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Warren Platts
2007-Oct-30, 06:42 PM
Say you have some metallic hydrogen under the pressures and temperatures characteristic of the interior of gas giant planets. If you raise the metallic hydrogen to a zone where the PT is low enough (or conversely lower the zone as the planet cools), the metallic hydrogen atoms will recombine to form H2, releasing energy according the following formula:



H + H → H2 + energy

So the question is, how many Joules will be released when two moles of H combine to form one mole of H2?

At standard temperatures and pressures, my old organic chemistry textbook says that the energy of dissociation of the H-H bond is 104 kcal per mole. At 4.1868 kJ per kcal, this works out to 435.4 kJ/mol. This agrees with my CRC Handbook (85th edition) where it says (e.g. p. 9-55) that the bond strength (Do298) at 298oK of H-H is 435.990 kJ/mole.

Of course the temperatures typical of the interior of gas giants are in the thousands of degrees K. So, I would like to know Do6000. The CRC Handbook (p. 9-52) gives the following approximate relation:



Do298 = Do0 + (3/2)RT

where R is the gas constant. Since I the Handbook gives me Do298, I can solve for Do0 (the bond strength at absolute zero), which is 432 kJ/mole. Thus, by resubstituting the 432 kJ/mole for Do0 into the above equation, and solving for 6,000o K, I get a value of:



Do6000 ~ 500 kJ/mole

So far so good. The problem is I can't reconcile this figure with the following cryptic reference in my Jupiter book (Bagenal et al., Jupiter: The Planet, Satellites, and Magnetosphere [2004], p. 41):


[A]s the planet cools, a fraction of the mass of the envelope is converted from one phase to the other with an associated latent heat release (or absorption). The effect on the evolution is not very pronounced for a laten heat of ~0.5kB (Saumon et al. 1992 (http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1992ApJ...391..827S&data_type=PDF_H IGH&whole_paper=YES&type=PRINTER&filetype=.pdf)).

where kB is Boltzmann's constant. The Saumon et al. (1992, 828) paper says that the latent heat figure is actually an "entropy jump . . . with the entropy of the metallic phase being higher than that of the molecular phase." If you work that figure out for one mole of H2, it's just Boltzman's constant times Avogadro's number which is 8.314 J mole-1 K-1, and that's just the ordinary gas constant R. The CRC Handbook (6-19) standard S for dihydrogen is 130.7 J mole-1, which is quite a bit higher than 8.314.

So which figure should a person interested in modeling the metallic-molecular phase transition use?

papageno
2007-Oct-30, 08:18 PM
In the metallic phase there is a bond between the H atoms. In the OP you neglected this binding energy.

Warren Platts
2007-Oct-30, 08:45 PM
In the metallic phase there is a bond between the H atoms. In the OP you neglected this binding energy.:confused:

So the answer is . . . . 8.314 J s-1 mole-1 ???

If so, why is it equal to Boltzmann's constant?

papageno
2007-Oct-30, 09:07 PM
:confused:

So the answer is . . . . 8.314 J s-1 mole-1 ???

If so, why is it equal to Boltzmann's constant?

The units of the Boltzmann constant are Energy/Temperature, just like specific heat or latent heat, or entropy...

Oh, and the paper says it is about 0.5 kB per proton (if you look into kinetic theory of gases, you find often 0.5 kB of energy per degree of freedom... or look Dulong-Petit law for specific heat of solids...).

Warren Platts
2007-Oct-30, 09:18 PM
The units of the Boltzmann constant are Energy/Temperature, just like specific heat or latent heat, or entropy...:cry:

Come on! Help me out here--please!

So to figure out the latent heat released when one mole of H2 is formed from metallic hydrogen, I should multiply 8.314 J K-1 mole-1 by 6,000o K??? If so, that would work out to about 50 kJ mole-1. Right? :confused:


Oh, and the paper says it is about 0.5 kB per proton.Exactly: 0.5 * 1.3806506(24) X 10-23 J K-1 per proton (it takes two protons to make one H2 molecule). But what does that mean? And what's the deal with Boltzmann's constant and the gas constant?

papageno
2007-Oct-30, 09:26 PM
So to figure out the latent heat released when one mole of H2 is formed from metallic hydrogen, I should multiply 8.314 J K-1 by 6,000o K??? If so, that would work out to about 50 J mole-1. Right?

Actually, you should get the equation of state and simply take the difference between final and initial state. Then you get amount of heat released (unless the planet is doing some mechanical work).

Did I mention that I am not an astrophysicist?

papageno
2007-Oct-30, 09:27 PM
Boltzmann constant: energy/temperature per particle or degree of freedom

=

Gas constant: energy/temperature per mole

grant hutchison
2007-Oct-30, 10:04 PM
Warren Platts, I have no idea of the latent heat associated with the phase change from molecular to metallic hydrogen, and you must feel free to ignore what follows.

But the pressure has been growing in me for some days now to ask you why, given your philosophy of science, you don't just ask yourself what the hydrogen "wants to do" or is "designed to do". Surely you should be able to cut through all this detail with a design stance or an intentional stance?

Grant Hutchison

Warren Platts
2007-Oct-30, 10:11 PM
Warren Platts, I have no idea of the latent heat associated with the phase change from molecular to metallic hydrogen, and you must feel free to ignore what follows.

But the pressure has been growing in me for some days now to ask you why, given your philosophy of science, you don't just ask yourself what the hydrogen "wants to do" or is "designed to do". Surely you should be able to cut through all this detail with a design stance or an intentional stance?

Grant Hutchison
Well, obviously, hydrogen atoms prefer monogamous relationships to loneliness. We humans could learn a few lessons from our humble precursors. So, they're sort of like us. You know how it is: when the relationship first starts, there's lots of heat generated, and then things cool to an equilibrium level. After a while, things get boring, so they'll do stupid things like forming a threesome with a passing oxygen atom. But that never works out for very long.

So my question is, what is the hormone level for a hydrogen atom at 6,000 degrees and a megabar of pressure?

Kaptain K
2007-Oct-31, 01:10 AM
Warren likes to play mind games and considers us to be perfect foils. Aside from the fact that he sets his "scenarios" in a Jupiter-type planet, his little exercises have little or nothing to do with astronomy!

Warren Platts
2007-Oct-31, 10:06 AM
Warren likes to play mind games and considers us to be perfect foils. Aside from the fact that he sets his "scenarios" in a Jupiter-type planet, his little exercises have little or nothing to do with astronomy!Sir, I assure you my little joke was intended in all good nature. I further assure you that the problem of equations of state for the conditions inside gas giants is not trivial, just because--unlike the interior of stars--the ideal gas laws do not apply. Yet the behavior of hydrogen under such conditions is crucial for our understanding of the gas giants. As things stand now, gas giants are the only extrasolar planets that have been directly observed through occultations that allow accurate estimates of their radius as well as their mass. This is one of the most active and exciting areas of astronomical research. Therefore, questions about the behavior of hydrogen in the interiors of the gas giants should not be unexpected here.

As for design stances, the whole epistemological point of taking a design stance toward x is that it speeds up the search for a mechanistic explanation of x by constraining the search space of plausible mechanisms to just those that satisfy the design criteria.

To return to the subject at hand, I have a range of values for the latent heat released by two moles of protons when they chemically fuse to form H2 that only differ by five orders of magnitude. I don't think it was unreasonable to hope that one of the experts here might have been able to narrow down the choices without much effort (there's nothing obvious that leaps out of Google or Wikipedia--believe me). Yet I've even managed to stump the great Dr. Hutchison.

At any rate, papageno gave me a clue, at least, on how to proceed. After dredging up old memories from chemistry class long ago, the energy of the phase transition should just be the enthalpy change (or so I am thinking):



∆H = T∆S + V∆P

So if the "entropy jump" mentioned in Saumon et al. (1992) of 0.5kB J K-1 per proton is equivalent to ∆S in the above and ∆P is assumed to be zero, then, at T = 6,000o K, ∆H would be:



6,000o K * 8.314 J mole-1 = 49.9 kJ mole(H2)-1

Considering that the heat of vaporization of water is 40.68 kJ mole-1, I'm thinking my figure of 49.9 kJ mole-1 might not be unreasonable as a back of the envelope estimate. But as a sanity check, my original question remains open and I would appreciate it if someone could verify my reasoning.

papageno
2007-Oct-31, 10:34 AM
To return to the subject at hand, I have a range of values for the latent heat released by two moles of protons when they chemically fuse to form H2 that only differ by five orders of magnitude.


The phase transition from metallic to molecular Hydrogen is not "chemical fusion" of two H atoms.




I don't think it was unreasonable to hope that one of the experts here might have been able to narrow down the choices without much effort (there's nothing obvious that leaps out of Google or Wikipedia--believe me). Yet I've even managed to stump the great Dr. Hutchison.


Have you looked up the equation of state and the phase diagram?

grant hutchison
2007-Oct-31, 10:42 AM
As for design stances, the whole epistemological point of taking a design stance toward x is that it speeds up the search for a mechanistic explanation of x by constraining the search space of plausible mechanisms to just those that satisfy the design criteria.Also known as "reasoning by false analogy", in the trade. :)

Grant Hutchison

tusenfem
2007-Oct-31, 12:53 PM
Let's go to the definition of Latent Heat:



Symbol L
The quantity of heat absorbed or released when a substance changes its physical phase at constant temperature (e.g. from solid to liquid at the melting point of from liquid to gas at the boiling point). *skip*
In thermodynamic terms the latent heat is the enthalpy of evaporation (ΔH), i.e. L = ΔH = ΔU + p ΔV, where ΔU is the change in the internal energy, p is the pressure and ΔV is the change in volume.

The specific latent heat (symbol l) is the heat absorbed or released per unit mass of a substance in the course of its isothermal change of phase. The molar latent heat is the heat absorbed or released per unit amount of substance during an isothermal change of state.


So, we have to be careful of which latent heat we are talking about, and most likely different kinds are compared here in the OP.
Also in a later post there is mention of a V Δp, where in the definition there is a p ΔV. You cannot just throw away that last part of the equation.

Warren Platts
2007-Oct-31, 03:30 PM
Let's go to the definition of Latent Heat:

Originally Posted by Oxford Consise Science Dictionary
Symbol L
The quantity of heat absorbed or released when a substance changes its physical phase at constant temperature (e.g. from solid to liquid at the melting point of from liquid to gas at the boiling point). *skip*
In thermodynamic terms the latent heat is the enthalpy of evaporation (ΔH), i.e. L = ΔH = ΔU + p ΔV, where ΔU is the change in the internal energy, p is the pressure and ΔV is the change in volume.

The specific latent heat (symbol l) is the heat absorbed or released per unit mass of a substance in the course of its isothermal change of phase. The molar latent heat is the heat absorbed or released per unit amount of substance during an isothermal change of state.
So, we have to be careful of which latent heat we are talking about, and most likely different kinds are compared here in the OP.
Also in a later post there is mention of a V Δp, where in the definition there is a p ΔV. You cannot just throw away that last part of the equation.

From the wiki on enthalpy (http://en.wikipedia.org/wiki/Enthalpy):



ΔH = ΔU + PΔV + VΔP = TΔS + VΔP

Therefore:



ΔU + PΔV = TΔS

So, if you set ΔP = 0, and let the volume change adiabatically, then all you have to worry about is TΔS--or so I'm thinking.

Is there a chemist in the house?

Warren Platts
2007-Oct-31, 04:41 PM
The phase transition from metallic to molecular Hydrogen is not "chemical fusion" of two H atoms.Then how would you describe it?


Have you looked up the equation of state and the phase diagram?It's not that easy--there isn't a nice little formula that one can apply. So a lot of ad hoc considerations are built into it, and the results are usually given in tabular form. However, I found this paper by Saumon et al. (1995) (http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1995ApJS...99..713S&data_type=PDF_H IGH&whole_paper=YES&type=PRINTER&filetype=.pdf) that's very helpful. It links to the tables in tabular form that give the EOS's for H and He (http://www.journals.uchicago.edu/AAS/cdrom/volume5/doc/files5.htm#1995ApJS...99..713S). It's table 1 lists the ∆S across the PPT for various pressures and temperatures.

I'm interested in 6,000 K (log[6000]=3.78), so that's the second line down, where it says the ∆S is 0.590kB. Multiplying by the gas constant R (which is equal to the product of Boltzmann's and Avogadro's constants) converts this figure into SI units, which equals 4.91 J mole-1 K-1. So, going back to my formula that no one has said I either should or should not use:



∆H = T∆S

yeilds



6000o K * 4.91 J mole-1 K-1 = 29.5 kJ mole-1

And so doubling this figure to get the enthalpy of formation of H2 gives ~ 60 kJ per mole of H2, which is fairly close to my earlier estimate of 50 kJ per mole of H2.

So the question is: Did I figure this out correctly? :neutral:

Warren Platts
2007-Oct-31, 04:45 PM
Also known as "reasoning by false analogy", in the trade. :)Ah, but it's only a false analogy if one is metaphysically commited to a strong realism with regard to minds, purposes, and designs of humans and their artifacts.

grant hutchison
2007-Oct-31, 04:52 PM
Ah, but it's only a false analogy if one is metaphysically commited to a strong realism with regard to minds, purposes, and designs of humans and their artifacts.Bluff. :)
One can choose many different metaphors which allow one to reason in many different ways about the same thing. A maximum of one will produce useful results when referred to the behaviour of the Universe. The remainder are false analogies. There are therefore more false analogies than true ones. Reasoning by analogy is therefore a minefield, however you care to dress it up.

Grant Hutchison

Warren Platts
2007-Oct-31, 04:59 PM
Bluff. :)
One can choose many different metaphors which allow one to reason in many different ways about the same thing. A maximum of one will produce useful results when referred to the behaviour of the Universe. The remainder are false analogies. There are therefore more false analogies than true ones. Reasoning by analogy is therefore a minefield, however you care to dress it up.Heh! I guess Kroto and Smalley just got lucky. :D

(It's better to be lucky. . . .)

grant hutchison
2007-Oct-31, 05:11 PM
Heh! I guess Kroto and Smalley just got lucky. :DThat's absolutely correct. If enough people try reasoning by dumb analogy, some of them get lucky. We don't hear from the rest of them.
So you have a numerator of one, and no idea what the denominator is.

Grant Hutchison

Warren Platts
2007-Oct-31, 06:09 PM
Well, I'm working on numerator number two. That's why I need to know about the latent heat of the phase transition.

papageno
2007-Oct-31, 09:23 PM
The phase transition from metallic to molecular Hydrogen is not "chemical fusion" of two H atoms.

Then how would you describe it?


Phase transition is the technical term.
You are going from one state with well-defined thermodynamic properties to another with well-defined but different thermodynamic properties.

By calling it "chemical fusion of two H atoms" one is not describing the actual phase transition, but simply one of the steps that can be used to calculate the binding energy. The step that is left out is the separation of the H atoms from the metallic state.






Have you looked up the equation of state and the phase diagram?

It's not that easy--there isn't a nice little formula that one can apply. So a lot of ad hoc considerations are built into it, and the results are usually given in tabular form. However, I found this paper by Saumon et al. (1995) (http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1995ApJS...99..713S&data_type=PDF_H IGH&whole_paper=YES&type=PRINTER&filetype=.pdf) that's very helpful. It links to the tables in tabular form that give the EOS's for H and He (http://www.journals.uchicago.edu/AAS/cdrom/volume5/doc/files5.htm#1995ApJS...99..713S). It's table 1 lists the ∆S across the PPT for various pressures and temperatures.


A quick glance at the paper shows that they used the Helmotz free energy for the equation of state:

F = U - TS

which implies that the processes occur at constant pressure and temperature (that is, determined by the environment, whatever that is in the specific case of the planet).

If there is no (mechanical) work done by the system in the phase transition, the change in free energy is zero, therefore:

Delta(U) = - T Delta(S).

The only thing exchanged with the environment is heat.





I'm interested in 6,000 K (log[6000]=3.78), so that's the second line down, where it says the ∆S is 0.590kB. Multiplying by the gas constant R (which is equal to the product of Boltzmann's and Avogadro's constants) converts this figure into SI units, which equals 4.91 J mole-1 K-1. So, going back to my formula that no one has said I either should or should not use:



∆H = T∆S

yeilds



6000o K * 4.91 J mole-1 K-1 = 29.5 kJ mole-1


This is all you need to do, as far as I can tell.




And so doubling this figure to get the enthalpy of formation of H2 gives ~ 60 kJ per mole of H2, which is fairly close to my earlier estimate of 50 kJ per mole of H2.


You don't need to double anything: the tabulated figures should take the molecules of Hydrogen already into account.

Warren Platts
2007-Nov-01, 12:18 AM
Phase transition is the technical term.
You are going from one state with well-defined thermodynamic properties to another with well-defined but different thermodynamic properties.

By calling it "chemical fusion of two H atoms" one is not describing the actual phase transition, but simply one of the steps that can be used to calculate the binding energy. The step that is left out is the separation of the H atoms from the metallic state. Good point.


A quick glance at the paper shows that they used the Helmotz free energy for the equation of state:

F = U - TS

which implies that the processes occur at constant pressure and temperature (that is, determined by the environment, whatever that is in the specific case of the planet).

If there is no (mechanical) work done by the system in the phase transition, the change in free energy is zero, therefore:

Delta(U) = - T Delta(S).

The only thing exchanged with the environment is heat.
If the latent heat causes thermal instabilities that result in convection, does that count as mechanical work? Indeed, couldn't the whole amount of latent heat be used to drive convection?


You don't need to double anything: the tabulated figures should take the molecules of Hydrogen already into account.It would probably be less confusing to say 29.9 kJ per gram (since there is one mole of hydrogen atoms in one gram).

papageno
2007-Nov-01, 02:41 PM
If the latent heat causes thermal instabilities that result in convection, does that count as mechanical work? Indeed, couldn't the whole amount of latent heat be used to drive convection?


I thought that the latent heat released in that transition is negligibly small compared to the energies involved.

Warren Platts
2007-Nov-01, 02:43 PM
Originally Posted by Warren Platts
Ah, but it's only a false analogy if one is metaphysically commited to a strong realism with regard to minds, purposes, and designs of humans and their artifacts.
Bluff. :)
A bluff is like in poker where one misrepresents a weak hand by overbetting. But I'm not misrepresenting my hand here. I'm dead serious. Whether the analogy is false or not depends on one's theory of human purposes and minds. I've consistently found that the most obstinate resistance to my ideas stems from unexamined assumptions about human minds and purposes--these are taken as a given by the vast majority of scientists who indeed practically define the physical domain as that which exists outside of the human mind.

But unless one is clear about what human minds, purposes, and designs really are and is prepared to defend that position, then one really isn't qualified to comment on the alledged presense or absence of purpose in the physical domain, or so I would think. :)

But be that as it may, why the sectarianism when it comes to mere epistemology? It seems to me scientists should be happy to use whatever techniques generate true results. Is a result somehow tainted by the use of an unorthodox epistemology--even after that result has withstood every single test thrown at it? If someone gets something right for the "wrong" reason, who cares? Think about all the times people have got things wrong for all the "right" reasons! :D

grant hutchison
2007-Nov-04, 02:31 PM
A bluff is like in poker where one misrepresents a weak hand by overbetting. But I'm not misrepresenting my hand here. I'm dead serious.In which case you are simply betting badly. I had thought better of you. :)


But be that as it may, why the sectarianism when it comes to mere epistemology? It seems to me scientists should be happy to use whatever techniques generate true results. Is a result somehow tainted by the use of an unorthodox epistemology--even after that result has withstood every single test thrown at it?Already answered. The generation of sporadic true results by repeated bad bets is not a methodology worthy of the name. It's a lot of monkeys and a lot of typewriters.

Grant Hutchison

Warren Platts
2007-Nov-05, 01:18 PM
In which case you are simply betting badly. I had thought better of you. :)Believing that human "minds" are not real is no sucker bet at all. It's the consensus view in neurology and the philosophy of mind these days. You're behind the times old man! :razz:

grant hutchison
2007-Nov-05, 01:56 PM
Believing that human "minds" are not real is no sucker bet at all. It's the consensus view in neurology and the philosophy of mind these days. You're behind the times old man! :razz:I do have some formal training in these matters (in contrast to my knowledge of thermodynamics!) and I very much doubt that "consensus" is the correct word for anything to do with consciousness. :)

But no matter. It has no impact, one way or the other, on the logical error of false analogy.

Grant Hutchison

Warren Platts
2007-Nov-07, 05:20 PM
Here's the latest from Dennett. (http://ase.tufts.edu/cogstud/papers/intentionalsystems.pdf) It will get you back up to speed. :D

grant hutchison
2007-Nov-07, 06:14 PM
Here's the latest from Dennett. (http://ase.tufts.edu/cogstud/papers/intentionalsystems.pdf) It will get you back up to speed. :DComfortably up to speed on Dennett, thanks, as I've told you before.
And (as I've mentioned before) I doubt you'd get much support from Dennett for what you're up to with his toolkit. :)

Grant Hutchison

Warren Platts
2007-Nov-07, 09:54 PM
Well, we might just have to ask him. He's a nice guy--the last time I wrote to him prospecting for philosophy Ph.D. programs, he wrote me back himself and kindly informed me that Tufts does not offer the Ph.D. in philosophy. But that's just because at Tufts, like Colorado State, they do philosophy, rather than just study it. If Icarus accepts my paper, I'll send him a copy and we'll see what he has to say.

grant hutchison
2007-Nov-07, 10:04 PM
Sounds good to me. :)

Grant Hutchison