1 / 61

Chapter 5 Continued : More Topics in Classical Thermodynamics

Chapter 5 Continued : More Topics in Classical Thermodynamics. Free Expansion (  The Joule Effect ). A type of Adiabatic Process is the FREE EXPANSION in which a gas is allowed to expand in volume adiabatically without doing any work. Free Expansion (  The Joule Effect ).

michaelcbaker
Télécharger la présentation

Chapter 5 Continued : More Topics in Classical Thermodynamics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 5 Continued: More Topics in Classical Thermodynamics

  2. Free Expansion ( The Joule Effect) • A type of Adiabatic Process is the FREE EXPANSION in which a gas is allowed to expand in volume adiabatically without doing any work.

  3. Free Expansion ( The Joule Effect) • A type of Adiabatic Process is the FREE EXPANSION in which a gas is allowed to expand in volume adiabatically without doing any work. • It is adiabatic, so by definition, no heat flows in or out (Q = 0). Also no work is done because the gas does not move any other object, so W = 0. The 1st Law is: Q = ΔE + W

  4. The 1st Law:Q = ΔE + W • So, since Q = W = 0, the 1st Law says that ΔE = 0. • Thus this is a very peculiar type of expansion and In a Free Expansion The Internal Energy of a Gas Does Not Change!

  5. Free Expansion Experiment • Experimentally, an Adiabatic Free Expansionof a gas into a vacuum • cools a real(non-ideal) gas. • Temperature is unchanged for an Ideal Gas. • Since Q = W = 0, the 1st Law says that • ΔE = 0.

  6. Free Expansion • For an Ideal GasE = E(T) = CTn • (C = constant, n > 0) • So, for Adiabatic Free Expansion of an Ideal Gassince ΔE = 0, ΔT = 0!! • Doing an adiabatic free expansion experiment • on a gas gives a means of determining • experimentally how close (or not) the gas is to • being ideal.

  7. T = 0 in the free expansion of an ideal gas. But, for the free expansion of Real Gases, T depends on V. • So, to analyze the free expansion of real gases, its convenient to • DefineThe Joule Coefficient • αJ (∂T/∂V)E(= 0 for an ideal gas)

  8. T = 0 in the free expansion of an ideal gas. But, for the free expansion of Real Gases, T depends on V. • So, to analyze the free expansion of real gases, its convenient to • DefineThe Joule Coefficient • αJ (∂T/∂V)E(= 0 for an ideal gas) • Some useful manipulation: • αJ (∂T/∂V)E = – (∂T/∂E)V(∂E/∂V)T •  – (∂E/∂V)T/CV • The Combined 1st & 2nd Laws: • dE = T dS – pdV.

  9. Joule Coefficient:αJ= – (∂E/∂V)T/CV • 1st & 2nd Laws: dE = T dS – pdV. • So (∂E/∂V)T = T(∂S/∂V)T – p.

  10. Joule Coefficient:αJ= – (∂E/∂V)T/CV • 1st & 2nd Laws: dE = T dS – pdV. • So (∂E/∂V)T = T(∂S/∂V)T – p. • A Maxwell Relation is • (∂S/∂V)T = (∂p/∂T)V, • so that the Joule coefficient can be written: • αJ = (∂T/∂V)E =– [T(∂P/∂T)T – p]/CV Obtained from the gas Equation of State αJ is a measure of how close to “ideal” a real gas is!

  11. Joule-Thompson(“Throttling”) Effect Also Known as the Joule-Kelvin Effect! (WHY??)

  12. Joule-Thompson(“Throttling”) Effect Also Known as the Joule-Kelvin Effect! (WHY??) • An experiment by Joule & Thompson showed that the enthalpy H of a real gas is not only a function of the temperature T, but it is also a function of the pressure p. See figure on the next slide.

  13. Thermometers Adiabatic Wall p2,V2,T2 p1,V1,T1 Porous Plug “Throttling” Expansionp1 > p2 Joule-Thompson Effect Sketch of the experiment by Joule & Thompson

  14. The Joule-Thompson Effect: A continuous, adiabatic process in which the wall temperatures remain constant after equilibrium is reached. For a given mass of gas, the work done is: W = p2V2 – p1V1.

  15. For a given mass the work is: W = p2V2 – p1V1. 1st Law: ΔE = E2 - E1 = Q – W. Adiabatic Process: Q = 0 So, E2 – E1 = – (p2V2 – p1V1). This gives E2 + p2V2 = E1 + p1V1.

  16. The Joule-Thompson Process: Results in the fact that E2 + p2V2 = E1 + p1V1. Recall the definition of Enthalpy: H  E+ pV . So in the Joule-Thompson process, the Enthalpy H stays constant (is conserved!) H2 = H1orΔH = 0.

  17. To analyze the Joule-Thompson Effect it is convenient to Define: The Joule-Thompson Coefficient μ (∂T/∂p)H (μ > 0for cooling. μ < 0 for heating) 17

  18. To analyze the Joule-Thompson Effect it is convenient to Define: The Joule-Thompson Coefficient μ (∂T/∂p)H (μ > 0for cooling. μ < 0 for heating) Some useful manipulation: μ= (∂T/∂p)H = – (∂T/∂H)P (∂H/∂p)T = – (∂H/∂p)T/CP. 18

  19. To analyze the Joule-Thompson Effect it is convenient to Define: The Joule-Thompson Coefficient μ (∂T/∂p)H (μ > 0for cooling. μ < 0 for heating) Some useful manipulation: μ= (∂T/∂p)H = – (∂T/∂H)P (∂H/∂p)T = – (∂H/∂p)T/CP. 1st & 2nd Laws: dH = TdS + V dp. 19

  20. Joule-Thompson Coefficient:μ (∂T/∂p)H (μ > 0for cooling. μ < 0 for heating). Manipulation: μ= (∂T/∂p)H = – (∂T/∂H)P (∂H/∂p)T = – (∂H/∂p)T/CP. The 1st Law:dH = TdS + Vdp. So, (∂H/∂p)T = T(∂S/∂p)T + V.

  21. Joule-Thompson Coefficient:μ (∂T/∂p)H (μ > 0for cooling. μ < 0 for heating). Manipulation: μ= (∂T/∂p)H = – (∂T/∂H)P (∂H/∂p)T = – (∂H/∂p)T/CP. The 1st Law:dH = TdS + Vdp. So, (∂H/∂p)T = T(∂S/∂p)T + V. A Maxwell Relation: (∂S/∂p)T = – (∂V/∂T)p So the Joule-Thompson Coefficient can be written: μ = (∂T/∂p)H = [T(∂V/∂T)T – V]/CP Obtained from the gas Equation of State

  22. More on the Joule-Thompson Coefficient • The temperature behavior of a substance • during a throttling (H = constant) • process is described by the Joule • Thompson Coefficient, defined as

  23. The Joule-Thompson Coefficient • The Joule-Thompson CoefficientJTis clearly a • measure of the change in temperature of a • substance with pressure during a constant enthalpy • process, & we have shown that it can also be • expressed as

  24. ThrottlingA Constant Enthalpy Process H = E + PV = Constant • Characterized by the Joule-Thomson Coefficient, • which can be written as shown below:

  25. ThrottlingA Constant Enthalpy Process H = E + PV = Constant • Characterized by the Joule-Thomson Coefficient:

  26. Another Kind of Throttling Process!(From American slang!)

  27. Throttling Processes:Typical T vs. p Curves Family of Curves of Constant H(Reif’s Fig. 5.10.3)

  28. Now, a brief, hopefully useful, interlude from macroscopic physics: A discussion of a microscopic Physics model of a gas. Let the system of interest be a real (non-ideal) gas. An early empirical model developed for such a gas is the Van der Waals’ Equation of State This is a relatively simple Empirical Modelwhich attempts to make corrections to the Ideal Gas Law. Recall the Ideal Gas Law: pV = NkBT = nRT

  29. Van der Waals Equation of State: (P + a/v2)(v – b) = RT v  molar volume = (V/n), n  # of moles This model reproduces the behavior of real gases more accurately than the ideal gas equation through the empirical parametersa & b. Their physical interpretation is discussed next. 29

  30. Van der Waals Equation of State: (P + a/v2)(v – b) = RT v  molar volume = (V/n), n  # of moles This model reproduces the behavior of real gases more accurately than the ideal gas equation through the empirical parametersa & b. Their physical interpretation is discussed next. 30

  31. Van der Waals Equation of State: (P + a/v2)(v – b) = RT v  molar volume = (V/n), n  # of moles This model reproduces the behavior of real gases more accurately than the ideal gas equation through the empirical parametersa & b. Their physical interpretation is discussed next. 31

  32. Van der Waals Equation of State: (P + a/v2)(v – b) = RT v  molar volume = (V/n) n  # of moles The term a/v2 : Represents the effects of the attractive intermolecular forces which reduce the pressure at the walls compared to that (P) within the gas. 32

  33. Van der Waals Equation of State: (P + a/v2)(v – b) = RT v  molar volume = (V/n) n  # of moles The term –b: Represents the effects of the molecular volume occupied by a kilomole of gas, & which is therefore unavailable to other molecules. 33

  34. Van der Waals Equation of State: (P + a/v2)(v – b) = RT v  molar volume = (V/n) n  # of moles As a & b become smaller, or as T becomes larger, the equation approaches ideal gas equation Pv = RT. 34

  35. Van der Waals Equation of State: Typical P vs V curves for Different T(isotherms) P Isotherms for different T • Below a critical temperature Tc, the curves show maxima & minima. C is a critical point. • A vapor, which occurs below Tc, differs from a gasin that it may be liquefied by applying pressure at constant temperature. C V Tc

  36. Van der Waals Equation of State: More P vs V isotherms for various T(isotherms) Isotherms for different T P vapor • An inflection point, which occurs on the curve at the • Critical Temperature • Tc, gives the Critical • Point: (Tc,Pc). C gas V Tc

  37. A New Topic!Adiabatic Processes in an Ideal Gas 37

  38. A New Topic!Adiabatic Processes in an Ideal Gas Ratio of Specific Heats: γ cP/cV = CP/CV 38

  39. A New Topic!Adiabatic Processes in an Ideal Gas Ratio of Specific Heats: γ cP/cV = CP/CV For a reversible quasi-static process: dE = dQ– PdV. 39

  40. A New Topic!Adiabatic Processes in an Ideal Gas Ratio of Specific Heats: γ cP/cV = CP/CV For a reversible quasi-static process: dE = dQ– PdV. For an adiabatic process: dQ = 0, so that dE = – P dV. 40

  41. A New Topic!Adiabatic Processes in an Ideal Gas Ratio of Specific Heats: γ cP/cV = CP/CV For a reversible quasi-static process: dE = dQ– PdV. For an adiabatic process: dQ = 0, so that dE = – P dV. For an ideal gas, E = E(T), so that CV = (dE/dT). 41

  42. A New Topic!Adiabatic Processes in an Ideal Gas Ratio of Specific Heats: γ cP/cV = CP/CV For a reversible quasi-static process: dE = dQ– PdV. For an adiabatic process: dQ = 0, so that dE = – P dV. For an ideal gas, E = E(T), so that CV = (dE/dT). Also, PV = nRT and H = E + PV. 42

  43. A New Topic!Adiabatic Processes in an Ideal Gas Ratio of Specific Heats: γ cP/cV = CP/CV For a reversible quasi-static process: dE = dQ– PdV. For an adiabatic process: dQ = 0, so that dE = – P dV. For an ideal gas, E = E(T), so that CV = (dE/dT). Also, PV = nRT and H = E + PV. So, H = H(T) and CP = (dH/dT). 43

  44. Ratio of Specific Heats(Ideal Gas) γ cP/cV = CP/CV. 44

  45. Ratio of Specific Heats(Ideal Gas) γ cP/cV = CP/CV. PV = nRT and H = E + PV. So, H = H(T) and CP = (dH/dT).And: CP – CV = (dH/dT) – (dE/dT) = d(PV)/dT = nR. 45

  46. Ratio of Specific Heats(Ideal Gas) γ cP/cV = CP/CV. PV = nRT and H = E + PV. So, H = H(T) and CP = (dH/dT).And: CP – CV = (dH/dT) – (dE/dT) = d(PV)/dT = nR. So,for an ideal gas ONLY, CP – CV = nR. 46

  47. Ratio of Specific Heats(Ideal Gas) γcP/cV= CP/CV. PV = nRT and H = E + PV. So, H = H(T) and CP = (dH/dT).And: CP – CV = (dH/dT) – (dE/dT) = d(PV)/dT = nR. So,for an ideal gas ONLY, CP – CV = nR. This is sometimes known as Mayer’s Equation, & it holds for ideal gases only. 47

  48. Ratio of Specific Heats(Ideal Gas) γcP/cV= CP/CV. PV = nRT and H = E + PV. So, H = H(T) and CP = (dH/dT).And: CP – CV = (dH/dT) – (dE/dT) = d(PV)/dT = nR. So,for an ideal gas ONLY, CP – CV = nR. This is sometimes known as Mayer’s Equation, & it holds for ideal gases only. For 1 kmole, cP–cV= R. cP& cVare specific heats. 48

  49. We had, CV(dP/P) + (CV + nR) (dV/V) = 0. For an ideal gas ONLY, CP – CV = nR 49

  50. We had, CV(dP/P) + (CV + nR) (dV/V) = 0. For an ideal gas ONLY, CP – CV = nR so that CV (dP/P) + CP (dV/V) = 0 or (dP/P) + γ(dV/V) = 0. 50

More Related