x1=5/12 x1=1/3 x1'=8/21 x2 X 7/12 4/21 4/21 (8m1苯」+|9mol甲苯+4mol二甲苯)-(5分) △S2=△S3-AS1=- Rong In xb-[- R2nB In xB 4 5 8314(8n1n+9nn+4lnn;)-(5ln;+7ln)-(3ln+2lna+4na) 35.6J·K 26.10分(0982) 0982 [答 因为dU= TdS-pd 所以(oUar=T(0p/0Tn-p T([RT/(Vm-b))aTr- RT/(m-b=0 故 △U ∫(oUad △Hm=△Um+△(plm=(p2m2Pp1m) b(P2-P1) 39×102dm3mol×(1-1000)×101325kPa △Gn ∫"(RT/p+b RTIn(pa/p1)+b(P2-P1) 8314JKmo1×500K×lnp531000p5 +39×102dm3mor4×(1-1000)×101.325kPa △Fa1-pur=∫a-b (积分区间:Imn1到Vm,2) RTIn((m, 2-b)(m, 2-b) RTIn(ppi) 72 kNoll △Sm=(△H△Gm)7=5742 J-K.mol 27.5分(1641)ln [ ln ] 2 3 1 B B B B ∆S = ∆S − ∆S = −RΣn′ x′ − −RΣn x = −8 314 8 + + − + − + + 8 12 9 3 7 4 4 21 5 5 12 7 7 12 3 1 3 2 2 9 4 4 9 . [( ln ln ln ) ( ln ln ) ( ln ln ln )] （5 分） 1 35.6 J K− = ⋅ 26. 10 分 (0982) 0982 [答] 因为 dU= TdS-pdV 所以 ( ∂ U/ ∂ V)T = T( ∂ p/ ∂ T)V- p = T{ ∂ [RT/(Vm-b)]/ ∂ T}V- RT/(Vm-b)=0 故 ΔUm=∫( ∂ U/ ∂ V)TdV=0 ΔHm=ΔUm+Δ(pVm)=(p2Vm,2- p1Vm,1) =b(p2-p1) =3.9×10-2 dm3 ·mol-1×(1-1000)×101.325 kPa = -3.948 kJ·mol-1 ΔGm 2 2 1 1 m d ( / ) p p p p = = V p RT p + b p ∫ ∫ d = RTln(p2/p1)+b(p2-p1) = 8.314 J·K-1 ·mol-1×500K×ln[ p$ /(1000× p$ )] +3.9×10-2 dm3 ·mol-1×(1-1000)×101.325kPa = -32.66 kJ·mol -1 ΔFm= 2 1 m m d ( )d( V V RT p V ) V b − = − − ∫ ∫ V b (积分区间：Vm,1到Vm,2) = -RTln[(Vm,2-b)/(Vm,2-b)] = RTln(p2/p1) = -28.72 kJ·mol -1 ΔSm= (ΔHm-ΔGm)/T= 57.42 J·K-1 ·mol-1 27. 5 分 (1641)