# Talk:Henry's law/Draft

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 Definition:  The relationship between the amount of gas dissolved in a liquid and the partial pressure of that gas above the liquid. [d] [e]
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## Wikipedia has a similar article

I was a significant contributor to the Wikipedia article with the same name as this CZ article. I have reworded and rearranged some it it somewhat, added some references and conformed it to a CZ article. - Milton Beychok 00:35, 18 February 2008 (CST)

## Approval of Version 1

Congratulations on approval of version 1. Please keep discussions concerning further developments below this section. D. Matt Innis 02:32, 7 November 2008 (UTC)

## Approval of Version 1.1

The Draft has been re-approved to version 1.1. D. Matt Innis 23:15, 15 December 2008 (UTC)

## Re-approval is again needed

I made some signicant changes in the "Temperature dependence of Henry's law constant" to correct the formatting of the equations and their parameter definitions in that section. It needs re-approval again. Milton Beychok 20:54, 15 September 2009 (UTC)

## A mistake?

"${\displaystyle e^{p}=e^{k_{\rm {H}}}\;c}$" probably should be "${\displaystyle e^{p}=e^{k_{\rm {H}}c}}$". --Boris Tsirelson 21:32, 24 September 2012 (UTC)

Thanks for finding that typo. I have now fixed it. Milton Beychok 17:23, 25 September 2012 (UTC)
Nice. I also bother about dimensions: are these ${\displaystyle p}$ and ${\displaystyle k_{\rm {H}}c}$ dimensionless? --Boris Tsirelson 16:35, 26 September 2012 (UTC)
Boris, I am not sure that I understand your question about dimensions. Just below the equation, the parameters are listed and noted that they have many different sets of dimensions. Then Table 1 lists the dimensions very often used .... but there are many others to be found in the technical literature.
${\displaystyle p}$ and ${\displaystyle k_{\rm {H}}}$ and ${\displaystyle c}$ are definitely not dimensionless. Shalom and good yomtov (a day late).Milton Beychok 17:16, 26 September 2012 (UTC)
If ${\displaystyle p}$ is not dimensionless then ${\displaystyle e^{p}=1+p+p^{2}/2+p^{3}/6+\dots }$ is a sum of quantities of different dimensions. Does it make sense? --Boris Tsirelson 19:32, 26 September 2012 (UTC)
Boris, I am not a mathematician and you have just lost me completely. In this article, p is the symbol for pressure and it may be expressed in measurement units (i.e., dimensions) of pascals or atmospheres or torrs or bars or kg-force per square centimeter, and many others. That is what I meant when I said it was not dimensionless. And that makes perfect sense to me.
Also, in this article, c is used as the symbol for concentration and it may be expressed in measurement units of mols of dissolved gas per liter of solution, or mols of dissolved gas per mol of solution, or mols of dissolved gas per cubic meter of solution, or cubic centimeters of dissolved gas per cubic centimeter of solution, or many other methods of expressing concentrations. And that also makes perfect sense to me.
Furthermore, since p = kH · c, the dimensions of kH depend on the dimensions used for expressing p and c. In other words, the dimensions of kH will be the dimensions of p divided by the dimensions of c.
Beyond explaining what I meant by dimensionless what the word "dimensions" means to me in this article's context, I cannot help you any further. Perhaps, reading some of the references (especially, reference 7) might solve your concerns. Or perhaps you could consult one of the chemical engineering professors at the University of Tel-Aviv. Milton Beychok 21:31, 26 September 2012 (UTC)
Boris, I realize that the word "dimensions" to you probably means the basic physical dimensions such as mass, length, time, and temperature (M, L, T, and Θ). But there are many equations in chemistry and in engineering whose parameters are expressed in units of measurement rather than in those basic physical dimensions ... and I "think" in those terms of those measurement units. Therefore, my answer to your question was that the parameters in Henry's Law are not dimensionless. However, I do not know if they are or are not dimensionless in terms of the basic physical dimensions. In my opinion, it really doesn't matter for equations such as Henry's Law which have been in use for over 200 years now. Milton Beychok 02:03, 27 September 2012 (UTC)

[Unindent]

Boris is right in that it would be more elegant to write

${\displaystyle e^{p/p_{0}}=e^{kc/(kc_{0})}=e^{c/c_{0}}}$

where p0 and c0 are the respective quantities for the same reference state. However, if one makes the assumption (which is implicit in Milton's reasoning):

${\displaystyle \,p_{0}=kc_{0}\Longrightarrow p_{0}/(kc_{0})=1}$

then taking the natural logarithm on both sides

${\displaystyle p/p_{0}=kc/(kc_{0})\Longrightarrow p=kc\;(p_{0}/kc_{0})=kc}$

Exponentiating

${\displaystyle p=kc\Longrightarrow e^{p}=e^{kc}}$

which is Milton's expression. Chemical engineers are not always aware of the reference state, usually they simply assume p0 = kc0 and say: take p0 = 1 in the appropriate pressure unit. --Paul Wormer 09:21, 13 October 2012 (UTC)

Thank you Paul; now I see. The key is, "p0 = 1 in the appropriate pressure unit". But probably it should be stated in the article. Also, is it clear to non-experts, why these exponentials? The formula "${\displaystyle p=k_{\rm {H}}\;c}$" is rather understandable, but the meaning of "${\displaystyle e^{p}=e^{k_{\rm {H}}c}}$" is utterly unclear for me. Is it clear to other readers? --Boris Tsirelson 09:40, 13 October 2012 (UTC)
I also note that "${\displaystyle k_{\rm {H,cp}}=k_{\rm {H,cp}}^{\Theta }\;\exp \left[-C\cdot \left({\frac {1}{T}}-{\frac {1}{T^{\Theta }}}\right)\right]}$ where ${\displaystyle C}$ is a constant with dimension of kelvins". In this case one hesitates to violate dimension rules writing (disturbingly) "${\displaystyle k_{\rm {H,cp}}=k_{\rm {H,cp}}^{\Theta }\;\exp \left[{\frac {1}{T}}-{\frac {1}{T^{\Theta }}}\right]}$" and assuming (implicitly) that C = 1 in the appropriate unit system! --Boris Tsirelson 09:53, 13 October 2012 (UTC)
As far as I understand it (but I may be mistaken) one introduces the exponential equation because it is assumed to be applicable for large c. In any case the converse is true: if the equation exp[p] = exp[kc] holds for all c then it follows for small c that the linear equation p = kc is true.
Setting p0 = 1 for a standard state (say atmospheric pressure at sea level) is a common step, but setting to 1 the rather esoteric constant C (that happens to have the dimension of temperature) is a big step. One would change the temperature scale, which is an unusual thing to do. If one would like to do something similar, one would introduce C′ = RC where R is the gas constant and RC has dimension of energy. Then in
${\displaystyle \exp \left[-C'\cdot \left({\frac {1}{RT}}-{\frac {1}{RT^{\Theta }}}\right)\right]}$
one would set C′ = 1 in appropriate unit of energy. Don't ask me why I would scale energy and not temperature, this is just my intuition/experience.
--Paul Wormer 12:27, 13 October 2012 (UTC)
Thank you very much for helping out, Paul. As for including the subject matter discussed by you and Boris about Henry's law and about the van't Hoff equation in the article, I would disagree with that. This is an encyclopedia article and, in my opinion, we should not turn it into a postgrad textbook. Milton Beychok 15:55, 13 October 2012 (UTC)
Well, about violating dimension rules, I understand the answer (frankly, I stay a bit disturbed, but let it be my problem). But the question "why these exponentials" remains. Paul wrote: "if the equation exp[p] = exp[kc] holds for all c then it follows for small c that the linear equation p = kc is true." No, Paul, think again: these two equations are exactly equivalent, not approximately for small c! Note that "exp" is present on both sides!
Here is a parody. Hope the pattern of the parody differs from the pattern in the article; but please explain me, what is the distinction between them!
Parody. The commulativity of addition law states that ${\displaystyle e^{x+y}=e^{y+x}}$ for all x, y. Taking the natural logarithm of the formula, gives us the more commonly used formula: ${\displaystyle x+y=y+x.}$
--Boris Tsirelson 16:32, 13 October 2012 (UTC)
Boris says: "these two equations are exactly equivalent, not approximately for small c! Note that "exp" is present on both sides!" Boris is completely right. (It occurred to me, too, this morning in bed just after I woke up). Yesterday I was mistaken, meaning that I also don't know why one takes exponentials on both sides of p=kc. It is not wrong, but unnecessarily complicated. --Paul Wormer 05:36, 14 October 2012 (UTC)
"Inelegant", "unnecessarily complicated"... This is not the point. Not my point, I mean. Either the exponential has a (physical? chemical?) meaning, or not. In the former case the meaning should be mentioned (at least). In the latter case the article starts (!) with a utterly puzzling claim, as nonsensical as this one: "A formula for Henry's Law is: ${\displaystyle \arctan {\,p}=\arctan {\,kc};}$ taking the tangent function of the formula, gives us the more commonly used formula: ${\displaystyle p=kc.}$" What does the reader think, if not "WTF?!" --Boris Tsirelson 06:40, 14 October 2012 (UTC)

(Unindent)Paul is right, the exponential form is probably unnecessarily complicated ... so I removed it. Boris, "WTF" and pejorative words like "nonsensical" were completely uncalled for!! That is the sort of language I expect of Wikipedia but not of Citizendium. Milton Beychok 18:19, 14 October 2012 (UTC)

I am sorry for that. But probably it happened because this sort of errors in articles (and this sort of unwilling to fix them, still calling...you know what...just "unnecessarily complicated") I expect of Wikipedia but not of Citizendium (the more so, not on approved articles written by CZ editors). Why did not you finalize the matter in September?
Happy editing. --Boris Tsirelson 19:00, 14 October 2012 (UTC)
Boris, it might well have been resolved in September. When you first brought up your concern about "dimensions" on Sept. 26th, I promptly replied to the best of my ability on that same day and followed up with more explanation on Sept. 27th. However, you then waited two weeks until October 12th before you raised the question of whether the exponential form was needed at all. I then deleted the the exponential form on October 14th.
Had you made yourself perfectly clear as to what you wanted back in September, rather than waiting two weeks to do so, the deletion of the exponential form might well have been resolved sooner. Milton Beychok 20:31, 14 October 2012 (UTC)