Thermochemical Properties - PDF Free Download

Thermochemical Properties Materials respond to Thermal stimuli (temperature) Chemical stimuli (composition or environment) Electromagnetic stimuli (electric or magnetic fields) Mechanical stimuli (mechanical forces) Consider the first two together Response to thermal or chemical stimuli defines thermochemical properties

Thermochemical Properties Essential features: Thermodynamics: what material wants to do (forces) Kinetics: what it can do, and how quickly Study Thermodynamics Properties Equilibrium phase diagrams Kinetics Continuous: heat and mass diffusion Structural phase transitions Environmental interactions Wetting and catalysis Corrosion

Thermodynamics The conditions of equilibrium and stability Equilibrium no desire for change Deviation from equilibrium driving force for change Beyond limits of stability must change Internal equilibrium T, P, {µ} are constant Deviation drives heat and mass diffusion Global equilibrium Thermodynamic potential is minimum Deviation drives structural phase transformations

First Law of Thermodynamics Defines internal energy, E Energy is conserved de = dw + dq dw = work done (chemical + mechanical +electromagnetic) dq = heat transferred (thermal work) Energy transferred to one material is taken from another

Second Law of Thermodynamics Defines entropy : S Entropy is associated with Evolutionary time (most fundamental) Heat Randomness (information) When a system is isolated, S can only increase Any system is isolated when its surroundings are included

A Simple Adiabatic System Simple system in thermally insulated container Can do mechanical work Reversibly, with a frictionless piston Irreversibly, with a paddle wheel But no thermal interaction because of insulation This is called an adiabatic system An isolated system is one example of an adiabatic system

Change of State in an Adiabatic System Moving piston generates E-V curve Turning paddle wheel Raises E at constant V Changes the reversible E-V curve E Irreversible change from paddle wheel Reversible curve from moving piston Paddle work is irreversible System moves to new E-V curve System can never return V

The Measure of Time in an Adiabatic System E Future The E-V curve divides states into Past (below current curve) Present (on current curve) Future (above current curve) E Past V Irreversible change from paddle wheel Present Reversible curve from moving piston States (E,V) below are the past System cannot do work on paddle wheel These states are unattainable States (E,V) above are in the future Can be reached by paddle + piston But system can never return The current (E,V) curve is the present System can sample these states at will V

Entropy = Time (State of Evolution) E curves of constant entropy S States (E,V) on a reversible curve have a common property: call it entropy (S) Assign a numerical value of S to each curve such that S is continuous S Then S = S(E,V) measures the evolutionary time of the state (E,V) S can only increase S divides past (S <S) from future (S >S) V

Entropy and Heat S = S(E,V ) ds(e,v ) = S E V de + S V E dv We can choose the measure of S (the numbers we assign to S) so that Then S E S V V E = 1 T = P T T = absolute temperature P = pressure ds = 1 T de + P T dv

Entropy and Heat ds = 1 T de + P T dv de = TdS PdV If the change of state is reversible (no friction, etc.) - dw = PdV dq = de dw = TdS (reversible thermal work ) When heat is added reversibly - ds = dq T Can measure S from heat added in a change of state

Statistical Entropy E,V Isolated system with given E,V Let Ω(E,V) be the number of distinguishable states for given (E,V) Distinguishable ways of arranging atoms Distinguishable ways of distributing energy or momentum among atoms Can show that a good measure of the entropy is S(E,V ) = k ln[ Ω(E,V )] Can calculate entropy from a knowledge of the system

The Fundamental Equation The entropy function ( fundamental equation ) S = S(E,V,{N}) ds = 1 T de + P T dv The energy function (alternate form of the fundamental equation ) The quantities S, V, E, {N} are fixed by the function, the intensities (forces: T, P and {µ}) are given by its derivatives n k=1 E = E(S,V,{N}) de = TdS PdV + n µ k T k=1 dn k µ k dn k

The Fundamental Equation E = E(S,V,{N}) = TS PV + µ k N k de = TdS PdV + n k=1 k µ k dn k ( Integrated form ) Thermodynamic forces from derivatives of FE T, P, {µ} These can be controlled independently Thermodynamic properties from second derivatives Specific heat, compressibility, CTE These are material properties

The Conditions of Equilibrium metastable X S unstable stable S stable unstable metastable X The entropy of an isolated system can only increase A stable state must have a maximal value of the entropy Conditions of equilibrium: Local: S is not increased by infinitesimal changes Global: maxiumum with respect to all possible changes Conditions of stability S is not increased by small, finite changes Stable vs. metastable equilibrium

The Conditions of Equilibrium and Stability (Gibbs) metastable S unstable stable Necessary conditions for equilibrium (δs) E,V,{N} 0 (δe) S,V,{N} 0 Necessary conditions for stability (δ 2 S) E,V,{N} 0 (δ 2 E) S,V,{N} 0 Sufficient conditions for stable equilibrium X (ΔS) E,V,{N} 0 (ΔE) S,V,{N} 0

The Necessary Conditions of Equilibrium metastable S unstable stable X For an isolated system in equilibrium, there must be no small change of state that increases its entropy: δs(e,v,{n}) 0 (for every possible change) Any part of it is in equilibrium with respect to internal changes That is, any subvolume is in equilibrium when regarded as an isolated system The inequality is always a necessary condition for equilibrium

Internal Equilibrium: Thermal The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. 1 2 Let two small volumes exchange energy at constant total energy,volume and composition: Equilibrium requires: δs 0 = δe 1 + δe 2 1 = δe 1 1 T 1 T 2 T 1 T 2 T 1 = T 2 (temperature is constant)

Internal Equilibrium: Mechanical The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. 1 2 Two small volumes in thermal equilibrium exchange volume at constant composition: Equilibrium requires: δs 0 = P 1 T δv 1 + P 2 T δv 2 = δv 1 [ T P 1 P 2 ] P 1 = P 2 (pressure is constant)

Internal Equilibrium: Chemical The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file 1 2 Two small volumes in thermal and mechanical equilibrium exchange one component: Equilibrium requires: δs 0 = µ 1 T δn 1 µ 2 T δn 2 = δn 1 [ T µ 2 µ 1 ] For every component µ 1 = µ 2 (chemical potential is constant)

Necessary Conditions for Internal Equilibrium Thermal equilibrium: Temperature is constant (T) Mechanical equilibrium (for a fluid): Pressure is constant (P) Chemical equilibrium: Chemical potential of each component is constant (µ k ) Disclaimer: the conditions of mechanical and chemical equilibrium change slightly in electrical or strong gravitational fields, but we won t care much

Equilibrium When the System is not Isolated: Thermodynamic Potentials Consider three cases: Material interacts thermally with environment: Equilibrium requires that the Helmholtz free energy be a minimum F = E - TS Material interacts thermally and mechanically with environment: Equilibrium requires that the Gibbs free energy be minimum G = E - TS + PV Material interacts thermally and chemically with environment Equilibrium requires that the work function be minimum Ω = E TS µ k N k k

Thermal Interaction: The Helmholtz Free Energy System (S) The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been Reservoir corrupted. Restart your computer, and then (R) open the file again. If the red x still appears, you may have to delete the image and then insert it again. T {N},V de T, V, {N} controlled Ex: solid in a furnace A system, S, is in contact with a reservoir, R, that fixes temperature, T The walls of S are rigid and impermeable The volume and content of S are fixed: {N}, V are constants S and R can exchange energy (de) in the form of heat R is so much bigger than S that de does not change its temperature S+R is an isolated system Its entropy must be maximum at equilibrium: for any change of state δs T = δs + δs R 0

Thermal Interaction: The Helmholtz Free Energy System (S) The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been Reservoir corrupted. Restart your computer, and then (R) open the file again. If the red x still appears, you may have to delete the image and then insert it again. T {N},V de T, V, {N} controlled δs T ( 0) = δs + δs R δs R = δe R T TδS δe = δ(ts E) 0 = δe T Define the Helmholtz free energy, F = E-TS The condition of equilibrium is that F be minimum for given T,V,{N}: δf(t,v,{n}) 0

Thermomechanical Interaction: The Gibbs Free Energy dv {N} de P, T T, P, {N} controlled Let a system with fixed {N} interact with a reservoir that fixes T and P: δs T ( 0) = δs + δs R δs R = δe R T + P T δv R = 1 T TδS δe PδV = δ(ts E PV ) 0 Define the Gibbs free energy, G = E-TS+PV The condition of equilibrium is that G be minimum for given T,P,{N}: δg(t,p,{n}) 0 ( δe + PδV )

Thermal Interaction: The Work Function T, {µ} V de dn T, V, {µ} controlled An open system with given V interacts with a reservoir that fixes T and {µ}: δs T ( 0) = δs + δs R δs R = δe R T 1 n µ k δn R k = 1 n T k=1 T δe µ kδn k k=1 TδS δe + µ k δn k = δ(ts E + µ k N k ) 0 Define the Work function, Ω = E TS µ k N k = PV 0 The condition of equilibrium is that Ω be minimum for given T,V,{µ}: k k δω(t,v,{µ}) 0 k

Summary: Thermodynamic Potentials and Conditions of Equilibrium Isolated system: E, V, {N} controlled Entropy, S(E,V{N}) = maximum Thermal contact: T, V, {N} controlled Helmholtz free energy, F(T,V,{N}) = E-TS = minimum Thermomechanical contact: T, P, {N} controlled Gibbs free energy, G(T,P,{N}) = E-TS+PV = minimum Open contact: T, V, {µ} controlled Work function, Ω(T,V,{µ}) = - PV = minimum Or, equivalently. P = maximum