1. Chapter 11 AP Packet

2. 1993 #5

3. (A) The temperature Tb of state b.

4. PaVa/Ta = PbVb/Tb 1. 2 x 105 x 17 x 10-3/250 = 1
PaVa/Ta = PbVb/Tb 1.2 x 105 x 17 x 10-3/250 = 1.2 x 105 x 51 x 10-3/Tb Tb = 750 K

5. (B) This problem is easier to do if you find (C) first.

6. (C) The change in internal energy Ub -Ua

7. ΔU = 3/2 nRΔT ΔU = 3/2 (1)8.31(500) ΔU = 6230 J

8. (B) The heat Qab added to the gas during the process ab

9. W = PΔV W = 1.2 x 105 (34 x 10-3) W = J (negative because the gas is expanding) ΔU = Q + W 6230 = Q + (-4080) Q = J

10. (D) The work Wbc done by the gas in its surroundings during process bc

11. Zero, there is no volume change so no work is done.

12. (E) The net work done by the gas on its surroundings for the entire cycle

13. In a cycle the total change in internal energy is zero
In a cycle the total change in internal energy is zero. ΔU = Q + W 0 = W W = J

14. (F) The efficiency of a Carnot engine that operates between the maximum and minimum temperatures in this cycle.

15. e = 1- (TC/TH) e = 1- (250/750) e = 2/3 or 66%

16. 1999 #6

17. (A) Determine the pressure of the gas at state C.

18. Temp is the same as A and C are on the isotherm
Temp is the same as A and C are on the isotherm. PAVA/TA = PCVC/TC 4 x 105 x 1 x 10-2/TA = PC x 2 x 10-2/TA PC = 2 x 105 K

19. (B) Sketch a graph of the complete cycle.

20. “A” is at point (1.0,4). The line moves horizontal to the right until it reaches “B”. The next line points down to point “C” which is at (2.0,2). The isotherm line moves from “C” to “A”. This line is curved toward the origin.

21. (C) State whether the net work done by the gas during the complete cycle is positive, negative, or zero. Justify your answer.

22. The net work done BY THE GAS is positive
The net work done BY THE GAS is positive. We would call the net work done negative because it is done BY THE GAS. We know this because it is a clockwise cycle.

23. (D) State whether this device is a refrigerator or a heat engine.

24. It is a heat engine because it is a clockwise cycle.

26. (A) Determine the change in internal energy, Ua – Uc, between the states a and c.

27. ΔUc-›a = Q + W ΔUc-›a = 685 + (-120) ΔUc-›a = 565 J

28. (B)(i) Is heat added to or removed from the gas when the gas is taken along the path abc?›

29. The internal energy decreases because a to c moves toward the origin
The internal energy decreases because a to c moves toward the origin. The work from b to c is positive. Since ΔU is negative and and W is positive, Q must be negative. Heat is removed from the gas. ›

30. (B)(ii) Calculate the amount added or removed.›

31. The work is listed: Wa-›b-›c = 75 J ΔU = Q + W -565 = Q + (75) Q = -490 J

32. (C) How much work is done on the gas in the process cda?›

33. No work is done from c to d, no change in volume
No work is done from c to d, no change in volume. From d to a: W = PΔV W = (6 x 105)( ) W = -150,000 J Negative because the gas is expanding.

34. (D) Is heat added to or removed from the gas when the gas is taken along the path cda?

35. ΔUc-›d-›a = Q + W ΔUc-›d-›a is an increase in U, and we just found that the W is negative. By ΔUc-›d-›a = Q + W, (+) = Q + (-), Q must be positive to raise U and do work to the environment, so heat is added to the gas.

36. 2004 #5

37. (A)(i) Calculate the work done by the gas as it expands.›

38. W = PΔV W = (600)(9 - 3) W = J We would normally call this negative work since the gas is expanding, but the question says by the gas so J is correct.

39. (A)(ii) Calculate the change in internal energy of the gas as it expands.›

40. PV = nRT 600(3) = 2(8.31)T TA = 108 K, at three times the volume TB = 324 K ΔU = 3/2 (2)8.31( ) ΔU = 5400 J

41. (A)(iii) Calculate the heat added to or removed from the gas during this expansion.›

42. ΔU = Q + W 5400 = Q + (-3600) Q = 9000 J (added)

43. (B) The pressure is then reduced to 200 N/m3 without changing the volume as the gas is taken from state B to state C. Label state C on the diagram and draw a line or curve to represent the process from state B to state C. ›

44. State c is at point (9,200). The line is drawn straight down from B to C.

45. (C)(i) The gas is then isothermally compressed back to state A
(C)(i) The gas is then isothermally compressed back to state A. Draw a line or curve on the diagram to represent this process. ›

46. The line curves from C to A because it is an isotherm
The line curves from C to A because it is an isotherm. The curve is bowed toward the origin.

47. (C)(ii) Is heat added to or removed from the gas during this isothermal expansion?›

48. Heat is removed from the gas. Since it is an isotherm, ΔU is zero
Heat is removed from the gas. Since it is an isotherm, ΔU is zero. Since the gas is compressed, work is positive. ΔU = Q + W, 0 = Q + (+), Q must be negative.

49. 2006 #5

50. (A) Calculate the temperature of the gas when it is in the following states. (i) State 2 (ii) State 3

51. (A)(i) P1V1/T1 = P2V2/T2 (1. 0 x 105)(0. 25)/373 = (1. 0 x 105)(0
(A)(i) P1V1/T1 = P2V2/T2 (1.0 x 105)(0.25)/373 = (1.0 x 105)(0.50)/T2 T2 = 746 K

52. (A)(ii) P1V1/T1 = P3V3/T3 (1. 0 x 105)(0. 25)/373 = (1. 5 x 105)(0
(A)(ii) P1V1/T1 = P3V3/T3 (1.0 x 105)(0.25)/373 = (1.5 x 105)(0.25)/T3 T3 = K

53. (B) Calculate the net work done on the gas during the cycle.

54. W1-›2 = PΔV W1-›2 = (1.0 x 105)(0.5 – 0.25) W1-›2 = J (expansion) W2-›3 = PΔV W2-›3 = (1.25 x 105)(0.5 – 0.25) W2-›3 = 31,250 J W3-›1 = PΔV W3-›1 = Zero, no volume change Wtotal = = 6250 J

55. (C) Was heat added to or removed from the gas during the cycle
(C) Was heat added to or removed from the gas during the cycle? Justify your answer.

56. Heat was removed from the gas during the cycle
Heat was removed from the gas during the cycle. This is a counter-clockwise cycle, so it is a refrigerator. Refrigerators add work to the gas to remove heat.

57. 2008 #5

58. (A) For each process in this cycle, indicate in the table below whether the quantities W, Q, and ΔU are positive (+), negative (-), or zero (0).

59. (B) Explain your response for the signs of the quantities for process A-›B.

60. A-›B (Next three slides)

61. Work (W) for A-›B is zero, no change in volume.

62. Change in internal energy (ΔU) for A-›B is positive; from A to B is a move away from the origin which is an increase in temperature.

63. Heat (Q) for A-›B is positive
Heat (Q) for A-›B is positive. Since W is zero and ΔU is positive, by ΔU = Q + W, heat must be added to the system: (+) = Q + (0).

64. B-›C (Next three slides)

65. Work (W) for B-›C is negative (-); an increase in volume is negative work.

66. Change in internal energy (ΔU) for B-›C is zero; an isotherm has no change in temperature.

67. Heat (Q) for B-›C is positive
Heat (Q) for B-›C is positive. Since W is negative and ΔU is zero, by ΔU = Q + W, heat must be added to the system to do the work: (0) = Q + (-).

68. C-›A (Next three slides)

69. Work (W) for C-›A is negative (+); a decrease in volume is positive work.

70. Change in internal energy (ΔU) for C-›A is negative; from C to A is a move toward from the origin which is a decrease in temperature.

71. Heat (Q) for C-›A is negative
Heat (Q) for C-›A is negative. Since W is positive and ΔU is negative, by ΔU = Q + W, enough heat must be removed from the system to make up for the work done to the system and to make up for the decrease in temperature: (-) = Q + (+).