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Unformatted text preview: m = (-M w c s , w T w )/( M m T m ) = -[(100.0g)(4.18 J K-1 g-1 )(6.39 C)]/[(61.0 g)(-93.61C)] = 0.468 J K-1 g-1 Chem 2B, Winter 2007 Professor Thuc-Quyen Nguyen 3) ( 10 points ) Imagine that 2.00 mol of argon, confined by a movable, frictionless piston in a cylinder at a pressure of 1.00 atm and a temperature of 398 K, is cooled to 298 K. Argon gas may be considered ideal, and its molar heat capacity at constant pressure is Cp = (5/2)R where R = 8.315 J/K.mol. Calculate: (a) The work done on the system, w. w = -nRT = -(2.00 mol)(8.3145 J K-1 mol-1 )(-100K) = +1.66x10 3 J (b) The heat absorbed by the system, q. This is a constant P process; q p = nc p T = (2.00 mol)(5/2)(8.3145 J K-1 mol-1 )(-100K) = -4.16x10 3 J (c) The energy change of the system, E. E = q + w = -4157 + 1663 = -2494 J = -2.494x10 3 J (d) The enthalpy change of the system, H. H of a system always equals to q p , hence H = -4.16 kJ...
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This note was uploaded on 08/06/2008 for the course CHEM 2B taught by Professor Nguyen during the Winter '07 term at UCSB.
- Winter '07