# hw 10 answers - Iype Shelby Homework 10 Due 6:00 pm Inst...

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Iype, Shelby – Homework 10 – Due: Nov 21 2007, 6:00 pm – Inst: Vandenbout 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points On January 1, 2001, a bookstore had 266,478 books in stock. On February 1, 2001, the same bookstore had 257,814 books in stock. What is the value of Δ(books) for the bookstore during that period? Correct answer: - 8664 books. Explanation: B f = 257 , 814 books B i = 266 , 478 books Δ B = B f - B i = 257 , 814 books - 266 , 478 books = - 8664 books Thus the bookstore had 8664 fewer books on February 1. 002 (part 1 of 1) 10 points Work is 1. chaotic molecular motion. 2. organized molecular motion. correct 3. heat. 4. kinetic energy. 5. potential energy. Explanation: Work is organized molecular motion since it is displacement of a solid against a force. 003 (part 1 of 1) 10 points A chemical reaction takes place in a container of cross-sectional area 100 cm 2 . As a result of the reaction, a piston is pushed out through 20 cm against an external pressure of 558 torr. What is the value for w for this reaction? (Sign does matter.) Correct answer: - 149 J. Explanation: 004 (part 1 of 1) 10 points A 100 W electric heater (1 W = 1 J / s) oper- ates for 14 min to heat the gas in a cylinder. At the same time, the gas expands from 5 L to 12 L against a constant atmospheric pressure of 4 . 332 atm. What is the change in internal energy of the gas? Correct answer: 80 . 9274 kJ. Explanation: P ext = 4 . 332 atm V ini = 5 L 1 L · atm = 101 . 325 J V final = 12 L If the heater operates as rated, then the to- tal amount of heat transferred to the cylinder will be q = (100 J / s) (14 min) (60 s / min) = 84000 J = 84 kJ Work will be given by w = - P ext Δ V in this case because it is an expansion against a constant opposing pressure: w = - (4 . 332 atm) (12 L - 5 L) = - 30 . 324 L · atm Convert to kilojoules (kJ) w = ( - 30 . 324 L · atm)(101 . 325J / L · atm) = - 3072 . 58 J = - 3 . 07258 kJ The internal energy change is Δ U = q + w = 84 kJ + ( - 3 . 07258 kJ) = 80 . 9274 kJ 005 (part 1 of 1) 10 points A system had 150 kJ of work done on it and its internal energy increased by 60 kJ. How much energy did the system gain or lose as heat? 1. The system gained 60 kJ of energy as heat.

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Iype, Shelby – Homework 10 – Due: Nov 21 2007, 6:00 pm – Inst: Vandenbout 2 2. The system lost 210 kJ of energy as heat. 3. The system gained 210 kJ of energy as heat. 4. The system lost 90 kJ of energy as heat. correct 5. The system gained 90 kJ of energy as heat. Explanation: 006 (part 1 of 1) 10 points When 2.00 kJ of energy is transferred as heat to nitrogen in a cylinder fitted with a piston at an external pressure of 2.00 atm, the nitro- gen gas expands from 2.00 to 5.00 L against this constant pressure. What is Δ U for the process? 1. +1 . 39 kJ correct 2. 0 3. +2 . 61 kJ 4. - 0 . 608 kJ 5. - 2 . 61 kJ Explanation: 007 (part 1 of 1) 10 points A 1.00 g sample of n -hexane (C 6 H 14 ) under- goes complete combustion with excess O 2 in a bomb calorimeter. The temperature of the 1502 g of water surrounding the bomb rises from 22.64 C to 29.30 C. The heat capacity of the calorimeter is 4042 J/ C. What is Δ U for the combustion of n -C 6 H 14 ? One mole of n -C 6 H 14 is 86.1 g. The specific heat of water is 4.184 J / g ·
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• Fall '07
• Fakhreddine/Lyon

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