Re: AMBER: Explanation of the energy unit ?

From: Marek Malý <maly.sci.ujep.cz>
Date: Mon, 08 Sep 2008 17:01:10 +0200

Dear Prof. Case,

thank you very much for your answer which
made me sure that my interpretation of
the [kcal/mol] unit is OK.

The second part of my question:

"The sense (background) of this "strange" unit" is unfortunately still
nonanswered.

I have to say that in relevant Wikipedia page:


http://en.wikipedia.org/wiki/Units_of_energy


is zero information about this unit which is very strange to me
since in computational(and probably not only computational) chemistry is
this (at least it seems to me)
energy unit standard. So maybe it would be worth to spend some time
to amend information about this unit in above Wikipedia page include the
right
interpretation and some background. Of course I do not mean that it is
task for you
personally :)), but for anybody who knows the right answers.

I think that could be interesting to answer question:
"What was the reason for introducing energy unit which gives me energy of
NA my systems in [kcal] or in another
words what was the reason to introduce energy unit which is NA times
smaller than [kcal] ?"

Is it because the relevant energies expressed using this small unit are
usually "nice" numbers like 30.56, 400.78 and
not numbers like 30.56 * 10^-n ... which should appear using any other
energetical unit or is there some
deeper background ? I think that there should be some deeper background
since there is use precisely Avogadro
constant in relationship with [kcal].


    Marek




Dne Mon, 08 Sep 2008 00:12:22 +0200 David A. Case
<case.biomaps.rutgers.edu> napsal/-a:

> On Sat, Sep 06, 2008, Marek Malý wrote:
>>
>> Let's assume that we have some molecular system Y and we for example
>> calculate the total energy of this system. Lets say that result is
>> X[kcal/mol].
>>
>> My only interpretation is that X is the energy in [kcal] of NA systems
>> Y.
>> So if I want energy of one system Y in [kcal] it is X/NA. Where NA is
>> Avogadro constant.
>
> This is correct. 1 cal = 4.184 J (exactly). See wikipedia (or other
> places) for a discussion of its history and use.
>
> ...dac
>
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Received on Wed Sep 10 2008 - 06:07:24 PDT
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