ch14 - padilla(tp5647 HW14 Thermodynamics II shear(52375...

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padilla (tp5647) – HW14: Thermodynamics II – shear – (52375) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. DrRuth says: here are some useful conver- sion factors and constants: R = 8.314 J/mol K 1 L atm = 101.325 J 1 cal = 4.184 J Also, remember that internal energy is E in our textbook, but some older textbooks, and some Quest questions, use U instead of E. Be careful of units. S is usually J/K mol and Delta H is usually kJ/mol. Notice the differ- ence between J and kJ. 001 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 gained 210 kJ of energy as heat. 2. The system lost 90 kJ of energy as heat. correct 3. The system lost 210 kJ of energy as heat. 4. The system gained 90 kJ of energy as heat. 5. The system gained 60 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.) 002 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. 0 2. - 0 . 608 kJ 3. - 2 . 61 kJ 4. +1 . 39 kJ correct 5. +2 . 61 kJ Explanation: q = 2 kJ P = 2 atm V 0 = 2 L V f = 5 L For expansion against a constant external pressure, w = - P ext Δ V = ( - 2 atm)(5 L - 2 L) × (101 . 325 J · L 1 · atm 1 ) = - 607 . 95 J = - 0 . 60795 kJ . Δ U = q + w = 2 kJ + ( - 0 . 60795 kJ) = +1 . 39205 kJ . 003 10.0 points When a gas expands into a vacuum, w = 0. 1. True correct 2. False Explanation: No work is done when a gas expands into a vacuum since there is nothing to push out of the way. 004 10.0 points For the methanol combustion reaction 2 CH 3 OH( ) + 3 O 2 (g) -→ 2 CO 2 (g) + 4 H 2 O(g) estimate the amount of work done and tell whether the work was done on or by the sys- tem. Assume a temperature of 27 C. 1. 7.5 kJ, work done by the system correct
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padilla (tp5647) – HW14: Thermodynamics II – shear – (52375) 2 2. No work is done in this reaction. 3. 7.5 kJ, work done on the system 4. 2.5 kJ, work done on the system 5. 2.5 kJ, work done by the system Explanation: T = 27 C + 273 = 300 K Considering only moles of gas, Δ n = n f - n i = (2 + 4) - 3 = 3 . w = - Δ nRT = - (3 mol) (8 . 314 J / mol · K) (300 K) = - 7500 J = - 7 . 5 kJ The system expands because Δ n is positive, so the system does the work on the surround- ings. Also, when w is negative, work is done by the system. 005 10.0 points 1.95 mol of an ideal gas at 300 K and 3.00 atm expands from 16 L to 28 L and a final pressure of 1.20 atm in two steps: (1) the gas is cooled at constant volume until its pressure has fallen to 1.20 atm, and (2) it is heated and allowed to expand against a constant pressure of 1.20 atm un- til its volume reaches 28 L. Which of the following is CORRECT? 1. w = - 6 . 03 kJ for the overall process 2. w = 0 for the overall process 3. w = 0 for (1) and w = - 1 . 46 kJ for (2) correct 4. w = - 4 . 57 kJ for (1) and w = - 1 . 46 kJ for (2) 5. w = - 4 . 57 kJ for the overall process Explanation: For step (1): If there is no change in volume, w = 0.
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