T=73℃,P=101058Pa P=2809Pap=0.95g·mo 17(g) 求:K=? 解:k=t’/t0=45-0.5/0.5=80 F。 f Po-Pm=101058-2809/101058=0972 Pc T=Tc/Tr=373/296=126 ∴FCo=80×0.972×1,26=9798(ml·min-) 3(pi/p0)2-13.(162920/101058) J(pilp)-1-2162920101058)-1 32.599-1 =0.752 24.1900-1 Fco=Fco·J=9788×00752=78.7(mlmn Vn=Fco·l0=73.7×0.5=36.85(m =4.2/0.95=442 K=k·Vn/V=8.0×3685/442=66.1 16.15已知:t=72(cm),t1正已烷=85(cm),t烷=146(cm) 15.5(cm),甲苯=18.7(cm) 315(cm) 求 环乙烷 解:I g(187-22)-g(159-22) 7 lg(315-2.2)-lg(15.9-2.2) 环已=100×9(146-22)-g(85-22) lg(15.9-22)-lg(8.5-2.2)Tr = 73 ℃， Pc =101058Pa P Pa g mol w = 2809  = 0.95  w用 = 4.7(g) 求： K = ? 解： 0 ' / 4.5 0.5/0.5 8.0 r k t t = = − = Fco = Fco  J Fco Fo wf Tf =   101058 2809 /101058 0.972 0 0 = − = − = p p p w w f Tf = Tc /Tr = 373/ 296 = 1.26 ∴ 80 0.972 1.26 97.98( min ) −1 Fco =   = ml  (162920 /101058) 1 (162920 /101058) 1 2 3 ( / ) 1 ( / ) 1 2 3 3 2 3 2 − − =  − − =  pi po pi po J 0.752 4.1900 1 2.599 1 2 3 = − − =  97.88 0.0752 78.7( min ) −1 Fco = Fco J =  = ml  73.7 0.5 36.85( ) V Fco t 0 ml n =  =  = Vs = 4.2/ 0.95 = 4.42 K = k Vm /Vs = 8.036.85/ 4.42 = 66.1 16. 15 已知： t 7.2(cm) c = ， 85( ) 1 t 正已烷 = cm ，t 146(cm) r环烷 = t 15.5(cm) r正庚烷 = ， 18.7( ) 1 t 甲苯 = cm ，t 315(cm) r正辛烷 = 求： I甲苯 = ? I环乙烷 = ? 解: 7] 724 lg( 31.5 2.2) lg(15.9 2.2) log(18.7 2.2) lg(15.9 2.2) 100 [ + = − − − − − − I甲苯 =  6] 687 lg(15.9 2.2) lg(8.5 2.2) lg(14.6 2.2) lg(8.5 2.2) 100 [ + = − − − − − − I环已烷 = 