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温故知新:熵乘以温度项

已有 18204 次阅读 2015-2-4 14:32 |个人分类:热力学性质|系统分类:科研笔记

关注:

1) 不同温度下气-固反应焓变和熵变的精确计算方法

2) 热力学公式:熵乘以温度项

3) 问题摘录


自由能F=E+PV-TS,S为熵entropy,E为结合能,一般算的应该是能量最低点,此时P~0.

所以energy without entropy应该就是酸的结合能吧。当ISMEAR=-5时,就是自由能。我是这么理解的。


问题摘录:

1) VASP计算过程中发现:做的sigma优化,各种sigma值得到的结果都是
entropy T*S    EENTRO =         0.00000000


http://emuch.net/html/201007/2228034.html

    sigma测试应该挺有意思,看看对各种性质的影响。


as if you set
ismear=-5
no ohter else sigma = whatever values
but the energy of entropy =0 in the case
Originally posted by Prettyswan at 2010-08-07 21:57:12:
我改了ISMEAR=0,做sigma优化,所有的sigma的取值都是  entropy T*S    EENTRO =        -0.26359441。请问这是怎么回事?
如果是体系是有 band gap的,那有可能是正常的。如果是金属的,就不正常了。
Originally posted by zhang_jaj at 2011-04-25 17:58:15:
楼主,把ismear改成0后,结果也是恒定值吗,最后搞定了吗,我也碰到问题了。ismear为0但是每个sigma对应的OUTCAR里的eentro的值都是0
搞定了,就是这样的

2)energy without entropy是什么意思?物理意义是什么呢?entropy什么是啊???
什么时候用TOTEN,什么时候用EENTRO,这两者有什么区别呢??????
也有人用tail -1 OSZICAR,这又是什么呢???
TOTEN和EENTRO各是什么意思呢?有什么物理意义呢?


http://emuch.net/html/201011/2618408.html

 

你所说的就是能量和没有熵的能量:
当你的ISMEAR=-5时,这2个是相等的TOTEN不就是tot energy吗?你可以用命令:grep ‘TOTEN’ OUTCAR查看你得到的能量。当ISMEAR不是等于-5时,Free energy TOTEN 与他就有差别。

  在计算体系的结合能时,体系的总能取为energy without entropy后的值。

查看体系的费米能级,用命令:


grep ‘Fermi energy: |tail -1


而OSZICAR是每次迭代或离子移动情况的简单汇总。


   

Originally posted by ustbwgs at 2010-11-21 20:11:50:
energy without entropy是什么意思?物理意义是什么呢?entropy什么是啊???
什么时候用TOTEN,什么时候用EENTRO,这两者有什么区别呢??????
也有人用tail -1 OSZICAR,这又是什么呢???
TOTEN和E ...

你所说的就是能量和没有熵的能量
当你的ISMEAR=-5时,这2个是相等的TOTEN不就是tot energy吗?你可以用命令:grep ‘TOTEN’ OUTCAR查看你得到的能量。

当ISMEAR不是等于-5时,Free energy TOTEN 与他就有差别。在计算体系的结合能时,体系的总能取为energy without entropy后的值。查看体系的费米能级,用命令:
grep ‘Fermi energy: |tail -1
而OSZICAR是每次迭代或离子移动情况的简单汇总。

给你举个例子吧:
Free energy of the ion-electron system (eV)
  ---------------------------------------------------
 alpha Z        PSCENC =        84.97356597
  Ewald energy   TEWEN  =     -1024.81602307      #Ewald energy   ?
  -1/2 Hartree   DENC   =      -517.90692606
  -V(xc)+E(xc)   XCENC  =      -130.78301910
  PAW double counting   =         0.00000000        0.00000000
   entropy T*S    EENTRO =         0.00000000                    #E-entropy
  eigenvalues    EBANDS =      -165.59366427
  atomic energy  EATOM  =      1717.99243751
  ---------------------------------------------------
 free energy    TOTEN  =       -36.13362901 eV
  energy without entropy =      -36.13362901  energy(sigma->0) =      -36.13362901
分割线中间的各行给出了各个分项的能量值,总能(TOTEN,即Total Energy)就是各个项的总和,倒数第二行给出的值。这里的总能,应该是指自由能(Free energy),

    自由能的公式是F=H-TS,H是焓(Enthalpy),T是温度,S是熵(Entropy),

     而EENTRO就是指-TS项的值。energy without entropy就是TOTEN-EENTRO,应该是焓值H。
这是我的理解,如有错请各位指正。
PS:楼上说的完全正确~~

自由能F=E+PV-TS,S为熵entropy,E为结合能,一般算的应该是能量最低点,此时P~0.

所以energy without entropy应该就是酸的结合能吧。当ISMEAR=-5时,就是自由能。我是这么理解的。

谢谢你的指点,如下是我的一点小小的疑问,望兄弟赐教:
VASP里算得都是绝对0度的情况,那么T=0,而EENTRO就是指-TS项的值话,它是不是应该总是0呢,实际不是呀,energy without entropy就是TOTEN-EENTRO,应该是含,这样的TOTEN恒等于energy without entropy,实际上像楼上说的一样,它们之间会因ISMEAR取值的不同会有差别啊。。。。
THANK YOU FOR YOUR CONSIDERRATION


Originally posted by SPRII at 2010-11-23 21:31:55:
建议你去读读vasp官方英文manual中第七章"Theoritical Background",那里对vasp中使用算法有较详细的说明.特别是针对你的这条疑问.先看看说明书!
I saw the Theoritical Background of the mannu, but I think  t is extremely profound.Thank you for your help.



 

引子:

Comment6: The calculated energy difference at 1 atmbetween the ScH3 (P63) and ScH2 (lowest energy one) + 1/2H2 equals -as authors state in p.20 - some 0.10 eV. Fromwhat is seen in Fig.10 the difference is around 0.15 eV, while ZPE for bothsystems are nearly the same (Tables 2 and 3), thus the E+ZPE should differ by ca. 0.15 and not 0.10, as authors write. Please double check.


Whichever of the two differences (0.15 or0.10 eV) is correct, the ST term for gaseous 1/2 H2 at 300 K equals 0.2 eV, and thus it is natural that ScH3 looses some H2 and becomes non-stoichiometric(H-deficient). It is worth to emphasize that.


解读:


As the reviewer pointed out to us, the ST term for gaseous 1/2 H2 at 300 K equals 0.2 eV

【熵增加了,氢分子能量增加了,ScH2 (Fm3m)+1/2H2(P63/m)变得能量更高?ScH3 (P63)变得更稳定,尤其是变成ScH2.9后?】, and thus it is natural that ScH3looses some H2 and becomes non-stoichiometric (H-deficient).Therefore, ScH2.9 is more stable than ScH2 (Fm3m)+1/2H2(P63/m) system. Actually, ScH2.9can be stored in liquid nitrogen and was experimentally observed to decomposeto ScH2+H2 at elevated temperatures at 1 atm1.




FREE ENERGIE OF THE ION-ELECTRON SYSTEM (eV)
 ---------------------------------------------------
 free  energy   TOTEN  =       -41.05379930 eV

 energy  without entropy=      -41.05060411  energy(sigma->0) =      -41.05273423
  enthalpy is  TOTEN    =      -33.47771040 eV   P V=        7.57608890



149.81    80.92   -33.47771040



高压下结构的变化规律,体积减小,键长缩短,晶格振动加剧;   G=H+TS, H=P+PV




获取焓值的脚本

#!/bin/sh

rm -f pv_et.txt

for file in OUTCAR_*4

do
pre=$(awk '/external pressure/{ print $4+$9 }' $file |tail -n 1)
vol=$(awk '/volume of cell/{print $5}' $file |tail -n 1)
et=$(awk '$3=="TOTEN"{print $5}' $file |tail -n 1)
#et=$(awk '/free  energy   TOTEN/{print $5}' $file |tail -n 1)
echo $pre \ \  $vol \ \  $et >> vv_et.txt
done
cat vv_et.txt|sort  -n -k 1 >pv_et.txt
rm vv_et.txt
a=$(awk '{print NR}' pv_et.txt)

echo $a
exit


问题参考:


















文献摘录:

Inclusion of the zero point energy of H2 molecules would tend to stabilize phases with high hydrogen stoichiometries overthose with low hydrogen stoichiometries, and thus would favor FeH2compared to FeH.

The enthalpy of FeH2 could also be lowered if this phase contains H vacancies, as discussed above.









参考网址:

http://webbook.nist.gov/cgi/cbook.cgi?ID=C1333740&Units=SI&Mask=1883#Thermo-Gas

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

http://butane.chem.uiuc.edu/pshapley/GenChem2/C7/1.html

 

 

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

 

http://casey.brown.edu/research/crp/Edu/Documents/00_Chem201/6_thermodyn_pot/6-thermodyn_pot-frames.htm

 

http://zh.wikipedia.org/zh-cn/%E5%90%89%E5%B8%83%E6%96%AF%E8%87%AA%E7%94%B1%E8%83%BD



吉布斯自由能英语Gibbs free energy),或称吉布斯函数Gibbs function)、自由Free Enthalpy)是热力学中描述等温等压过程的一个重要参量,常用G表示,它的定义是:

G = U - TS + pV = H - TS

其中U是系统的内能T温度Sp压强V体积H

吉布斯自由能的微分形式是:

.mathrm{d}G = - S.;.mathrm{d}T + V.;.mathrm{d}p + .mu.;.mathrm{d}N

其中.mu.,化学势。一个重要的推论是G=.mu N.,。也就是说每个粒子的平均吉布斯自由能等于化学势。



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

The Gibbs free energy is defined as:

G(p,T) = U + pV - TS

which is the same as:

G(p,T) = H - TS

where:

The expression for the infinitesimal reversible change in the Gibbs free energy as a function of its 'natural variables' p and T, for an open system, subjected to the operation of external forces (for instance electrical or magnetical) Xi, which cause the external parameters of the system ai to change by an amount dai, can be derived as follows from the First Law for reversible processes:

T.mathrm{d}S= .mathrm{d}U + p.mathrm{d}V-.sum_{i=1}^k .mu_i .,.mathrm{d}N_i + .sum_{i=1}^n X_i .,.mathrm{d}a_i + .cdots.mathrm{d}(TS) - S.mathrm{d}T= .mathrm{d}U + .mathrm{d}(pV) - V.mathrm{d}p-.sum_{i=1}^k .mu_i .,.mathrm{d}N_i + .sum_{i=1}^n X_i .,.mathrm{d}a_i + .cdots.mathrm{d}(U-TS+pV)=V.mathrm{d}p-S.mathrm{d}T+.sum_{i=1}^k .mu_i .,.mathrm{d}N_i - .sum_{i=1}^n X_i .,.mathrm{d}a_i + .cdots.mathrm{d}G =V.mathrm{d}p-S.mathrm{d}T+.sum_{i=1}^k .mu_i .,.mathrm{d}N_i - .sum_{i=1}^n X_i .,.mathrm{d}a_i + .cdots

where:

This is one form of Gibbs fundamental equation.[10] In the infinitesimal expression, the term involving the chemical potential accounts for changes in Gibbs free energy resulting from an influx or outflux of particles. In other words, it holds for an open system. For a closed system, this term may be dropped.

Any number of extra terms may be added, depending on the particular system being considered. Aside from mechanical work, a system may, in addition, perform numerous other types of work. For example, in the infinitesimal expression, the contractile work energy associated with a thermodynamic system that is a contractile fiber that shortens by an amount −dl under a force f would result in a term fdl being added. If a quantity of charge −de is acquired by a system at an electrical potential Ψ, the electrical work associated with this is −Ψde, which would be included in the infinitesimal expression. Other work terms are added on per system requirements.[11]

Each quantity in the equations above can be divided by the amount of substance, measured in moles, to form molar Gibbs free energy. The Gibbs free energy is one of the most important thermodynamic functions for the characterization of a system. It is a factor in determining outcomes such as the voltage of an electrochemical cell, and the equilibrium constant for a reversible reaction. In isothermal, isobaric systems, Gibbs free energy can be thought of as a "dynamic" quantity, in that it is a representative measure of the competing effects of the enthalpic and entropic driving forces involved in a thermodynamic process.

The temperature dependence of the Gibbs energy for an ideal gas is given by the Gibbs–Helmholtz equation and its pressure dependence is given by:

.frac{G}{N}  = .frac{G}{N}^.circ  + kT.ln .frac{p}{{p^.circ }}

if the volume is known rather than pressure then it becomes:

.frac{G}{N}  = .frac{G}{N}^.circ  + kT.ln .frac{V^.circ}{{V }}

or more conveniently as its chemical potential:

.frac{G}{N}  = .mu  = .mu^.circ  + kT.ln .frac{p}{{p^.circ }}.

In non-ideal systems, fugacity comes into play.

 


http://www.americanscientist.org/issues/pub/the-thermodynamic-sinks-of-this-world/4

Nothing Is Simple

Going back to the cartoon world of H2, Cl2, O2 and Na, you would get mostly NaCl (s) and H2O (l) at equilibrium at 298 kelvin. Not really—for mixing is most certainly allowed, and salt dissolves spontaneously in water, as you well know. And perhaps I should worry about other reactions among the six species. For instance Na with H2 (to give NaH [s]), Na with O2 (to give Na2O [s]), Cl2 with H2 (to give HCl [g]), Cl2 with O2 (to give Cl2O or ClO2, both known). And I haven’t started to be concerned about ternaries (compounds of three elements with each other); I’ve kept my world simple with binaries, isolated from each other.

Remember that other pyrotechnic lecture demonstration—dropping a chunk of sodium into water?

Na (s) + H2O (l) ↔ NaOH (aq) + ½ H2 (g)      ΔH = –183 kJ/mol

I write NaOH (aq), the “aq” standing for aqueous, because NaOH is very soluble in water (ΔG for solution is –42 kJ/mol). Now how about quaternaries?


If the temperature is very high, the entropy term in ΔG will steer things, so that reactions in which the entropy is increased are favored.

All solids will be spontaneously converted into gases. Yes, even solid NaCl (boiling point [b.p.] = 1,413 degrees Celsius) and NaOH (b.p. 1,388 degrees Celsius). All molecular gases will decompose into atoms, a reaction with a nice positive ΔS, as translational degrees of freedom are created. And some ions will be created (for example, H ↔ H+ + e-), depending on the temperature. From a chemist’s point of view, the surface or interior of a star (now that is high T!) is boring—there are no molecules there.

But from a nuclear physicist’s point of view, these are the greatest fun.

The low temperature limit, T approaching absolute zero, poses different problems. For then the history of the model world really, really matters. Let me explain. At very low temperatures, the atoms are not moving quickly enough to overcome in their collisions any barriers (remember the ubiquitous activation energies introduced above?).

So if you take that hydrogen balloon with its mixture of H2 and O2 down to close to absolute zero, you will have to wait a very, very long time, a time approaching infinity, to get any water. Here H2 and O2 are metastable relative to H2O, perfectly happy on their own. You can see what I mean by history—at low T what you get (in a human, finite observation time) depends on whether the reactant molecules were first heated to overcome activation barriers. Or if they were not, just allowed to cool.

The high vacuum interstellar medium poses another challenge. If a rare collision between two molecules or atoms that are prone to react were to take place—say H2 + O (atomic)—the reaction to H2O being highly exothermic (by 491 kJ/mol, gas phase) the product molecule is born with a large energy. In the absence of collisions, that energy will have nowhere to go, and the reaction, exothermic, will not happen

A Real Problem Lurking

I built my world as one of essentially limitless amounts of each element. If instead I began with equal and large finite masses, I would immediately run into the constraints of a limiting reagent. Consider for instance that simple water-forming reaction. If you have a finite amount (say 16 grams each) of H2 and O2, you will get at equilibrium 18 grams of H2O, with 14 grams of H2 left over and a miniscule amount of O2. Thirty-two grams we began with, 32 grams we have at the end—from Lavoisier’s time we knew that nothing is gained, nothing is lost. And let’s not worry about what mass/energy equivalence leads to in an exothermic reaction. There just wasn’t enough O2 there to react with the H2 supplied.

Here’s the nagging thought then about the utility of my Gedankenexperimentany real world must begin with pretty constrained if not fixed local concentrations of the elements. And so the outcome of all reactions, not just for the equal-mass example above, will be constrained by those initial conditions. We will come back to this.


 

 


摘录学习:


这是热力学里可能是最重要的问题之一。简单的说,熵和绝对温度是个捆绑的概念,0熵就是0温度,而且精确的温度定义是用熵来定义的,粗略的说熵和决定了我们所认为的温度,熵才是温度在物理学意义上的度量。
Thermodynamic beta 这条里面的推论基本是可以理解的,即{T} =.frac{.partial U}{.partial S} , U是物体的内能而S是熵。
另,不可能让物体保持绝对零度。绝对零度不可达到,这是我们目前物理学的界限之一,和普朗克时间以下是没有物理意义,光速不可超越是同样的概念。


要理解“熵”跟“温度”有没有关系,这个问题的理解类似于考察“体积”和“密度”有没有关系。为什么用这样的一个类比呢?因为对于“密度”这样的概念我们更熟悉一些,而其定义又可以类似于温度来定义,我们可以把它定义成质量对体积的导数,正如温度可以定义成内能对熵的导数。


首先,从热力学角度来说,熵的定义是dS=dQ/T,这里说一句,Q(系统从外界吸收的热量)是过程量,而T和S是状态量,这里的dQ一般都要在微分符号d上加一横表示不是完整微分,以示与普通微分的区别。从定义来看,熵与温度的关系并不明确,或者说,没有显式关系。热力学里对于一个系统存在三个基本的热力学函数,分别是物态方程(由实验确定),内能,熵。知道了这三者,就能够推导出其他的量。如果局限到理想气体,熵可以用T和另外一个状态参量表示出来。


关于储氢材料的计算











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