第8章 O=4E+n AEa=Qat-Wt=350-126=224(J) (1)∵内能是态函数,故△Ea=△Eab 故 Odh= Eadh +Wadh= 224+42= 266J) e+w 224-84=-308(J)放热 (3) Od=AEd +Wad=(ed-ea)+wadb 168+42=210 Qb=△Eb+=Eb-E =(E-E.)+(E-E (Eb-E(Ed-Ea=Ae =AEd-168=224-168=56(J) 8-21mol单原子理想气体r=3 3 AE=RAT=×8.31×(350-300)=62325(J) (1)等容Q=AE=62325(J),W=0 (2)等压gn=C1=1+2, RT=-×8.31×50 =1039(J) Wn=Q,-AE=1039-62325≈4158(J) 8-3P==latm,T=7=293K,V=V Tb=T。=80+353(K),V=Vd=210 ∵两过程的初末态相同,∴内能增量相同 4E=-RAT 8.31×(7。-7) 题8-3图
71 第 8 章 8-1 Q = E +W = − = 350 −126 = 224(J) Eacb Qacb Wacb （1）∵内能是态函数，故 Eabd = Eacb 故 = + = 224 + 42 = 266(J) Qadb Eadb Wadb （2） Qba = Eba +Wba = −Eacb +Wba = −224 − 84 = −308(J) 放热 （3） Qad = Ead +Wad = Ed − Ea +Wadb ( ) = 168 + 42 = 210(J) Qdb = Edb + O = Eb − Ed ( ) ( ) = Eb − Ea + Ea − Ed = (Eb − Ea ) − (Ed − Ea ) = Eab −168 = −168 = 224 −168 = 56(J) Eacb 8-2 1 mol 单原子理想气体 i=3 8.31 (350 300) 623.25(J) 2 3 2 = R T = − = i E （1）等容 Qv = E = 623.25(J),Wv = 0 （2）等压 8.31 50 2 5 2 2 = + = = R T i Qp CpT = 1039(J) W = Q − E = 1039 − 623.25 415.8(J) p p 8-3 0 0 0 Pa = P =1atm, Ta = T = 293K, Va =V 2 0 Tb = Tc = 80 + 353(K), Vc =Vd = V ∵ 两过程的初末态相同，∴ 内能增量相同 8.31 ( ) 2 5 2 R Tac Tc Ta i E = = − 题 8-3 图
×8.31×(353-293)=1246(J) (1)Wabc=Wh=rTh c)=kT h( =8.31×353×ln2=2033(J) Qab=Wak+AEa=2033+1246=3296(J) (2) Wadc =wad=StaIn( g)=KT, hn 2 831×293×h2=1687(J) Qak=△Ea+Wa=1246+1687=2933J) 8-4(1)N2体积 等温 PH=p22→V=P0.1×0.01 (m3) p, 绝热:BH=p=12=(By1=(01)y14×001=373×0-m3 (2)N2温度 等温:72=T=300K 绝热:p1T=p2T2 →T2=T()=1118K P (3)N2对外做功 等温:H=Qn="Rrn(P p2 又P1 m rt= v2 ∴W=DVmP1=1013×103×0.01×h()=-467×103J P 绝热:W,=-△E=-C,m(72-7)=(PV1-p2V2)=R
72 8.31 (353 293) 1246(J) 2 5 = − = （1） ) 2 ln( ) ln( 0 0 V V kT V V W W RT b b c abc = bc = b = = 8.31353 ln 2 = 2033(J) = + = 2033+1246 = 3296(J) Qabc Wabc Eabc （2） ln( ) a ln 2 a d adc ad a KT V V W = W = JT = = 8.31 293 ln 2 = 1687(J) = + =1246 +1687 = 2933(J) Qadc Eadc Wad 8-4 （1）N2 体积 等温： 10 (m ) 10 0.1 0.01 4 3 2 1 1 1 1 2 2 2 − = = = = p p V p V p V V 绝热： 1/1.4 4 3 1 1/ 2 1 1 1 2 2 2 ) 0.01 3.73 10 m 10 0.1 ( ) ( − = = V = = p p p V p V V r r r （2）N2 温度 等温： T2 = T1 = 300K 绝热： r r r r p T p T − − − − = 2 1 1 2 1 1 K p p T T r r ( ) 1118 1 1 2 2 = 1 = − （3）N2 对外做功 等温： ln( ) 2 1 p p RT M m WT = QT = 又 1 1 RT p2V2 M m pV = = ∴ ) 4.67 10 J 10 0.1 ln( ) 1.013 10 0.01 ln( 5 3 2 1 = 1 1 = = − p p WT p V 绝热： R i C T T p V p V M m Ws E v m 2 ( ) ( ) = − = − , 2 − 1 = 1 1 − 2 2
(1.013×103×0.01-100×1.013×105×373×10-4)××8.31 -6.9×103(J) 8-5 W 4E=m C,n(T-70)=xC,m(70-T i+2 +2 m,时=-m)=B=m 8-6由p=M RT→p=,R(),由图知为恒量,故a→b为一等压过 程.P R W=(2-V) 8-7设状态A的温为 AB为等容线,故Tn=PT7=27 题8-7图 V p, 2 AB为等温线,故T=7=TPVe3123 Q=△EAB+O=R(7B-7)==R(2T1-71)=3R71=3PV1吸热 QBC=△Ec+WB AEBC =R( -TB)=RT-21=-3RT=-3pVI WBC等于图中阴影部分的面积值 Bc=(p+P)x(2-H2)(B+2p)×2V1-) 2 PI @Bc=-3p, 1+P,=--pV1 放热 Oc=WC+O=RTahn( . )=rti hn=-p, V,n 放热
73 6.9 10 (J) 8.31 2 5 (1.013 10 0.01 100 1.013 10 3.73 10 ) 3 5 5 4 − = − − 8-5 ( ) ( ) , 0 C , T0 T M m C T T M m Ws = −E = − v m − = v m − ∵ 2 2 / 2 2 2 , , + = + = = i i i C C r v m p m ∴ 1 2 − = r i 又∵ RT M m pV = , ∴ 1 ( ) 1 1 0 0 0 − − − = − = r p V pV RT M m RT M m r Ws 8-6 由 ( ) V T R M m RT p M m pv = = , 由 图 知 V T 为 恒 量 ， 故 a →b 为一等压过 程. 0 0 2V RT P = 2 (2 ) 0 2 0 0 0 0 RT W V V v v = − = 8-7 设状态 A 的温为 TA=T1 ∵AB 为等容线，故 T1 2T1 p p T A B B = = AB 为等温线，故 1 1 1 1 1 3 2 3 / 2 P V p V V p V T T T p C A A C = A = C = = = 1 1 3 1 3 1 1 (2 ) 2 6 ( ) 2 R T T R T T RT PV i QAB = EAB + O = B − A = − = = 吸热 QBC = EBC +WBC 1 1 3 1 3 1 1 ( 2 ) 2 6 ( ) 2 R T T R T T RT p V i EBC = C − B = − = − = − WBC等于图中阴影部分的面积值. 2 ) 2 3 )( 3 2 (2 2 ( ) ( ) 1 1 1 1 2 p p V V p p V V W B C A BC + − = + − = 1 1 3 2 = p V ∴ 1 1 1 1 1 1 3 7 3 2 QBC = −3p V + p V = − p V 放热 2 3 ln 3 2 ln( ) RT1 ln p1V1 V V Q W O RT C A CA = CA + = A = = − 放热 题 8-7 图
g2=1-19mc|+|4 (+ln=)p1V1 ≈8.7% 3P1V1 8-8(1)AB等温膨胀AEA,=0W=O,>0 吸热 BC等容降温△EBC<O.WBC=0,QBc=AEC<0放热 (2)CA绝热过程TV=T2→Tc=7(y (3)不是卡诺循环 (4)AB过程 BC过程Qm=mcn(r-Tn)=-mc,n1 CA绝热过程Qc4=0 =1-9=1-g1,C-《/ g Q RIn(v,/v 9(1)1→2等温线的熵变为 As=3-1x9,RM(/)=R(/) =Rh2=8.31×0.693=5.76(J/k) (2)1→4绝对热线和4→2等压线 AS2=AS14+△S42=0+△S42= CdT RIn( 又pH=P2H4 P,I=P2V2 PI P RIn 2 (3)1→>3等压线,3→2等容线的过程
74 8.7% 3 ) 2 3 ln 3 7 ( 1 | | | | 1 1 1 1 1 1 1 2 + = − + = − = − p V p V Q Q Q Q Q AB BC CA 8-8 （1）AB 等温膨胀 EAB = 0,WAB = QAB 0 吸热 BC 等容降温 EBC 0,WBC = 0,QBC = EBC 0 放热 （2）CA 绝热过程 1 2 1 1 1 2 1 1 1 ( ) − − − = = r C r C r V V TV T V T T （3）不是卡诺循环 （4）AB 过程 ln( ) 1 2 1 V V RT M m QAB = WAB = BC 过程 = − = − − −1 2 1 , , 1 ( ) 1 ( ) r BC v m C B v m V V C T M m C T T M m Q CA 绝热过程 QCA = 0 ( ) ln( / ) [1 ] 1 | | 1 1 2 1 1 , 1 2 1 2 R V V C V V Q Q Q Q r v m AB BC − − = − = − = − 8-9 （1） 1→2 等温线的熵变为 ln( ) d ln( ) 2 1 2 1 1 1 2 1 1 12 2 1 R V V T RT V V T Q T Q S S S T = − = = = = Rln 2 = 8.31 0.693 = 5.76(J/K) （2） 1→4 绝对热线和 4 → 2 等压线 = + = + = 2 4 12 14 42 42 0 ( ) p T dQ S S S S ) 2 ln( 2 7 ln( ) ln( ) d 4 1 4 2 4 2 2 4 V V R V V C T T C T C T p p p = = = = 又 p1V1 p4V4 p1V1 p2V2 r r = = r r r r V V p p p p V V 1/ 1/ 2 1 1/ 1 2 1/ 1 4 4 1 2 1 = = = = ∴ ln 2 2 1 ln 2 2 7 1/1.4 S12 R = R = （3） 1→3 等压线， 3→ 2 等容线的过程
△S, CpdT, r 2 Cr dT C. In T1 ∵1>3等压过程,故T3/T1=V3/W=V2/V 3→2等容过程,故T2/73=P2/P3=P2/P1=V1/2 Rh(2)+-rhn(-)=rhn 2 AS12=AS12=4S12 因为熵是状态函数,熵变只与初末态有关,而这三个过程的初末态相同,所以自然熵变 也相同 8-10Q放=Q吸,设最后其同温度为r T1+ Cn(T1-T)=Cn(T-T2)→T AS=AS,+AS, Cdt cr CdT T ln()+1 T;72 4T, 8-11(1)冰的溶解热C=3352×103Jkg (水吸热,Q0) g=320690K Q-3352×103 0.5≈-571J/K 293 (3)△S=AS1+AS,=613-571=42J/K 8-12房间热量流失,则
75 = + = + 2 3 3 1 2 3 3 1 12 d T d ) d ) ( d ( T C T C T T Q T Q S p V p V + = 3 2 1 3 ln ln T T C T T Cp v ∵ 1→3 等压过程，故 T3 T1 =V3 V1 =V2 V1 3→ 2 等容过程，故 T2 T3 = p2 P3 = P2 P1 =V1 V2 ∴ ln( ) ln 2 2 5 ln( ) 2 7 2 1 1 2 12 R V V R V V S = R + = ∴ = S12 = S12 S12 因为熵是状态函数，熵变只与初末态有关，而这三个过程的初末态相同，所以自然熵变 也相同. 8-10 Q放 = Q吸 ，设最后其同温度为 T 2 ( ) ( ) 1 2 1 2 T T Cm T T Cm T T T + − = − = = + = + T T T T T Q T Q S S S 1 2 d d 1 2 = + T T m T T m T C T T C T 1 2 d d 1 2 2 1 2 1 2 2 1 2 4 ( ) ln( ) ln( ) ln ln T T T T C T T T C T T T T Cm m m + = = = + 8-11 （1）冰的溶解热 335.2 10 J/kg 3 C = (水吸热，Q>0) 0.5 613J/K 273 d 335.2 103 1 = = = = T C T Q S m （2） 0.5 571J/K 293 335.2 103 2 − − = − = T Q S （3） S = S1 + S2 = 613−571= 42J/K S 0 8-12 房间热量流失，则 T t Q t S d d d d 1 1 = −