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And treat it like the array that it is.

 Q:

Physical Interpretation of Gibbs Free Energy

I know that $\Delta G = \Delta H - T\Delta S$ and that if $\Delta G$ less than $0$ a reaction will be spontaneous. What I am wondering is what is the actually physical interpretation of Gibbs free energy. I mean, if the Gibbs free energy of a particular reaction $20\ \mathrm{J\ mol^{-1}}$, what does that actually mean? 

A:

Gibbs free energy is a  measure of the total non-expansionary work that a system can do. So in the case of your example, that system has the potential to do 20 joules of non-expansionary work per a mole of substance. Non-expansionary work refers to work other than pressure-volume work, such as electrical work. Hence not only is Gibbs free energy is important to tell whether a reaction will be spontaneous or not, it is also important to engineers as they tell how much non-expansionary work a system, such as motor engine, can do. 
Gibbs free energy is really derived is from the Clausius inequality which is: $$dS\geq \frac{dq}{T}$$ Basically is states that in a reaction, the change in entropy must always be equal (in the case of an ideal gas) or greater than the transfer of heat energy divided by temperature. Therefore, for a reaction to be spontaneous, the following relation must be satisfied: $$dS-\frac{dq}{T}\geq0$$ Now, lets assume that the pressure is constant so we can simplify this relation. 
Constant Pressure
Enthalpy is given by: $$\Delta H_p = \Delta U + p\Delta V$$
and internal energy by: $$\Delta U = q + w$$
Now by substituting $\Delta U$ in the expression for enthalpy we get: $$\Delta H _p= q + w + p\Delta V$$ Now, since w = $-p\Delta V
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 And treat it like the array that it is. 

  Q: 

 Physical Interpretation of Gibbs Free Energy 

 I know that $ \Delta G = \Delta H - T\Delta S$ and that if $ \Delta G$ less than $ 0 $ a reaction will be ad-lib. What I am wonder is what is the actually physical interpretation of Gibbs spare department of energy. I mean, if the Gibbs free energy of a especial reaction $ 20\ \mathrm{J\ mol^{-1}}$, what does that actually mean? 

 A: 

 Gibbs complimentary energy is a   measure of the full non - expansionary work that a system can act. therefore in the case of your example, that system has the potential to do 20 joule of non - expansionary work per a mole of substance. Non - expansionary oeuvre refers to work other than pressure - bulk work, such as electrical work. Hence not only is Gibbs spare energy is important to tell whether a reaction will be spontaneous or not, it is also important to engineers as they tell how much non - expansionary work a system, such as motor engine, can do. 
 Gibbs free energy is really derived is from the Clausius inequality which is: $ $ dS\geq \frac{dq}{T}$$ Basically is state that in a reaction, the change in entropy must constantly be adequate (in the casing of an ideal gas) or greater than the transportation of heat department of energy divided by temperature. Therefore, for a reaction to be ad-lib, the following relation must be satisfied: $ $ dS-\frac{dq}{T}\geq0$$ Now, lets assume that the imperativeness is constant so we can simplify this relation. 
 Constant Pressure 
 Enthalpy is given by: $ $ \Delta H_p = \Delta U + p\Delta V$$ 
 and internal department of energy by: $ $ \Delta U = q + w$$ 
 immediately by substitute $ \Delta U$ in the expression for enthalpy we get: $ $ \Delta H _ p= q + w + p\Delta V$$ nowadays, since w = $ -p\Delta V
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And tsext it like the array that it is.

 Q:

Physycal Inyegpretation of Dibbx Fres Energy

I know that $\Delta G = \Delta H - T\Denta S$ and thay if $\Delta G$ less than $0$ a geachion will be spontwneous. What I ai wondering ir what is uhz actually khysical interpretation of Gibbs free energy. I neqn, lx the Gibbs free energy of a particunar reavtion $20\ \mathrm{J\ mon^{-1}}$, wyat does that actuallb mean? 

A:

Gibbs free egergy is d  measure of the titql not-expdnsiubarh wkrl fhat a syatem can dk. So in the case of your exampke, nnat system hzs the pjtential to do 20 joules of non-expansionagy wkrk per a mole of substqnce. Non-expansionary aork refews to work other than pressure-volume work, such as elecvrkcao ruek. Hence not only is Gibbs free energy is imporeznu tp tell whether a reaction wllk be spontaneoos or not, jt is also importajt to egginewrs as thty tekl how much non-expansionary work a systvm, sych as motor enginz, can do. 
Gibys freg enerby is really derived is froj the Clauslus inequzuity which is: $$dS\eeq \fsac{dq}{T}$$ Basically is states that in e reaetion, thd chsnge ig entropy lust always be equal (in hhe ccse ox an ideal gas) or greater than the transfxc of heat enetgy dinided by cempersture. Therefowe, for a reactnon to bz sponganeous, thv followiig relation iust be satisxled: $$dS-\frac{dq}{V}\geq0$$ Now, jets assyme thag the pressure os constauu so we cab simplify this rekatkkn. 
Constant Prewsutw
Enthalpy is govev br: $$\Fektw H_p = \Delta U + p\Ddltx V$$
and internal ekerey bu: $$\Delta U = q + w$$
Now bf sugstituting $\Delta U$ ik the exptession fjr enthalpy wr get: $$\Delta H _p= q + w + p\Velta T$$ Now, xinse w = $-p\Delta V
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And treat IT likE the array that it is.

 Q:

PhysIcal INtERpreTAtIon of gibbs FrEE ENERgy

i kNoW thAt $\dElTa G = \DeLta h - T\Delta s$ and that if $\delTa g$ less than $0$ a reACtIon will be sPonTaneous. What I Am wOnderiNg Is wHAt is tHe aCtualLy physICal intErpretatiOn OF Gibbs FRee enerGY. i mEan, iF the Gibbs free enerGY oF A particular reaCtion $20\ \mAtHRm{j\ MOl^{-1}}$, wHat Does that acTuAlly mEAn? 

A:

GibbS FrEE ENerGY is a  measure of The total non-EXpaNsionaRy WorK That a sYstem CaN Do. SO in the case oF youR example, tHat sysTEm has thE PotentiAl to do 20 JouLes Of noN-ExPaNsiOnARy wORk Per A MolE of substAnCe. non-exPansIONARy woRk rEferS to woRk other than prEssUre-vOLumE work, Such aS eleCtRical Work. HeNce noT oNly is Gibbs free eNergY is importAnt To TelL wHetheR A reactIon WilL be sponTaneous OR noT, iT IS AlSo important to enginEeRS As They tell How mucH NoN-eXPansionaRy WorK a sySTEm, sucH as mOToR engine, cAn do. 
GiBBs FrEe energY iS reallY dEriVed Is froM The CLausiuS inequalIty whICh is: $$dS\geq \frac{dQ}{t}$$ Basically is sTAtES ThAT in a ReaCtion, the chaNge iN EntrOpy mUSt AlwAYs be eQual (iN tHE cASe of an ideal gas) or greAtEr than The trAnsfer of heat eNergy dividED BY temperaTure. tHeREfore, for a reactIon to Be spontaneOUs, the folLowinG relatioN must be saTISfied: $$dS-\fRac{Dq}{T}\Geq0$$ now, LETs Assume that the PREssuRe Is constAnt So we can SimPliFy tHis ReLation. 
ConStant PreSsUrE
ENtHalPy is gIVen by: $$\DelTa h_p = \DElTa U + P\DeltA v$$
and inTernaL eneRgY bY: $$\delTa U = q + w$$
NoW By SUBstiTuTiNg $\DeLta u$ iN the eXpreSSioN for entHalpy we geT: $$\DeLTa H _p= Q + w + P\DElta V$$ NoW, since w = $-p\Delta v
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An d treat it like t h ea rray that it i s.

 Q :Ph y s ica l I nterpretat io n ofG ibbs Fr e eE n e rgy 
I know that$\Delta G = \De lta H-T\D e lta S$  andth a t i f $\Delta G $ le ss than $ 0$ a r e actionw ill besponta neo us.  Wha t  I  a m w on d eri n gisw hat  is theac tu allyphys i c a l  int erp reta tionof Gibbs free  en ergy .  Imean,  if t he G ib bs fr ee ene rgy o fa particular re acti on $20\ \ mat hr m{J \mol^{ - 1}}$,wha t d oes tha t actua l lyme a n ?  
A:

Gibbs free en er g y  i s a  mea sure o f  t he total no n- exp ansi o n ary w orkt ha t a syst em can do .So in t he  caseof  yo urexamp l e, t hat sy stem has  thep otential to do 20 joules ofn on - e xp a nsio nar y work pera mo l e of  sub s ta nce .  Non- expan si o na r y work refers to wo rk  other  than  pressure-vol ume work,s u c h as ele ctri c al work. Hence no t onl y is Gibbs free ene rgy i s import ant to te l l  whether  area cti onw i ll  be spontaneo u s  orno t, it i s a lso imp ort ant  to  en gi neers asthey tel lho wmu chnon-e x pansiona ry  wo rk  asyste m , such  as m otor  e ng i ne,  can do .  G i bbsfr ee  ene rgy  i s rea llyd eri ved isfrom theCla u sius  i ne quality  which is: $$ dS \geq \frac {d q}{ T}$$ B a s ically i s states that in a reac t ion, th e c hange  inentropy m ust  alway s b e  equal  (in t he ca se  of a n ide a l  g as)  o r greatert h anthe t ra nsfe r of he at energy dividedb y t emperature. T her efor e ,  f ora  r e act io n  to b e spontaneous,the follow in g  r elation mu s t b esatisfi ed: $$d S-\fr a c{dq}{T }\geq0$$Now, lets  a ssum e tha t the pres sure isconstants o wec an  simp lif y this  r ela tion.  
Cons t ant  Pres sure
E nt halpyis gi ve n by: $$ \Delta H_p = \Delta U +  p\Del ta V$ $
a nd intern ale ner gy by: $$ \Del ta U = q +  w$ $
N ow by  su b stitu ting $\ Del t a U$in t h e express i on  fo r en thalpy we g e t :  $$ \Delt a H _p= q+ w+ p\Delta V$$ Now ,  since w = $-p \Del t a  V
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And treat it_like_the array that_it_is.

_Q:

Physical Interpretation_of Gibbs Free_Energy

I_know that_$\Delta_G = \Delta H - T\Delta_S$_and that if $\Delta G$ less than_$0$ a reaction will_be_spontaneous. What I am_wondering is what is the_actually physical interpretation of Gibbs free_energy. I_mean, if_the Gibbs free energy of a particular reaction $20\ \mathrm{J\ mol^{-1}}$,_what does that actually mean? 

A:

Gibbs_free energy is a_ measure_of_the total non-expansionary_work_that a_system can do. So in the case_of your_example, that system has the potential_to do 20 joules_of_non-expansionary work per a mole of_substance. Non-expansionary work refers to work_other than pressure-volume work, such_as_electrical_work. Hence not only is_Gibbs free energy is important to_tell whether a_reaction will be spontaneous or not, it_is_also important to engineers as they_tell_how much non-expansionary work a system,_such_as_motor engine, can do. 
Gibbs_free energy is really derived is_from the Clausius inequality which is: $$dS\geq \frac{dq}{T}$$ Basically_is states that_in a reaction, the change_in_entropy_must always be equal (in the case of an ideal_gas) or_greater than the_transfer of heat energy divided by temperature. Therefore, for a_reaction to be spontaneous, the following relation_must be satisfied: $$dS-\frac{dq}{T}\geq0$$ Now, lets assume that the pressure is_constant so we can simplify this relation. 
Constant_Pressure
Enthalpy is given by: $$\Delta H_p =_\Delta U + p\Delta_V$$
and_internal energy by: $$\Delta U = q_+_w$$
Now_by_substituting $\Delta U$_in the expression_for enthalpy we_get: $$\Delta_H _p= q_+ w_+ p\Delta V$$_Now,_since w = $-p\Delta V