[試題] 101-2 高振宏 材料熱力學 期中考
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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