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 chemistr Physical Chemistry Cheng Xuan February 2004, Spring Semester

Physical Chemistry Cheng Xuan February 2004, Spring Semester Physical Chemistry

Physical Chemistry Material Equilibrium Review from the last class → Basic equations du tds- pdv closed syst, rev. proc., P-v (425)* work only H≡U+PV (426) A≡U-Ts (427)* G≡H-Ts U closed syst in equi lib., P-v aT work only (4.29) OH closed syst., in equi lib., P-v P aT ly (4.30) P aS aS OT ar/closed syst., in equilib.(4.31)* P

Physical Chemistry Review from the last class Material Equilibrium Basic Equations H  U + PV (4.26)* A  U – TS (4.27)* G  H - TS (4.28)* dU = TdS - PdV (4.25)* closed syst., rev. proc., P-V work only V V T U C         = (4.29)* closed syst., in equilib., P-V work only P P T H C         = (4.30)* closed syst., in equilib., P-V work only V V T S C T         = P P T S C T         = closed syst., in equilib. (4.31)*

Physical Chemistry Material Equilibrium Review from the last class H H=U+pv U G=H-TS TS G FA+pv TS pV A=U-TS Relations among different functions

Physical Chemistry Review from the last class Material Equilibrium Relations among different functions

Physical Chemistry Material Equilibrium The Gibbs equations du e tds- pdv (4.33)2 dh = tas t vdP closed syst, rev 4.34) proc., P-v work da=-sdT- Pdv only (4.35) dg=-sdt+ vdp. (4.36)

The Gibbs Equations Physical Chemistry Material Equilibrium dG = -SdT + VdP (4.36)* dA = -SdT - PdV (4.35) dH = TdS + VdP (4.34) dU = TdS - PdV (4.33)* closed syst., rev. proc., P-V work only

Physical Chemistry Material Equilibrium aM aN、02za ON Oy croy axa du e tds- pdv S dh= tds t vdP aP S a示 P C da=-sdt- pdv dg=-sdt+ dP T P

Physical Chemistry Material Equilibrium dG = -SdT + VdP dA = -SdT - PdV dH = TdS + VdP dU = TdS - PdV S S P V P T        =          V V S T S P        =      −     T T V P V S        =          T T P V P S        = −          ( ) ( ) x y M N y x   =   2 2 ( ) , ( ) x y M z N z y x y x x y     = =      

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