[gmx-users] Enthalpy of Vaporization

[Fabian Casteblanco]

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.casteblanco at gmail.com