厦门大学：《物理化学 Physical Chemistry》课程电子教案（PPT教学课件，英文版）chapter04-2 Material Equilibrium

Physical Chemistry Material Equilibrium Maxwell relations a丿( (4.44) C as( ap C aP (4.45) T T

Maxwell Relations T T V P V S        =          T T P V P S        = −          (4.45) S S P V P T        =          V V S T S P        =      −     (4.44) Physical Chemistry Material Equilibrium

Physical Chemistry Material Equilibrium Isobaric thermalexpansivit 1(oⅣ a(1,P) (1.43) VaT P Isothermal compressibility (4.39) 1(a K(T, P) (1.4) V aP 2 C (453) K aT JT (a-1)(452) aP (264)*1m H P

T P V V T P          1 ( , ) P T V V T P          − 1 ( , ) (4.39)* Isobaric thermal expansivity Physical Chemistry Material Equilibrium Isothermal compressibility (1.43)* (1.44)* (4.53)   2 TV CP −CV = ( −1)         = T C V P JT  (4.52) H JT P T           (2.64)*

Physical Chemistry Material Equilibrium Application of Maxwell Relations Example 1: Prove that the internal energy of ideal gas is a function of temperature only Answers. du=tds-Pdr O S b For ideal gas PV=nRT P=nRT aP nR aT aU aP nR P P=0 aT

Application of Maxwell Relations V nRT PV = nRT P = Physical Chemistry Material Equilibrium Example 1: Prove that the internal energy of ideal gas is a function of temperature only. Answers. For ideal gas V nR T P V  =        P T P T V U T V  −         =        P V nR = T − = 0 dU =TdS −PdVP V S T V U T T  −         =        T T V P V S        =         

Physical Chemistry Material Equilibrium Application of Maxwell Relations Example 2: Calculate 4U when an ideal gas changes from PI,V, T, to P2,2, T2 (Hint: applying/ou Answers U=U(7,V) aU aP P aT aU T+aU aT aP dT+ P ldv aT △U=∫Gd+ aP P aT

Physical Chemistry Material Equilibrium Application of Maxwell Relations Example 2: Calculate U when an ideal gas changes from P1 , V1 , T1 to P2 , V2 , T2 . (Hint: applying ) V T U         U U T V = ( , ) Answers. dV V U dT T U dU V T          +        = P dV T P C dT T V V        −        = + P dV T P U C dT T V  V         −         = + P T P T V U T V  −         =       

Physical Chemistry Material Equilibrium e Chapter 4 Material Equilibrium Calculation of changes in state functions ● Calculation of as aS dT+ dP OT' aP PdT-avdP(4.59) aS (449) P O(4.50 aT TaP OT' △S=S2-S dT-aldP (460

(4.59) Calculation of S dT VdP T C dP P S dT T S dS P P T  = −         +        = Calculation of changes in state functions    = − = − 2 1 2 1 2 1 dT VdP T C S S S P  (4.60) V T V P S T P  = −         = −        (4.50) T C T S P P  =        (4.49) Physical Chemistry Chapter 4 Material Equilibrium Material Equilibrium