H,0(1D (1) H20(g) (2）H,0(g) 101325Pa相变 一101325Pa 恒温 一31025.7Pa 理想气体恒温过程△=0，△U=0 AhAH+Ah=△H=40.59kJ AU-AU+AUh=△U=△H-Ap=△H-RT=37.47kJ 因对整个容器系统△=0，故W-0,Q-△U=37.47k △S△S+△S2=△h/T-nRIn(p2lp)=118.60J·K-1 4M=4U-74S-37.49J-118.60x373.15J=-6.771kJ △G-△H-T△S=40.59kJ-118.60×373.15J=-3.672kJ 六、 解：)求p(苯)和p(甲)，可由克克方程 In P2=AL (T -T) RTT m101325Pa83145J.K.moJ'x35315Kx373号 得 30.03×10J-m011373.15K-353.15=0.5482 p'(*)=175.30kPa 后保惠-e0架-w p'(甲苯)=76.20kPa (2)+液相组成及气相组成可由拉乌尔定律求得： p(总)=卫（苯）x(苯)什p(甲苯){1x(苯)} x(苯)={p(总)-p(甲苯)}/{p(苯)-p(甲苯)} =(101.325-76.20)kPa/175.30-76.20Pa =0.2535 x(甲苯)=1-x(苯)=1-0.2535=0.7465 (苯)p(苯)x(*)/p(总)=175.30kPa×0.2535/101.325kPa=0.4386 甲苯)=1-（苯）=1.0.4386=0.5614 理想气体恒温过程 H2=0， U2=0 H= H1+ H2= H1=40.59 kJ U= U1+ U2= U1= H1–(pV)= H1–RT=37.47 kJ 因对整个容器系统 V=0，故 W=0，Q= U = 37.47 kJ S= S1+ S2= H1/T – nRln(p2/p1) =118.60 J·K–1 A= U –TS=37.49kJ –118.60373.15 J = –6.771 kJ G= H –TS=40.59kJ –118.60373.15 J = –3.672 kJ 六