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Thermodynamics Problem: Gas Cycle Calculations and Analysis

This text discusses the calculations and analysis of a gas cycle, including temperature, internal energy, work, heat, and efficiency.

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Thermodynamics Problem: Gas Cycle Calculations and Analysis

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  1. Chapter 11AP Packet

  2. 1993#5

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

  4. PaVa/Ta = PbVb/Tb1.2 x 105 x 17 x 10-3/250 = 1.2 x 105 x 51 x 10-3/TbTb = 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 nRΔTΔU = 3/2 (1)8.31(500)ΔU = 6230 J

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

  9. W = PΔVW = 1.2 x 105 (34 x 10-3)W = -4080 J (negative because the gas is expanding)ΔU = Q + W6230 = Q + (-4080)Q = 10310 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.ΔU = Q + W0 = 1800 + WW = -1800 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.PAVA/TA = PCVC/TC4 x 105 x 1 x 10-2/TA = PC x 2 x 10-2/TAPC = 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. 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.

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

  26. ΔUc-›a = Q + WΔUc-›a = 685 + (-120)ΔUc-›a = 565 J

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

  28. 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. ›

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

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

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

  32. No work is done from c to d, no change in volume. From d to a:W = PΔV W = (6 x 105)(1- 0.75)W = -150,000 JNegative because the gas is expanding.

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

  34. ΔUc-›d-›a = Q + WΔUc-›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.

  35. 2004#5

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

  37. W = PΔV W = (600)(9 - 3)W = +3600 JWe would normally call this negative work since the gas is expanding, but the question says by the gasso +3600 J is correct.

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

  39. PV = nRT600(3) = 2(8.31)TTA = 108 K, at three times the volume TB = 324 KΔU = 3/2 (2)8.31(324-108)ΔU = 5400 J

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

  41. ΔU = Q + W5400 = Q + (-3600)Q = 9000 J (added)

  42. (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. ›

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

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

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

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

  47. 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.

  48. 2006#5

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

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