Care este diferența dintre W în energia internă și PV în entalpie?


Răspunsul 1:

Luați în considerare ce se întâmplă cu fiecare când presiunea și volumul sunt modificate cu o cantitate mică.

ΔW=pΔV\Delta W = p \Delta V

Δ(pV)=pΔV+VΔp=ΔW+VΔp\Delta(pV) = p\Delta V + V\Delta p = \Delta W + V\Delta p

SotheamountofworkgeneratedisnotthesameastheamountofchangeinpV,ingeneral.So the amount of work generated is not the same as the amount of change in pV, in general.

ThetwoareequalwhenthepressureisconstantandthusΔp=0.Soatconstantpressurethechangein[math]pV[/math]isthesameasthework.The two are equal when the pressure is constant and thus \Delta p = 0. So at constant pressure the change in [math]pV[/math] is the same as the work.

Din acest motiv, entalpia găsește utilizări pentru calcule în medii de presiune constantă, cum ar fi reacțiile chimice în aer liber.

Thesechemicalreactionsmayresultinexpansionorcontraction.Sothereactiondoessomeamountofworkonthesurroundingatmosphere.Becausethepressureisbasicallyconstantforreactionsintheopenair,thechangeinenthalpyH=U+pVisequaltothechangeininternalenergyUplustheworkdone[math]W=pΔV[/math].These chemical reactions may result in expansion or contraction. So the reaction does some amount of work on the surrounding atmosphere. Because the pressure is basically constant for reactions in the open air, the change in enthalpy H = U + pV is equal to the change in internal energy U plus the work done [math]W= p\Delta V[/math].

ΔH=ΔU+ΔW+VΔp\Delta H = \Delta U + \Delta W + V\Delta p

Atât timp cât presiunea este constantă

ΔH=ΔU+ΔW\Delta H = \Delta U + \Delta W

Așadar, pentru presiune constantă, putem folosi H pentru a urmări schimbările energetice totale, inclusiv lucrările efectuate în împrejurimi.

În situațiile în care presiunea nu este constantă, entalpia nu corespunde la fel de simplu cu energia.


Răspunsul 2:

Ijustwanttoanswerthisquestionasachemistbecausethesignconventionforusisoppositewhatanengineerwoulduse.manengineerisinterestedintheworkproducedbutachemistisinterestedinthesystemthatdidthework.midthesystemdidworkthenithaslessinternalenergythatitdidbeforetheworkwasdonesothevalueofw(notΔW)isnegative.So,achemistwillwrite[math]w=pΔV[/math].Thenegativesignisbecauseforapositivevolumechange,[math]ΔV[/math],ofthesystemtheworkisnegativeandviceversa.Also,atconstantpressure,theheatexchangedistheenthalpychange[math]q=ΔH.[/math]thus,internalenergyisdefinedasthesumofthechangeinheat,q,andthechangeinwork,w:I just want to answer this question as a chemist because the sign convention for us is opposite what an engineer would use. man engineer is interested in the work produced but a chemist is interested in the system that did the work. mid the system did work then it has less internal energy that it did before the work was done so the value of w (not \Delta W) is negative. So, a chemist will write [math]w = -p\Delta V[/math]. The negative sign is because for a positive volume change, [math]\Delta V[/math], of the system the work is negative and vice versa. Also, at constant pressure, the heat exchanged is the enthalpy change [math]q = \Delta H. [/math]thus, internal energy is defined as the sum of the change in heat, q, and the change in work, w:

ΔU=q+w\Delta U = q + w

or,substitutinginthevalueforworkatconstantpressureandtheheatchangeisenthalpychangeso.ΔU=ΔHpΔVor, substituting in the value for work at constant pressure and the heat change is enthalpy change so. \Delta U = \Delta H - p\Delta V

diferitele convenții privind semnele fac ca totul să pară greșit, dar este important să definim ce convenție folosim înainte de a scrie ecuațiile.


Răspunsul 3:

Ijustwanttoanswerthisquestionasachemistbecausethesignconventionforusisoppositewhatanengineerwoulduse.manengineerisinterestedintheworkproducedbutachemistisinterestedinthesystemthatdidthework.midthesystemdidworkthenithaslessinternalenergythatitdidbeforetheworkwasdonesothevalueofw(notΔW)isnegative.So,achemistwillwrite[math]w=pΔV[/math].Thenegativesignisbecauseforapositivevolumechange,[math]ΔV[/math],ofthesystemtheworkisnegativeandviceversa.Also,atconstantpressure,theheatexchangedistheenthalpychange[math]q=ΔH.[/math]thus,internalenergyisdefinedasthesumofthechangeinheat,q,andthechangeinwork,w:I just want to answer this question as a chemist because the sign convention for us is opposite what an engineer would use. man engineer is interested in the work produced but a chemist is interested in the system that did the work. mid the system did work then it has less internal energy that it did before the work was done so the value of w (not \Delta W) is negative. So, a chemist will write [math]w = -p\Delta V[/math]. The negative sign is because for a positive volume change, [math]\Delta V[/math], of the system the work is negative and vice versa. Also, at constant pressure, the heat exchanged is the enthalpy change [math]q = \Delta H. [/math]thus, internal energy is defined as the sum of the change in heat, q, and the change in work, w:

ΔU=q+w\Delta U = q + w

or,substitutinginthevalueforworkatconstantpressureandtheheatchangeisenthalpychangeso.ΔU=ΔHpΔVor, substituting in the value for work at constant pressure and the heat change is enthalpy change so. \Delta U = \Delta H - p\Delta V

diferitele convenții privind semnele fac ca totul să pară greșit, dar este important să definim ce convenție folosim înainte de a scrie ecuațiile.