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CHE 344 Chemical Engineering Thermodynamics, Spring 1996 Exam I, February 2, 1996 Each question 15 points, total 45 points Attempt ALL Q’s. Setting it up right is more important . Using the pressure explicit (equation 3-3, page 41) and the volume explicit (equation 3- 3a, page 41) virial equation of state, show that B’ = #& and C’ = Sane (Express the pressure explicit equation in terms of volume, compare terms, or express the volume explicit equation in terms of pressure and compare terms) . Acetonitrile, C2H3N, is stored at a temperature of 550°F and a pressure of 4500 psia in a high pressure vessel with a volume of 0.2 cubic feet. For safety, the 0.2 cubic feet vessel, A, is surrounded by a second pressure vessel. The volume of the second pressure vessel is 2.0 cubic feet (exclusive of the volume occupied by the vessel A). The second vessel is filled with inert nitrogen at 10.0 atm. The entire assembly is maintained at 550°F. If the inner vessel ruptures, what will be the final pressure? Assume a simple n mixing rule for this problem: Zmizture = )|(x:Z;) where n is the total number of components, 2; is the mole fraction of component t and Z; is the compressibility factor of i in the mixture, under mixture T and P. (Note: I have greatly simplified the mixture rule for this problem) Here is some data, state any additional assumptions you may have to make to solve the problem. N2 126.2 33.5 CyH3N | 548.0 47.7 . Let C, for Nitrogen be 5.0 cal/gmol.K. Two pound moles of nitrogen at 350°F are contained in a piston-cylinder arrangement, kept at a constant pressure. Calculate the heat that must be extracted from the system to cool it to 60°F if you can neglect the heat capacity of the piston and the cylinder. Assume that nitrogen behaves as an ideal gas for this problem. Values of the Universal Gas Constant R=8.314J-mol'-K~! =8.314kJ-kmol~'- K~'=8.314 m?- Pa-mol™'-K7! = 0.008 314 kJ -mol~* = 0.083 14 L- bar-mol™'-K~' . = 83.14cm?-bar-mol7!- K7! = 8314cm?- kPa-mol”'-K7' 2.06 cm? - atm: mol~!- K~! = 0.082 06 L- atm-mol™'-K7! = 62356 cm’ - torr- mol”! - K~! = 62.356 L- torr- mol~'-K~! = 1.987 cal- mol”! - K~! = 1.986 Btu - Ib mol”'- R7* .7302 ft? - atm -1b mol! R7! = 10.73 ft? - psia “Ib mol”! R~! = 1545 ft-Ib,-Ib mol”'-R7! 
