[gmx-users] Enthalpy of Vaporization

[Justin A. Lemkul]

Fabian Casteblanco wrote:
> 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?
> 

No, run both at the same temperature.  Kinetic energy doesn't enter the equation 
here.  Also, you need temperature to be constant in the equation below.  If you 
run two simulations at two different temperatures, which T do you use.  DHvap 
and other thermodynamic quantities are generally obtained at standard state, 
i.e. 298.15 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?
> 

No, that's why you use g_energy -nmol for the liquid system.  The output of 
g_energy is in kJ/mol of equivalent systems.  Dividing by the value given in 
-nmol is energy in kJ/mol of molecules.  That way, you compare the same 
quantities.  Otherwise, you've got the energy of one molecule in the gas phase 
(kJ/mol of molecules) minus the energy of one mole of a system of molecules in 
the liquid phase (kJ/mol of systems).  You'd get a nonsensical result.

-Justin

-- 
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Justin A. Lemkul
Ph.D. Candidate
ICTAS Doctoral Scholar
MILES-IGERT Trainee
Department of Biochemistry
Virginia Tech
Blacksburg, VA
jalemkul[at]vt.edu | (540) 231-9080
http://www.bevanlab.biochem.vt.edu/Pages/Personal/justin

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