mississippichem Posted November 29, 2010 Share Posted November 29, 2010 The Van der Waals equation of state for gases: [math]\left(p + \frac{n^2 a}{V^2}\right)\left(V-nb\right) = nRT[/math] where "a-prime" is the attractive force between gas particles. My question is, does "a-prime" already account for any paramagnetic interaction that may arise in gases with un-paried electrons like triplet [ce] O_2[/ce]? Or, does "a-prime" only factor in the Van der Waals (induced dipole-induced dipole) interactions present meaning that one would have to account for paramagnetic interactions separately? I realize one could neglect the paramagnetic interactions entirely for most purposes, obviously they shouldn't add up to much in the gas phase. This is sheer morbid curiosity. *I'm familiar with the standard derivation of the Van der Waals equation, but I've never had an opportunity to use it in real life because [math] PV=nRT [/math] has always been sufficient. I never work with the mega-pressures, micro-volumes, or small experimental margins where any of this really matters. Link to comment Share on other sites More sharing options...
Mr Skeptic Posted November 29, 2010 Share Posted November 29, 2010 I think it is supposed to account for the pressure increase due to particle collisions/interactions with each other. That said, it can only really be an estimate, as it accounts for all sorts of interactions simultaneously as a single value. 1 Link to comment Share on other sites More sharing options...
mississippichem Posted November 30, 2010 Author Share Posted November 30, 2010 I think it is supposed to account for the pressure increase due to particle collisions/interactions with each other. That said, it can only really be an estimate, as it accounts for all sorts of interactions simultaneously as a single value. Oh, so "a" counts for all intermolecular interactions. Thanks Link to comment Share on other sites More sharing options...
Mr Skeptic Posted November 30, 2010 Share Posted November 30, 2010 Yup, but I can't really tell you whether the "b" value would also partially account for those. The "b" value corresponds to a lower effective volume due to molecules having some "size", but interactions between the molecules could also contribute to their effective size. I suppose a more complicated formula could be made to account even more accurately for the different types of interactions, but if you've never even had to use this one then I doubt a more complicated one would be of much use, just yet more accurate and yet less used. Link to comment Share on other sites More sharing options...
mississippichem Posted November 30, 2010 Author Share Posted November 30, 2010 (edited) Yup, but I can't really tell you whether the "b" value would also partially account for those. The "b" value corresponds to a lower effective volume due to molecules having some "size", but interactions between the molecules could also contribute to their effective size. I suppose a more complicated formula could be made to account even more accurately for the different types of interactions, but if you've never even had to use this one then I doubt a more complicated one would be of much use, just yet more accurate and yet less used. I don't think "b" accounts for changes in molecular size that stem from dipoles and induced dipoles in question. I think it uses the volume extrapolated from the regular covalent radii. I'm sure the equation accounting or all those various "psuedo-coulombic" interactions would become a nightmare because polarizability is different for every effective nuclear charge and therefore every element in every compound would have some different factor making the formula different for every substance. I imagine this equation could be expanded in many interesting and useless ways . Edited November 30, 2010 by mississippichem Link to comment Share on other sites More sharing options...
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