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Unformatted text preview: elfenbein (ee4265) Homework 9 sutcliffe (50985) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Whats the chemical formula of water? HIJKLMNO (a little geek humor for you) 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, and sometimes we switch our viewpoint of what the system is once we have finished a calculation. This cov- ers Ch. 9. The rest will be on HW 10, the last HW. 001 10.0 points When a hydrocarbon is burned in the pres- ence of oxygen, the amount of energy in the universe 1. stays the same. correct 2. decreases. 3. increases. Explanation: 002 10.0 points A sample of an ideal gas having a volume of 0.125 L at 298 K and 10.0 atm pressure is al- lowed to expand against a constant opposing pressure of 0.341 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:- 30 . 2254 J. Explanation: P 1 = 10.0 atm P 2 = 0.341 atm V 1 = 0.125 L V 2 = 1.000 L T 1 = 298 K T 2 = 200 K In this case, the P is the constant opposing pressure 0.341 atm. V is the change in volume you observe, from 0.125 L to 1.000 L. Plugging these values into the equation, you get w =- P V w =- (0 . 341 atm)(1 . 000 L- . 125 L) =- (0 . 341 atm)(0 . 875 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 . 341 atm)(0 . 875 L) parenleftbigg 1000 cm 3 1 L parenrightbiggparenleftbigg 101325 N / m 2 1 atm parenrightbigg parenleftbigg 1 m 3 10 6 cm 3 parenrightbigg =- 30 . 2 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. +450 kJ 2.- 550 kJ 3.- 50 kJ correct 4.- 950 kJ 5.- 800 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 . elfenbein (ee4265) Homework 9 sutcliffe (50985) 2 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 player does is irrelevant, unless it involves the...
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This note was uploaded on 11/28/2010 for the course CH 50985 taught by Professor Sarasutcliffe during the Spring '10 term at University of Texas at Austin.
- Spring '10