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)/Jx CrxFe (2510411.7152T ); Gex(fcc)/Jx CrxFe (1310831.823T2.748T logeT)For the bcc phase, please do the following calculations using one ca
2、lculator.(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 (aCr and aFe) as a function of composition at 873K(d) What are the Henrys law constants for Fe and Cr?For the fcc pha
3、se, 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)/Jx CrxFe (2510411.7152T); Gex(fcc)/Jx CrxFe (1310831.823 T2.748T lnT) Gex/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
4、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。（a）由 exGj = exGm + exGm/ xj - xi exGm/ xi 可得 exGFe=XcrX FeexG(bcc)+XCr exGm(bcc)-XFeXCr exG+XCrXFe exG=XcrX Fe (2510411.7152T) +XCr (2510411.7152T) -XFeXCr(2510411.7152 T) +XCrXFe (2510411.7152T) =X2Cr (2510411.7152 T)同理；可得;exGCr=X2Fe(25
5、10411.7152 T)(b)当 T=873K 时，Gex(bcc)x CrxFe (2510411.7152T)= xCrxFe 14876.6304 J设 xCr =X，则 XFe=1-XexGFe=X214876.6304 J （T=873K）exGCr=（1-X） 214876.6304 J （T=873K）0. 0.20.40.60.81.0-2002040608010120140160exGFe (J)XexGFe0. 0.20.40.60.81.0-2002040608010120140160图一 exGFe-X 图0. 0.20.40.60.81.0-200204060801
6、0120140160exGFe (J)XexGFe0. 0.20.40.60.81.0-2002040608010120140160图二 exGcr-X 图0. 0.3 0.6 0.91203040exGxexG.00.20.40.60.81.01015020250303504图三 exG-X 图（C）a m=Xmfm aB = xB expX2oL /(RT) exG (bcc)/Jx CrxFe oLoL=2510411.7152T因而aFe= (1-X) expX2(2510411.7152T) /(RT)（T=873K）aCr= X exp(1-X)2(2510411.7152T )
7、/(RT) （T=873K）0. 0.5 1.00.0.51.0aFexaFe00图三 aCr X 图0. 0.5 1.00.80.40.aCrxaCr00图五 aFeX 图(d)fb = expoL /RT所以:fFe =fCr = exp2510411.7152T /RTfcc:exGFe = X2Cr (1310831.823T2.748T InT)exGCr = X2Fe (1310831.823T2.748 T lnT)设 xCr =X，则 XFe=1-XexGFe = X2 (1310831.823T 2.748T InT)exGCr = (1-X) 2(1310831.823T2.748T lnT)
