I am confused by the entropy.

Queries about input and output files, running specific calculations, etc.


Moderators: Global Moderator, Moderator

Post Reply
Message
Author
dyc_2008
Newbie
Newbie
Posts: 49
Joined: Tue May 25, 2010 9:03 am

I am confused by the entropy.

#1 Post by dyc_2008 » Tue Nov 15, 2011 8:51 am

dear all,

I substitute an atom in my system with alloying element.
In the OUTCAR, I find the energy of the system is lower than the un-substitute one.
Base on thermodynamics, G=H-TS, it is possible to achieve a lower energy if H and S increase simultaneously.
But, in VASP, the temperature is 0k, then TS=0. So entropy dose not contribute any effect in the equation G=H-TS.

please Somebody tell me how to explain this.
thank you so much.
Last edited by dyc_2008 on Tue Nov 15, 2011 8:51 am, edited 1 time in total.

alex
Hero Member
Hero Member
Posts: 586
Joined: Tue Nov 16, 2004 2:21 pm
License Nr.: 5-67
Location: Germany

I am confused by the entropy.

#2 Post by alex » Tue Nov 15, 2011 12:16 pm

Dear dyc_2008,

could you clarify, if you are talking about the total energy of your doped system or energy differences between undoped and doped systems?

Thx,

alex
Last edited by alex on Tue Nov 15, 2011 12:16 pm, edited 1 time in total.

dyc_2008
Newbie
Newbie
Posts: 49
Joined: Tue May 25, 2010 9:03 am

I am confused by the entropy.

#3 Post by dyc_2008 » Tue Nov 15, 2011 12:31 pm

dear alex,
differences between undoped and doped systems
thank you
Last edited by dyc_2008 on Tue Nov 15, 2011 12:31 pm, edited 1 time in total.

tlchan

I am confused by the entropy.

#4 Post by tlchan » Wed Nov 16, 2011 10:18 am

I don't think you should directly compare the total energies of the doped and undoped system. You can calculate the formation energy. For your case, the formation energy is the energy release to substitute a host atom by a dopant, where the dopant is extracted from a source, and the substituted host atom should go to a sink. Energy is required to extract the dopant from the source, and there is energy gain by depositing a host atom to the sink. The two energies correspond to the chemical potentials, which can be determined by the experimental conditions or your project.

As you mentioned, the total energies from VASP correspond to T = 0, so the contribution from entropy is not included. If you think entropy is important for your project, you have to separately add in the configurational entropy yourself.
Last edited by tlchan on Wed Nov 16, 2011 10:18 am, edited 1 time in total.

zzhlax
Newbie
Newbie
Posts: 28
Joined: Mon Jun 29, 2009 6:41 am
License Nr.: Medea via Materials Design

I am confused by the entropy.

#5 Post by zzhlax » Wed Nov 16, 2011 1:04 pm

What is "the configurational entropy"? Is it different from the entropy due to temperature?
thanks
Last edited by zzhlax on Wed Nov 16, 2011 1:04 pm, edited 1 time in total.

vistawanted

I am confused by the entropy.

#6 Post by vistawanted » Fri Nov 18, 2011 1:12 pm

Well, based on thermodynamics, G=H-TS, when T=0, G=H, especially when P=0, G=H=U=A, where A and U are Helmholtz Free Energy and internal energy, respectively.
In your system, we may only think about U, which was contributed by chemical bonding and configuration.
Then, all the substitutiaonal process has two steps, from un-sub one -> point defect one -> substituated one. Since doping element and replaced element, let us say element 1 was replaced by element 2, are definitely different, the chemical potential of them should be considered, too.
Therefore, the change of free energy, or internal energy, should be:
DeltaU = (Usub - U2) - (Uun - U1)
Where, Usub and Uun are the total energy of substituted and un-sub models, U2 and U1 are chemical potential of element 2 and 1, which should be obtained from their corresponding most stable phase.
Last edited by vistawanted on Fri Nov 18, 2011 1:12 pm, edited 1 time in total.

Post Reply