I think Axel made the same point as me clearer.
The default output in real units for energies and enthalpy is a sum of energy terms, the unit of which is scaled by \mathcal{N} for convenience. That’s all there is.
Okay, perfect: @m.adibi
Short terms answer to the question you were posting, regardless of getting into the specific science of your experimental enthalpy thingy: divide by total number of superatoms (since your entire CH4 is coarsened into one bead).
You can take my explanation or @Germain or the one in LAMMPS page
And yes Germain, I think we all have different ways of understanding something: as long as we reach the final result (which we do), it’s okay.
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That has now been explained multiple times by different people in different way. One of those explanations should be sufficient. Otherwise, I doubt that anybody will be able to provide an explanation that will satisfy you.
Can any of us verify independently what you said about the enthalpy not matching the data?
Axel I highly doubt if you even fully read the replies. There was a comment that agreed with what i said and another that disagreed. So there is no such a thing like different people said the same thing! I asked a simple question and you keep referring to different irrelevant issues!
I agreed with the procedure of normalizing by the atoms to get a per atom quantity. I was only speaking with regard to how lammps handles units with regard to common use of moles as a normalizing factor. Which was about answering the specifics of your question.
Axel is totally right concerning the fact that enthalpy measurements are defined with regard to processes. “Enthalpy of methane” means nothing. I don’t know to which paper you are referring, but I doubt they define such a thing. My best guess is that it is defined with regard to formation or condensation (or anything, I do not know) and normalized to the quantity of matter involved through moles of molecules (or atoms, again, can’t tell). Since you do not tell to which data you are referring too, I have no opinion on how correct your comparison with the simulation is.
We agreed with Cecilia that we were trying to say the same thing in different forms.
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Here is the paper:
https://pubs.acs.org/doi/pdf/10.1021/jp1013576
Look for table 3 in the paper. However, the issue of calculating h or dh is way beyond my question! I never asked such a thing. I simply asked about what does that mole in the denominator of kcal/mole refer to.
hey @m.adibi , dont worry. I think at some point we were all a bit pissed, just bashing at one another. Being human is too difficult - next life I will be a dolphin
The important thing is that now you understood (I promise you that mol as if it as 1 mol of simulation box and then take a look of how many atoms you will have in 1 mol of simulation box in order to convert to per mol of substance it works just fine).
Sorry if I confused you with the reasoning of the 5000 molecules: I had not realised as I was writing it that in your case it would ultimately converge to dividing by N
.
I think that reasoning I made can still serve to you in some other contexts: e.g. if you were simulating SiO2 that division to transform it to “per mol of SiO2” would not be by N, and you would not have molecules either. So I think my reasoning can allow you to understand the meaning of the operations you are making, even if by the end of everything, the 6.02E23 will appear in the numerator and denominator and look stupid.
Best,
Cecilia
Being human is not difficult, but apparently, being human and endeavoring to answer people’s queries without judging their personalities or predicting their satisfaction capabilities is really hard for some people! Of course, I am not talking about you, if you know what I mean 
I already knew those high-school concepts. My problem was that I presumed you knew that I knew 
so I assumed you know that when I say atoms, my system is coarse-grained so atoms=molecules. That is where the misunderstanding roots!
Anyway, Good luck @ceciliaalvares, and thanks for your helpful reply!
2 things here:
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The paper makes it explicit that the experimental data used to adjust the parameters of one of the models (M_{sw}) was the enthalpy of vaporization for methane, and that the difference was calculated using differences of the ensemble averages of the enthalpy between gas and liquid states. This is what you were asked to provide when you said you compared with “experimental enthalpy”. Also notice how they always mention to which experimental enthalpy they refer to, that they are all noted \Delta{}H and the variety of values they mention. This is not a trivial point and it is not being picky. This precision is necessary to understand what you were doing and without it, the conversation is confusing, circles around and easily leads to nowhere.
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Table 3 of the paper compares ensemble averages between two models (M_{sw} and OPLS-UA) at several pressure values. These are not experimental data, hence telling us to look at table 3, while we ask you which experimental data you refer to, is incorrect. These are not the same thing. An ensemble average at given P and T is not an experimental value, but can be compared between models for proof checking.
Anyway that was outside of the scope of the pure usage of LAMMPS. But this point on what “experimental enthalpy” meant is the point that might have heated up (no pun intended) the conversation, and understand that if people answer that they cannot help you because you are not precise enough or that what you say appears to make no sense, it might be more appropriate to refer directly to the source material you have in mind or try to understand why they do not understand what you say instead of repeating that what you say was very simple. We are not in your shoes, we do not have all the same infos as you do.
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