Presentation is loading. Please wait.

Presentation is loading. Please wait.

Byeong-Joo Lee Byeong-Joo Lee POSTECH - MSE Thermodynamic Criterions, Functions and Relations.

Similar presentations


Presentation on theme: "Byeong-Joo Lee Byeong-Joo Lee POSTECH - MSE Thermodynamic Criterions, Functions and Relations."— Presentation transcript:

1 Byeong-Joo Lee http://cmse.postech.ac.kr Byeong-Joo Lee POSTECH - MSE calphad@postech.ac.kr Thermodynamic Criterions, Functions and Relations

2 Byeong-Joo Lee http://cmse.postech.ac.kr 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 는 control 이 쉽지 않은 변수. 따라서 control 이 쉬운 변수를 택하고 그때 평형을 가늠할 수 있는 척도 ( 새로운 state function) 를 개발할 필요. → F, G

3 Byeong-Joo Lee http://cmse.postech.ac.kr dF ≡ dU – TdS – SdT For a reversible process dF = [TdS – PdV – δw’] – TdS – SdT = – SdT – PdV – δw’ dF T = – PdV – δw’ = – δw T.Total ▷ Constant T 조건에서 system 이 상태를 변화해 가면서 할 수 있는 최대 work 은 -ΔF. For an irreversible isothermal process △ F T = [q – w] – T △ S T △ S = q + T △ S irr w = P △ V + w’ = w’ For constant V △ F T,V + w’ + T △ S irr = 0 ▷ 최대의 일을 하지 못한다면 이는 ΔS irr 의 발생 때문. Constant T, V 조건일 때 평형은 system 의 maximum (w’+ ΔS irr ) 또는 system 의 minimum F 에서 얻어짐. Helmholtz Free Energy - Work Function, F ≡ U ST Helmholtz Free Energy - Work Function, F ≡ U – ST

4 Byeong-Joo Lee http://cmse.postech.ac.kr Helmhomlz Free Energy - Example 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

5 Byeong-Joo Lee http://cmse.postech.ac.kr dG ≡ dU + PdV + VdP – TdS – SdT For a reversible process dG = [TdS – PdV – δw’] + PdV + VdP – TdS – SdT = – SdT + VdP – δw’ dG T,P = – δw’ ▷ Constant T, P 조건에서 system 이 상태를 변화해 가면서 할 수 있는 최대 non-PV work 은 -ΔG. For an irreversible isothermal process △ G T,P = [q – w] + P △ V – T △ S T △ S = q + T △ S irr w = P △ V + w’ △ G T,P + w’ + T △ S irr = 0 ▷ 최대의 일을 하지 못한다면 이는 ΔS irr 의 발생 때문. Constant T, P 조건일 때 평형은 system 의 maximum (w’+ΔS irr ) 또는 system 의 minimum G 에서 얻어짐. Gibbs Free Energy - Gibbs Function, G ≡ U + PV ST Gibbs Free Energy - Gibbs Function, G ≡ U + PV – ST

6 Byeong-Joo Lee http://cmse.postech.ac.kr Thermodynamic Relations - For a closed system dU = TdS – PdV dH = TdS + VdP dF = – SdT – PdV dG = – SdT + VdP

7 Byeong-Joo Lee http://cmse.postech.ac.kr Thermodynamic Relations - For a multicomponent system is the chemical work done by the system ▷ ▷ Chemical Potential

8 Byeong-Joo Lee http://cmse.postech.ac.kr Thermodynamic Relations

9 Byeong-Joo Lee http://cmse.postech.ac.kr Thermodynamic Relations - Maxwell Relation

10 Byeong-Joo Lee http://cmse.postech.ac.kr Thermodynamic Relations - Transformation Formula For an incremental change of state at constant x

11 Byeong-Joo Lee http://cmse.postech.ac.kr Thermodynamic Relations - Application ▷ Heat capacity equation ▷ Gibbs-Helmholtz equation

12 Byeong-Joo Lee http://cmse.postech.ac.kr Thermodynamic Relations 1.Gibbs Free Energy, Helmholtz Free Energy 에서 “Free” 라는 단어의 의미는 무엇일까 ? 2. 수 많은 thermodynamic relation 들이 존재한다. 그 relation 들의 중요성은 무엇일까 ? 3.Gibbs-Helmholtz equation 을 어떻게 활용할 수 있는지 용도를 조사하시오.


Download ppt "Byeong-Joo Lee Byeong-Joo Lee POSTECH - MSE Thermodynamic Criterions, Functions and Relations."

Similar presentations


Ads by Google