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Chapter 5 auxiliary functions

Chapter 5 auxiliary functions. 5.1 Introduction. The power of thermodynamics lies in its provision of the criteria for

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Chapter 5 auxiliary functions

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  1. Chapter 5auxiliary functions

  2. 5.1 Introduction

  3. The power of thermodynamics lies in its provision of the criteria for • equilibrium within a system and its ability to facilitate determination of the effect on the equilibrium state of change in the external influences which can be brought to bear on the system. • the practical usefulness of this power is consequently determined by the practicality of the equations of state of the system ί.e ,the relationships among the state functions which can be established .

  4. * The combination of the first and second laws of thermodynamics leads to the following equation: du = Tds - pdVˇ Or u = f ( s , Vˇ ) This equation of state gives the relationship between the dependent variables u and the independent variables s and v for a closed system of fixed composition which is in states of equilibrium and is undergoing a process involving volume change against external pressure as the only form of work performed in or by the system (p – v work)

  5. * the combination of the first and the second laws of thermodynamic provides also the following criteria for the equilibrium : 1. For closed system of constant energy and constant volume the entropy is a maximum. 2. for closed system of constant entropy and volume the energy a minimum

  6. * Since the entropy is an inconvenient choice of independent variable from the point of view of experimental measurement of control; it is desirable to develop a simple equation similar to the previous one which contains a more convenient choice of independent variable. * from experimental point of view the most convenient pair of independent variables would be temperature and pressure because they are the most easily measured and controlled parameters in a practical experiment.

  7. * from theoretical point of view the most convenient pair of independent variable would be volume and temperature when they are fixed fore closed system the Eίs level values ; and hence the Boltzman , factor ( exp (- Eί / kT ) and the partition function are fixed this will ease the theoritical calculations using the methods of statistical mechanism * thus , in this chapter to meet the previously discussed points the enthalpy function , H , the helmholtz free energy function ( the work function ) , A , the Gibbs free energy function , G , and the chemical potential of species ί , μί , are introduced

  8. 5.2 The enthalpy H

  9. * The enthalpy function H is defined as: H = υ + pv Thus : dH = dυ + pdv+ vdp = Tds – pdv + vdp + pdv Therefore : dH = Tds + vdp . ί . e, H = ƒ ( s , p ) * The dependent variable, in this case, is enthalpy while the pair of independent variables are the entropy and the pressure.

  10. * Since the enthalpy is state function, we have: Δ H = H2 – H1 = ( υ2 + p2v2 ) – (u1+ p1v1 ) Thus :under constant pressure: Δ H = ( υ2 – υ1 ) + p ( v2 – v1 ) = Δ υ + Wp-v = qp

  11. * Therefore, the equation of state, dH = Tds +vdp , gives the relationship between the dependent variable H and independent variables S and P for a closed system of fixed composition whish is in states of equilibrium and is undergoing a process involving volume change against external pressure as the only form of work performed on , or by , the system ; the enthalpy change of the system equals the heat leaving or entering the system .

  12. 5.3 Helmholtz Free Energy Function A

  13. * Helmholtz Free Energy Function A is defined as : A = υ – Ts Thuse : dA = dυ – Tds – SdT = Tds – pdv – Tds – SdT Therefore dA = – sdT – pdv = ƒ (v , T ) * the dependent variable in this case is A and the pair of independent variables are volume and temperature .

  14. * Since the Helmholtz free energy function is a state function, thus : Δ A = A2 – A1 = (U2 – T2S2) – (U1 – T1S1) = (U2 – U1) – (T2S2 –T1S1) . ί . e, Δ A = q + w – ( T2S2 – T1S1) ThusΔ A – w = q – ( T2S2 – T1S1)

  15. * in describing the work in the previous equation , a positive sign is assigned to work done on the system and a negative sign is assigned to work done by the system . if the apposite convention is used . ί . e, the work done by the system is assigned a positive value , the previous equation become w + ΔA = q – ( T2S2 –T1S1) * If the process is isothermal; that is T2 = T1 = T,then : W + Δ A = q – T ( S2 – S1)

  16. * from the second law of thermodynamics : q ≤ T ( S2 – S1), Thus w ≤ –Δ A * Therefore, for reversible isothermal process : wmax = – Δ A . ί . e tha maximum amount of work done by system equal the decrease in the work function * Since=wmax–wdeg; thusw = –Δ A–TΔsirr= – (ΔA + T Δsirr ) Thus for process which occurs at constant V and T , we can write : Δ A + T Δ sirr = 0 For an infinitesrnal increment of such on process : SA +TdS = 0

  17. * For spontaneous processes dsirr is a positive value , thus processes occur at constant T and V will be spontaneous if d A is negative value ; ί . e, for spontaneous processes that occur at constant T and V : dA < 0 * since the condition for thermodynamic equilibrium is that dsisr = 0 then with respect to the described process . ί . e , at constant V and T equilibrium is defended by the condition that dA = 0

  18. * Thus in closed system , held at constant T and V Helmholtz free energy can only decrease , for spontaneous process or remain constant , at equilibrium ; the attainment of equilibrium in the system coincides with the system having a minimum value of A constant with the fixed values of V and T

  19. * consideration of A thus provides a criterion of equilibrium for a closed system with fixed composition at constant value V and T : * consider for example m n atoms of seme element accusing in both asides crystalline phase and a vapor phase contained in a constant – volume vessel which is immersed in constant – temperature heat reservoir . the point now is to determine the equilibrium of the n atoms between the sold phase and the vapor phase .

  20. * At constant V and T this distribution occurs at the minimum value of A, and hence with low values with U and high value of S since A = U – TS * The two extreme states of existence of this system are : 1- all n atoms are in the sold phase and none occur in the vapor phase . 2- all n atoms are in the vapor phase and the solid phase is absent .

  21. * starting with system occurring in the first of these two states , ί . e ,the solid crystalline state ; the atom in such case are held together by interatomic force ; thus , if an atom to be removed from the crystal surface and placed in vapor phase ( the first atom is placed in vacuum) , energy is a boarded as heat from the heat reservoir to the system to break the interatomic bonds to increase the internal energy , υ , of the system and its randomness ί. e , the system entropy as shown in figure ( figure 5.1 page 97 ) which chins the variation of internal energy and entropy with the number of atom in in the vapor phase of the closed solid vapor system at constant V and T .

  22. *this figure shows that U increase linearly with nυ while the entropy increase is nonlinear figure (1a) the note of deincrease of S with nυ increase . * The saturated vapor pressure is calculated as : P = [ nv( q,T) kT ] ÷ [ V – Vs ] Where V is the volume ob the containing vessel, Vs is the volume of solid phase present, and nυ ( eq,T) is the number of atoms in the vapor phase at the equilibrium point which correspond to the minimum value of A as shown in figure (2) (figure 5.2 page 98 ) this minimum value is obtained by adding the values of U to the corresponding values of (–Ts) and thus having a curve that represents the variation of A with nυ .

  23. * As the magnitude of the entropy contribution the value of A , –Ts , is temperature dependent and the internal energy contribution is independent of temperature , the entropy contribution becomes incuriously predominate as temperature is increased and the compromise between U and (– Ts) which minimize A occurs at layer of nυ . * This is illustrated in figure ( figure 5.3 page 99 ) which is draw for T1 and T2 where T1 < T2 * As increase in the temperature from T1 to T2 increase the saturated vapor pressure from p(T1) = nv (eq.T1) le T1 [v–vs(T1)] to p(T2) = nv (eq.T2) le T2 [v– vs(T2)] the saturated vapor pressure increase exponentially with increasing the temperature .

  24. * for the constant – volume system , the maximum temperature at which both solid vapor phase occur in the temperature at which minimization of A occurs at nυ (eq.T) = n, above this temperature , the entropy contribution overwhelms the enternal energy contribution and hence all n atomsoccur in the solid phase * Conversely, as T decrease, then nυ (eq.T) decrease and, in the limit that T = ok, the entropy contribution to AS vanishes and minimization of A coincides with minimization o U , that is all n atoms occur in the solid phase .

  25. * Now considers that the constant temperature heat reservoir containing the constant-volume system, is of constant-volume and is adiabatically continued , then the cornbraid system , the particle containing system and heat reservoir , is one o constant U and V ; accordingly the combined system attains equilibrium at its maximum point of entropy .

  26. * if nυ > nυ (eq.T) , the evaporating process stimulating occur , this process is accompanied by transfer of heat q , from the heat reservoir to the particles containing system , thus the entropy chang of the combined system is giving by : Δ s combined system = Δs heat reservoir +Δs particles system = – q/T + ( q/T + Δsirr ) = Δsirr Thus : Δ A = – TΔsirr Hence minimization of A correspond to maximization of entropy

  27. * also if nυ > nυ (eq.T) condensation will occur ; the entropy change the combined system , in this case , is given by : Δ s combined system = Δsheat reservoir+Δsparticlessystem = q/T + (– q/T + Δsirr ) = Δsirr Thus : Δ A = – TΔsirr AT equilibrium Δsirr = 0thusΔA = 0

  28. * It should be point that at , or near , equilibrium the probability that nυ deviates bu even the smallest amount from the value nυ (eq.T) exceedingly small . this probability is small enough that the practical terms it corresponds to the thermodynamic statement that spontaneous deviation of asystem from its equilibrium state is impossible .

  29. 5.4 The Gibbs Free Energy G

  30. The Gibbs Free Energy G is defined as : G = H – TS Thus G = U + PV – TS And dG = du + pdv + vdp – Tds – sdT = Tds – pdv + pdv + vdp – Tds – sdT = – sdT = Vdp =ƒ ( p , T ) * the dependent variable in this case G and the pair of independent variable are the pressure and temperature .

  31. *since the Gibbs Free energy function is state property : Thus ΔG = G2 – G1 = ( H2 – T2S2 ) – ( H1 – T1S1 ) = ( υ 2 + P2V2 – T2S2 ) – ( υ1 + P1V1 – T1S1 ) = ( υ2 – υ1 ) + ( P2V2 – P1V1 ) – ( T2S2 – T1S1 ) • If the process is carried out under constant temperature and pressure . • Thus ΔG = Δ υ + P(V2 –V1) – T(S2 –S1) • = q + w + P(V2 –V1) – T(S2 –S1)

  32. Since w = wp –v +wَ where wp –v is the work carried out due to volume change and wَ is the sum of all forms of work other than the p – v work : ΔG = q + wp –v + wَ + P(V2 –V1) – T(S2 –S1) = q – P(V2 –V1) + wَ + P(V2 –V1) – T(S2 –S1) = q + wَ – T(S2 –S1) * in describing the work in the pervious equation , appositive sign is assigned to work done on the system and negative sign is assigned to work done by the system IF the opposite convention is used ί . e. the work done by the system is assigned appositive value , the pervious equation because : ΔG = q + wَ – T(S2 –S1)

  33. * From the second law of thermodynamics : q ≤ T(S2 –S1) wَ ≤ – ΔG therefore for reversible processes that occur at constant temperature and pressure ; the maximum amount of work , other than the p – v work is given by equation : wَ max = – ΔG * again the pervious inequality can b written as ; wَ = – (ΔG + T Δsirr )

  34. * in the case of an isothermal , isobaric process during which no work other thann the p – v work is done , that is w َ = 0 , then : ΔG + TΔsirr = 0 For infinitesimal changes , the pervious equation because : dG + Tdsirr = 0 such process can only spontaneously If The Gibbs Free Energy decrease since in any spontaneously process dsirr > 0 * As the condition for the thermodynamic equilibrium is that d sirr = 0 then with respect to isothermal and isobanic processes , equilibrium is defined by the coordination that : d G = 0

  35. * thus for a closed system under going process at constant T and P The Gibbs Free Energy can only decrease or remain constant , and hence the attainment of equilibrium in the system coincides with the system having the minimum value of G constant with the fixed value of P and T .

  36. 5.5 Function Thermodynamic equation For a closed system

  37. * The Function Thermodynamic equation are foni Thermodynamic equation of states that related thermodynamic state function as dependent variable to pair of independent thermodynamic state function for closed system of fixed composition which is in statue of equilibrium and the under going a process involving a change in the two independent variable . these equation are : du = Tds – pdv thus , U = ƒ ( S , V ) dH = Tds + vdp thus , H = ƒ ( S , P ) dA = – sdT – pdv thus , A = ƒ ( T , V ) dG = – sdT + vdp thus , G = ƒ ( T , V )

  38. * applying the partial differentiation principles , the following relationship are obtained : 1- T = ( ∂ u \ ∂ s )υ = ( ∂ H \ ∂ s )p p = ( ∂ u \ ∂ v )s = ( ∂ A \ ∂ v )T v = ( ∂ H \ ∂ p )s = ( ∂ G \ ∂ p )T T = ( ∂ A \ ∂ T )v = ( ∂ G \ ∂ T )p 2 - ( ∂ u \ ∂ s )υ ( ∂ s \ ∂ v )u ( ∂ v \ ∂ u )s = –1 ( ∂ H \ ∂ s )p ( ∂ s \ ∂ p) H ( ∂ p \ ∂ H )s = –1 ( ∂ A \ ∂ T )υ ( ∂ T \ ∂ A )v ( ∂ T \ ∂ v )A = –1 ( ∂ G \ ∂ T )p ( ∂ T \ ∂ p )G ( ∂ p \ ∂ G )T = –1

  39. 3- Maxwell's Relations: ( ∂ T \ ∂ v )s = ( ∂ p \ ∂ s )υ ( ∂ T \ ∂ P )s = ( ∂ v \ ∂ s )p ( ∂ s \ ∂ v )T = ( ∂ p \ ∂ T )υ ( ∂ s \ ∂ p )T = ( ∂ v \ ∂ T )p

  40. 5.6 examples of the use of thermodynamic relations

  41. 1- Equation of State Relating the Eternal Energy of a closed one-component system to the experimentally measurable Quantities T , P and v . Since du = Tds – pdv Thus ( ∂ u \ ∂ v )T = T ( ∂ s \ ∂ v )T – P Use the Maxwell's relation : ( ∂ s \ ∂ v )T = ( ∂ p \ ∂ T ) v ( ∂ u \ ∂ T )t = T( ∂ p \ ∂ T ) v – P It can be shown that ideal gases which obeys the equation of state pv = nRt , ( ∂ u \ ∂ v )T = 0 ; ί.e the enternal energy of the as volume .

  42. 2- Equation of State Relating the Eternal Energy of a closed one-component system to the experimentally measurable Quantities T , P and . since : dH = Tds + Vdp thus : ( ∂ H \ ∂ P )T = T ( ∂ s \ ∂ p )T + V Use the Maxwell's relation : ( ∂ s \ ∂ p )T = – ( ∂ v \ ∂ T )p yields ( ∂ H \ ∂ P )T = – T ( ∂ v \ ∂ T )p + V If again the system is a fixed quantity o an ideal gas , the equation of statue indicates that the enthalpy o an ideal as is independent of its pressure .

  43. 3. The Gribs-Helmholz Equation : Since : G = H – Ts And : ( ∂ G \ ∂ T )p = – S Thus : G = H + T ( ∂ G \ ∂ T )p Therefore for closed system of fixed composition undergoing processes at constant pressure : G = H + T ( dG / dT ) Thus : GdT = HdT + TdG Or : TdG – HdT = – HdT Then : (TdG – HdT) / T2 = – HdT / T2 Or : d ( G / T ) = – ( H / T2 ) dT Thus : [ d ( G / T ) ]/dT = – H / T2

  44. Also we can write : [ d ( ΔG / T ) ]/dT = –Δ H / T2 Similarly : [ d ( A / T ) ]/dT = – U / T2 And : [ d (ΔA / T ) ]/dT = – Δ U / T2 The preview two equation are applicable to closed system of fixed composition under going process at constant volume

  45. 4. The relationship between cp and cv The difference between cp and cv is given by : cp – cv = ( ∂ H \ ∂ T )p – ( ∂ u \ ∂ T ) υ = [ ∂ ( u + pv ) \ ∂ T ]p – ( ∂ u \ ∂ T ) υ = ( ∂ u \ ∂ T ) p + p ( ∂ v \ ∂ T ) p – ( ∂ u \ ∂ T ) υ But : U = ƒ ( υ , T ) Thus : du = ( ∂ u \ ∂ v )T dv + ( ∂ u \ ∂ T )υ dT There : ( ∂ u \ ∂ T ) p = ( ∂ u \ ∂ v )T ( ∂ v \ ∂ T ) p + ( ∂ u \ ∂ T ) υ Hence : cp – cv = ( ∂ u \ ∂ T ) υ + ( ∂ u \ ∂ v )T ( ∂ v \ ∂ T ) p + p ( ∂ v \ ∂ T ) p – ( ∂ u \ ∂ T ) υ = ( ∂ v \ ∂ T ) p [ p + ( ∂ u \ ∂ v )T ] = ( ∂ v \ ∂ T ) p [( ∂ u \ ∂ v )T + ( ∂ u \ ∂ v )T ] = T( ∂ v \ ∂ T ) p ( ∂ s \ ∂ v )T

  46. The T Maxwell's relation: ( ∂ s \ ∂ v )T = ( ∂ p \ ∂ T ) υ leads to : Cp – Cυ = ( ∂ v \ ∂ T ) p ( ∂ p \ ∂ T ) υ Since : ( ∂ p \ ∂ T )υ ( ∂ T \ ∂ v ) p ( ∂ p \ ∂ T ) υ = – 1 Then : ( ∂ p \ ∂ T ) υ = – ( ∂ v \ ∂ T ) p ( ∂ p \ ∂ v )T Thus : cp – cv = – T ( ∂ v \ ∂ T ) p ( ∂ v \ ∂ T ) p ( ∂ p \ ∂ v )T Since : α = 1/v ( ∂ v \ ∂ T ) p And : β = – 1/v ( ∂ v \ ∂ p ) T Thus : cp – cv = ( V T α2 ) / β ί.e. cp / cv > 1 Where α is the coefficient of thermal expensive at constant pressure , and β is the compreslility factor at constant temperature .

  47. 5.7 The Gibbs Free Energy and the composition of the system

  48. * thus for the discussion has been restricted to closed system of fixed composition ; as an example , the system containing fixed number of moles of component , etc . in such cases the system has to independent variable which when fixed uniquely fix the state of the problem . * How ever , if the composition of the system varies during the process for example , if the system contains the gaseous co, co2 and O2 , then at constant T and P minimization of G would occur when the equilibrium o the reaction Co + 1/2 O2 is established .

  49. * Thus , G is a function of T, P and the number of moles of all species present in the system i,e . G = G ( T , P , and n1, n2 , n3 , …. , nj , …. nr ) When n1, n2 , …. Are the number of moles o species 1, 2 , ….. * Differentiation of the pervious equation yields : dG = (∂G /∂T)p dT + (∂G /∂p)T dp + ∑ ί =1 ί=r (∂G /∂nί) t.p all nj=1.2. dni = – SdT + Vdp + ∑ ί =1 ί=k ί (∂G /∂nί) t.p all nj=1.2. dni Where ∑ ί =1 ί=r (∂G /∂nί) t.p all nj=1.2. dni Is the sum of r terms , one of the ί species .

  50. 5.8 The chemical potential

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