njit-etd1961-002 - New Jersey Institute of Technology

c =p13The critical properties were calculated as follows:(A) Critical pressure P c by Lydersen's method8, 10Mc o + I. 2where P c = critical pressure in atmosphereM = molecular weight = 30.03= atomic and structural constant= 0.33P 30.03c = 2 - 66.897 atm.(0.33 + 0.34)(B) Critical temperature Tc by Lydersen's method8,¹0TbT c(εΔT)²T c =where T c = critical temperature in °KTb = normal boiling point = 253.96 ° KT= structural contribution from-CHO group = 0.048253.96414.49°K0.567 + 0.048 - (0.048)²(C) Critical volume by Lydersen's method 8V c = 40 +where V c = critical volume in cc/g-moleΔv = atomic and atomic group contributionto V c = 73 for -CHOgroupV c= 40 + 73 = 113 cc/gm-mole