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Unformatted text preview: dewitt (aad777) – Homework 9 – sutcliffe – (52380) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. NOTE: Read the questions carefully. If it asks for a CHANGE (Delta) in a quantity or the VALUE of that quantity, then you must be sure to include the correct sign. Remem- ber that the quantities are for the SYSTEM, unless stated differently. This covers Ch. 9. The rest will be on HW 10, the last HW. As always, the deadline is 11pm, the due date is shown on Quest. 001 10.0 points When a hydrocarbon is burned in the pres- ence of oxygen, the amount of energy in the universe 1. increases. 2. decreases. 3. stays the same. correct Explanation: 002 10.0 points A sample of an ideal gas having a volume of 0.413 L at 298 K and 10.0 atm pressure is al- lowed to expand against a constant opposing pressure of 0.753 atm until it has a volume of 1.000 L at 200 K and the pressure equals the opposing pressure. What is the work for the system? Correct answer:- 44 . 7757 J. Explanation: P 1 = 10.0 atm P 2 = 0.753 atm V 1 = 0.413 L V 2 = 1.000 L T 1 = 298 K T 2 = 200 K In this case, the P is the constant opposing pressure 0.753 atm. Δ V is the change in volume you observe, from 0.413 L to 1.000 L. Plugging these values into the equation, you get w =- P Δ V w =- (0 . 753 atm)(1 . 000 L- . 413 L) =- (0 . 753 atm)(0 . 587 L) However, some conversion factors will have to be applied. To get to Joules, which is the same as N · m, two conversions will be needed: L to m 3 and atm to unit of pressure which includes newtons (which is Pa, defined as N/m 2 ). Recalling that 1 mL = 1 cm 3 , and 100 cm = 1 m, the equation should be w =- (0 . 753 atm)(0 . 587 L) × parenleftbigg 1000 cm 3 1 L parenrightbiggparenleftbigg 101325 N / m 2 1 atm parenrightbigg × parenleftbigg 1 m 3 10 6 cm 3 parenrightbigg =- 44 . 8 J 003 10.0 points A CD player and its battery together do 500 kJ of work, and the battery also releases 250 kJ of energy as heat and the CD player re- leases 50 kJ as heat due to friction from spin- ning. What is the change in internal energy of the system, with the system regarded as the CD player alone? Assume that the bat- tery does 500 kJ of work on the CD player, which then does the same amount of work on the surroundings. 1.- 550 kJ 2. +450 kJ 3.- 800 kJ 4.- 50 kJ correct 5.- 950 kJ Explanation: Heat from the CD player is- 50 kJ. Battery (part of the surroundings ) does- 500 kJ work on the CD player. CD player does- 500 kJ work on some other part of the surroundings . This question is testing your ability to see what the system is, and then look at ONLY the energy flow for the system. Here the sys- tem is just the CD player. What the battery dewitt (aad777) – Homework 9 – sutcliffe – (52380) 2 player does is irrelevant, unless it involves the CD player: Δ U = q + w =- 50 kJ + [500 kJ + (- 500 kJ)] =- 50 kJ 004 10.0 points Calculate the energy required to heat 128 g...
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This note was uploaded on 10/20/2011 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.
- Fall '07