上海交通大学：《材料热力学》教学资源_2018 lecture 7 equilibrium

Contents of Today S.J.T.U. Phase Transformation and Applications Review previous Condition of equilibrium Phase equilibrium Clapeyron equation in vapor equilibria Variation of vapor pressure with particle size Second-order transition SJTU Thermodynamics of Materials Fall 2018 ©X.J.Jin Lecture 7 equilibrium I

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Fall 2018 © X. J. Jin Lecture 7 equilibrium I Contents of Today Review previous Condition of equilibrium Phase equilibrium Clapeyron equation in vapor equilibria Variation of vapor pressure with particle size Second-order transition

Property Relation S.J.T.U. Phase Transformation and Applications aU dU-Tds-Pdv =T =-P dF=-SdT-Pav =-S 8) dG=-SdT+Vdp =-S =V dH=TaS+Vap -T =V 二 as ap) H as aP as ap SJTU Thermodynamics of Materials Fall 2018 ©X.J.Jin Lecture 7 equilibrium I

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Fall 2018 © X. J. Jin Lecture 7 equilibrium I Property Relation S S V P V T          = −        S S P V P T          =        T T V P V S          =        T T P V P S          =        − dU =TdS−PdV dF = −SdT−PdV dG= −SdT+VdP dH =TdS+VdP P V U T S U V S  = −         =        P V F S T F V T  = −         = −        V P G S T G P T  =         = −        V P H T S H P S  =         =       

Review S.J.T.U. Phase Transformation and Applications Properties relations Entropy of Mixing,Partial Molar Quantities 热力学关系式的应用 -) p.65..13 a△G =△V ap ap (av 0s OH aH S,o SJTU Thermodynamics of Materials Fall 2018 ©X.J.Jin Lecture 7 equilibrium I

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Fall 2018 © X. J. Jin Lecture 7 equilibrium I Review Properties relations Entropy of Mixing，Partial Molar Quantities 热力学关系式的应用 T T P V P S          =        − = 0           T V U p.65, 2.13 V p G T =            T p p C           S p T           S T          T p V V           − 1 H S T           H S,        

Quiz Q3 Answer S.J.T.U. Phase Transformation and Applications 15.进行下述过程时，系统的△U、△H、△S和△G何者为零？ (1)非理想气体的卡诺循环； (2)隔离系统中的任意过程： (3)在100℃，01325Pa下1mol水蒸发成水蒸气； (4)绝热可逆过程。 15题：1)内能变化为0，焓变为0，熵变为0，自由能变化为0。 2):内能变化为0， 焓变不一定为0，熵变不一定为0，自由能变化不 一定为0。 3)自由能变化为0。错，主要认为相变过程中内能为0，焓变也为0。 4)只有熵变为0。错误很多。 SJTU Thermodynamics of Materials Fall 2018 ©X.J.Jin Lecture 7 equilibrium I

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Fall 2018 © X. J. Jin Lecture 7 equilibrium I Quiz Q3 Answer 15题：1）内能变化为0，焓变为0，熵变为0，自由能变化为0。 2）：内能变化为0，焓变不一定为0，熵变不一定为0，自由能变化不 一定为0。 3）自由能变化为0。错，主要认为相变过程中内能为0，焓变也为0。 4）只有熵变为0。错误很多

Problem 3.2 S.J.T.U. Phase Transformation and Applications At-5 C,the vapor pressure of ice is 3.012 mmHg and that of supercooled liquid water is 3.163 mmHg.Tha latent heat of fusion of ice is 5.85 kJ/mol at -5C.Calculate AG and AS per mole for the transition from water to ice at-5°C. △G1-5C Vapor pressure △G=0 △G=-0 Ice Water △Gand△Sat-5C SJTU Thermodynamics of Materials Fall 2018 ©X.J.Jin Lecture 7 equilibrium I

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Fall 2018 © X. J. Jin Lecture 7 equilibrium I Problem 3.2 At -5 C, the vapor pressure of ice is 3.012 mmHg and that of supercooled liquid water is 3.163 mmHg. Tha latent heat of fusion of ice is 5.85 kJ/mol at -5 C. Calculate G and S per mole for the transition from water to ice at -5C. Ice Water G and S at -5 C G1 -5 C G=0 G=0 Vapor pressure