HW12_Solutions (1)

# HW12_Solutions (1) - triesault(njt298 H12 Thermo 1st Law...

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triesault (njt298) – H12: Thermo - 1st Law – McCord – (53740) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Remember that internal energy is E in our book, but is U in many other books. Quest (and this assignment) has questions with both symbols. 001 10.0 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 10.0 points Work is 1. organized molecular motion. correct 2. chaotic molecular motion. 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 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 21 cm against an external pressure of 519 torr. What is the value for w for this reaction? (Sign does matter.) Correct answer: - 145 J. Explanation: 004 10.0 points A 100 W electric heater (1 W = 1 J / s) oper- ates for 12 . 5 min to heat the gas in a cylinder. At the same time, the gas expands from 3 L to 15 L against a constant atmospheric pressure of 3 . 742 atm. What is the change in internal energy of the gas? Correct answer: 70 . 4501 kJ. Explanation: P ext = 3 . 742 atm V ini = 3 L 1 L · atm = 101 . 325 J V final = 15 L If the heater operates as rated, then the to- tal amount of heat transferred to the cylinder will be q = (100 J / s) (12 . 5 min) (60 s / min) = 75000 J = 75 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 = - (3 . 742 atm) (15 L - 3 L) = - 44 . 904 L · atm Convert to kilojoules (kJ) w = ( - 44 . 904 L · atm)(101 . 325J / L · atm) = - 4549 . 9 J = - 4 . 5499 kJ The internal energy change is Δ U = q + w = 75 kJ + ( - 4 . 5499 kJ) = 70 . 4501 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?

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triesault (njt298) – H12: Thermo - 1st Law – McCord – (53740) 2 1. The system gained 90 kJ of energy as heat. 2. The system lost 90 kJ of energy as heat. correct 3. The system gained 210 kJ of energy as heat. 4. The system gained 60 kJ of energy as heat. 5. The system lost 210 kJ of energy as heat. Explanation: Δ U = q + w +60 kJ = q + 150 kJ q = - 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 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. +2 . 61 kJ 2. +1 . 39 kJ correct 3. 0 4. - 2 . 61 kJ 5. - 0 . 608 kJ Explanation: Δ U = q + w We know q , we need w : For expansion against a constant external pressure w = - P ext Δ V w = ( - 2 atm)(5 L - 2 L) × (101 . 325 J · L 1 · atm 1 ) = - 607 . 95 J = - 0 . 60795 kJ .
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