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1、2.7 综合练习二:磁性转变对热力学性质的影响 (1)热容的非磁性部分,并画草图;Gmno-mag = -5179+117.85 T 22.1T ln(T) 0.0048 T2S=-G/TS=- (117.85-22.1( ln (T)+1)-0.0096T) H=G+TS=-5179+117.85T22.1Tln(T)0.0048T2 - (117.85-22.1( ln (T)+1)-0.0096T)TH =Cp n0-magdTCpno-mag = 22.1 +0.0096*T 在OriginPro 8中输入此公式得出下图:(2)热容的磁性部分,并画出草图;Cpmag = R*c()*l
2、n(1+Bo)p=0.28 =T/T* T*=633 K Bo=0.52 R=8.31451 当298 KT633 K时Cpmag=8.31451*(ln(1.52)*(474/479)(1/0.28)-1)(23+2/39+2/515)/2.342457在origin中输入下列函数8.31451*(ln(1.52)*(2.452428*(2*(x/633)3+(2/3)*(x/633)9+(2/5)*(x/633)15)/2.3424当633 KT900 K时Cpmag = 8.31451*(ln(1.52)* 2-5+2/3-15+2/5-25/ 2.342457在origin中输入下列函
3、数8.31451*(ln(1.52)*(2*(x/633)(-5)+(2/3)*(x/633)(-15)+(2/5)*(x/633)(-25)/2.3424得出下图:(3)热容的非磁性部分和磁性部分之和,并画草图。将(1)和(2)的Cpno-mag和Cpmag相加得出下列公式当298 KT633 K时Cp=22.1 +0.0096*T + 8.31451*(ln(1.52)* (474/479)(1/0.28)-1)(23+2/39+2/515)/ 2.342457在origin中输入下列函数:22.1+0.0096*x+8.31451*(ln(1.52)*(2.452428*(2*(x/633)3+(2/3)*(x/633)9+(2/5)*(x/633)15)/2.3424当633 KT900 K时Cp=22.1+0.0096*T+8.31451*(ln(1.52)*2-5+2/3-15+2/5-25/2.342457在origin中输入下列函数:22.1+0.0096*x+8.31451*(ln(1.52)*(2*(x/633)(-5)+(2/3)*(x/633)(-15)+(2/5)*(x/633)(-25)/2.3424得到下图: