Talk:Henry's law/Draft

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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)

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Approval of Version 1.1

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

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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?

"$e^{p}=e^{k_{\rm {H}}}\;c$" probably should be "$e^{p}=e^{k_{\rm {H}}c}$". --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 $p$ and $k_{\rm {H}}c$ 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.

$p$ and $k_{\rm {H}}$ and $c$ are definitely not dimensionless. Shalom and good yomtov (a day late).Milton Beychok 17:16, 26 September 2012 (UTC)

If $p$ is not dimensionless then $e^{p}=1+p+p^{2}/2+p^{3}/6+\dots$ 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 = k_H · c, the dimensions of k_H depend on the dimensions used for expressing p and c. In other words, the dimensions of k_H 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

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

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):

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

then taking the natural logarithm on both sides

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

Exponentiating

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

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 "$p=k_{\rm {H}}\;c$" is rather understandable, but the meaning of "$e^{p}=e^{k_{\rm {H}}c}$" is utterly unclear for me. Is it clear to other readers? --Boris Tsirelson 09:40, 13 October 2012 (UTC)

I also note that "$k_{\rm {H,cp}}=k_{\rm {H,cp}}^{\Theta }\;\exp \left[-C\cdot \left({\frac {1}{T}}-{\frac {1}{T^{\Theta }}}\right)\right]$ where $C$ is a constant with dimension of kelvins". In this case one hesitates to violate dimension rules writing (disturbingly) "$k_{\rm {H,cp}}=k_{\rm {H,cp}}^{\Theta }\;\exp \left[{\frac {1}{T}}-{\frac {1}{T^{\Theta }}}\right]$" 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

$\exp \left[-C'\cdot \left({\frac {1}{RT}}-{\frac {1}{RT^{\Theta }}}\right)\right]$

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)