材料热力学练习三_各种热力学性质的计算

1、.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:Gex(bcc)/J xCr xFe (25104 11.7152 T);Gex(fcc)/J xCrxFe (13108 31.823 T 2.748 T log eT)For the bcc phase, please do the following calculations using

2、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 slaw constants for Fe and Cr?For the

3、 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 ：Gex (bcc)/J xCrx Fe (25104 11.7152 T);Gex (fcc)/J xCr x Fe (13108 31.823 T2.748 T 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

4、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 ） Fe 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 。.学习参考.（a ）由 exGj = ex Gm + ex Gm / x j -xiex Gm / x i 可得exGFe= XcrX FeexG(bcc)+X Cr ex Gm(bcc)-XFeX Cr ex G+X CrXFe ex G= XcrX Fe (25104 11.7152T)+X Cr (25104 11.7152 T) -X FeXCr (25104 11.7152 T) +X CrXFe

5、 (25104 11.7152T) =X 2 Cr (25104 11.7152 T)同理 ；可得 ;exGCr =X 2Fe (25104 11.7152 T)(b) 当 T=873K时，Gex (bcc) xCr xFe (25104 11.7152 T)= x Cr xFe 14876.6304J设 xCr =X ，则 XFe=1-XexGFe= X2·14876.6304 J（T=873K ）ex GCr = （1-X ） 2·14876.6304 J（ T=873K）.学习参考.0.00.20.40.60.8160001400012000)10000J(e8000

6、FG6000xe400020000-20000.00.20.40.60.8X图一ex GFe-X 图0.00.20.40.60.8160001400012000)10000J(8000eFG6000xe400020000-20000.00.20.40.60.8X图二ex Gcr-X 图.exGFe1.01600014000120001000080006000400020000-20001.0exGFe1.01600014000120001000080006000400020000-20001.0.学习参考.0.00.20.40.60.8exG1.040004000350030003000G25

7、00xe200020001500100010000.00.30.60.9x图三ex G-X 图2 o（C） am =X m ·fma B = x B expX·L /(RT)exG (bcc)/J xCrxFe o Lo L=25104 11.7152 T因而a Fe= (1-X) ·expX 2·(25104 11.7152 T) /(RT) （ T=873K ）aCr= X ·exp(1-X) 2·(25104 11.7152 T) /(RT)（T=873K ）0aFe1.0e0.5a F0.000.00.51.0x.学习参考.图三aCr X 图0aCr0.00.4ra C0.800.00.51.0x图五a FeX 图(d)f b =exp o L /RT所以 :fFe =f Cr = exp25104 11.7152 T /RTfcc:ex GFe = X 2Cr(13108 31.823 T2.748 T In T)ex GCr = X 2 Fe (13108 31.823 T