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 solidsolutionof (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 xCr xFe (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 (aCr and a Fe) as a function of composition at 873K(d) What are the Henryslaw constants for Fe and Cr?For the fc

3、c 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 x Cr xFe (25104 11.7152 T);Gex (fcc)/J x Cr x Fe (13108 31.823 T 2.748 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體

4、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 。（ a）由 exGj = ex G m + ex Gm/x j - xiex Gm /x i 可得exGFe = XcrX Feex G(bcc)+XCr ex Gm (bcc)-XFe X Cr ex G+X Cr XFe ex G=XcrX Fe (25104 11.7152 T)+X Cr (25104 11.7152 T) -X Fe X Cr (25104 11.7152 T) +X Cr X Fe (

5、25104 11.7152 T) =X 2Cr (25104 11.7152 T)同理；可得 ;ex GCr =X 2Fe (25104 11.7152 T)(b) 當(dāng) T=873K時(shí)，Gex (bcc) x Crx Fe (25104 11.7152 T)= x Cr xFe 14876.6304J設(shè) xCr =X ，則 X Fe =1-Xex GFe= X 2 14876.6304 J（ T=873K ）ex GCr =（ 1-X ） 2 14876.6304 J（ T=873K ）0.00.20.40.60.8exGFe1.0160001600014000140001200012000)

6、1000010000J(80008000eFGx60006000e400040002000200000-2000-20000.00.20.40.60.81.0X圖一exGFe-X 圖)J(eFGxeGxe0.00.20.40.60.8exGFe1.016000160001400014000120001200010000100008000800060006000400040002000200000-2000-20000.00.20.40.60.81.0X圖二 exGcr-X圖0.00.20.40.60.8exG1.04000400035003000300025002000200015001000

7、10000.00.30.60.9x圖三exG-X 圖（C ） am =X m fmaB = x B expX 2 oL /(RT)exG (bcc)/J xCr xFe oLo L=25104 11.7152 T因而2a Fe = (1-X) expX (25104 11.7152 T) /(RT) （ T=873K ）2a Cr= X exp(1-X) (25104 11.7152 T) /(RT)（T=873K ）0aFe1.0e0.5a F0.000.00.51.0x圖三aCr X 圖0aCr0.00.4rCa0.800.00.51.0x圖五aFeX 圖(d)f b = expoL /RT所以 :f Fe =fCr = exp25104 11.7152 T /RTfcc:ex G Fe = X 2Cr (13108 31.823 T 2.748 T In T)ex G Cr = X 2Fe (13108 31.823 T 2.748