Problems and Short Answer Questions: Show your work where appropriate. 8.) A steel thick walled container contains 5 ft of butane at 20 atm and 300 F. (0F = 460R, 1 atm = 14.696 lb/in, R= 1545.3 ft lb/lbmol-'R, Tc = 765 R, Pc= 37.5 atm, MW = 78.11 lb/lbmol) (a) Determine the compressibility factor of the butane. (6 points) Z = PV RT (20) (14.696) 5 (1545-3) (300) (460) R = 6.89106 (b) Determine the mass of butane in the container. (4 points) Convert psi (in 381) M = PV (20) (14.696)5 = 6.89 -6 (1545-3)(300)(460) R = 1.0002 lb 3.50 V-0 Compressibility factor, Z = PvIRT 1.20 1.10 1.00 0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20 (M2 P P K 1.1.1 XX Ph 1 7 1.0 L 10 L 1.05 2.0 Solid lines = TR Dotted lines = v 5.0 3.0 4.0 Reduced pressure, PR 6.0 7.0