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Unformatted text preview: reyes (mr29667) Homework 6: Thermodynamics II Shear (52375) 1 This print-out should have 12 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 lost 210 kJ of energy as heat. 2. The system gained 60 kJ of energy as heat. 3. The system gained 210 kJ of energy as heat. 4. The system gained 90 kJ of energy as heat. 5. The system lost 90 kJ of energy as heat. correct Explanation: U = q + w q = U- w = 60 kJ- (+150 kJ) =- 90 kJ (Negative means the system lost energy as heat.) 002 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 = 0 for (1) and w =- 1 . 46 kJ for (2) correct 2. w =- 4 . 57 kJ for (1) and w =- 1 . 46 kJ for (2) 3. w =- 6 . 03 kJ for the overall process 4. w =- 4 . 57 kJ for the overall process 5. w = 0 for the overall process Explanation: For step (1): If there is no change in volume, w = 0. For step (2): For expansion against a con- stant external pressure, w =- P ext V = (- 1 . 2 atm)(18 L- 6 L) (101 . 325 J L 1 atm 1 ) =- 1 . 45908 kJ . The total work done by the system would be the sum of the work for each step....
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- Spring '09