相图的用处3 S.J.T.0. Phase Transformation and Applications Page 6/34 估算热力学数据，如 熔化热 -从相图上查出一定成 分的合金系统中，作 LT-T) 为溶剂的金属熔点降 XB 低了多少度，然后用 RTR 稀溶液凝固点下降的 依数性公式，即可估 算溶剂金属的摩尔熔 化焓 SJTU Thermodynamics of Materials Fall 2011 @X.J.Jin Lecture 17 Phase Diagram IV
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Fall 2011 © X. J. Jin Lecture 17 Phase Diagram IV Page 6/34 相图的用处 3 • 估算热力学数据，如 熔化热 – 从相图上查出一定成 分的合金系统中，作 为溶剂的金属熔点降 低了多少度，然后用 稀溶液凝固点下降的 依数性公式，即可估 算溶剂金属的摩尔熔 化焓 2 ( ) m m B RT L T T x − =
相图的用处4 S.J.T.0. Phase Transformation and Applications Page 7/34 Weight Percent Cedmium ·求出活度数据 列442℃ 一在两相共存区，某组 分在这两个共存相中 的化学势应相等 (cd) 4an（sln)=as ALomie Percent Cadmium Lp(sIn)=+RTIn am =M+RTIn ap AG=LBx(D)-H(s) T→Tm Solid->liquid AG=△Hm-TASm=△Hm(I- T SJTU Thermodynamics of Materials Fall 2011 ©X.J.Jin Lecture 17 Phase Diagram IV
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Fall 2011 © X. J. Jin Lecture 17 Phase Diagram IV Page 7/34 相图的用处 4 • 求出活度数据 – 在两相共存区，某组 分在这两个共存相中 的化学势应相等 D * ( ) ( ln) Bi Bi s μ s = μ Bi Bi Bi Bi l Bi (sln) RT ln a RT ln a * ( ) 0 μ = μ + = μ + ΔG = Bi l − Bi s T →Tm * ( ) * μ ( ) μ (1 ) m m m m T T ΔG = ΔH −TΔS = ΔH − Solid -> liquid
二元例题1 ww S.J.T.0. Phase Transformation and Applications Page 8/34 10.用相律判断下列相图是否正确？说出理由。 B B B B UB● (4) (1) (2) (3) 图中A及B是纯物质，l指液态，a、B及Y均是固熔体。 SJTU Thermodynamics of Materials Fall 2011 @X.J.Jin Lecture 17 Phase Diagram IV
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Fall 2011 © X. J. Jin Lecture 17 Phase Diagram IV Page 8/34 二元例题 1
Liquid Liquid S.J.T.0. Phase Transformation and A L+a L+B 29/34 L+a L+B B a+B a+B B A B A Mole fraction of B,xB→ Mole fraction of B,xB-> (a) (b) Liquid Liquid L+B 个 L+x L+Y a+L B+Y +B a+B B A B Mole fraction of B,xB→ Mole fraction of B,xB-> SJTU (c) (d)
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Fall 2011 © X. J. Jin Lecture 17 Phase Diagram IV Page 9/34
二元例题2b S.J.T.0. Phase Transformation and Applications Page 10/34 Molar Gibbs free energy of What is the most stable state solution,drawn at T and P of the system at an average constant for an a phase and composition of Xo? B phase. · Suppose the a-phase was somehow prevented from forming (i.e., consider the B-phase equilibrium under the a-phase constraint that the a-phase is absent).In this case,what would be the equilibrium state of the system with average composition X,？ SJTU Thermodynamics of Materials Fall 2011 @X.J.Jin Lecture 17 Phase Diagram IV
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Fall 2011 © X. J. Jin Lecture 17 Phase Diagram IV Page 10/34 二元例题 2b • What is the most stable state of the system at an average composition of X0? • Suppose the α-phase was somehow prevented from forming (i.e., consider the equilibrium under the constraint that the α-phase is absent). In this case, what would be the equilibrium state of the system with average composition X0? Molar Gibbs free energy of solution, drawn at T and P constant for an α phase and β phase