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Transcript
ME 083 Thermodynamic Aside: Gibbs Free Energy Professor David M. Stepp Mechanical Engineering and Materials Science 189 Hudson Annex [email protected] 549-4329 or 660-5325 24 February 2003 From Last Time…. Gibbs Free Energy: G = H – T*S Recall our Arrhenius relationship for the equilibrium number of vacancies (defects): Nv Ntot e Q v kB T Frenkel defects (vacancy plus interstitial defect) THERMODYNAMIC ASIDE: GIBBS FREE ENERGY Energy of a System: A State Function (Depends only on the current condition of the system) Three Categories of Energy: Kinetic (motion), Potential (position), and INTERNAL (molecular motions) State Function Internal Energy: – How can we change a material’s (or system’s) internal energy? ∆U = Q + W + W’ State Function Internal Energy (U): ∆U = Q + W + W’ (The First Law of Thermodynamics) Q: Heat flow (into system) W: P-V work (on system) W’: Other work (on system) Energy Conservation Processes in nature have a natural direction of change SPONTANEOUS DOES NOT OCCUR State Function Entropy (S) ∆SP ≥ 0 The Entropy of a system may increase or decrease during a process. The Entropy of the universe, taken as a system plus surroundings, can only increase. (The Second Law of Thermodynamics) “Entropy is Time’s Arrow” Note: The laws of thermodynamics are empirical, based on considerable experimental evidence. One Can Show That: For Reversible Processes QREV = TdS – QREV denotes the maximum heat absorbed – Note the cyclic path integral (reversibility) With this, the combined Notation for the First and Second Law can be expressed: dU =TdS + dWREV + dW’ BUT dWREV = F*dx = F*(A/A)*dx = F/A*(-dV) = -PdV dU = TdS – PdV + dW’ Combined statement of First and Second Laws The Third Law of Thermodynamics: The Entropy of all substances is the same at 0 K Both Entropy and Temperature Have Absolute Values (Both have an empirically observed zero point) State Function Enthalpy (H) H = U + P*V So: dH = dU + PdV + VdP Consider the special case where dP = 0 and dW’ = 0: dHP = TdSP = dQREV,P State Function Gibbs Free Energy G = H – TS = U + PV - TS dG = dU + PdV + VdP – TdS – SdT = (TdS – PdV + dW’) + PdV + VdP – TdS – SdT = VdP – SdT + dW’ Special Case of dT = 0 and dP = 0: dGT,P = dW’T,P Or, if W’ = 0: dGP = -SdT At constant Temperature and Pressure, the (change in) Gibbs Free Energy reflects all non-mechanical work done on the system. Back to Crystals… Consider Frenkel Defects (vacancy + interstitial) The Free Energy of the Crystal can be written as – the Free Energy of the perfect crystal (G0) – plus the free energy change necessary to create n interstitials and vacancies (n*∆g) – minus the entropy increase that derives from the different possible ways in which defects can be arranged (∆SC) ∆G = G-G0 = n*∆g - T∆SC n = the number of defects ∆g = ∆h - T∆S (the energy to create defects) The Configurational Entropy, ∆SC, is proportional to the number of ways in which the defects can be arranged (W). ∆SC = kB* ln(W) (The Boltzmann Equation) Note: this constitutes a connection between atomistic and phenomenological descriptions (thru statistics). Perfect Crystal: N atoms which are indistinguishable can only be placed in one way on the N lattice sites: ∆SC = k*ln(N!/N!) = k*ln(1) = 0 Crystal with Vacancies and Interstitials With N normal lattice sites (and equal interstitial sites), ni (number of interstitial atoms) can be arranged in: N! (N ni )!ni! distinct ways Similarly, the vacant sites can be arranged in: N! distinct ways (N nv )! nv ! Recalling that the equilibrium number of vacancies obeys an Arrhenius relationship: n e N g 2kT e s 2k e h 2kT Remember: ∆g = ∆h - T∆s n e Therefore: N h 2kT Assuming configurational entropy in creating defects is negligible Now compare this with our equation for Frenkel Defects: NV NI e NT NT Q sV / I 2kT The Configurational Entropy of these noninteracting defects is: N! N! SC k ln N n ! n ! N n ! n ! i i v v Probability theory for statistically independent events: PAB = PA * PB Stirling’s Approximation for Large Numbers: ln(x!) ≈ x * ln(x) - x Recalling that ni = nv = n (for Frenkel defects): N N n SC 2 k N ln n ln Nn n At equilibrium, the free energy is a minimum with respect to the number of defects Note: Entropy (per defect) is a maximum at this point Recall: ∆G = G – G0 = n * ∆g – T∆SC G = G0 + n * (∆h – T∆s) - T∆SC Next Time… The Statistical Interpretation of Entropy ∆SC = k * ln(W) Configurational Entropy is proportional to the number of ways in which defects can be arranged