Discussion problems 1 - KEY

Discussion problems 1 - KEY - In a constant volume process:...

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Chemistry 152 Discussion Section Problems 1. The volume of a system changes from 20.0 to 30.0 L in a process against a 15.0 atm pressure by adsorbing 215 J heat from the surroundings. Determine the ∆E (in J) of the system (1 atm∙L = 101.3 J). This is a constant pressure process: ∆V = 30.0 – 20.0 = 10.0 L; P = 15.0 atm; w = -P∆V = -150 atm L = -150 * 101.3 = -15.2 kJ q = 215 J (q > 0 since the system adsorbs energy from the surroundings) ∆E = q + w = 215 J - 15195 J = -150 kJ 2. 2.00 mole monatomic ideal gas argon at 2.00 atm and 25.0 o C. What will the final temperature be if 3.00 kJ heat is transferred at: a) constant volume; b) constant pressure. Calculate w , E, and H for each case.
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Unformatted text preview: In a constant volume process: w = 0 (because this is a constant volume process) q v = E = nC v T = 3000 J = E = (2.00 mol)(12.47 J.K-1 mol-1 ) T T = 3000 J/(2.00mol)(12.47 J K-1 mol-1 ) = 120 K T final = 298 K + 120 K = 418 K H = nC p T = (2.00 mol)(20.79 J.K-1 mol-1 ) 120 K = 4.99 kJ In a constant pressure process: q p = H = nC p T = 3000 J = H = (2.00 mol)(20.79 J.K-1 mol-1 ) T T = 3000 J/(2.00mol)(20.79 J K-1 mol-1 ) = 72.2 K T final = 298 K + 72.2 K = 370 K p V = nR T w = -p V = -(2.00 mol)(8.314 J.K-1 mol-1 )72.2 K = -1.20 kJ E = nC v T = (2.00 mol)(12.47 J.K-1 mol-1 ) 72.2 K = 1.80 kJ...
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This note was uploaded on 01/18/2012 for the course CHEM 152B taught by Professor Zhang during the Winter '12 term at University of Washington.

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