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Unformatted text preview: shah (rps587) HW 14 Kleinman (58225) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. This HW covers Chs. 20 and 21. 001 10.0 points During a certain thermodynamic process a sample of gas expands and cools, reducing its internal energy by 2380 J, while no heat is added or taken away. How much work is done during this process? Correct answer: 2380 J. Explanation: Let : E = 2380 J and Q = 0 J . Q = E + W W = Q E = 0 J ( 2380 J) = 2380 J . 002 5.0 points Suppose you squeeze an airfilled hollow rub ber ball in your hand. Assuming no heat escapes, what happens to the internal energy of the air inside? 1. It increases. correct 2. Unable to determine 3. It decreases. 4. It doesnt change. Explanation: You do work on the air by decreasing its volume while no heat escapes, so by the first law of thermodynamics, the internal energy increases. 003 (part 1 of 2) 10.0 points A gas is compressed at a constant pressure of 0 . 586 atm from 5 . 26 L to 2 . 75 L. In the process, 344 J of energy leaves the gas by heat. What is the work done on the gas? Correct answer: 148 . 998 J. Explanation: Given : P = 0 . 586 atm = 59361 . 8 Pa , V 1 = 5 . 26 L = 0 . 00526 m 3 , and V 2 = 2 . 75 L = 0 . 00275 m 3 . The work done on the gas is W = P V = (59361 . 8 Pa) ( . 00275 m 3 . 00526 m 3 ) = 148 . 998 J . 004 (part 2 of 2) 10.0 points What is the change in its internal energy? Correct answer: 195 . 002 J. Explanation: Given : Q = 344 J . The change in the internal energy is U = Q + W = 344 J + 148 . 998 J = 195 . 002 J . 005 (part 1 of 3) 10.0 points Hint: look at equation (20.23) for part 2. Given: R = 8 . 31451 J / K mol . Two moles of helium gas initially at 366 K and 0 . 21 atm are compressed isothermally to . 74 atm. Find the final volume of the gas. Assume the helium to behave as an ideal gas. Correct answer: 0 . 0811908 m 3 . Explanation: Given : n = 2 mol , R = 8 . 31451 J / K mol , T f = 366 K , and P f = 0 . 74 atm . shah (rps587) HW 14 Kleinman (58225) 2 From the ideal gas law, P f V f = nRT V f = nRT P f = (2 mol)(8 . 31451 J / K mol)(366 K) (0 . 74 atm)(1 . 013 10 5 Pa / atm) = 0 . 0811908 m 3 . 006 (part 2 of 3) 10.0 points Find the work done by the gas. Correct answer: 7 . 66586 kJ. Explanation: Given : P i = 0 . 21 atm and T i = 366 K . The initial volume was V i = nRT P i = (2 mol)(8 . 31451 J / K mol)(366 K) (0 . 21 atm)(1 . 013 10 5 Pa / atm) = 0 . 286101 m 3 and the work done was W = integraldisplay P dV = nRT ln parenleftbigg V f V i parenrightbigg = (2 mol)(8 . 31451 J / K mol)(366 K) ln parenleftbigg . 0811908 m 3 . 286101 m 3 parenrightbigg 1 kJ 1000 J = 7 . 66586 kJ ....
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This note was uploaded on 05/07/2009 for the course PHY 58225 taught by Professor Klienman during the Spring '09 term at University of Texas at Austin.
 Spring '09
 KLIENMAN

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