Byeong-Joo Lee 이 병 주 포항공과대학교 신소재공학과

Byeong-Joo Lee Gibbs Energy for a Unary System - from dG = SdT + VdP Gibbs Energy for a Unary System - from dG = – SdT + VdP Gibbs Energy as a function of T and constant constant T

Byeong-Joo Lee Gibbs Energy for a Unary System - from G = H ST Gibbs Energy for a Unary System - from G = H – ST Gibbs Energy as a function of T and constant constant T

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

Byeong-Joo Lee Gibbs Energy for a Unary System - Effect of Pressure Molar volume of Fe = 7.1 cm3 Expansivity = 0.3 × 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 cm 3 Expansivity = 0.69 × 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) = e.u. for Fe 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 Gibbs Energy for a Unary System - absolute value available ? V(T,P) based on expansivity and compressibility C p (T) S 298 : by integrating C p /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) H 298 : from first principles calculations, but generally unknown ※ H 298 becomes a reference value for G T ※ Introduction of Standard State

Byeong-Joo Lee Temperature and Pressure Dependence of Molar Volume of Fe

Byeong-Joo Lee Temperature dependence of Specific Heat of Fe

Byeong-Joo Lee Temperature dependence of molar Enthalpy of Fe

Byeong-Joo Lee

Byeong-Joo Lee 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 a condensed phase, the entropy change is that corresponding to the disappearance of the gas. disappearance of the gas.

Byeong-Joo Lee First Approximations Richards’ rule: Trouton’s rule: cal/degree cal/degree

Byeong-Joo Lee 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 T m = 505 K 505 K 495 K 1 mole of liquid x moles of solid (1-x) moles of liquid

Byeong-Joo Lee

Byeong-Joo Lee First Law of thermodynamics  Microscopic vs. Macroscopic View Point 의 이해  State function vs. Process variable, 기타 용어의 이해  열역학 1 법칙의 탄생 과정, 1 법칙 중요성의 이해  Special processes 의 중요성 이해, 응용력 1.Constant-Volume Process: ΔU ＝ q v 2.Constant-Pressure Process: ΔH ＝ q p 3. Reversible Adiabatic Process: q ＝ 0 4. Reversible Isothermal Process: ΔU ＝ ΔH ＝ 0

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

Byeong-Joo Lee Statistical Thermodynamics  Statistical Thermodynamics 의 개본 개념 이해  Ideal Gas 에 대한 Statistical Thermodynamics 의 응용력  통계열역학 개념을 통한 엔트로피의 이해  Heat capacity 계산에의 응용  Heat capacity at low temperature

Byeong-Joo Lee Criterion of Thermodynamic Equilibrium, Thermodynamic Relations  Helmholtz Free Energy, Gibbs Free Energy 의 탄생 배경  Free energy minimum 과 equilibrium 간의 상관관계  Chemical Potential 의 정의, Gibbs energy 와의 관계  Chemical Work 으로서의 term  Thermodynamic Relation 들의 응용력, 중요성 이해

Byeong-Joo Lee Application of Criterion 1.1 기압 하 Pb 의 melting point 는 600K 이다. 1 기압 하 590K 로 과냉된 액상 Pb 가 응고하는 것은 자발적인 반응이라는 것을 (1) maximum-entropy criterion 과 (2) minimum-Gibbs-Energy criterion 을 이용하여 보이시오 번 문제에서의 Pb 가 단열된 용기에 보관되어 있었다면 용기 내부는 결국 어떠한 ( 평형 ) 상태가 될 것인지 예측하시오.

Byeong-Joo Lee 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 T m = 505 K 505 K 495 K 1 mole of liquid x moles of solid (1-x) moles of liquid

Byeong-Joo Lee Example - Phase Transformation of Graphite to Diamond Calculate graphite→diamond transformation pressure at 298 K, given H 298,gra – H 298,dia = J S 298,gra = 5.74 J/K S 298,dia = 2.37 J/K density of graphite at 298 K = 2.22 g/cm 3 density of diamond at 298 K = g/cm 3

Byeong-Joo Lee Gibbs Energy for a Unary System V(T,P) based on expansivity and compressibility C p (T) S 298 : by integrating C p /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) H 298 : from first principles calculations, but generally unknown ※ H 298 becomes a reference value for G T ※ Introduction of Standard State