材料熱力學(xué)練習(xí)三_各種熱力學(xué)性質(zhì)的計(jì)算

1、.新型材料設(shè)計(jì)及其熱力學(xué)與動(dòng)力學(xué)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翻譯 ：BCC（ Fe， Cr ）固溶體的過(guò)剩吉布斯自由能和fcc固溶體 （ Fe， Cr）的吉布斯自由能表達(dá)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對(duì)于體心立方相，請(qǐng)使用計(jì)算器做下面的計(jì)算。（ a） 計(jì)算 Fe 和 Cr 的局部吉布斯能量表達(dá)式；（ b ）畫(huà)出 873K 時(shí)局部吉布斯自由能和整

4、體吉布斯自由能的復(fù)合函數(shù)圖。（ c） 畫(huà)出 873K 時(shí) Fe 和 Cr 反應(yīng)的活度圖 。（ d ） Fe 和 Cr 亨利定律常數(shù)是什么 ？對(duì)于 fcc, 請(qǐng)用你自己的符號(hào)計(jì)算a 和 b 。.學(xué)習(xí)參考.（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) 當(dāng) T=873K時(shí)，Gex (bcc) xCr xFe (25104 11.7152 T)= x Cr xFe 14876.6304J設(shè) xCr =X ，則 XFe=1-XexGFe= X2·14876.6304 J（T=873K ）ex GCr = （1-X ） 2·14876.6304 J（ T=873K）.學(xué)習(xí)參考.0.00.20.40.60.8160001400012000)10000J(e8000

6、FG6000xe400020000-20000.00.20.40.60.8X圖一ex GFe-X 圖0.00.20.40.60.8160001400012000)10000J(8000eFG6000xe400020000-20000.00.20.40.60.8X圖二ex Gcr-X 圖.exGFe1.01600014000120001000080006000400020000-20001.0exGFe1.01600014000120001000080006000400020000-20001.0.學(xué)習(xí)參考.0.00.20.40.60.8exG1.040004000350030003000G25

7、00xe200020001500100010000.00.30.60.9x圖三ex G-X 圖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.學(xué)習(xí)參考.圖三aCr X 圖0aCr0.00.4ra C0.800.00.51.0x圖五a FeX 圖(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