rather than making a point about the Clausius-Clapeyron equation.
Let the facts speak for themselves, and let the reader draw their own OMG! conclusions if they wish.
All the best,
Ξxtreme Unction
13:15, 22 October 2006 (UTC)

[edit] Question about derivation

I post a question about the derivation here: [5]. Because I want some quick answer I decide to post on that page instead of this formal one. -- 131.111.164.228 11:18, 26 October 2006 (UTC)

Let me put it here for further references. -- 131.111.164.110 13:10, 30 October 2006 (UTC)

Hi everyone. From the article Clausius-Clapeyron relation, I'm confusing about its derivation. It says that

Along the coexistence curve, we also have dμI = dμII.

And after that we use the dμ = − sdT + vdP relation to give the final formula. My question is that we use two points in the coexistence (phase-equilibrium) curve to derive the formula. Those two points describe the same situations where there are two phase appeared in equilibrium in the system. Hence, the Clausius-Clapeyron equation should be used only for describing a change between two points along the phase-equilibrium curve.

But commonly we use it to refer about latent heat for describing phase transition. We try to describe two situations between a point at equilibrium (two phase) and a point outside the equilibrium curve (only one phase). So why is the equation valid in this case? I know I'm missing some important concept here. Thanks everyone for your attention. -- 131.111.164.228 11:13, 26 October 2006 (UTC)

It's derived at equillibrium between the two phases and describes the relation between the latent heat of the phase change and the associated changes in temperature, pressure and volume. The latent heat of the phase change is what it's all about. If you want to study the change between states that aren't in equillibrium, then you set up a thermodynamic cycle. --BluePlatypus 14:00, 26 October 2006 (UTC)
As far as I know, the Clausius-Clapeyron equation gives an approximation to the slope of the coexistence curve--that's it. ("Equillibrium" is a tricky word because thermodynamic equillibrium is a separate concept from phase equillibrium.) It says nothing about states of single phase, nor does it even describe the coexistence curve properly--firstly, it's a linear equation (which most coexistence curves are not); secondly, it only describes the transition between two out of three possible phases. In other words, I am questioning the validity of your last paragraph. Ckerr
It is not an approximation at all. It is exact. Nor is it a linear equation. It's a PDE. Now, if you chose to integrate that as if it wasn't, which some people certainly do, then it becomes an approximation. But it's not one present in the equation itself. --BluePlatypus 21:00, 26 October 2006 (UTC)
The points I and II are actually the same point on a PT diagram; they are different points on a VT, or PS, or VS diagrams. You do not take two different points on a melting curve on a PT diagram, you just take one point (which gives you the temperature at which your element would melt under given pressure, provided the pressure does not change in the process). Now you would like to know what is the melting curve slope at that point. Here the Clausius-Clapeyron equation becomes useful. You see, that single point in a PT diagram becomes an infinite number of points when you go to the VT diagram. Indeed, under given P and T, when you are on the melting curve, you can vary the fraction of phase I in the mixture continuously between 0 and 100%. This changes V (as the specific volumes of the two phases differ, generally speaking), without changing either P or T or mju (but changing S). The Clausius-Clapeyron equation is then derived from the fact that mju stays constant. The two points used in the derivation are at the same P and T; simply, one is 100% phase I, 0% phase II; and the second one is 0% phase I, 100% phase II. The slope is thus defined at a point (PT) proper, not at two close points. Hope this helps. Dr_Dima
Thank you all very much -- 131.111.164.110 09:24, 30 October 2006 (UTC)
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