Hello Justin, Thank you for your response. I just wanted to make sure I understand what you meant. I'm assuming that you want the one molecule in the gas phase at its boiling point and you are doing this because you want the molecule alone with no intermolecular interactions? (since heat of vaporization is the energy required to break those interactions amoung liquid molecules) Is that why we ignore kinetic energy? Is the liquid alcohol that I already simulated (liquid methanol, 1 bar, 298 K) also suppose to be at the 338 K boiling temperature? Would I simply run the npt equilibrium again but simply change the temperature from 298 to 338 K?
Also, you stated the equation: DHvap = <Epot(gas)> - <Epot(liq)> + RT The first Epot(gas) would be the potential energy for 1 molecule but the Epot(liq) would be the potential energy for 1 mole? Thanks for your help Justin. -- Best regards, Fabian F. Casteblanco Rutgers University -- Chemical Engineering PhD Student C: +908 917 0723 E: fabian.castebla...@gmail.com -- gmx-users mailing list gmx-users@gromacs.org http://lists.gromacs.org/mailman/listinfo/gmx-users Please search the archive at http://www.gromacs.org/Support/Mailing_Lists/Search before posting! Please don't post (un)subscribe requests to the list. Use the www interface or send it to gmx-users-requ...@gromacs.org. Can't post? Read http://www.gromacs.org/Support/Mailing_Lists
