2. Physics 1710—Warm-up Quiz A jar is sealed when it is at a temperature of 20 C then placed in an oven where it is heated to a temperature of 200 C. What is the pressure inside? A.About 101 kPa B.About 1010 kPa C.About 1.61 kPa D.About 163 kPa E.Must know the volume

3. Physics 1710 Chapter 21 Kinetic theory of Gases Solution: P 1 V = nRT 1 P 1 V = nRT 1 P 2 V = nRT 2 P 2 V = nRT 2 P 2 = P 1 (T 2 /T 1 ) [Charles’ Law] T 2 = 200 C +273 C = 473 K T 1 = 20 C +273 C = 293 K P 2 = 101 kPa (473 /293) = 163 kPa

4. No Talking! Think! Confer! Peer Instruction Time What factors are important in explaining why one feels cold when one first enters a swimming pool but later feels warmer? Physics 1710 Chapter 21 Kinetic theory of Gases

5. What factors are important in explaining why one feels cold when one first enters a swimming pool but later feels warmer? Physics 1710 — e-Quiz Answer Now ! 1.Your skin temperature drops. 2.The water warms near your skin. 3.Radiation into the water decreases with time. 4.Your skin hair insulates you. 5.Something else.

6. How does heat get from one place to another? Physics 1710 C hapter 20 Heat & 1 st Law of Thermo Conduction Convection Radiation Examples: Touching a hot stove Feeling the air rising from it Feeling the glow

7. Conduction: P = kA |dT/dx | P = kA |dT/dx | Examples: Examples: – Thermos bottles – Blankets – Double pane windows – Newton’s law of cooling P = h A(T 2 – T 1 ) – Pans – R factor or R value P = A(T 2 – T 1 )/∑ i R i Physics 1710 C hapter 20 Heat & 1 st Law of Thermo

8. Convection: Heat transfer by material transfer Heat transfer by material transfer Forced convection (fluids) Forced convection (fluids) – External force produces material transfer Natural Convection Natural Convection – Buoyancy-driven flow – Newton’s law of cooling applied P = h A(T 2 – T 1 ) h depends on flow conditions Physics 1710 C hapter 20 Heat & 1 st Law of Thermo

9. Radiation: Stefan-Boltzmann Law Stefan-Boltzmann Law P = εσ AT 4 Wien’s LawWien’s Law P ∝T4P ∝T4P ∝T4P ∝T4 σ = 5.6696 x 10 -8 W/m 2 ‧K 4 σ = 5.6696 x 10 -8 W/m 2 ‧K 4 Emissivity 0< ε <1; ε ~ ½ Emissivity 0< ε <1; ε ~ ½ Reflectivity (albedo) R = (1- ε) Reflectivity (albedo) R = (1- ε) Energy balance Energy balance P in - εσ A(T ave ) 4 = 0 Physics 1710 C hapter 20 Heat & 1 st Law of Thermo

10. Skin Physics 1710 Chapter 21 Kinetic theory of Gases Observations: Cold sensation on first entering poolCold sensation on first entering pool Sensation lost after a few minutes in still waterSensation lost after a few minutes in still water Sensation returns with fluid motionSensation returns with fluid motionModel: Conduction: P = kA |dT/dx | Skin Water

11. 1′ Lecture: Energy average distributes equally (is equipartitioned) into all available states. The distribution of particles among available energy states obeys the Boltzmann distribution law. The distribution of particles among available energy states obeys the Boltzmann distribution law. n V = n o e –E/kT Physics 1710 Chapter 21 Kinetic theory of Gases

12. (Molar) Specific Heat of an Ideal Gas: ∆Q = n C V ∆T (constant volume) ∆Q = n C P ∆T (constant pressure) W = ∫ P dV; at constant volume W = 0. ∆E int = ∆Q = n C V ∆T E int = n C V T C V = (1/n) d E int /dT C V = 3/2 R = 3/2 N o kT = 12.5 J/mol‧K Physics 1710 Chapter 21 Kinetic theory of Gases

13. ∆E int = ∆Q –W = nC P ∆T - P∆V nC V ∆T= nC P ∆T – n R ∆T C P - C V = R C P = 5/2 R γ = C P / C V = (5/2 R)/(3/2 R) = 5/3 γ = 5/3 Physics 1710 Chapter 21 Kinetic theory of Gases

14. Adiabatic Expansion of an Ideal Gas: For adiabatic case: dE int = n C V dT = - PdV So that dT = -P dV /(nC V ) Also: PV = nRT PdV + VdP = nR dT PdV + VdP = -RP /(nC V ) dV Physics 1710 Chapter 21 Kinetic theory of Gases

15. PdV + VdP = -R/(nC V ) PdV Rearranging: dP/P = [1 – R/(nCV)] dV/V dP/P = - γ dV/V ln P = - γ lnV + ln K PV γ = constant Physics 1710 Chapter 21 Kinetic theory of Gases

16. Bulk Modulus of an Ideal Gas: B = -∆P/ (∆V/V) P= K V - γ dP = - γ KV - γ - 1 dV B = -dP/(dV/V) B= ( γ KV 1 dV)/(dV/V) B= γ KV - γ B = γ P Physics 1710 Chapter 21 Kinetic theory of Gases

17. Velocity of Sound in Ideal Gas: v = √(B/ρ) v = √[ γ P/(ρ o P/ P o )] v = √[ γ P o /ρ o ] N.B.: no pressure dependence v = √[ γ P o /ρ o ] N.B.: no pressure dependence For air (ideal) v = √[ γ P o /ρ o ] v = √[(5/3)(101 kPa/1.29 kg/m 3 )] v = 362 m/s (Cf 343 m/s) Physics 1710 Chapter 21 Kinetic theory of Gases

18. Velocity of Molecule in Ideal Gas (N 2 ): ½ m = 3/2 kT v rms = √ = √3kT/m v rms = √3(1.38 x10 –23 J/K)(273 K)/(28 x 1.66 x10 –27 kg) v rms = 493 m/s = 1.36 v sound Physics 1710 Chapter 21 Kinetic theory of Gases

19. Law of Atmospheres dP = -mg n V dy P = n V kT dP = kT dn V kT dn V = -mg n V dy dn V / n V = -(mg/kT) dy n V = n o e –(mgy/kT) Physics 1710 Chapter 21 Kinetic theory of Gases

20. Boltzmann Distribution Function n V = n o e –(mgy/kT) n V = n o e –U/kT n V (E) = n o e –E/kT Physics 1710 Chapter 21 Kinetic theory of Gases

21. Summary: γ = C P / C V γ = C P / C V PV γ = constant B = γ P The distribution of particles among available energy states obeys the Boltzmann distribution law. The distribution of particles among available energy states obeys the Boltzmann distribution law. n V = n o e –E/kT Physics 1710 Chapter 21 Kinetic theory of Gases

22. Physics 1710 Chapter 22 heat Engines etc 1′ Lecture: The work done by a heat engine is equal to the difference in the heat absorbed at the high temperature and expelled at the low. The work done by a heat engine is equal to the difference in the heat absorbed at the high temperature and expelled at the low. The thermal efficiency is the work done divided by the heat absorbed. The thermal efficiency is the work done divided by the heat absorbed. The second law of thermodynamics states that (1) no heat can be 100% efficient; (2) heat flows from hot to cold unless work is expended. The second law of thermodynamics states that (1) no heat can be 100% efficient; (2) heat flows from hot to cold unless work is expended. The maximum efficiency is obtained via a Carnot cycle and is equal to the temperature difference divided by the high temperature. The maximum efficiency is obtained via a Carnot cycle and is equal to the temperature difference divided by the high temperature.

23. 1′ Lecture: Entropy S is a measure of the disorder of a system. Entropy S is a measure of the disorder of a system. ∆S = ∫dQ/T ∆S = ∫dQ/T S≡ k ln N S≡ k ln N Physics 1710 Chapter 22 heat Engines etc

24. Observation: No irreversible process runs backward. Heat flows from hot to cold. Energy moves from organized to disorganized forms. Physics 1710 Chapter 22 heat Engines etc

25. If ∆E int = ∆Q – W Then ∆W = ∆Q h – ∆Q c The net work done by a heat engine is equal to the change in the heat content of its heat reservoir. Physics 1710 Chapter 22 heat Engines etc

26. Thermal Efficiency e = ∆W/ ∆Q h = [∆Q h - ∆Q c ] / ∆Q h e = 1 - ∆Q c / ∆Q h Physics 1710 Chapter 22 heat Engines etc

27. Kelvin-Planck form of 2 nd Law of Thermodynamics Since e = 1 - ∆Q c / ∆Q h It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the absorption of energy from a reservoir and the performance of an equal amount of work. [ie no 100% efficient heat engine.] Physics 1710 Chapter 22 heat Engines etc

28. Clausius Form of 2 nd Law of Thermodynamics Run heat engine in reverse = heat pump It is impossible to construct a cyclical machine whose sole effect is the continuous transfer of energy from one object to another at a higher temperature without the input of work. Physics 1710 Chapter 22 heat Engines etc

29. Carnot Cycle Named for Sadi Carnot (1796- 1832) (1)Isothermal expansion (2)Adiabatic expansion (3)Isothermal compression (4)Adiabatic compression Physics 1710 Chapter 22 heat Engines etc

30. Carnot’s Theorem No real heat engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs. All real heat engines are less efficient than the Carnot engine because they do not operate through a reversible cycle. Physics 1710 Chapter 22 heat Engines etc

31. Efficiency of Carnot Cycle e Carnot = 1 - ∆Q c / ∆Q h ∆Q ab = W ab = nRT h ln (V b /V a ) ∆Q cd = W cd = nRT c ln (V d /V c ) But (V b /V a )=(V d /V c ) ∆Q c / ∆Q h =T c / T h Thus e Carnot = 1 - T c / T h Physics 1710 Chapter 22 heat Engines etc

32. Entropy (macroscopic definition) dS = dQ / T The change in entropy during a process depends only on the end points and therefore is independent of the actual path followed. Isolated systems tend toward disorder and entropy is a measure of that disorder. Physics 1710 Chapter 22 heat Engines etc

33. Free Energy E dE = dH – TdS dH = dQ – dW Physics 1710 Chapter 22 heat Engines etc

34. Examples of Entropy Changes Reversible change in volume and temperature of ideal gas: ∆S = ∫ i f dQ/T ∆S = ∫ i f dQ/T ∆S = ∫ i f {C v dT + nR(dV/V)} ∆S = nC v T f /T i + nR ln (V f /V i ) Ice melting: ∆S = ∫ i f dQ/T= Q/T m = mL f /T m Physics 1710 Chapter 22 heat Engines etc

35. Entropy (Microscopic Definition) S = k ln N Entropy is a measure of the disorder of a system. N is the number of of occupied microstates. k is the Boltzmann constant. (k = 1.38 x 10 –23 J/K) Physics 1710 Chapter 22 heat Engines etc

36. Summary: The work done by a heat engine is equal to the difference in the heat absorbed at the high temperature and expelled at the low.The work done by a heat engine is equal to the difference in the heat absorbed at the high temperature and expelled at the low. ∆W = ∆Q h – ∆Q c The thermal efficiency is the work done divided by the heat absorbed. The thermal efficiency is the work done divided by the heat absorbed. e = 1 - ∆Q c / ∆Q h Physics 1710 Chapter 22 heat Engines etc

37. Summary: Kelvin-Planck form of 2 nd Law of Thermo: It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the absorption of energy from a reservoir and the performance of an equal amount of work. Kelvin-Planck form of 2 nd Law of Thermo: It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the absorption of energy from a reservoir and the performance of an equal amount of work. Clausius Form of 2 nd Law of Thermo: It is impossible to construct a cyclical machine whose sole effect is the continuous transfer of energy from one object to another at a higher temperature without the input of work.Clausius Form of 2 nd Law of Thermo: It is impossible to construct a cyclical machine whose sole effect is the continuous transfer of energy from one object to another at a higher temperature without the input of work. Physics 1710 Chapter 22 heat Engines etc

38. Summary: The maximum efficiency is obtained via a Carnot cycle and is equal to the temperature difference divided by the high temperature.The maximum efficiency is obtained via a Carnot cycle and is equal to the temperature difference divided by the high temperature. e Carnot = 1 - T c / T h Entropy S is a measure of the disorder of a system. Entropy S is a measure of the disorder of a system. ∆S = ∫dQ/T ∆S = ∫dQ/T S≡ k ln N S≡ k ln N Physics 1710 Chapter 22 heat Engines etc