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이 병 주 포항공과대학교 신소재공학과 calphad@postech.ac.kr

이 병 주 포항공과대학교 신소재공학과 calphad@postech.ac.kr. Thermodynamics. Gibbs energy, Enthalpy. Gibbs Energy for a Unary System - from dG = – SdT + VdP. Gibbs Energy as a function of T and P. @ constant P. @ constant T. Gibbs Energy for a Unary System - from G = H – ST.

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이 병 주 포항공과대학교 신소재공학과 calphad@postech.ac.kr

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  1. 이 병 주 포항공과대학교 신소재공학과 calphad@postech.ac.kr Thermodynamics Gibbs energy, Enthalpy ...

  2. Gibbs Energy for a Unary System - from dG = –SdT + VdP Gibbs Energy as a function of T and P @ constant P @ constant T

  3. Gibbs Energy for a Unary System - from G = H – ST Gibbs Energy as a function of T and P @ constant P @ constant T

  4. Gibbs Energy for a Unary System - Temperature Dependency 서로 다른 출발점에서 유도된 위의 두 식은 같은 식인가? Empirical Representation of Heat Capacities 를 이용하여 위의 두 식이 동일한 것임을 증명하라.

  5. Gibbs Energy for a Unary System - Effect of Pressure Molar volume of Fe = 7.1 cm3 Expansivity = 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 = 10 cm3 Expansivity = 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

  6. Gibbs Energy for a Unary System - absolute value available ? • 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

  7. Temperature and Pressure Dependence of Molar Volume of Fe

  8. Temperature dependence of Specific Heat of Fe

  9. Temperature dependence of molar Enthalpy of Fe

  10. Gibbs Energy change of a reaction • Neumann-Kopp rule: heat capacity of a solid compound is equal to the • sum of the heat capacities of its constituent elements. • Richards’ rule: cal/degree • Trouton’s rule: cal/degree • In reactions in which a gas reacts with a condensed phase to produce • a condensed phase, the entropy change is that corresponding to the • disappearance of the gas.

  11. First Approximations Richards’ rule: Trouton’s rule: cal/degree cal/degree

  12. Numerical Example • A quantity of supercooled liquid Tin is adiabatically contained at 495 K. • Calculate the fraction of the Tin which spontaneously freezes. Given J at Tm = 505 K 505 K 495 K x moles of solid (1-x) moles of liquid 1 mole of liquid

  13. Thermodynamics 교과과정 이해에 대한 Check Point - I

  14. First Law of thermodynamics • Microscopic vs. Macroscopic View Point의 이해 • State function vs. Process variable, 기타 용어의 이해 • 열역학 1법칙의 탄생 과정, 1법칙 중요성의 이해 • Special processes의 중요성 이해, 응용력 • Constant-Volume Process: ΔU ＝ qv • Constant-Pressure Process: ΔH ＝ qp • 3. Reversible Adiabatic Process: q ＝ 0 • 4. Reversible Isothermal Process: ΔU ＝ ΔH ＝ 0

  15. Second Law of thermodynamics • (Mechanical, Thermal, Chemical) Irreversibility • Irreversibility vs. Creation of Irreversible Entropy • Maximum Entropy, Minimum Internal Energy as a Criterion of Equilibrium

  16. Statistical Thermodynamics • Statistical Thermodynamics 의 개본 개념 이해 • Ideal Gas에 대한 Statistical Thermodynamics의 응용력 • 통계열역학 개념을 통한 엔트로피의 이해 • Heat capacity 계산에의 응용 • Heat capacity at low temperature

  17. Criterion of Thermodynamic Equilibrium, Thermodynamic Relations • Helmholtz Free Energy, Gibbs Free Energy의 탄생 배경 • Free energy minimum과 equilibrium 간의 상관관계 • Chemical Potential의 정의, Gibbs energy와의 관계 • Chemical Work으로서의 term • Thermodynamic Relation 들의 응용력, 중요성 이해

  18. Application of Criterion • 1기압하Pb의melting point 는600K이다. • 1기압하590K로과냉된액상Pb가응고하는것은 • 자발적인반응이라는것을 • (1) maximum-entropy criterion과 • (2) minimum-Gibbs-Energy criterion을이용하여보이시오. 2. 1번문제에서의Pb가단열된용기에보관되어있었다면 용기내부는결국어떠한(평형)상태가될것인지예측하시오.

  19. Numerical Example • A quantity of supercooled liquid Tin is adiabatically contained at 495 K. • Calculate the fraction of the Tin which spontaneously freezes. Given J at Tm = 505 K 505 K 495 K x moles of solid (1-x) moles of liquid 1 mole of liquid

  20. Example - Phase Transformation of Graphite to Diamond • Calculate graphite→diamond transformation pressure at 298 K, given • H298,gra – H298,dia = -1900 J • S298,gra = 5.74 J/K • S298,dia = 2.37 J/K • density of graphite at 298 K = 2.22 g/cm3 • density of diamond at 298 K = 3.515 g/cm3

  21. Gibbs Energy for a Unary System • 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

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