材料热力学试三:各种热力学性质的计算
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材料热力学试三:各种热力学性质的计算
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新型材料设计及其热力学与动力学
The excess Gibbs energies of bcc solid solution of (Fe,Cr) and fcc solid solution of (Fe,Cr) is represented by the following expressions:
G ex(bcc)/J=x Cr x Fe (25104-11.7152T);
G ex(fcc)/J=x Cr x Fe (13108-31.823T+2.748T log e T)
For the bcc phase, please do the following calculations using one calculator.
(a) Calculate the partial Gibbs energy expressions for Fe and Cr
(b) Plot the integral and partial Gibbs energies as a function of composition at 873 K
(c) Plot the activities (a Cr and a Fe) as a function of composition at 873K
(d) What are the Henry’s law constants for Fe and Cr?
For the fcc phase, please do the calculations (a) to (b) by using your own code
翻译:
BCC(Fe,Cr)固溶体的过剩吉布斯自由能和fcc固溶体(Fe,Cr)的吉布斯自由能表达式如下:
G ex(bcc)/J=x Cr x Fe (25104-11.7152T);
G ex(fcc)/J=x Cr x Fe (13108-31.823T+2.748T ln T) G ex/J
对于体心立方相,请使用计算器做下面的计算。
(a)计算Fe和Cr的局部吉布斯能量表达式;
(b)画出873K时局部吉布斯自由能和整体吉布斯自由能的复合函数图。
(c)画出873K时Fe和Cr反应的活度图。
(d)F e和Cr亨利定律常数是什么?
对于fcc,请用你自己的符号计算a和b。
(a )由ex G j = ex G m + ∂ex G m / ∂ x j - ∑ x i ∂ex G m / ∂ x i 可得
ex
G Fe =Xc r X Fe ex G (bcc)+X Cr ex G m (bcc)-[X Fe X Cr ex G+X Cr X Fe ex G ]
=Xc r X Fe
(25104-11.7152T ) +X Cr (25104-11.7152T ) -[X Fe X Cr (25104-
11.7152T ) +X Cr X Fe (25104-11.7152T ) ]
=X 2Cr (25104-11.7152T ) 同理;可得;
ex G Cr =X 2Fe
(25104-11.7152T ) (b)当T=873K 时,
G ex (bcc)=x Cr x Fe (25104-11.7152T )= x Cr x Fe 14876.6304 J 设x Cr =X ,则X Fe =1-X
ex G Fe=X 2
·14876.6304 J
(T=873K )
ex G Cr
=(1-X )2·14876.6304 J (T=873K )
0.0
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40006000800010000
120001400016000e x G F e (J )
X
exGFe
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0200040006000800010000120001400016000
图一 ex G Fe -X 图
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120001400016000e x G F e (J )
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exGFe
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0200040006000800010000120001400016000
图二 ex G cr -X 图
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e x G
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图三 ex G-X 图
(C )a m =X m ·f m a B = x B exp[X 2·o L /(RT)]
ex
G (bcc)/J =x Cr x Fe o L
o L=25104-11.7152T
因而
a Fe = (1-X)· exp[X 2·(25104-11.7152T ) /(RT)](T=873K ) a Cr = X · exp[(1-X)2·(25104-11.7152T ) /(RT)] (T=873K )