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IV. Kinetic theory (continued – see previous lecture)

IV. Kinetic theory (continued – see previous lecture). 5. Heat capacitance a) Monoatomic gas. }. b) Equipartition principle. Degrees of freedom. Diatomic gas, including rotations:. Diatomic gas, including rotations and vibrations:.

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IV. Kinetic theory (continued – see previous lecture)

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  1. IV. Kinetic theory (continued – see previous lecture) 5. Heat capacitance a) Monoatomic gas } b) Equipartition principle. Degrees of freedom Diatomic gas, including rotations: Diatomic gas, including rotations and vibrations:

  2. Molar Specific Heats of Gases at Constant Volume CV At Room Temperature CV (J/mol·K) Monatomic He 12.47 Ar 12.47 Diatomic H2 20.42 N2 20.76 O2 21.10 Polyatomic CO2 28.46 SO2 31.39 3R/2 = 12.47 J/mol·K 5R/2 = 20.79 J/mol·K 7R/2 = 29.10 J/mol·K

  3. How many rotational degrees of freedom does the water molecule have? Question • 1 • 2 • 3 • 4

  4. c) A hint of quantum theory b) Heat capacitance of solids 5. Distribution of molecular speeds (Maxwell distribution)

  5. Molar Specific Heats of Elemental Solids at Constant Volume due to Lattice Vibrations The rule of Dulong and Petit only holds at high temperatures. At low T, quantum mechanical effects reduce CV.

  6. V. The first low of Thermodynamics (conservation of energy) ΔU = ΔQ - ΔW dU = dQ - dW • 1. • Macro- a micro- parameters • Equation of state • for monatomic ideal gas: • Thermodynamic equilibrium • 2. Internal energy (kinetic plus potential energy of particles) dU = dQ + dWext dWext = -dW dW = P dV dQ = T dS dU = dQ - dW = T dS - P dV

  7. 3. Work P P 1 2 1 2 ΔW = P(V2 - V1)>0 ΔW V V V1 V2 V1 V2 P P 3 2 1 2 ΔW ΔW = P(V2 - V1)<0 4 1 V V V2 V1 V1 V2 • Work is path dependent • Heat is path dependent • Internal energy is path independent ΔW12 = 0 ΔW23 > 0 ΔW34 > 0 ΔW42 < 0

  8. Example1:A quantity of air is taken from state a to state b along a path that is a straight line in the PV-diagram, where Va=0.070 m3, Vb=0.110 m3, Pa=1.00*105 Pa, Pb=1.40*105 Pa. Assume that the gas is ideal. a) What is the work, W, done by the gas in this process? W = ½ (Pa + Pb )(Vb - Va ) W = ½ (1.00 + 1.40)* 105 Pa*(0.1100- 0.0700) m3 W = 4.8* 103 J P b a V Va Vb b) What happen with temperature and internal energy of this gas? For ideal gas:

  9. Example2: This P-V diagram represents a system consisting of a fixed amount of ideal gas that undergoes three different processes in going from state A to state B. Rank work, heat transfer, change in internal, kinetic, and potential energy for each process. P State A 3 ΔW1 <ΔW2 <ΔW3 ΔU1 =ΔU2 =ΔU3 = UB - UA ΔT1 =ΔT2 =ΔT3 = TB - TA ΔU = ΔQ - ΔW ΔQ1 <ΔQ2 <ΔQ3 For ideal gas: PE=0 andΔ PE=0 T~ K=U ΔK1 =ΔK2 =ΔK3 = KB - KA 2 I State B V

  10. Example3:A system consisting of a quantity of ideal gas is in equilibrium state A. It is slowly heated and as it expands, its pressure varies. It ends up in equilibrium state B. Now suppose that the same quantity of ideal gas again starts in state A, but undergoes a different thermodynamic process (i.e., follows a different path on a P-V diagram), only to end up again in the same state B as before. Consider the net work done by the system and the net heat absorbed by the system during these two different processes. Which of these statements is true? 1. The work done may be different in the two processes, but the heat absorbed must be the same. 2. The work done must be the same in the two processes, but the heat absorbed may be different. 3. The work done may be different in the two processes, and the heat absorbed may be different in the two processes. 4. Both the work done and the heat absorbed must be the same in the two processes, but are not equal to zero. 5. Both the work done and the heat absorbed by the system must be equal to zero in both processes. Note: See example 2

  11. Final Exam: Tues, Dec. 16, 2008, 7:00-9:00 p.m. • ~1/3 of questions test understanding of concepts/principles • ~2/3 of questions test numerical application of concepts/principles • Bring soft (#2) pencils, erasers, and your scientific calculator • Studying: • Review during lecture: Friday, Dec. 12 • Practice problems and solutions for Ch. 20 are posted • Compare your homework solutions with the posted solutions • Review the lecture questions and the text supplements • Review and solve the example problems in the text • Solve problems on the practice exams • -- Seven Phys 221 practice Final Exams are posted plus practice • problems on waves and thermodynamics (Ch. 15-20) • The formula sheet that will be provided with the exam will be posted • Meet with me before the exam to clear up any serious problems you are having with the course material

  12. Physics 221 Final Exam Tuesday, Dec. 16, 7:00-9:00 p.m. 20 2-point problems + 20 4-point problems (multiple choice) Total: 40 problems worth 120 course points ~ 1/3 Ch 17-20 ~ 1/2 Comprehensive (prior to Ch. 17) ~ 1/5 Laboratory Final Exam Physics 221 Final Exam Conflicts If you are enrolled in any of the three following courses you are entitled to take a Physics 221 Final Exam Make-up at a time and place to be arranged. Math 195, Math 196, Acct 215 Prior to final exam week, students must request to take the make-up exam.

  13. Question 1 One cm3 of liquid water at 100 ˚C is boiled off at 1 atm pressure and is converted to 1670 cm3 of steam at 100 ˚C. The “system” is the water. The work done by the system when it is converted to steam is __ L·atm. (1 L = 1000 cm3) • 0.17 • 1.7 • 17 • 1700 Question 2:The work done by the system in going from point 1 to point 2 is ___ L·atm. • 1 • 2 • 3 • 3

  14. ΔU = ΔQ - ΔW 4. Thermodynamic processes ΔW = P ΔV • Adiabatic process: ΔQ = 0 ΔU = - ΔW • Isochoric process: ΔV = 0 ΔW = 0 ΔU = ΔQ • Isobaric process: ΔP = 0 ΔW = PΔV • Isothermal process: ΔT = 0 • For ideal gas (!) only:ΔT = 0 ΔU = 0 ΔQ = ΔW • Closed cycle process:ΔT = ΔP = ΔV = ΔU = 0 ΔQ = ΔW

  15. The processes on a PV diagram (ideal gas)

  16. The work done by the gas during this process is ___ L·atm. 1.5 2.7 3.9 5.4 Question An ideal gas undergoes an isothermal expansion from V1 = 1.00 L to V2 = 2.72 L and p2 = 1.00 atm as shown in the figure.

  17. The net heat absorbed by the system during one cycle is ___ L·atm Question • 2 • -2 • 4 • -4

  18. Example:A monatomic ideal gas undergoes an increase in pressure from p1 = 1.00 atm to p2 = 3.00 atm at V = 24.0 L as shown in the figure. The heat absorbed by the gas during this process is ??? L·atm. For Ideal monatomic gas:

  19. Heat capacitance of an ideal gas For an ideal gas: at constant volume at constant pressure

  20. Relating heat capacities at constant volume and pressure

  21. Adiabatic processes for an ideal gas dU = dQ -PdV For an adiabatic process: dQ = 0 For an ideal gas: PV=nRT

  22. Example: P 1 2 V V1 V2 (a) (b)

  23. The final pressure of the air is ___ atm. ( = 1.40 for air) Example The volume of the air inside the cylinder of an engine decreases from 1.00 L to 0.100 L during adiabatic compression. The initial pressure of the air is 1.00 atm and initial temperature is 27 ˚C. The final temperature of the air is ___ ˚C.

  24. Adiabatic processes for an ideal gas (2) Work PV=nRT

  25. The work done by the gas during this process is ___ L·atm. ( = 1.40 for air) -8 -4 2 6 Question The volume of the air inside the cylinder of an engine decreases from 1.00 L to 0.100 L during adiabatic compression. The initial pressure of the air is 1.00 atm and the final pressure is 25.1 atm.

  26. The heat absorbed by the gas during this process is ___ L·atm. Question Some air undergoes an isobaric expansion at 1.00 atm pressure from V1 = 5.00 L to V2 = 9.00 L. (CV = 5R/2)

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