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HW12 solution_pdf-1

# HW12 solution_pdf-1 - tan(jt27678 H12 Thermo 1 mccord(50970...

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tan (jt27678) – H12: Thermo 1 – mccord – (50970) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points On January 1, 2001, a bookstore had 266,478 books in stock. On ±ebruary 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 ±ebruary 1. 002 10.0 points Work is 1. kinetic energy. 2. chaotic molecular motion. 3. organized molecular motion. correct 4. potential energy. 5. heat. Explanation: Work is organized molecular motion since it is displacement oF a solid against a Force. 003 10.0 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 22 cm against an external pressure oF 554 torr. What is the value For w For this reaction? (Sign does matter.) Correct answer: - 162 J. Explanation: 004 10.0 points A 100 W electric heater (1 W = 1 J / s) oper- ates For 6 . 5 min to heat the gas in a cylinder. At the same time, the gas expands From 2 L to 9 L against a constant atmospheric pressure oF 4 . 077 atm. What is the change in internal energy oF the gas? Correct answer: 36 . 1083 kJ. Explanation: P ext = 4 . 077 atm V ini = 2 L 1 L · atm = 101 . 325 J V Fnal = 9 L IF the heater operates as rated, then the to- tal amount oF heat transFerred to the cylinder will be q = (100 J / s) (6 . 5 min) (60 s / min) = 39000 J = 39 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 . 077 atm) (9 L - 2 L) = - 28 . 539 L · atm Convert to kilojoules (kJ) w = ( - 28 . 539 L · atm)(101 . 325J / L · atm) = - 2891 . 71 J = - 2 . 89171 kJ The internal energy change is Δ U = q + w = 39 kJ + ( - 2 . 89171 kJ) = 36 . 1083 kJ 005 10.0 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 lost 90 kJ oF energy as heat. correct

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tan (jt27678) – H12: Thermo 1 – mccord – (50970) 2 2. The system gained 60 kJ of energy as heat. 3. The system gained 90 kJ of energy as heat. 4. The system gained 210 kJ of energy as heat. 5. The system lost 210 kJ of energy as heat. Explanation: Δ U = q + w q = Δ U - w = 60 kJ - (+150 kJ) = - 90 kJ (Negative means the system lost energy as heat.) 006 10.0 points When 2.00 kJ of energy is transferred as heat to nitrogen in a cylinder Ftted 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?
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HW12 solution_pdf-1 - tan(jt27678 H12 Thermo 1 mccord(50970...

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