Gibbs Free Energy, Enthalpy, … Thermodynamics Gibbs Free Energy, Enthalpy, … Byeong-Joo Lee POSTECH - MSE calphad@postech.ac.kr

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 Gibbs Energy as a function of T and P @ constant P @ constant T

서로 다른 출발점에서 유도된 위의 두 식은 같은 식인가? 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.

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

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

Temperature and Pressure Dependence of Molar Volume of Fe

Temperature dependence of Specific Heat of Fe

Temperature dependence of molar Enthalpy of Fe

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.

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

Calculate the fraction of the Tin which spontaneously freezes. Given 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 1 mole of liquid x moles of solid (1-x) moles of liquid

교과과정 이해에 대한 Check Point - I Thermodynamics 교과과정 이해에 대한 Check Point - I

Microscopic vs. Macroscopic View Point의 이해 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

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

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

Helmholtz Free Energy, Gibbs Free Energy의 탄생 배경 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

Application of Criterion 1기압 하 Pb의 melting point 는 600K이다. 1기압 하 590K로 과냉된 액상 Pb가 응고하는 것은 자발적인 반응이라는 것을 (1) maximum-entropy criterion과 (2) minimum-Gibbs-Energy criterion을 이용하여 보이시오. 2. 1번 문제에서의 Pb가 단열된 용기에 보관되어 있었다면 용기 내부는 결국 어떠한 (평형)상태가 될 것인지 예측하시오.

Calculate the fraction of the Tin which spontaneously freezes. Given 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 1 mole of liquid x moles of solid (1-x) moles of liquid

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

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