Talk:Enthalpy of vaporization

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About the section "physical model for vaporization"[edit]

Please note, that the same expression for enthalpy of vaporization was already derived in the article in Russian Physical Chemistry Journal in 1983. You can see this article on http://commons.wikimedia.org/wiki/File:Surface_tension_.pdf. There r is the latent heat of vaporization, σ is the surface tension. In fact, the only difference between these two expressions is use of 3 instead of π in the Russian paper.

Please, edit this section accordingly.

Water values[edit]

There are two values for water posted on this page, 40.8 kj/mol and 40.65 kj/mol. Could someone verify the correct value, or explain the differences?

Webelements says for Radium:[edit]

Enthalpy of vaporization [/kJ mol-1]: about 125 and not 37! doesn't that sound more plausible? Malbi

I think water should be included too. Andries 15:27, 31 May 2004 (UTC)

Yes, go ahead, please; also in the water article.--Patrick 21:15, 31 May 2004 (UTC)
done, although I dunno much about the subject and feel uncertain. Andries 20:42, 3 Jun 2004 (UTC)

Enthalpy and heat[edit]

I think the word "heat" should be replaced by "enthalpy". It is more suitable and more accurate.

I think Don Anonymous is tilting at windmills. Gene Nygaard 13:43, 23 Apr 2005 (UTC)

Definition[edit]

Heat of vaporization is a quantity that exists at conditions other than only standard pressure.--Renier Maritz 17:13, 13 October 2005 (UTC)

Title change[edit]

Physchim62 has changed the title of this article without discussion here, and despite the fact that opposition to a change from the most common English name of this quantity by changing "heat" to "enthalpy".

Physchim62 has also ignored the fact pointed out by Renier Maritz above, and has created a title of more limited scope by adding the word "standard" to the title. The article, however, remains written in the more general sense.

This name change needs to be undone. I don't have time to take care of the redirects right now, but intend to do it later unless someone else does it before I get to it. Of course, the same problem applies to heat of fusion. Gene Nygaard 14:54, 27 October 2005 (UTC)

Proportional to melting point?[edit]

Just looking at the list it seems that the heat of fusion is correlated with the melting or boiling point. Is there a theory/formula connecting the two? njh 11:20, 25 November 2005 (UTC)


Units Consistency[edit]

Wikipedia should make it's units consistent for physical properties. The pages on specific heat, heat of vaporization, and heat of fusion all have different units. That is, grams, kilograms, and moles all used which makes comparisons all the more difficult.

The units in the table of latent heat in the latent heat article are J/g. The units in the table for latent heat of vaporization kJ/kg. However the values for water are 2272 and 2260 respectively. So, according to wikipedia the latent heat of vaporization is both 2272 J/g and 2260 kJ/kg. How? The true value is 2259.36 J/g.Antinoah (talk) 02:53, 24 February 2008 (UTC)

N/A's[edit]

Some of the elements marked N/A (e.g. Actinium) and ??? (e.g., Terbium) have heats of vaporization added in these infoboxes. Why are they listed there and not here? --Geoffrey 21:21, 18 April 2006 (UTC)

Value for Hafnium not consistent[edit]

Although the difference is minor, there is a discrepency between this page and the information page of Hafnium about the enthalpy of vaporization (571 vs 575 kJ/mol). Also webelements gives a complete other value: 630 kJ/mol. I don't know which is the better value, I just noticed the difference. --Paul 14:55, 17 July 2006 (UTC)

heat is no source of energy[edit]

energy could be gotten from the mass of an atom.thus ,mass of an atom of an element is the determinant of energy content of any substance.the more an atom of an element the more the collision which then increase the kinetic energy of an atom. —Preceding unsigned comment added by 41.219.244.62 (talk) 18:16, 17 May 2008 (UTC)

Determination of the latent heat[edit]

I'm preparing a work for the university (i'm a student) about how to obtain the variation of enthalpy of vaporization of liquids like water and alcohols, if somebody thinks this interesting, i can explain three easy and cheap methods for that. Tell me what you think! --Pedvi (talk) 16:50, 12 June 2008 (UTC)

Inconsistencies and ambiguities[edit]

In the periodic table data, the definition of a mole is unclear. The diatomic gases, for example, have values that are half those one finds for the same gases on their respective pages. For hydrogen, the table gives 449.36 J/mol while the hydrogen page gives (slightly more than) twice that, 904 J/mol. (Which is the more accurate value?) Similarly, nitrogen in the periodic table shows 2.7928 kJ/mol and the nitrogen page gives 5.56 kJ/mol. Since vaporization doesn't rupture the diatomics' bonds, it seems odd to put this data in this way, but I can see the difficulty in trying to make the entire table consistent. At least it should be made prominently clear that the data in the table is for one mole of atoms, not molecules. Some data in the table clearly needs to be corrected, the value for sulfur given is dramatically low, 1.7175 kJ/mol, as compared with 45 kJ/mol on the sulfur page. I assume that the value given for phosphorus is for white phosphorus (P₄) - which should be noted since that isn't the thermodynamically stable allotrope. Gadolinist (talk) 19:14, 19 June 2009 (UTC) Why doesn't the data in this table agree with that in http://en.wikipedia.org/wiki/Heats_of_vaporization_of_the_elements_%28data_page%29? --206.55.183.156 (talk) 12:49, 6 December 2009 (UTC)

The value for methane is inconsistent: 8.19 kJ/mol / .016042 kg/mol = 510.53 kJ/kg, but the value listed is 760 kJ/kg! where are the references? -- 99.233.186.4 (talk) 00:16, 19 May 2010 (UTC)

Why energy as function of T discontinuous?[edit]

I asked a similar question on the enthalpy of fusion talk page, but got no answer. Can someone explain why at the point of vaporization, the kinetic energy of the molecules stop increasing (assuming T ∝ KE) and only potential energy w.r.t to intermolecular forces increases?

In most classic mechanics, textbook cases, the 2 quantities change continuously, so is this some quantum effect? --UncleJoe1985 (talk) 00:48, 23 June 2009 (UTC)

If you have a liquid and vapor in equilibrium at the boiling point, then they are at the same temperature. Hence, as you say, T ∝ KE, and so the kinetic energies are equal. Dezaxa (talk) 05:56, 5 January 2014 (UTC)

Reference for "Enthalpy of Vaporization"[edit]

Can someone provide a reliable source to the use of the term "Enthalpy of Vaporization" over "Heat of Vapourization". I've gone through all of my thermodynamic texts and I can't find any that use this term. There may be reason for referring to the term in the body of the article, but to have the title of the article use a term that no one in the real world uses seems . . . odd. At this point, without a reliable source, this nomenclature would seem to be a personal point of view and might even be classified as original research. I am not going to get involved in an edit war, so I will not make any change.John G Eggert (talk) 13:09, 12 November 2011 (UTC)

The standard chemical engineering textbook (Poling et al. "The properties of liquids and gases")uses Enthalpy all the time. "Heat" is a familiar everyday word, but it is not thermodynamically precise. I'm too new here to have an opinion whether to go for precision or familiarity... AlanParkerFrance (talk) 12:19, 17 October 2012 (UTC)

Confusion between values Fusion and Vaporization[edit]

Hello. English is not my mother tongue, so be indulgent.

Some of values in table is for fusion, not for vaporization. Example for S : 1,7175 = fusion, 45 is for vaporization like write in sulfur under "Heat of vaporization". But since I'm not sure, I prefer not to correct. For exemple, i'm not sure that "enthalpy" is same that "heat". Jerome66 (talk) 13:48, 22 February 2012 (UTC)

explicitize symbol meanings and units[edit]

Reader comment: the background section is hard to follow since the symbols in the formula are not explained nor are their units given. — Preceding unsigned comment added by 137.224.252.10 (talk) 13:47, 29 May 2012 (UTC)