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Basic Review of Thermodynami cs Byeong-Joo Lee POSTECH - MSE [email protected] Byeong-Joo Lee www.postech.ac.kr/~calphad Objective Understanding and Utilizing Thermodynamic Laws State function Thermodynamic Laws Statistical thermodynamics Gibbs energy Extension of Thermodynamics Multi-Phase System Multi-Component System Partial Molar Quantities Utilization of Thermodynamics Phase Diagrams Defect Thermodynamics Byeong-Joo Lee www.postech.ac.kr/~calphad 1-1. Understanding and Utilizing Thermodynamic Laws Why define state function ? Why Thermodynamic Laws are important ? Statistical vs. classical thermodynamics Why use Gibbs energy ? Byeong-Joo Lee www.postech.ac.kr/~calphad Fundamentals State function vs. Process variable First Law of Thermodynamics “The change of a body inside an adiabatic enclosure from a given initial state to a given final state involves the same amount of work by whatever means the process is carried out” It was necessary to define some function which depends only on the internal state of a system – Internal Energy. For adiabatic process: UB – UA = -w Generally: UB – UA = q - w dU = δq – δw dU 0 : as a state function Special processes 1.Constant-Volume Process: ΔU ＝ qv 2.Constant-Pressure Process: ΔH ＝ qp 3.Reversible Adiabatic Process: q＝0 4.Reversible Isothermal Process: ΔU ＝ ΔH ＝ 0 Byeong-Joo Lee www.postech.ac.kr/~calphad Second Law of thermodynamics - Reversible vs. Irreversible △S = measurable quantity + un-measurable quantity = q/T + △Sirr = qrev/T Byeong-Joo Lee www.postech.ac.kr/~calphad Second Law of thermodynamics - Maximum Work UB U A q w q dU system w dS system dS system q T dS irr dU system w T dS irr w TdS system dU system TdS irr w wmax TS system U system Byeong-Joo Lee www.postech.ac.kr/~calphad Second Law of thermodynamics - Entropy as a Criterion of Equilibrium ※ for an isolated system of constant U and constant V, (adiabatically contained system of constant volume) equilibrium is attained when the entropy of the system is maximum. ※ for a closed system which does no work other than work of volume expansion, dU = T dS – P dV (valid for reversible process) U is thus the natural choice of dependent variable for S and V as the independent variables. ※ for a system of constant entropy and volume, equilibrium is attained when the internal energy is minimized. w TdS system dU system TdS irr PdV TdS system dU system TdS irr 0 dU system TdS irr Byeong-Joo Lee www.postech.ac.kr/~calphad New Thermodynamic Functions – Reason for the necessity ※ Further development of Classical Thermodynamics results from the fact that S and V are an inconvenient pair of independent variables. + need to include composition variables in any equation of state and in any criterion of equilibrium + need to deal with non P-V work (e.g., electric work performed by a galvanic cell) dU = TdS – PdV S, V are not easy to control. Need to find new state functions which are easy to control and can be used to estimate equilibrium → F, G Byeong-Joo Lee www.postech.ac.kr/~calphad Helmholtz Free Energy - Work Function, F ≡ U – ST dF ≡ dU – TdS – SdT For a reversible process dF = [TdS – PdV – δw’] – TdS – SdT = – SdT – PdV – δw’ dFT = – PdV – δw’ = – δwT.Total ▷ Maximum work that the system can do by changing its state at Constant T, V = -ΔF. For a irreversible isothermal process △FT = [q – w] – T△S T△S = q + T△Sirr w = P△V + w’ = w’ For constant V △FT,V + w’ + T△Sirr = 0 ▷ If cannot do a maximum work, it’s due to the creation of Δsirr. Under a Constant T, V, an equilibrium is obtained when the system has maximum (w’+ Δsirr) or minimum F. Byeong-Joo Lee www.postech.ac.kr/~calphad Helmholtz Free Energy - Example of Application Equilibrium between condensed phase and gas phase. Use Helmholtz Free Energy Criterion to determine equilibrium amount of gaseous phase at a given temperature, and how it changes with changing temperature Byeong-Joo Lee www.postech.ac.kr/~calphad Gibbs Free Energy - Gibbs Function, G ≡ U + PV – ST dG ≡ dU + PdV + VdP – TdS – SdT For a reversible process dG = [TdS – PdV – δw’] + PdV + VdP – TdS – SdT = – SdT + VdP – δw’ dGT,P = – δw’ ▷ Maximum work that the system can do by changing its state at Constant T, P = -ΔG. For a irreversible isothermal process △GT,P = [q – w] + P△V – T△S T△S = q + T△Sirr w = P△V + w’ △GT,P + w’ + T△Sirr = 0 ▷ If cannot do a maximum work, it’s due to the creation of Δsirr. Under a Constant T, P, an equilibrium is obtained when the system has maximum (w’+ Δsirr) or minimum G. Byeong-Joo Lee www.postech.ac.kr/~calphad Thermodynamic Relations - For a closed system dU = TdS – PdV dH = TdS + VdP dF = – SdT – PdV dG = – SdT + VdP Byeong-Joo Lee www.postech.ac.kr/~calphad Thermodynamic Relations - For a multicomponent system ▷ G G (T , P, ni , n j , n k , ) G G G G dG dT dP dni n T P ,ni ,n j , P T ,ni ,n j , ni T , P ,n j ,nk , j ▷ Chemical Potential dn j T , P ,ni ,nk , G i n i T , P ,n j ,nk , n dG SdT VdP i dni w PdV i dni i dni 1 is the chemical work done by the system Byeong-Joo Lee www.postech.ac.kr/~calphad Gibbs Energy ▷ Effect of Temperature on G ▷ Effect of Pressure on G Byeong-Joo Lee www.postech.ac.kr/~calphad Gibbs Energy for a Unary System - from dG = –SdT + VdP Gibbs Energy as a function of T and P dG S dT V dP T P To Po G(T , P) G(To , Po ) S dT V dP @ constant P C P (T ) dT T T S (T ) S o To T G(T ) G(To ) To T C (T ) P S dT o To dT T P @ constant T G( P) G( Po ) V dP Po Byeong-Joo Lee www.postech.ac.kr/~calphad Gibbs Energy for a Unary System - from G = H - ST Gibbs Energy as a function of T and P G H ST G(T , P) ( H (T , P) H (To ,1)) H o (S (T , P) S (To ,1) So )T H H o (S So )T H o SoT H TS @ constant P T S (T ) S o T H (T ) H o C P (T ) dT To To T T To To G(T ) H o S oT C P (T ) dT T C P (T ) dT T C P (T ) dT T @ constant T H P Po P H dP P o P T P P S V V dP P V (1 T ) dP T V dP Po T o P T T T P S P V P S dP dP V dP Po P P o o P T T T Byeong-Joo Lee www.postech.ac.kr/~calphad Gibbs Energy for a Unary System - Temperature Dependency T G(T ) G(To ) To T C (T ) P S dT dT o To T T T To To G(T ) H o S oT C P (T ) dT T C P (T ) dT T Are both of above expressions the same? Use Empirical Representation of Heat Capacities cP a bT cT 2 to verify that the above expressions are exactly the same. Byeong-Joo Lee www.postech.ac.kr/~calphad Gibbs Energy for a Unary System - Effect of Pressure Molar volume of Fe Expansivity = 7.1 cm3 = 0.3 × 10-4 K-1 ∆H(1→100atm,298) = 17 cal ※ The same enthalpy increase is obtained by heating from 298 to 301 K at 1atm Molar volume of Al Expansivity = 10 cm3 = 0.69 × 10-4 K-1 ∆H(1→100atm,298) = 23.7 cal ※ The same enthalpy increase is obtained by heating from 298 to 302 K at 1atm ∆S(1→100atm,298) = -0.00052 e.u. for Fe -0.00167 e.u. for Al ※ The same entropy decrease is obtained by lowering the temperature from 298 by 0.27 and 0.09 K at 1 atm. ※ The molar enthalpies and entropies of condensed phases are relatively insensitive to pressure change Byeong-Joo Lee www.postech.ac.kr/~calphad Gibbs Energy for a Unary System T T To To G(T ) H o S oT C P (T ) dT T P C P (T ) dT V dP Po T • V(T,P) based on expansivity and compressibility • Cp(T) • S298: by integrating Cp/T from 0 to 298 K and using 3rd law of thermodynamics (the entropy of any homogeneous substance in complete internal equilibrium may be taken as zero at 0 K) • H298: from first principles calculations, but generally unknown ※ H298 becomes a reference value for GT ※ Introduction of Standard State Byeong-Joo Lee www.postech.ac.kr/~calphad Statistical Thermodynamics Basic Concept of Statistical Thermodynamics Application of Statistical Thermodynamics to Ideal Gas Understanding Entropy through the Concept of the Statistical Thermodynamics Heat capacity Heat capacity at low temperature Byeong-Joo Lee www.postech.ac.kr/~calphad Numerical Examples 1. The melting point of Pb is 600K at 1 atm. Show that solidification of supercooled liquid Pb at 590K and 1 atm is a natural process, based on (1) maximum-entropy criterion (2) minimum-Gibbs-Energy criterion. H melting 4810 J / mole C p (l ) 32.4 3.1 103 T J / mol K C p ( s ) 9.75 103 T J / mol K 2. Estimate what will eventually happen (equilibrium state) if the Pb in problem #1 is contained in an adiabatic container. 3. A quantity of supercooled liquid Tin is adiabatically contained at 495 K. Calculate the fraction of the Tin which spontaneously freezes. Given H mSn 7070 J at Tm = 505 K C p,Sn(l ) 34.7 9.2 103T J / K C p,Sn( s ) 18.5 26 103T J / K Byeong-Joo Lee www.postech.ac.kr/~calphad