ME 083 Thermodynamic Aside: Gibbs Free Energy Professor David M. Stepp Mechanical Engineering and Materials Science 189 Hudson Annex or February 2003

From Last Time…. Gibbs Free Energy: G = H – T*S Recall our Arrhenius relationship for the equilibrium number of vacancies (defects): Frenkel defects (vacancy plus interstitial defect)

Energy of a System: Three Categories of Energy: State Function Internal Energy: A State Function A State Function (Depends only on the current condition of the system) Kinetic (motion), Potential (position), and INTERNAL (molecular motions) –How can we change a material’s (or system’s) internal energy? ∆U = Q + W + W’ THERMODYNAMIC ASIDE: GIBBS FREE ENERGY

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 DOES NOT SPONTANEOUSOCCUR

State Function Entropy (S) ∆S P ≥ 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 Q REV = TdS –Q REV 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 + dW REV + dW’ BUT dW REV = 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: dH P = TdS P = dQ REV,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 = (TdS – PdV + dW’) + PdV + VdP – TdS – SdT = VdP – SdT + dW’ = VdP – SdT + dW’ Special Case of dT = 0 and dP = 0: dG T,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 (G 0 ) –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 (∆S C ) ∆G = G-G 0 = n*∆g - T∆S C n = the number of defects ∆g = ∆h - T∆S (the energy to create defects)

The Configurational Entropy, ∆S C, is proportional to the number of ways in which the defects can be arranged (W). ∆S C = k B * 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: ∆S C = k*ln(N!/N!) = k*ln(1) = 0

Crystal with Vacancies and Interstitials With N normal lattice sites (and equal interstitial sites), n i (number of interstitial atoms) can be arranged in: distinct ways distinct ways Similarly, the vacant sites can be arranged in: distinct ways distinct ways

Recalling that the equilibrium number of vacancies obeys an Arrhenius relationship: Remember: ∆g = ∆h - T∆s Remember: ∆g = ∆h - T∆s Therefore: Assuming configurational entropy in creating defects is negligible Now compare this with our equation for Frenkel Defects:

The Configurational Entropy of these non- interacting defects is: Probability theory for statistically independent events: P AB = P A * P B Stirling’s Approximation for Large Numbers: ln(x!) ≈ x * ln(x) - x Recalling that n i = n v = n (for Frenkel defects):

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 – G 0 = n * ∆g – T∆S C G = G 0 + n * (∆h – T∆s) - T∆S C

Next Time… The Statistical Interpretation of Entropy ∆S C = k * ln(W) Configurational Entropy is proportional to the number of ways in which defects can be arranged