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IV. Kinetic theory (continued – see previous lecture) 5. Heat capacitance a) Monoatomic gas 3 E K tr nRT 2 3 nRdT 2 dQ nCV dT dK tr } 3 CV R 2 1 dQ CV - molar specific heat at constant v olume n dT b) Equipartition principle. Degrees of freedom f CV R 2 f E nRT 2 Diatomic gas, including rotations: 5 CV R 2 Diatomic gas, including rotations and vibrations: 7 CV R 2 Molar Specific Heats of Gases at Constant Volume CV At Room Temperature Monatomic He Ar CV (J/mol·K) 12.47 12.47 Diatomic H2 N2 O2 20.42 20.76 21.10 Polyatomic CO2 SO2 28.46 31.39 3R/2 = 12.47 J/mol·K 5R/2 = 20.79 J/mol·K 7R/2 = 29.10 J/mol·K Question How many rotational degrees of freedom does the water molecule have? 1. 1 2. 2 3. 3 4. 4 c) A hint of quantum theory b) Heat capacitance of solids E 3nRT CV 3R 5. Distribution of molecular speeds (Maxwell distribution) 3 dN Nf (v)dv m 2 2 mv2 / 2 k BT v e f (v) 4 2k BT 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. V. The first low of Thermodynamics (conservation of energy) ΔU = ΔQ - ΔW dU = dQ - dW 1. •Macro- a micro- parameters •Equation of state m for monatomic ideal gas: PV RT •Thermodynamic equilibrium 2. Internal energy (kinetic plus potential energy of particles) dU = dQ + dWext dWext = -dW dW = P dV dU = dQ - dW = T dS - P dV dQ = T dS 3. Work dW Fdx PAdx PdV dW PdV •Work is path dependent •Heat is path dependent •Internal energy is path independent V2 W P(V )dV V1 P P 1 2 1 ΔW = P(V2 - V1)>0 V1 V2 ΔW V P 2 V1 V2 P 2 1 V1 3 2 ΔW = P(V2 - V1)<0 V2 V ΔW 4 1 V V1 V2 V ΔW12 = 0 ΔW23 > 0 ΔW34 > 0 ΔW42 < 0 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 ) P b W = ½ (1.00 + 1.40)* 105 Pa*(0.1100 - 0.0700 ) m3 a W = 4.8* 103 J V Va Vb b) What happen with temperature and internal energy of this gas? For ideal gas: PaVa PbVb Ta Tb K ~T Tb PbVb 1.40 105 Pa 0.1100m3 2.2 5 3 Ta PaVa 1.00 10 Pa 0.0700m K b Tb 2 .2 K a Ta 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 ΔW1 < ΔW2 < ΔW3 2 ΔU1 = ΔU2 = ΔU3 = UB - UA ΔT1 = ΔT2 = ΔT3 = TB - TA ΔU = ΔQ - ΔW ΔQ1 <ΔQ2 <Δ Q3 For ideal gas: PE=0 and Δ PE=0 T~ K=U ΔK1 = ΔK2 = ΔK3 = KB - KA 3 I State B V 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 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 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. 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) 1. 0.17 2. 1.7 3. 17 4. 1700 Question 2: The work done by the system in going from point 1 to point 2 is ___ L·atm. 1. 2. 3. 4. 1 2 3 3 ΔU = ΔQ - ΔW 4. Thermodynamic processes ΔW = P ΔV •Adiabatic process: ΔQ = 0 ΔU = - ΔW •Isochoric process: ΔV = 0 ΔW = 0 •Isobaric process: ΔW = PΔV ΔP = 0 ΔU = ΔQ •Isothermal process: ΔT = 0 •For ideal gas (!) only: ΔT = 0 PV P0V0 ΔU = 0 V2 V2 V1 V1 W PdV P0V0 •Closed cycle process: ΔT = ΔP = ΔV = ΔU = 0 ΔQ = ΔW V dV P0V0 ln 2 V V1 ΔQ = ΔW The processes on a PV diagram (ideal gas) Question 3.0 The work done by the gas during this process is ___ L·atm. 1 2.5 p (atm) 2.0 1. 2. 3. 4. 1.5 2 1.0 0.5 0.0 0.0 0.5 1.0 1.5 2.0 V (L) 2.5 3.0 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. V2 V2 V1 V1 W PdV P0V0 V dV P0V0 ln 2 V V1 1.5 2.7 3.9 5.4 Question The net heat absorbed by the system during one cycle is ___ L·atm 1. 2 2. -2 3. 4 4. -4 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: U K 32 nRT 32 PV V 0 Q U U 3 2 3 2 P V 2atm24L 72L atm Heat capacitance of an ideal gas 1 dQ CV n dT / V 1 dQ CP n dT at constant volume / For an ideal gas: P at constant pressure dU dT dU / V dT dU / P dT PV nRT dU dQ dW dQ dU dW dQ dU PdV CV 1 dQ 1 dU n dT / V n dT CP 1 dQ 1 dU 1 PdV CV R / P n dT n dT n dT CP CV CV f R 2 CV CP CV R R 1 CP R 1 monoatomic gas : f 3 C v 32 R, C P 52 R, 53 1.67 diatomic gas at room temperatu re : f 5 C v 52 R, C P 72 R, 75 1.40 Relating heat capacities at constant volume and pressure U ~T U ~ T Adiabatic processes for an ideal gas dU = dQ -PdV For an adiabatic process: dQ = 0 nRT nCV dT dV V For an ideal gas: PV=nRT CV 1 dU dU nCV dT n dT C CV R P ( 1) CV CV dT dV - - 1 T V ln T 1 ln V const ln T ln V 1 const ln TV 1 const PV nRT TV 1 const T1V1 1 T2V2 1 PV const P1V1 P2V2 Example: P 1 P1 2atm V1 1l 2 T1 60 C V2 2l a T1 T2 b Q 0 P2 ? V1 (a) V2 P1V1 P2V2 T1 T2 V V1 P2 P1 V2 P1V1 P2V2 T1 T2 P2 2atm (b) P1V1 P2V2 T1 T2 P1V1 P2V2 V1 P2 P1 V2 V T2 T1 1 V2 1 1 P2 2 atm 2 1 T2 60 C 2 1l 1atm 2l 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. Given: p1 1.00 atm; V1 1.00 L; V2 0.100 L; r V1 / V2 10.0; T1 27 °C = 300 K. Adiabatic compression: p1V1 p2V2 V1 V1 p2 p1 p1 p1r V2 V2 The final pressure of the air is ___ atm. ( = 1.40 for air) = (1.00 atm)(10.0)1.40 25.1 atm. T1V1 1 T2V2 1 The final temperature of the air is ___ ˚C. V1 T2 T1 V 1 T1r 1 2 = (300 K)(10.0)1.40 1 754 K = 481 °C. Adiabatic processes for an ideal gas (2) TV 1 const T1V1 1 T2V2 1 PV const P1V1 P2V2 Work V2 W PdV P1V1 V1 P1V1 V2 V1 dV V 1 V21 V11 1 1 ( P1V1 V21 - P1V1 ) 1 dU dW nCV dT PV=nRT 1 W (P1V1 P2V2 ) 1 W nCV (T1 T2 ) CV 1 W (P1V1 P2V2 ) (P1V1 P2V2 ) R 1 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. The work done by the gas during this process is ___ L·atm. ( = 1.40 for air) 1. 2. 3. 4. -8 -4 2 6 W 1 (P1V1 P2V2 ) 1 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) The heat absorbed by the gas during this process is ___ L·atm. pV nRT Q nC p T C p R PV C p CV R Q C p R PdV (7 / 2)1.00atm(4.00 L) Q 14 L atm