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Physics 101: Chapter 15 Thermodynamics, Part I

Physics 101: Chapter 15 Thermodynamics, Part I. Textbook Sections 15.1 – 15.5. The Ideal Gas Law (review). pV = Nk B T N = number of molecules N = number of moles (n) x N A molecules/mole k B = Boltzmann’s constant = 1.38 x 10 -23 J/K pV = nRT

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Physics 101: Chapter 15 Thermodynamics, Part I

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  1. Physics 101: Chapter 15Thermodynamics, Part I • Textbook Sections 15.1 – 15.5

  2. The Ideal Gas Law(review) • pV = NkBT • N = number of molecules • N = number of moles (n) x NA molecules/mole • kB = Boltzmann’s constant = 1.38 x 10-23 J/K • pV = nRT • R = ideal gas constant = NAkB = 8.31 J/mol/K = .0823 atm-l/mol/K

  3. Careful with units:R = 8.31 J/mol/Km = molar mass in kgvrms = speed in m/s Summary of Kinetic Theory:The relationship between energy and temperature(for monatomic ideal gas) • Internal Energy U • = number of molecules x ave KE/molecule • = N (3/2)kBT • = (3/2)nRT = (3/2)pV

  4. Thermodynamic Systems and p-V Diagrams • ideal gas law: pV = nRT • for n fixed, p and V determine “state” of system • T = pV/nR • U = (3/2)nRT = (3/2)pV • examples: • which point has highest T? • 2 • which point has lowest U? • 3 • to change the system from 3 to 2, energy must be added to system P 1 2 P1 P3 3 V V1 V2

  5. y W W Work Done by a System • W = p V • W > 0 if V > 0 • expanding system does positive work • W < 0 if V < 0 • contracting system does negative work • W = 0 if V = 0 • system with constant volume does no work

  6. Problem: Thermo I 1 P 2 4 3 P P W = PDV (>0) W = PDV = 0 V 1 1 2 2 1 P 2 4 4 3 3 Wtot > 0 4 3 DV > 0 V DV = 0 V V P P W = PDV (<0) W = PDV = 0 1 1 2 2 1 P 2 If we gothe other way thenWtot < 0 4 4 3 3 4 3 V DV = 0 V DV < 0 V Wtot = ??

  7. Now try this: P 1 P1 P3 2 Area = (V2-V1)x(P1-P3)/2 3 V V1 V2

  8. P Work done by system 1 P1 P3 2 3 V V1 V2 First Law of ThermodynamicsEnergy Conservation DU = Q- W Increase in internalenergy of system Heat flow into system Equivalent ways of writing 1st Law: Q = DU + W Q = DU + p DV

  9. P 1 P 2 V V1 V2 First Law of ThermodynamicsExample 2 moles of monatomic ideal gas is taken from state 1 to state 2 at constant pressure p=1000 Pa, where V1 =2m3 and V2 =3m3. Find T1, T2, DU, W, Q. 1. pV1 = nRT1 T1 = pV1/nR = 120K 2. pV2 = nRT2 T2 = pV2/nR = 180K 3. DU = (3/2) nR DT = 1500 J DU = (3/2) p DV = 1500 J (has to be the same) 4. W = p DV = 1000 J 5. Q = DU + W = 1500 + 1000 = 2500 J

  10. First Law of ThermodynamicsExample 2 moles of monatomic ideal gas is taken from state 1 to state 2 at constant volume V=2m3, where T1=120K and T2 =180K. Find Q. P P2 P1 2 1. pV1 = nRT1 p1 = nRT1/V = 1000 Pa 2. pV2 = nRT1 p2 = nRT2/V = 1500 Pa 3. DU = (3/2) nR DT = 1500 J 4. W = p DV = 0 J 5. Q = DU + W = 0 + 1500 = 1500 J requires less heat to raise T at const. volume than at const. pressure 1 V V

  11. P(atm) P(atm) A B 4 4 A B 2 2 Case 1 Case 2 correct 3 9 3 9 V(m3) V(m3) Preflight Shown in the picture below are the pressure versus volume graphs for two thermal processes, in each case moving a system from state A to state B along the straight line shown. In which case is the work done by the system the biggest? 1. Case 1 2. Case 2 3. Same Net Work = area under P-V curve Area the same in both cases!

  12. P(atm) P(atm) A B 4 4 A B 2 2 Case 1 Case 2 correct 3 9 3 9 V(m3) V(m3) Preflight Shown in the picture below are the pressure versus volume graphs for two thermal processes, in each case moving a system from state A to state B along the straight line shown. In which case is the change in internal energy of the system the biggest? 1. Case 1 2. Case 2 3. Same U = 3/2 (pV) Case 1: (pV) = 4x9-2x3=30 atm-m3 Case 2: (pV) = 2x9-4x3= 6 atm-m3

  13. P(atm) P(atm) A B 4 4 A B 2 2 Case 1 Case 2 correct 3 9 3 9 V(m3) V(m3) Preflight (followup) Shown in the picture below are the pressure versus volume graphs for two thermal processes, in each case moving a system from state A to state B along the straight line shown. In which case is the heat added to the system the biggest? 1. Case 1 2. Case 2 3. Same Q = U + W W is same for both U is larger for Case 1 Therefore, Q is larger for Case 1

  14. P Work done by system 1 P1 P3 2 Q= DU + W Increase in internalenergy of system 3 Heat flow into system V V1 V2 DEMO Some questions: • Which part of cycle has largest change in internal energy, DU ? • 2  3 (since U = 3/2 pV) • Which part of cycle involves the least work W ? • 3  1 (since W = pDV) • What is change in internal energy for full cycle? • U = 0 for closed cycle (since both p & V are back where they started) • What is net heat into system for full cycle? • U = 0  Q = W = area of triangle (>0)

  15. Work done by system Increase in internalenergy of system Heat flow into system Recap: • 1st Law of Thermodynamics: Energy Conservation Q= DU + W P • point on p-V plot completely specifies state of system (pV = nRT) • work done is area under curve • U depends only on T (U = 3nRT/2 = 3pV/2) • for a complete cycle DU=0  Q=W V

  16. Thermodynamics, part 2 • engines and refrigerators • 2nd law of thermodynamics

  17. HEAT ENGINE REFRIGERATOR system TH TH QH QH W W QC QC TC TC Engines and Refrigerators • system taken in closed cycle  Usystem = 0 • therefore, net heat absorbed = work done • QH - QC = W (engine) • QC - QH = -W (refrigerator) • energy into green blob = energy leaving green blob

  18. HEAT ENGINE TH QH W QC TC The objective: turn heat from hot reservoir into work The cost: “waste heat” 1st Law: QH -QC = W efficiency e  W/QH =W/QH = (QH-QC)/QH = 1-QC/QH

  19. REFRIGERATOR TH QH W QC TC The objective: remove heat from cold reservoir The cost: work 1st Law: QH = W + QC coeff of performance CPr QC/W = QC/W = QC/(QH - QC)

  20. correct Preflights Consider a hypothetical device that takes 1000 J of heat from a hot reservoir at 300K, ejects 200 J of heat to a cold reservoir at 100K, and produces 800 J of work. Does this device violate the first law of thermodynamics ? 1. Yes 2. No This device doesn't violate the first law of thermodynamics because no energy is being created nor destroyed. All the energy is conserved. • W (800) = Qhot (1000) - Qcold (200) • Efficiency = W/Qhot = 800/1000 = 80% 80% efficient

  21. New concept: Entropy (S) • A measure of “disorder” • A property of a system (just like p, V, T, U) • related to number of different “states” of system • Examples of increasing entropy: • ice cube melts • gases expand into vacuum • Change in entropy: • S = Q/T • >0 if heat flows into system (Q>0) • <0 if heat flows out of system (Q<0)

  22. Second Law of Thermodynamics • The entropy change (Q/T) of the system+environment  0 • never < 0 • order to disorder • Consequences • A “disordered” state cannot spontaneously transform into an “ordered” state • No engine operating between two reservoirs can be more efficient than one that produces 0 change in entropy. This is called a “Carnot engine”

  23. HEAT ENGINE TH QH W QC TC Engines and the 2nd Law The objective: turn heat from hot reservoir into work The cost: “waste heat” 1st Law: QH -QC = W efficiency e  W/QH =W/QH = 1-QC/QH S = QC/TC - QH/TH  0 S = 0 for Carnot Therefore, QC/QH  TC/ TH QC/QH = TC/ TH for Carnot Therefore e = 1 - QC/QH  1 - TC/ TH e = 1 - TC/ TH for Carnot e = 1 is forbidden! e largest if TC << TH

  24. correct Preflight Consider a hypothetical device that takes 1000 J of heat from a hot reservoir at 300K, ejects 200 J of heat to a cold reservoir at 100K, and produces 800 J of work. Does this device violate the second law of thermodynamics ? 1. Yes 2. No • W (800) = Qhot (1000) - Qcold (200) • Efficiency = W/Qhot = 800/1000 = 80% • Max eff = 1 - 100/300 = 67% .

  25. Second Law of Thermodynamics • The entropy change (Q/T) of the system+environment  0 • never < 0 • order to disorder • Consequences • A “disordered” state cannot spontaneously transform into an “ordered” state • No engine operating between two reservoirs is more efficient than one that produces 0 change in entropy(“Carnot engine”) • Heat cannot be transferred spontaneously from cold to hot

  26. TH QH W QC TC Preflight Consider a hypothetical refrigerator that takes 1000 J of heat from a cold reservoir at 100K and ejects 1200 J of heat to a hot reservoir at 300K. 1. How much work does the refrigerator do? 2. What happens to the entropy of the universe? 3. Does this violate the 2nd law of thermodynamics? Answers: 200 J Decreases yes QC = 1000 J QH = 1200 J Since QC + W = QH, W = 200 J DSH = QH/TH = (1200 J) / (300 K) = 4 J/K DSC = -QC/TC = (-1000 J) / (100 K) = -10 J/K DSTOTAL = DSH + DSC = -6 J/K  decreases (violates 2nd law)

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