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Chapter 11 Thermodynamics Worksheet

Chapter 11 Thermodynamics Worksheet. Section 11.1 The Ideal Gas Law. Problem 1 : A two mole sample of a gas has a temperature of 500 kelvins and a volume of 3 m 3 , what is the pressure on the gas?. PV = nRT P x 3 = 2 x 8.31x 500 P = 2770 Pa.

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Chapter 11 Thermodynamics Worksheet

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  1. Chapter 11Thermodynamics Worksheet

  2. Section 11.1 The Ideal Gas Law

  3. Problem 1: A two mole sample of a gas has a temperature of 500 kelvins and a volume of 3 m3, what is the pressure on the gas?

  4. PV = nRTP x 3 = 2 x 8.31x 500P = 2770 Pa

  5. Problem 2: A three mole sample of a gas has a temperature of 500 kelvins and a pressure of 3000 pascals, what is the volume of the gas?

  6. PV = nRT3000 x V = 3 x 8.31x 500V = 4.155 m3

  7. Problem 3: If a 10 m3 sample of a gas at a temperature of 200 K and a pressure of 200000 pascals is compressed to 6 m3 while the pressure increases to 300000 pascals, what is the resulting temperature?

  8. P1V1/T1 = P2V2/T2200000 x 10/200 = 300000 x 6/T2T2 = 180 K

  9. Problem 4: If a 2 m3 sample of a gas at a temperature of 200 K and a pressure of 400000 pascals is allowed to expand to 8 m3 at constant temperature, what is the resulting pressure of the gas?

  10. P1V1/T1 = P2V2/T2400000 x 2/200 = P2 x 8/200P2 = 100000 Pa

  11. Section 11.2 Thermodynamics

  12. Problem 5: A gas is allowed to expand from 4 m3 to 10 m3 while under a constant pressure of 3000 Pa. What amount of work is done in this process?

  13. W = P DVW = 3000 x (10-6)W = 12000 JSince the gas is expanding, the gas did work on the environment, so we call this negative work: -12000 J.

  14. Problem 6: A gas is compressed from a volume of 11 m3 to 6 m3 while under a constant pressure of 15000 Pa. What amount of work is done in this process?

  15. W = P DVW = 15000 x (11-6)W = 75000 JSince the gas is conpressed, the environment did work on the gas, so we call this positive work: +75000 J.

  16. Problem 6: Which of the graphs shown above depicts an isothermal change?

  17. Isothermal changes are curved toward the origin on a PV diagram, so “E” and “K”.

  18. Problem 7: Which of the graphs shown above depicts an isobaric change?

  19. Since pressure is the y-axis, isobaric changes are horizontal lines on a PV diagram, so “B”, “H”, and “J”.

  20. Problem 8: Which of the graphs shown above depicts an isovolumetric change?

  21. Since volume is the x-axis, isovolumetric changes are vertical lines on a PV diagram, so “D”, and “G”.

  22. Problem 9: Which of the graphs shown above represents an increase in temperature?

  23. Temperature increases if pressure or volume or both increase, which means the arrow must move away from the origin, so “A”, “B”, “F”, “G”, “J”, and “L”.

  24. Problem 10: Which of the graphs shown above represents a decrease in temperature?

  25. Temperature decreases if pressure or volume or both decrease, which means the arrow must move toward the origin, so “C”, “D”, “H”, and “I”.

  26. Problem 11: Which of the graphs shown above represents an increase in internal energy?

  27. An increase in internal energy occurs when the temperature increases, so the answers are the same as problem 9: “A”, “B”, “F”, “G”, “J”, and “L”.

  28. Problem 12: Which of the graphs shown above represents work done by the system?

  29. Work is done by the system if the gas expands, so the arrows must point to the right. Therefore, the answers are : “B”, “E”, “F”, “J”, and “L”.

  30. Problem 13: Which of the graphs shown above represents work done on the system?

  31. Work is done on the system if the gas is compressed, so the arrows must point to the left. Therefore, the answers are : “C”, “H”, “I”, “K”.

  32. Section 11.3 The First Law of Thermodynamics

  33. Problem 14: A gas in a container is held at a temperature of 1000 kelvins. 3000 J of work is done by a force that slowly compresses the gas. A) Is this process isovolumetric, adiabatic, or isothermal?B)How much energy is transferred as heat?C) Is this heat gained or lost by the gas?

  34. 14A.Since the temperature is held constant, it is isothermic.

  35. 14B.Since the gas is held at a constant temperature, it is isothermal.

  36. 14C.DU = Q + WSince the temperature is held constant DU is zero. The gas is compressed so the work is positive.DU = Q + W0 = Q + 3000Q = -3000 J

  37. Problem 15: A gas is compressed as 257 joules of work is done to the gas. During this change, the internal energy of the gas increases 225 joules. What is the total heat transferred in this change? Is the heat added to the gas or removed from it?

  38. DU = Q + WThe gas is compressed so the work is positive.DU = Q + W225 = Q + 257Q = -32 J Since Q is negative, heat is removed.

  39. Section 11.3 Refrigeration Systems and Heat Engines

  40. Problem 17: Which of the graphs shown above represents a heat engine?

  41. A clockwise cycle represents a heat engine: “A”, “D”, “F”, “H”, “K”, and “L”.

  42. Problem 18: Which of the graphs shown above represents a refrigeration system?

  43. A counter-clockwise cycle represents a refrigerator: “B”, “C”, “E”, “G”, “I”, and “J”.

  44. Problem 19: A gas is taken through a cyclic change during which 4400 joules of heat energy is added to the gas. How much work is done on the system in this case? Is this a heat engine or a refrigerator?

  45. DU = Q + WIn a cyclic change, DU = zero. 0 = 4400 + WW = -4400 JIf heat is added and work comes out of the system, it is a heat engine.

  46. Problem 20: A gas is taken through a cyclic change during which 3400 joules of heat energy is removed from the gas. How much work is done on the system in this case? Is this a heat engine or a refrigerator?

  47. DU = Q + WIn a cyclic change, DU = zero. 0 = -3400 + WW = +3400 JIf heat is removed and work is done on the system, it is a refrigerator.

  48. Section 11.4 The Second Law

  49. Problem 21: Find the efficiency of a steam engine that, during one cycle, receives 304 J of energy from steam and loses 221 J as heat to the exhaust.

  50. W = QH - QCW = 304 – 221 = 83e = W/QHe = 83/304e = 0.27 or 27%

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