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. 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

The Internal Energy of a Gas Does Not Change! 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!

Free Expansion Experiment Experimentally, an Adiabatic Free Expansion of 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.

Free Expansion Doing an adiabatic free expansion For an Ideal Gas E = E(T) = CTn (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.

αJ  (∂T/∂V)E = – (∂T/∂E)V(∂E/∂V)T The Combined 1st & 2nd Laws: 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 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. 6

αJ = (∂T/∂V)E = – [T(∂P/∂T)T – p]/CV 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! 7

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.

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

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. 10

Adiabatic Process:  Q = 0 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. 11

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: H2 = H1 or ΔH = 0. 12

μ = (∂T/∂p)H = – (∂T/∂H)P (∂H/∂p)T To analyze the Joule-Thompson Effect it is 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/CP. 1st & 2nd Laws: dH = TdS + V dp. 13 13

μ = (∂T/∂p)H = – (∂T/∂H)P (∂H/∂p)T Joule-Thompson Coefficient: μ  (∂T/∂p)H (μ > 0 for 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. 14

μ = (∂T/∂p)H = [T(∂V/∂T)T – V]/CP The 1st Law: dH = TdS + Vdp. Joule-Thompson Coefficient: μ  (∂T/∂p)H (μ > 0 for 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 15

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

The Joule-Thompson Coefficient The Joule-Thompson Coefficient JT 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

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

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

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

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

Van der Waals’ Equation of State 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 Model which attempts to make corrections to the Ideal Gas Law. Recall the Ideal Gas Law: pV = nRT 22

(P + a/v2)(v – b) = RT Van der Waals Equation of State: 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. Their physical interpretation is discussed on the next slide. 23 23

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 attractive intermolecular forces, which reduce the pressure at the walls compared to that (P) 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. 24 24

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

Van der Waals Equation of State: More Typical P vs V curves for various T (isotherms) P Isotherms An inflection point, which occurs on the curve at the critical temperature Tc, gives the critical point (Tc,Pc). vapor Tc gas C V

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). 27 27

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). 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 & cV are specific heats. 28 28

P dV +V dP = nR dT = – (nRP/CV) dV 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 29 29

PVγ = constant. For an ideal gas ONLY, CP – CV = nR 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. Simple Kinetic Theory for a monatomic ideal gas (Ch. 6) gets E = (3/2)nRT so Cv = (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. 30 30

Method 1: Direct Integration W = –(3/2)] [P2V2 – P1V1]. 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]. 31 31

W = – ncV (T2 – T1). Work Done on an Ideal Gas in a Reversible Adiabatic Process Method 2: From the 1st Law For a reversible process, W = Qr – ΔE. So, for a reversible adiabatic process: W = – ΔE. For an ideal gas, ΔE = CV ΔT = ncV ΔT = ncV (T2 – T1). So, for a reversible adiabatic process in an ideal gas: W = – ncV (T2 – T1). 32 32

Work Done on an Ideal Gas in a Reversible Adiabatic Process 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]. 33 33

Summary: Reversible Processes for an Ideal Gas Adiabatic Process Isothermal Process Isobaric Process Isochoric Process PVγ = K γ = CP/CV T = constant P = constant V = constant W = – [1/(γ - 1)] [P2V2 – P1V1] W = nRT  ln(V2 /V1) W = P V W = 0 ΔE = CV ΔT ΔE = 0 ΔE = CP ΔT PV = nRT, E = ncVT, cP – cV = R, γ = cP/cV Monatomic ideal gas cV = (3/2)R, γ = 5/3 34 34