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9 th Week Chap(12-13) Thermodynamics and Spontaneous Processes

9 th Week Chap(12-13) Thermodynamics and Spontaneous Processes. • Standard State: The Thermodynamically stable state for pure liquids(Hg) and solid(graphite), for gases ideal gas behavior, for solutions 1.0 molar concentration of the dissolved species at

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9 th Week Chap(12-13) Thermodynamics and Spontaneous Processes

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  1. 9th Week Chap(12-13) Thermodynamics and Spontaneous Processes • Standard State: The Thermodynamically stable state for pure liquids(Hg) and solid(graphite), for gases ideal gas behavior, for solutions 1.0 molar concentration of the dissolved species at P= 1.0 atm and some specified T in each case • Reversible and Irreversible processes: Processes that occur through a series of equilibrium states are reversible. Adiabatic(q=0) paths are reversible • Thermodynamic Universe= System + Surrounding Isolated : No Energy and Matter can go in or out Adiabatic: No heat goes in or out • Entropy (S): measure of disorder Absolute Entropy S=kBln Number of Available Microstate ~ # of quantum states~size • Second Law of Thermodynamics: Heat cannot be transferred from cold to hot without work S ≥ q/T In-quality of Clausius S = q/T for reversible (Isothermal) processes S > q/T for irreversible processes • Third Law of Thermodynamics: S as T  for a pure substance • Reversible Processes: S =∫qrev/T but not Isothermal From the 2nd LawST=0∫T qrev/T Absolute Entropy

  2. Thermodynamic Processes no reactions/phase Transitions T>T Pext isotherm Isotherms T>T>T P = nRT/V for T =const T T<T Isothermal Paths are always reversible, e.g. AB and BA Whereas ABC and ADB are not reversible since they occur in a finite number of steps. Though both paths must result in the same changes in U, DU=UB – UA, since U is a state function.

  3. Thermodynamic Processes no reactions/phase Transitions T>T qin>0 Pext isotherm T qout<0 T<T UAC = qin + wAC qin= n cP(TC– TA) > 0 and wAC = - PextV UCB = qout + wCB qout= n cV(TC– TB) < 0 and wCB = - PV=0 UAB = UAC + UCB = n cP(TC– TA) - PextV + n cV(TC – TB)

  4. Thermodynamic Processes no reactions/phase Transitions wAC= - PextVAC work=-(area) Under PV curve qin>0 Pext isotherm qout<0 Area under A-C-B (-)work done by the system or (+)work done on the system

  5. Thermodynamic Processes no reactions/phase Transitions qin>0 Pext isotherm wDB= - PextVAC work=-(area) Under PV curve qout<0 Area under A-D-C (-)work done by the system or (+)work done on the system In the irreversible path A-D-B-C-A more work is done on the system than by the system! Whereas the reversible path AB the work is the same

  6. Hess’s Law applies to all State Function DH A D • AD DH • AB DH1 • BC DH2 • CD DH3 DH1 DH3 C B DH2 DH= DH1 + DH2 + DH3 Fig. 12-14, p. 506

  7. Example of Hess’s Law C(s,G) + O2(g)  CO2(g) DH1 = -393.5 kJ CO2((g ) CO(g) + ½O2(g) DH2 = + 283 kJ C(s,G) + O2(g)  CO(g) + ½ O2(g) DH = ?

  8. Example of Hess’s Law C(s,G) + O2(g)  CO2(g) DH1 = -393.5 kJ CO(g) + ½O2(g)  CO2((g) DH2 = - 283 kJ =========================== C(s,G) + O2(g)  CO(g) + ½ O2(g) DH =DH1+DH2= -110.5 kJ

  9. DH°f = 0 The Standard State: Elements in their most stables form are assigned a standard heat of Formation DH°f= 0 The standard state of solids/liquids is the pure phase @1 atm Gases it’s the ideal gas @ 1 atm For species in solution the Standard State is the concentration of 1.0 Molar @ P=1.0 atm DH°f= 0 for H+ ions solution For compounds the DH°f is defined by formation of one mole of the compound from its elements in their Standard States @ 1 atm, 1 molar and some T C(s,G) + O2(g)  CO2(g) DH°f = -393 kJmol-1 @25 °C

  10. In General for a reaction, with all reactants and products at a partial pressure of one 1 atm and/or concentration of 1 Molar aA + bB  fF + eE The Standard Enthalpy Change at some specified Temperature DH°(rxn) = DH°f(prod) - DH°f(react) DH°(rxn) = f DH°f(F) + eDH°f(E) – {aDH°f(A) + bDH°f(B)} energy Elements in their standard states DH°f(elements)=0 DH°f(reactants) DH°f(product) DH°(rxn

  11. For Example consider the reaction: 1/2 O2(g) O(g) DH°f[O(g)] (atomization energy/bond enthalpy) the formation of Ozone 3/2O2(g)  O3(g) DH°f[O3(g)] energy O(g) DH°f[O(g)]>0 DH°f(reactants) DH°f[O2(g)]=0 DH°f[O3] DH°(rxn) O3 Standard Heat of rxn for O(g) + O2(g)  O3(g)DH°=DH°f[O3] - DH°f[O] - DH°f[O2]

  12. For example consider the combustion rxn(not balanced): CH4(g) + O2(g) CO2(g) + H2O(l) DH°= standard DH? DH°= DH°f[CO2] +DH°f[H2O] - DH°f[CH4] - DH°f[O2] energy DH°f[O2(g)]=0 DH°f[CH4] + DH°f[O2] CH4(g) + O2(g) DH°f[CO2] +DH°f[H2O] DH°(rxn) CO2(g) + H2O(l)

  13. The combustion rxn(balanced): CH4(g) + 2O2(g) CO2(g) +2H2O(l) DH°= standard DH? DH°= DH°f[CO2] +2DH°f[H2O] - DH°f[CH4] - 2DH°f[O2] energy DH°f[O2(g)]=0 DH°f[CH4] + DH°f[O2] CH4(g) + O2(g) DH°f[CO2] +DH°f[H2O] DH°(rxn) CO2(g) + H2O(l)

  14. For example consider the combustion rxn: CH4(g) + O2(g) CO2(g) + H2O(l) DH°= standard DH? DH°= DH°f[CO2] +DH°f[H2O] - DH°f[CH4] - DH°f[O2] energy DH°f[O2(g)]=0 DH°f[CH4] + DH°f[O2] CH4(g) + O2(g) DH°f[CO2] +DH°f[H2O] DH°(rxn) CO2(g) + H2O(l)

  15. Consider the Reversible Isothermal Expansion of an Ideal Gas: So T=const for the system. U = q + w = 0 q = -w = (PextV) Heat must be transferred from the Surroundings to the System And the System does work on the surroundings If the process is reversible, Pext ~ P=nRT/V Which means done slowly and changes are infinitesimal V dV: The gas expands from V1 to V2 then q ~ PdV= (nRT/V)dV = (nRT) dlnV since dlnV=dV/V Integrating from V1 to V2q = nRT[lnV2 - lnV1]=nRTln(V2/V1) (q) Heat transferred in the Reversible Isothermal Expansion of an Ideal Gas q = nRTln(V2/V1)

  16. Reversible Irreversible q = nRTln(V2/V1)

  17. Carnot will use both Isothermal and Adiabatic q = 0 processes to define the entropy change S For an Ideal gas U = q + w =w U=ncV T = -Pext V Reversible Process Pext ~ P=nRT/V T dT and VdV ncV dT = - PdV

  18. Carnot will use both Isothermal and Adiabatic q = 0 processes to define the entropy change S T1 Reversible Process Pext ~ P=nRT/V T dT and VdV ncv dT = - PdV ncv(dT/T)=-nRdV/V cvdln(T) =-R dln(V) cvln(T2/T1)=Rln(V1/V2) T2 V1 V2

  19. cvln(T2/T1)=-Rln(V1/V2) A simpler form cab be obtained cvln(T2/T1)= ln(T2/T1)Cv =Rln(V1/V2)= ln(V1/V2)R (T2/T1)Cv = (V1/V2)R Recall that cp = cv + R (T2/T1)Cv = (V1/V2)R = (V1/V2)Cp-Cv let =cp/cv (T2/T1) = (V1/V2)-1 this relationship is very important for adiabatic processes in engines

  20. Carnot considered a cyclical process involving the adiabatic/Isothermal Compression/Expansion of an Ideal Gas: The so called Carnot cycle Th qh;Th ad ad ql::Tl Tl Which Yields (qh/Th) –(ql/Tl) = 0 and therefore S = q/T and that S is a State Function Since q = nRTln(V2/V1) then S = q/T= nRln(V2/V1)

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