Thank you very much.
David A. Case wrote:
> I think you should do this: take the average total potential energy for the
> liquid, and divide by the number of molecules. Then, you will have do a
> separate calculation for a _single_, gas-phase molecule. Subtracting these
> should give you an estimate of the enthalphy of vaporization (if you include
> some corrections for pressure-volume terms). A detailed description of this
> process is given here (among many other places, I am sure):
>
> %A H.W. Horn
> %A W.C. Swope
> %A J.W. Pitera
> %A J.D. Madura
> %A T.J. Dick
> %A G.L. Hura
> %A T. Head-Gordon
> %T Development of an improved four-site water model for biomolecular
> simulations: TIP4P-Ew
> %J J. Chem. Phys.
> %V 120
> %P 9665-9678
> %D 2004
>
> (See esp. Eq. 19. If you don't have a polarizable force field, the correction
> term "C" is pretty small.) Then you can roughly say:
>
> Delta H(vap) = <Ugas> - <Uliquid>/N + RT
>
> (But see the above reference for additional corrections). If you need a free
> energy of transfer from liquid to vapor, (and not just an enthalpy), you will
> need to carry out much more extensive calculations.
>
> ...hope this helps...dac
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Received on Tue Sep 28 2004 - 18:53:00 PDT
