mississippichem Posted November 29, 2010 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.
Mr Skeptic Posted November 29, 2010 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
mississippichem Posted November 30, 2010 Author 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
Mr Skeptic Posted November 30, 2010 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.
mississippichem Posted November 30, 2010 Author 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
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now