厦门大学：《物理化学 Physical Chemistry》课程电子教案（PPT教学课件，英文版）chapter03 The Second Law of Thermodynamics

Physical Chemistry Chapter 3 Chapter 3 The Second Law of Thermodynamics Reaction The first law The second law N2+3H2→>2NH3Ess+Em (with a catalyst)constant Provide such information 2leq INH ? througl th s ec

1 Chapter 3 The Second Law of Thermodynamics Reaction The first law The second law N2+3H2 2NH3 (with a catalyst) Esys + Esurr =constant [N2]eq =? [H2]eq =? [NH3]eq =? Provide such information through S Physical Chemistry Chapter 3

Physical Chemistry Chapter 3 The Statements of second law a It is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a heat reservoir and the performance of an equivalent amount of work by the system on the surroundings Kelvin-Plank statement Heat g Cyclic Heat inmachine b wOrk output =q reservoir (system) Fig3.1

2 The Statements of Second Law § It is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a heat reservoir and the performance of an equivalent amount of work by the system on the surroundings. Kelvin-Plank statement Physical Chemistry Chapter 3 Heat reservoir Cyclic machine (system) Work output = q Fig. 3.1 Heat q

Physical Chemistry Chapter 3 The Statements of second law It is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a cold (heat) reservoir and the flow(performance )of an equal equivalent amount of heat(work)out of(by the) system into a hot reservoir (on the surroundings. Clausius statement (Kelvin-Plank statement)

3 The Statements of Second Law § It is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a cold (heat) reservoir and the flow (performance) of an equal equivalent amount of heat (work) out of (by the) system into a hot reservoir. (on the surroundings.) Clausius statement (Kelvin-Plank statement) Physical Chemistry Chapter 3

Physical Chemistry Converts some of the random Chapter 3 molecular energy of heat flow into HEAT ENGINES macroscopic mechanical energy ( work) a heat engine operating Heat reservoir at T e between two temperatures The heat and work quantities q are for one cycle Engin ●●● (3.1) C Cold reservoir TC gH t qc (32) Fg.32 qH e +gC=l qc <1 (3.3) qH qH

4 HEAT ENGINES H C H H C q q q q q e     1 (3.3) Heat reservoir at TH Cold reservoir at TC qH -qC -w Fig. 3.2 Heat Engine A heat engine operating between two temperatures. The heat and work quantities are for one cycle. H qH w q w e | |    (3.1) -w = qH + qC (3.2) < 1 Physical Chemistry C Chapter 3 onverts some of the random molecular energy of heat flow into macroscopic mechanical energy (work)

Physical Chemistry Chapter 3 Carnot's Principle Carnot Principle: No heat engine can be more efficient than a reversible heat engine when both engines work between the same pair of temperatures The maximum amount of work from a given supply of heat is obtained with a reversible engine

5 Carnot’s Principle Carnot Principle: No heat engine can be more efficient than a reversible heat engine when both engine s work be twe en the s ame pa ir of temperatures. The maximum amount of work from a given supply of heat is obtained with a reversible engine. Physical Chemistry Chapter 3

Physical Chemistry Chapter 3 Carnot's Principle A Carnot cycle: a reversible cycle that consists of two isothermal steps at different temperatures and two adiabatic ster iSotherm 2 diabat adiabat isotherm (b) Fig 3.4

6 Carnot’s Principle A Carnot cycle: a reversible cycle that consists of two isothermal steps at different temperatures and two adiabatic steps. isotherm isotherm 1 2 4 3 V P     (b) adiabat adiabat Fig. 3.4 1 2 4 3 V P     (a) TH TC Physical Chemistry Chapter 3

Physical Chemistry Chapter 3 aB q1=nITIn - B CD q2=nITIn D A(PA,VA,) BC:7y-=1 B q Be D DA TVTYY-I (PD, V, T2) C 四l(Pc) From(iii), (iv) get So q1+q2=nR(71-72)

7 AB： (i) A B V V q1  nRT1 ln CD ： (ii) C D V V q2  nRT2 ln BC ： (iii) 1 2 1 1      T VB T VC V P C C   DA ： (iv) 1 2 1 1      TVA T VD From (iii), (iv) get D C A B V V V V  So (v) A B V V q1  q2  nR(T1 -T2 )ln T2 T1 q1 q2 A(PA,VA,T1) B(PB,VB,T1) D (PD,VD,T2) C (PC,VC,T2) {V} {P}     Physical Chemistry Chapter 3

Chapter 3 A(PA VA,T) D B(PB, VB, T1) q1=nRTIn-'B D q2 I72 (Pc,Vc,T2) q2=nRT)InD q1+q2=nR(Ti-72) V (33) q1+q2 From(3.3),(1),(V) (3.15 → ysical Chemistry

8 (i) A B V V q1  nRT1 ln (ii) C D V V q2  nRT2 ln (v) A B V V q1  q2  nR(T1 -T2 )ln From (3.3), (i)，(v), （3.15） 1 1 2 1 1 2 1 T T T q q q q w e       1 1 2 1 q q q q w e     （3.3） T2 T1 q1 q2 A(PA,VA,T1) B(PB,VB,T1) D (PD,VD,T2) C (PC,VC,T2) {V} {P}     Physical Chemistry Chapter 3 D C A B V V V V 

Chapter 3 、 isotherm Carnot's Principle adiabat adiabat dq+dw=dq-pav isotherm·3 p nRT dU=Cr(r)dr Fg.34(b) cut= do g-nRT dT nRp (39) c1()m=cnam+了 dT+ Cr(T) dT Cr(T) (3.10 7 T T (Sec.18) ∮Cv(T dT cRo d T CyT) T,Cy t dT=o dT=∫ T T → hysical Chemistry TT

9 Carnot’s Principle V dV CV dT  dq  nRT          3 2 4 3 1 4 2 1 ( ) ( ) ( ) ( ) ( ) T T T T T T V V V T T V V dT T C T dT T C T dT T C T dT T C T T dT C T (3.10) isotherm isotherm 1 2 4 3 V P     Fig. 3.4 (b) adiabat adiabat      (3.9) V dV nR T dq T dT CV (T) dU  dq  dw  dq  PdV , V nRT P  dU  CV (T )dT Physical Chemistry Chapter 3         1 3 1 1 3 1 0 ( ) ( ) ( ) ( ) T T T T V V T T V V dT T C T dT T C T dT T C T T dT C T (Sec. 1.8)

→ Physical Chemistry Chapter 3 Carnot's Principle iSotherm 2 dT adiabat ∮C()m=0(311) adiabat isotherm·3 0 Carnot cycle, perf. gas(3. 12) Fg.34(b) 2 (3.13) 0 0 2 H fa=H+c=0 Carnot cycle, perf. gas(3. 14)10

10 Physical Chemistry Chapter 3  ( )  0 T dT CV T (3.11) Carnot’s Principle   0 T dq Carnot cycle, perf. gas (3.12)     2 1 2 1 1 H H H T q dq T T dq     4 3 4 3 1 C C C T q dq T T dq    0  C C H H T q T q T dq Carnot cycle, perf. gas (3.14)   3 2 0 T dq   1 4 0 T dq isotherm isotherm 1 2 4 3 V P     Fig. 3.4 (b) adiabat adiabat (3.13)      1 4 4 3 3 2 2 1 T dq T dq T dq T dq T dq