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Chapter 5 Continued : More Topics in Classical Thermodynamics

Chapter 5 Continued : More Topics in Classical Thermodynamics. Einstein on Thermodynamics (1910). “A theory is the more impressive the greater the simplicity of its premises , and the more extended its area of applicability .”

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Chapter 5 Continued : More Topics in Classical Thermodynamics

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  1. Chapter 5 Continued: More Topics in Classical Thermodynamics

  2. Einstein on Thermodynamics (1910) “A theory is the more impressivethe greater the simplicity of its premises, and the more extended its area of applicability.” “Classical Thermodynamics… isthe ONLY physical theory of universal contentwhich I am convinced that, within the applicability of its basic concepts,WILL NEVER BE overthrown.” 2

  3. Eddington on Thermodynamics (1929) “If someone points out to you that your pet theory of the universe isin disagreement with Maxwell’s Equations– then so much the worse for Maxwell’s equations. “But if your theory is found to beagainst the 2nd Law of Thermodynamics- I can offer you no hope; there is nothing for it but to collapse in deepest humiliation!” 3

  4. Free Expansion ( The Joule Effect) • A type of Adiabatic Process is the FREE EXPANSIONin 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 • 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. Figure for a Free Expansion Experiment • An Adiabatic Free Expansionof a gas into a vacuumcools a real(non-ideal) gas. • The temperature is unchanged for an Ideal Gas. • Since Q = W = 0, the 1st Law says that ΔE = 0. • For an Ideal Gas it is easily shown that E is independent of volume V, so that E = E(T). So, since ΔE = 0, ΔT = 0.

  6. Define:Joule Coefficient αJ (∂T/∂V)E (= 0 for an ideal gas) Some useful manipulation: (∂T/∂V)E = – (∂T/∂E)V(∂E/∂V)T – (∂E/∂V)T/CV 1st Law: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 αJ = (∂T/∂V)E =– [T(∂P/∂T)T – p]/CV 6

  7. Adiabatic Wall Thermometer p2,V2,T2 p1,V1,T1 Porous Plug Throttle Expansionp1 > p2 The Joule-Thompson or “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.

  8. The Joule-Thompson Effectis 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. 1st Law: ΔE = E2 - E1 = Q – W. Adiabatic Process: Q = 0 So, E2 – E1 = – (p2V2 – p1V1). This gives E2 + p2V2 = E1 + p1V1. Recall the definition of Enthalpy: H  pV . So in the Joule-Thompson process, the Enthalpy H stays constant: H2 = H1 or ΔH = 0.

  9. Define: Joule-Thompson Coefficientμ (∂T/∂p)H (μ > 0for cooling. μ < 0 for heating) Some useful manipulation: (∂T/∂p)H = – (∂T/∂H)P (∂H/∂p)T = – (∂H/∂p)T/CP. 1st Law:dH = TdS + V dp. So, (∂H/∂p)T = T(∂S/∂p)T + V. A Maxwell Relation is (∂S/∂p)T = – (∂V/∂T)p so that μ = (∂T/∂p)H = [T(∂V/∂T)T – V]/CP 9

  10. Throttling Processes:Typical T vs. p Curves

  11. Throttling Processes:Typical T vs. p Curves Family of Curves of Constant H(Like Fig. 5.10.3 in Reif)

  12. Now, for a brief, hopefully useful Discussion of a Microscopic Physics Model in this Macroscopic Thermodynamicschapter! 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 Model which attempts to make corrections to the Ideal Gas Law. Recall the Ideal Gas Law: pV = nRT

  13. The Van der Waals Equation of State has the form: (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 parameters a & b, which represent the following phenomena: 13

  14. The Van der Waals Equation of State has the form: (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 parameters a & b, which represent the following phenomena: The term a/v2represents the attractive intermolecular forces,which reduce the pressure at the walls compared to that within the gas. 14

  15. The Van der Waals Equation of State has the form: (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 parameters a & b, which represent the following phenomena: The term a/v2represents the attractive intermolecular forces,which reduce the pressure at the walls compared to that within the gas. The term – b represents the molecular volume occupied by a kilomole of gas, & which is therefore unavailable to other molecules. 15

  16. The Van der Waals Equation of State has the form: (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 parameters a & b, which represent the following phenomena: The term a/v2represents the attractive intermolecular forces,which reduce the pressure at the walls compared to that within the gas. The term – b represents the molecular volume occupied by a kilomole of gas, & which is therefore unavailable to other molecules. As a & b become smaller, or as T becomes larger, the equation approaches ideal gas equation Pv = RT. 16

  17. 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 that H =H(T). So, H = H(T) and CP = (dH/dT). Thus, CP – CV = (dH/dT) – (dE/dT) = d(PV)/dT = nR. So,CP – CV = nR. This is sometimes known as Mayer’s Equation, & it holds for ideal gases only. For 1 kmole, cP–cV= R, where cP& cVare specific heats. 17

  18. Since dQ = 0 for an adiabatic process: dE = – P dV & dE = CV dT, so dT = – (P/CV) dV . For an ideal gas, PV = nRT so that P dV +V dP = nR dT = – (nRP/CV) dV. & V dP + P (1 +nR/CV) dV = 0. This gives, 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 + γdV/V = 0. Simple Kinetic Theory for a monatomic ideal gas (Ch. 6) gets E = (3/2)nRTsoCv = (3/2)nR&γ = (Cp/Cv) = (5/3) Integration of the last equation in green gives: ln P + γ ln V = constant, so that PVγ = constant. 18

  19. Work Done on an Ideal Gas in a Reversible Adiabatic Process Method 1: Direct Integration For a reversible adiabatic process, PVγ = K. Since the process is reversible, W =  PdV, so that W = K  V–γ dV = – [K/(γ –1)] V–(γ–1) = – [1/(γ –1)] PV | (limits: P1V1 P2V2) So, W = – [1/(γ –1)] [P2V2 – P1V1]. For an ideal monatomic gas, γ = 5/3, so that W = –(3/2)] [P2V2 – P1V1]. 19

  20. Work Done on an Ideal Gas in a Reversible Adiabatic Process Method 3: From the 1st Law For a reversible process, W = Qr – ΔE. So, for a reversible adiabatic process: W = – ΔE. For an ideal gas, ΔE = CVΔT = ncV ΔT = ncV (T2 – T1). So, for a reversible adiabatic process in an ideal gas: W = – ncV (T2 – T1). For an ideal gas PV = nRT, so that W = – (cV/R)[P2V2 – P1V1]. But, Mayer’s relationship for an ideal gas gives: R = cP – cV so that W = – [cV/(cP – cV)][P2V2 – P1V1] or W = – [1/(γ –1)] [P2V2 – P1V1]. 20

  21. Summary: Reversible Processes for an Ideal Gas PV = nRT, E = ncVT, cP – cV = R, γ = cP/cV Monatomic ideal gas cV = (3/2)R, γ = 5/3 21

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