1. Chapter 5 Continued: More Topics in Classical Thermodynamics

2. 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.”

3. 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.” “Classical Thermodynamics… is the ONLY physical theory of universal content which I am convinced that, within the applicability of its basic concepts, WILL NEVER BE overthrown.”

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

5. 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 1 st Law is: Q = ΔE + W So, since Q = W = 0, the 1 st 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!

6. Free Expansion Experiment Experimentally, an Adiabatic Free Expansion of a gas into a vacuum cools a real (non-ideal) gas. The temperature is unchanged for an Ideal Gas. Since Q = W = 0, the 1 st 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) = CT n (C = constant, n > 0)

7. Free Expansion For an Ideal Gas E = E(T) = CT n (C = constant, n > 0) So, for Adiabatic Free Expansion of an Ideal Gas since 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.

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 Define The Joule Coefficient α J  (∂T/∂V) E (= 0 for an ideal gas) Some useful manipulation: α J  (∂T/∂V) E = – (∂T/∂E) V (∂E/∂V) T  – (∂E/∂V) T /C V Combined 1 st & 2 nd Laws: dE = T dS – pdV.

9. Joule Coefficient: α J = – (∂E/∂V) T /C V 1 st & 2 nd 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: α J = (∂T/∂V) E = – [T(∂P/∂T) T – p]/C V Obtained from the gas Equation of State α J is a measure of how close to “ideal” a real gas is!

10. 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. p 1,V 1,T 1 p 2,V 2,T 2 Porous Plug Adiabatic Wall Thermometers Throttling Expansion p 1 > p 2

11. The Joule-Thompson Effect is 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 = p 2 V 2 – p 1 V 1. 1 st Law: ΔE = E 2 - E 1 = Q – W. Adiabatic Process:  Q = 0 So, E 2 – E 1 = – (p 2 V 2 – p 1 V 1 ). This gives E 2 + p 2 V 2 = E 1 + p 1 V 1. Recall the definition of Enthalpy: H  pV. So in the Joule-Thompson process, the Enthalpy H stays constant: H 2 = H 1 or ΔH = 0.

12. 12 To analyze the Joule-Thompson Effect its convenient to Define: The Joule-Thompson Coefficient μ  (∂T/∂p) H (μ > 0 for cooling. μ < 0 for heating) Some useful manipulation: μ = (∂T/∂p) H = – (∂T/∂H) P (∂H/∂p) T = – (∂H/∂p) T /C P. 1 st & 2 nd Laws: dH = TdS + V dp.

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

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

15. Throttling  A Constant Enthalpy Process H = E + PV = Constant Characterized by the Joule-Thomson Coefficient, which can be expressed as

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

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

18. Now, for a brief, hopefully useful Discussion of a Microscopic Physics Model in this Macroscopic Thermodynamics chapter! 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

19. 19 The Van der Waals Equation of State has the form: (P + a/v 2 )(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/v 2 represents 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.

20. 20 Adiabatic Processes in an Ideal Gas Ratio of Specific Heats: γ  c P /c V = C P /C V. 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 C V = (dE/dT). Also, PV = nRT and H = E + PV, so that H =H(T). So, H = H(T) and C P = (dH/dT). Thus, C P – C V = (dH/dT) – (dE/dT) = d(PV)/dT = nR. So, C P – C V = nR. Mayer’s EquationThis is sometimes known as Mayer’s Equation, & it holds for ideal gases only. For 1 kmole, c P – c V = R, where c P & c V are specific heats.

21. 21 Since dQ = 0 for an adiabatic process: dE = – P dV & dE = C V dT, so dT = – (P/C V ) dV. For an ideal gas, PV = nRT so that P dV +V dP = nR dT = – (nRP/C V ) dV. & V dP + P (1 +nR/C V ) dV = 0. This gives, C V dP/P + (C V + nR) dV/V = 0. For an ideal gas ONLY, C P – C V = nR. so that C V dP/P + C P dV/V = 0, or dP/P + γ dV/V = 0. Simple Kinetic Theory for a monatomic ideal gas (Ch. 6) gets E = ( 3 / 2 )nRT so C v = ( 3 / 2 )nR & γ = (C p /C v ) = ( 5 / 3 ) Integration of the last equation in green gives: ln P + γ ln V = constant, so that PV γ = constant.

22. 22 Work Done on an Ideal Gas in a Reversible Adiabatic Process Method 1: Direct Integration For a reversible adiabatic process, PV γ = 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: P 1 V 1  P 2 V 2 ) So, W = – [1/(γ –1)] [P 2 V 2 – P 1 V 1 ]. For an ideal monatomic gas, γ = 5/3, so that W = –(3/2)] [P 2 V 2 – P 1 V 1 ].

23. 23 Work Done on an Ideal Gas in a Reversible Adiabatic Process Method 3: From the 1 st Law For a reversible process, W = Q r – ΔE. So, for a reversible adiabatic process: W = – ΔE. For an ideal gas, ΔE = C V ΔT = nc V ΔT = nc V (T 2 – T 1 ). So, for a reversible adiabatic process in an ideal gas: W = – nc V (T 2 – T 1 ). For an ideal gas PV = nRT, so that W = – (c V /R)[P 2 V 2 – P 1 V 1 ]. But, Mayer’s relationship for an ideal gas gives: R = c P – c V so that W = – [c V /(c P – c V )][P 2 V 2 – P 1 V 1 ] or W = – [1/(γ –1)] [P 2 V 2 – P 1 V 1 ].

24. 24 Summary: Reversible Processes for an Ideal Gas Adiabatic Process Isothermal Process Isobaric Process Isochoric Process PV γ = K γ = C P /C V T = constantP = constantV = constant W = – [1/(γ –1)] [P 2 V 2 – P 1 V 1 ] W = nRT ln(V 2 /V 1 ) W = P  V W = 0 ΔE = C V ΔTΔE = 0ΔE = C V ΔT PV = nRT, E = nc V T, c P – c V = R, γ = c P /c V Monatomic ideal gas c V = (3/2)R, γ = 5/3