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)/JxCrxFe (2510411.7152T); Gex(fcc)/JxCrxFe (1310831.823T2.748T logeT)For the bcc phase, please do the following calculations using one calc

2、ulator.(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 phas

3、e, 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)/JxCrxFe (2510411.7152T); Gex(fcc)/JxCrxFe (1310831.823T2.748T 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时Fe和Cr反应的活度图。（

4、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 - å xi ¶exGm/ ¶ xi可得 exGFe=XcrX FeexG(bcc)+XCr exGm(bcc)-XFeXCr exG+XCrXFe exG =XcrX Fe (2510411.7152T) +XCr (2510411.7152T) -XFeXCr(2510411.7152T) +XCrXFe (2510411.7152T) =X2Cr (2510411.7152T)同理；可得;exGCr=X2

5、Fe(2510411.7152T)(b)当T=873K时，Gex(bcc)xCrxFe (2510411.7152T)= xCrxFe 14876.6304 J设xCr =X，则XFe=1-XexGFe=X2·14876.6304 J （T=873K）exGCr=（1-X）2·14876.6304 J （T=873K）图一 exGFe-X图图二 exGcr-X图图三 exG-X图（C）am=Xm·fm aB = xB expX2·oL /(RT) exG (bcc)/JxCrxFe oL oL=2510411.7152T因而 aFe= (1-X)· expX2·(2510411.7152T) /(RT)（T=873K）aCr= X· exp(1-X)2·(2510411.7152T) /(RT) （T=873K）图三 aCr X图图五 aFeX图(d) fb = expoL /RT所以:fFe =fCr = exp2510411.7152T /RTfcc:exGFe = X2Cr (1310831.823T2.748T InT)exGCr = X2Fe (1310831.823T2.748T lnT