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CH_301-Sutcliffe-HW9 - romasko(qrr58 Homework 9...

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romasko (qrr58) – 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. What’s 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. increases. 2. decreases. 3. stays the same. correct Explanation: 002 10.0 points A sample of an ideal gas having a volume of 0.220 L at 298 K and 10.0 atm pressure is al- lowed to expand against a constant opposing pressure of 0.825 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: - 65 . 1865 J. Explanation: P 1 = 10.0 atm P 2 = 0.825 atm V 1 = 0.220 L V 2 = 1.000 L T 1 = 298 K T 2 = 200 K In this case, the P is the constant opposing pressure 0.825 atm. Δ V is the change in volume you observe, from 0.220 L to 1.000 L. Plugging these values into the equation, you get w = - P Δ V w = - (0 . 825 atm)(1 . 000 L - 0 . 220 L) = - (0 . 825 atm)(0 . 780 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 . 825 atm)(0 . 780 L) × parenleftbigg 1000 cm 3 1 L parenrightbigg parenleftbigg 101325 N / m 2 1 atm parenrightbigg × parenleftbigg 1 m 3 10 6 cm 3 parenrightbigg = - 65 . 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. - 50 kJ correct 3. - 550 kJ 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 .
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romasko (qrr58) – 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 CD player: Δ U = q + w = - 50 kJ + [500 kJ + ( - 500 kJ)] = - 50 kJ 004 10.0 points Calculate the energy required to heat 436 g of mercury from 37 C to 42 C. The specific heat capacity of Hg( ) is 0 . 138 J / g · C.
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