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# hw 14 - shah(rps587 HW 14 Kleinman(58225 This print-out...

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shah (rps587) – HW 14 – Kleinman – (58225) 1 This print-out should have 23 questions. Multiple-choice 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 : Δ E = - 2380 J and Q = 0 J . Q = Δ E + W W = Q - Δ E = 0 J - ( - 2380 J) = 2380 J . 002 5.0 points Suppose you squeeze an air-filled 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 doesn’t 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 Δ V = - (59361 . 8 Pa) ( 0 . 00275 m 3 - 0 . 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 Δ 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 0 . 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 .

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shah (rps587) – HW 14 – Kleinman – (58225) 2 From the ideal gas law, P f V f = n R T V f = n R T 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 = n R T 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 = n R T ln parenleftbigg V f V i parenrightbigg = (2 mol)(8 . 31451 J / K · mol)(366 K) ln parenleftbigg 0 . 0811908 m 3 0 . 286101 m 3 parenrightbigg 1 kJ 1000 J = - 7 . 66586 kJ . 007 (part 3 of 3) 10.0 points Find the thermal energy transferred. Correct answer: - 7 . 66586 kJ. Explanation: From the first law of thermodynamics, since Δ U = 0, Q = W = - 7 . 66586 kJ .
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hw 14 - shah(rps587 HW 14 Kleinman(58225 This print-out...

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