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Exploring Classical Thermodynamics: Free Expansion & Joule-Thompson Effect

Dive into the concepts of free expansion and Joule-Thompson effect in classical thermodynamics, including adiabatic processes and gas behavior. Learn how to analyze real gases and determine their "ideal" characteristics.

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Exploring Classical Thermodynamics: Free Expansion & Joule-Thompson Effect

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  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

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