Chem 2B-quiz 2-011707

# Chem 2B-quiz 2-011707 - m = (-M w c s , w T w )/( M m T m )...

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Chem 2B, Winter 2007 Professor Thuc-Quyen Nguyen Name: Quiz #2 01/12/07 1) ( 5 points ) The gas mixture inside one of the cylinders of an automobile engine expands against a constant external pressure of 0.98 atm, from an initial volume of 150 mL (at the end of compression stroke) to a final volume of 800 mL. Calculate the work done on the gas mixture during this process and express it in joules. The work done in a change of volume at constant pressure is; w = -P ext ΔV ΔV = V 2 -V 1 = 800-150 = 650 ml = 0.65 L w = -0.98 atm x 0.65 L = -0.64 L atm 1 L atm = 101.325 J, Therefore w = -65 J 2) ( 5 points ) Suppose 61g of hot metal, which is initially at 120 ° C, is plunged into 100 g of water that is initially at 20 ° C. The metal cools down and the water heats up until they reach a common temperature of 26.39 ° C. Calculate the specific heat capacity of the metal, using 4.18 J/g. K as the specific heat capacity of the water. q sys = 0 = q m + q w For both sub-systems; q m + q w = M w c s , w ΔT w + M m c s , m ΔT m = 0 c s ,

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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.

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Chem 2B-quiz 2-011707 - m = (-M w c s , w T w )/( M m T m )...

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