上海交通大学：《材料热力学》教学资源_2010课件_lecture 6 property relation II

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？ (])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 Spring2010©X.J.Jin Lecture 6 Property Relation ll

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2010 © X. J. Jin Lecture 6 Property Relation II 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 -5 C.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 Spring2010©X.J.Jin Lecture 6 Property Relation ll

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2010 © X. J. Jin Lecture 6 Property Relation II 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

Index of nomenclature S.J.T.U. Phase Transformation and Applications Entropy of Mixing混合熵 ·Partial Molar Quantities偏摩尔量 p65,213 O△G @Cp =△V ap ap G) 0s aH aH S,o SJTU Thermodynamics of Materials Spring 2010 ©X.J.Jin Lecture 6 Property Relation ll

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2010 © X. J. Jin Lecture 6 Property Relation II Index of nomenclature • Entropy of Mixing混合熵 • Partial Molar Quantities偏摩尔量 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 ,        

Partial Molar Quantities S.J.T.U. Phase Transformation and Applications The partial derivative of that quantity with respect to mass(number of moles)at constant temperature and constant pressure,and the mass of all other materials in the system The partial molar volume of material a in a solution. Vo= OnaT,P,nne… Pure a > The partial molar volume would be Slope-V =OV equal to the molar volume. Ona )T,Pno-ne" Constant T and P The chemical potential is the partial molar Gibbs free energy.(Only) a SJTU Thermodynamics of Materials Spring 2010 ©X.J.Jin Lecture 6 Property Relation ll

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2010 © X. J. Jin Lecture 6 Property Relation II Partial Molar Quantities The partial derivative of that quantity with respect to mass (number of moles) at constant temperature and constant pressure, and the mass of all other materials in the system. a T ,P,nb ,nc , a n V V            The partial molar volume of material a in a solution. a T ,P,nb ,nc , a n V Slope V             a n V Constant T and P Pure a The partial molar volume would be equal to the molar volume. The chemical potential is the partial molar Gibbs free energy. (Only)

Property Relations (1) S.J.T.U. Phase Transformation and Applications OM dz Mdx Ndy dU=TdS-Pav dH TdS +VaP dF =-SdT-Pav dG=-SaT+Vap as 〔) SJTU Thermodynamics of Materials Spring 2010 ©X.J.Jin Lecture 6 Property Relation ll

Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2010 © X. J. Jin Lecture 6 Property Relation II Property Relations (1) dz  Mdx  Ndy dU TdS  PdV dH TdS VdP dF  SdT  PdV dG  SdT VdP x y x N y M                    S S V P V T                   S S P V P T                  T T V P V S                  T T P V P S                  