課程名稱︰材料熱力學 課程性質︰必修 課程教師︰高振宏 開課學院：工學院 開課系所︰材料系 考試日期（年月日）︰101/4/19 考試時限（分鐘）：110分鐘 是否需發放獎勵金：是 （如未明確表示，則不予發放） 試題 : 1. Melts in the system Pb-Snexhibit regular solution behavior.At 473°C a = 0.057 in a liquid solution of X = 0.1 Calculate the value of Ω Pb Pb for the system and calculate the activity of Sn in the liquid solution of X = 0.4 at 500°C . Sn 2. The system A-B forms a regular solution with the heat of mixing given by: ΔH = -13500X X (J/mol) mix A B a.Derive expressions for the Henry's law constant for A as a solute in B and B as a solute in A. b.Calculate ΔS at 1000 K , X = 0.2 mix A c.ΔH = -13500X X is in a regular solution , but in a real mix A B situation , is ΔH be overestimate or underestimate? Please mix give your answer and explain why. d.那 P , P , P 在真實情況下跟 regular solution 比還要大還是小， AA AB BB 請寫答案並簡述原因。 (英文原文有點忘記QAQ) 3. a.Explain why μ = (∂H'/∂n ) = (∂G'/∂n ) k k S',P,nj k T,P,nj b.So we can say that the chemical potential of component k, μ , k equals to the partial molar enthalpy of component k. Is this correct? Explain your answer. c.So we can say that the chemical potential of component k, μ , k equals to the partial molar Gibbs free energy of component k. Is this correct? Explain your answer. 4. In closed system, when we make T and P in constant, which thermodynamic property will have maximum (or minimum).Please give your answer and reason. And what about S and V in constant? 5. a.Please derive the below equation: _ Y = Y + ( 1- X )(∂Y /∂X ) B mix B mix B _ Y is the partial molal property, such as H , U or S. B X is the molar fraction of component B. B And the equation is T , P in constant and closed system. b.Sketch representation of the relation between mixing properties, and partial molal properties of the components . Mark each of the factors in above equation, and derive above equation. c.為何此equation 要在 T , P constant 和 closed system 下才成立 ? 6. Use the Gibbs-Duhem equation to show that, if the activity coefficients of the components of a binary solution can be expressed as ln γ = α X + (1/2) x α X ^2 + (1/3) x α X ^3 +.... A 1 1 2 2 3 3 ln γ = β X + (1/2) x β X ^2 + (1/3) x β X ^3 +.... B 1 1 2 2 3 3 over the entire range of composition, then α = β = 0 1 1 --

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