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Physics II PHY 202/222 Thermodymanics

Physics II PHY 202/222 Thermodymanics. 452 South Anderson Road Rock Hill, SC 29730 www.yorktech.com. Thermodynamics– Test 3. Beiser Chapter 20 Multiple Choice – 1 to 7 Supplementary Problems – 6,7,8 Beiser Chapter 21 Multiple Choice – All Supplementary Problems - Odd

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Physics II PHY 202/222 Thermodymanics

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  1. Physics IIPHY 202/222Thermodymanics 452 South Anderson Road Rock Hill, SC 29730 www.yorktech.com

  2. Thermodynamics– Test 3 • Beiser Chapter 20 • Multiple Choice – 1 to 7 • Supplementary Problems – 6,7,8 • Beiser Chapter 21 • Multiple Choice – All • Supplementary Problems - Odd For PHY 222 Students • Browne Chapter 18 - 3,8,9 • Browne Chapter 19 – 2,3,12

  3. Beiser Chapter 20Kinetic Theory of Matter Beiser p.233

  4. Kinetic Theory of Gases Gas is made of particles that are: • Small • In constant random motion • With lots of empty space • That interact only in collisions Collisions with container are source of pressure. More volume = less collisions = less pressure (Boyle’s Law: ) Beiser p.233

  5. Molecular Energy Note: If T = 0 K (absolute zero) then KE = 0, so no motion. Beiser p.233

  6. Chapter 21 Thermodynamics Beiser p.245

  7. What is thermodynamics? “Thermodynamics is the branch of Physics that deals with the relations between heat and other forms of energy, and of the conversion of one into the other.” Specifically we are interested in how heat, temperature, pressure and volume interact.

  8. Zeroth Law of Thermodynamics Two objects, each in thermal equilibrium with a third object, are in thermal equilibrium with each other. Beiser p.245

  9. First Law of Thermodynamics Due to conservation of energy: Heat into a system either raises the energy or accomplishes work T1 Q1 Heat engine Work T2 Q2 Beiser p.245

  10. Heat Engines Any machine that takes heat and converts it into work is called a heat engine. Beiser p.245

  11. Chamber Piston can be raised up or down to change the internal volume, V. Changing the volume, of course, can change the pressure and/or temperature. Heat can be added or taken away, or the chamber can be insulated so that no heat escapes or enters. Changing the temperature will change the pressure and/or volume. A change in volume can do work. Beiser p.247

  12. pressure V1 V2 Volume Isobaric Processes Δp = 0 Add heat to a chamber. The gas will expand. If we allow the piston to rise we can keep the pressure constant. Since the pressure applies a force over a distance, we can find the amount of work done. The amount of work is equal to the area of the shaded region in the graph. Beiser p.246

  13. Add heat to a chamber. The gas will try to expand. If we do not allow the piston to rise there will be no change in volume, therefore no work gets done. However the pressure will go up. p2 pressure p1 Volume Isochoric Processes ΔV = 0 Beiser p.245 Beiser p.246

  14. Isothermal Processes ΔT = 0 Increase the volume of the cylinder, but maintain constant temperature, and the pressure will fall. The temperature can be maintained either by surrounding the chamber with a constant temperature heat reservoir, or by performing the process very slowly so that the change in temperature is negligible. pressure V1 V2 Volume Beiser p.246

  15. A generally rapid change in volume/pressure in which no heat is gained or lost. pressure V1 V2 Volume Adiabatic Processes Q = 0 Beiser p.245 Beiser p.246

  16. Summary of Processes

  17. Second Law of Thermodynamics • Clausius statement: Heat cannot, by itself, pass from a colder to a warmer body. • Kelvin-Planck statement: It is impossible for any system to undergo a cyclic process whose sole result is the absorption of heat from a single reservoir at a single temperature and the performance of an equivalent amount of work. • Beiser book: It is impossible to construct a continuously operating engine that takes heat from a source and performs an exactly equivalent amount of work. Beiser p.245

  18. Three Laws of Thermodynamics • 0th - Two objects, each in thermal equilibrium with a third object, are in thermal equilibrium with each other. • 1st - Heat into a system either raises the energy or accomplishes work. • 2nd - It is impossible to construct a continuously operating engine that takes heat from a source and performs an exactly equivalent amount of work.

  19. Poker Game Analogyof 3 laws of thermo. 0th Law – You can’t get out of the game. Everything in the universe is in “the system”, so all objects are in thermal equilibrium or are moving towards it - unless we do something about it. 1st Law – You can’t win. You can't get something for nothing.  If you want work out, you must put heat/energy in and vice versa 2nd Law – You always lose. Efficiency < 1, Entropy > 0

  20. b c 6x105 Pa Pressure 2x105 Pa d .02 m3 Volume Heat Cycle ab – Isochoric - add heat but don’t allow piston to move bc – Isobaric - add heat and allow piston to move a cd – Isochoric – cylinder cools and pressure drops .05 m3 da – Isobaric – compress cylinder back to starting point

  21. b c 6x105 Pa Pressure a d 2x105 Pa .02 m3 .05 m3 Volume Heat Cycle - Questions Consider a system that is taken along the paths shown on the P-V diagram. Assume Ua = 30,000 J. • Find the work done by the system in going from a to b. • Find the work done by the system in going from b to c. • If 20 kJ of heat enters the system along the path from a to b, what is the internal energy at point b? • If the internal energy at point c is 95 kJ, how much heat enters or leaves the system along the path from b to c? • Run it backwards: If 21 kJ of heat enters the system in going from a to d, what is internal energy at point d? • Run it backwards: Find the heat that enters the system along the path from d to c. • If the system is taken along the closed loop a→b→c→d→a, how much work is done? • Find the area of the rectangular path. • What is the net heat that enters the system? • What is the efficiency of the engine

  22. b c 6x105 Pa Pressure a d 2x105 Pa .02 m3 .05 m3 Volume Heat Cycle - Wab Find the work done by the system in going from a to b. Isochoric process

  23. b c 6x105 Pa Pressure a d 2x105 Pa .02 m3 .05 m3 Volume Heat Cycle - Wbc Find the work done by the system in going from b to c. Isobaric process

  24. b c 6x105 Pa Pressure a d 2x105 Pa .02 m3 .05 m3 Volume Heat Cycle - Ub Ub=50kJ Ua = 30,000 J. If 20 kJ of heat enters the system along the path from a to b, what is the internal energy at point b? Q=ΔU=20kJ W=0 Ua=30kJ

  25. b c 6x105 Pa Pressure a d 2x105 Pa .02 m3 .05 m3 Volume Heat Cycle - Qbc Q=63 kJ Ub=50kJ If the internal energy at point c is 95 kJ, how much heat enters or leaves the system along the path from b to c? Uc=95kJ

  26. b c 6x105 Pa Pressure a d 2x105 Pa .02 m3 .05 m3 Volume Heat Cycle - Ud Ua = 30,000 J. Run it backwards: If 21 kJ of heat enters the system in going from a to d, what is internal energy at point d? Ua=30kJ Ud=45kJ Going forward: the gas loses 15 kJ of internal energy from c to d as we do 6 kJ of work to compress the cylinder back to start over.

  27. b c 6x105 Pa Pressure a d 2x105 Pa .02 m3 .05 m3 Volume Heat Cycle - Qdc Run it backwards: Find the heat that enters the system along the path from d to c. Uc=95kJ Ud=45kJ Going forward: the system loses 50 kJ from c to d and does no work

  28. b c 6x105 Pa Pressure a d 2x105 Pa .02 m3 .05 m3 Volume Heat Cycle - WTotal If the system is taken along the closed loop a→b→c→d→a, how much work is done? W=18kJ W=0 W=0 W=-6kJ

  29. b c 6x105 Pa Pressure a d 2x105 Pa .02 m3 .05 m3 Volume Heat Cycle - Area Find the area of the rectange enclosed by the path of the cycle. Note: the enclosed area = amount of work performed = 12 kJ!!!

  30. b c 6x105 Pa Pressure a d 2x105 Pa .02 m3 .05 m3 Volume Heat Cycle - QTotal What is the net heat that enters the system? Qbc=63 kJ Qcd = -50kJ Qab=20kJ Or, since the the gas molecules are back to their original state there is no change in internal energy, so Qda= -21kJ

  31. b c 6x105 Pa Pressure a d 2x105 Pa .02 m3 .05 m3 Volume Heat Cycle - Efficiency W=18kJ An enviable result! W=0 W=0 W=-6kJ

  32. Carnot Cycle Here are some web animations of the Carnot cycle Ani1Ani 2Ani3 Check ‘em out. Beiser p.249

  33. Carnot Efficiency Beiser p.250

  34. T2 Q2 Work Refrigerator T1 Q1 Refrigeration 3 4 2 1 Beiser p.253

  35. √ √ Floor heats up ? ? Floor cools down Reversibility Arrow of time – forward only Not in Beiser Browne p.233

  36. Entropy A measure of the unavailability of thermal energy to do work. ΔS = 0 for a reversible cycle and ΔS > 0 for a irreversible cycle. All real processes are irreversible. Thus entropy always increases. Not in Beiser Browne p.236

  37. 21.2

  38. 21.4

  39. 21.6

  40. 21.8

  41. 21.10

  42. 21.14

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