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) • 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 st 1 Law is: Q = ΔE + W

The st 1 Law: 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!

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 1 st Law says that ΔE = 0.

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.

• 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)

• 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 1 st & 2 nd Laws: d. E = T d. S – pd. V.

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

• Joule Coefficient: αJ = – (∂E/∂V)T/CV • 1 st & 2 nd Laws: d. E = T d. S – pd. V. • 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!

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

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 Thermometers Adiabatic Wall p 1, V 1, T 1 p 2, V 2, T 2 Porous Plug “Throttling” Expansion p > p 1 2

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 = p 2 V 2 – p 1 V 1.

• For a given mass the work is: st 1 W = p 2 V 2 – p 1 V 1. 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.

The Joule-Thompson Process: • Results in the fact that E 2 + p 2 V 2 = E 1 + p 1 V 1. • Recall the definition of Enthalpy: H E+ p. V. • So in the Joule-Thompson process, the Enthalpy H stays constant (is conserved!) H 2 = H 1 or ΔH = 0.

• 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) 17

• 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. 18

• 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. st • 1 & nd 2 Laws: d. H = Td. S + V dp. 19

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. st 1 The Law: d. H = Td. S + Vdp. So, (∂H/∂p)T = T(∂S/∂p)T + V.

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. st 1 The Law: d. H = Td. S + 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

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)

• 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: p. V = Nk. BT = n. RT

Van der Waals Equation of State: (P + 2 a/v )(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. • Their physical interpretation is discussed next. 29

Van der Waals Equation of State: (P + 2 a/v )(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. • Their physical interpretation is discussed next. 30

Van der Waals Equation of State: (P + 2 a/v )(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. • Their physical interpretation is discussed next. 31

Van der Waals Equation of State: 2 (P + a/v )(v – b) = RT v molar volume = (V/n) n # of moles • The term : Represents the effects of the attractive intermolecular forces 2 a/v which reduce the pressure at the walls compared to that (P) within the gas. 32

Van der Waals Equation of State: 2 (P + a/v )(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

Van der Waals Equation of State: 2 (P + a/v )(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

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

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

A New Topic! Adiabatic Processes in an Ideal Gas 37

A New Topic! Adiabatic Processes in an Ideal Gas Ratio of Specific Heats: γ c. P/c. V = CP/CV 38

A New Topic! Adiabatic Processes in an Ideal Gas Ratio of Specific Heats: γ c. P/c. V = CP/CV • For a reversible quasi-static process: d. E = d. Q – Pd. V. 39

A New Topic! Adiabatic Processes in an Ideal Gas Ratio of Specific Heats: γ c. P/c. V = CP/CV • For a reversible quasi-static process: d. E = d. Q – Pd. V. • For an adiabatic process: d. Q = 0, so that d. E = – P d. V. 40

A New Topic! Adiabatic Processes in an Ideal Gas Ratio of Specific Heats: γ c. P/c. V = CP/CV • For a reversible quasi-static process: d. E = d. Q – Pd. V. • For an adiabatic process: d. Q = 0, so that d. E = – P d. V. • For an ideal gas, E = E(T), so that CV = (d. E/d. T). 41

A New Topic! Adiabatic Processes in an Ideal Gas Ratio of Specific Heats: γ c. P/c. V = CP/CV • For a reversible quasi-static process: d. E = d. Q – Pd. V. • For an adiabatic process: d. Q = 0, so that d. E = – P d. V. • For an ideal gas, E = E(T), so that CV = (d. E/d. T). • Also, PV = n. RT and H = E + PV. 42

A New Topic! Adiabatic Processes in an Ideal Gas Ratio of Specific Heats: γ c. P/c. V = CP/CV • For a reversible quasi-static process: d. E = d. Q – Pd. V. • For an adiabatic process: d. Q = 0, so that d. E = – P d. V. • For an ideal gas, E = E(T), so that CV = (d. E/d. T). • Also, PV = n. RT and H = E + PV. • So, H = H(T) and CP = (d. H/d. T). 43

Ratio of Specific Heats (Ideal Gas) γ c. P/c. V = CP/CV. 44

Ratio of Specific Heats (Ideal Gas) γ c. P/c. V = CP/CV. • PV = n. RT and H = E + PV. • So, H = H(T) and CP = (d. H/d. T). And: CP – CV = (d. H/d. T) – (d. E/d. T) = d(PV)/d. T = n. R. 45

Ratio of Specific Heats (Ideal Gas) γ c. P/c. V = CP/CV. • PV = n. RT and H = E + PV. • So, H = H(T) and CP = (d. H/d. T). And: CP – CV = (d. H/d. T) – (d. E/d. T) = d(PV)/d. T = n. R. So, for an ideal gas ONLY, CP – CV = n. R. 46

Ratio of Specific Heats (Ideal Gas) γ c. P/c. V = CP/CV. • PV = n. RT and H = E + PV. • So, H = H(T) and CP = (d. H/d. T). And: CP – CV = (d. H/d. T) – (d. E/d. T) = d(PV)/d. T = n. R. So, for an ideal gas ONLY, CP – CV = n. R. • This is sometimes known as Mayer’s Equation, Equation & it holds for ideal gases only. 47

Ratio of Specific Heats (Ideal Gas) γ c. P/c. V = CP/CV. • PV = n. RT and H = E + PV. • So, H = H(T) and CP = (d. H/d. T). And: CP – CV = (d. H/d. T) – (d. E/d. T) = d(PV)/d. T = n. R. So, for an ideal gas ONLY, CP – CV = n. R. • This is sometimes known as Mayer’s Equation, Equation & it holds for ideal gases only. • For 1 kmole, c. P – c. V = R. c. P & c. V are specific heats. 48

• We had, CV(d. P/P) + (CV + n. R) (d. V/V) = 0. For an ideal gas ONLY, CP – CV = n. R 49

• We had, CV(d. P/P) + (CV + n. R) (d. V/V) = 0. For an ideal gas ONLY, CP – CV = n. R so that CV (d. P/P) + CP (d. V/V) = 0 or (d. P/P) + γ(d. V/V) = 0. 50

• We had, CV(d. P/P) + (CV + n. R) (d. V/V) = 0. For an ideal gas ONLY, CP – CV = n. R so that CV (d. P/P) + CP (d. V/V) = 0 or (d. P/P) + γ(d. V/V) = 0. • Simple Kinetic Theory for a monatomic ideal gas (Ch. 6) gets E = (3/2)n. RT so Cv = (3/2)n. R & γ = (Cp/Cv) = (5/3) 51

• We had, CV(d. P/P) + (CV + n. R) (d. V/V) = 0. For an ideal gas ONLY, CP – CV = n. R so that CV (d. P/P) + CP (d. V/V) = 0 or (d. P/P) + γ(d. V/V) = 0. • Simple Kinetic Theory for a monatomic ideal gas (Ch. 6) gets E = (3/2)n. RT so Cv = (3/2)n. R & γ = (Cp/Cv) = (5/3) • Integration of the last equation in green gives: ln P + γ ln V = constant, so that 52

• We had, CV(d. P/P) + (CV + n. R) (d. V/V) = 0. For an ideal gas ONLY, CP – CV = n. R so that CV (d. P/P) + CP (d. V/V) = 0 or (d. P/P) + γ(d. V/V) = 0. • Simple Kinetic Theory for a monatomic ideal gas (Ch. 6) gets E = (3/2)n. RT so Cv = (3/2)n. R & γ = (Cp/Cv) = (5/3) • Integration of the last equation in green gives: ln P + γ ln V = constant, so that γ PV = constant. 53

Work Done on an Ideal Gas in a Reversible Adiabatic Process Method 1: Direct Integration • For a reversible adiabatic process, PVγ = K. 54

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 = Pd. V, so that W = K V–γ d. V = – [K/(γ – 1)] V–(γ– 1) = – [1/(γ – 1)] PV | (limits: P 1 V 1 P 2 V 2) 55

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 = Pd. V, so that W = K V–γ d. V = – [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]. 56

Work Done on an Ideal Gas in a Reversible Adiabatic Process Method 2: From the 1 st Law • For a reversible process, W = Qr – ΔE. 57

Work Done on an Ideal Gas in a Reversible Adiabatic Process Method 2: From the 1 st Law • For a reversible process, W = Qr – ΔE. • So, for a reversible adiabatic process: W = – ΔE. • For an ideal gas, ΔE = CV ΔT = nc. V (T 2 – T 1). 58

Work Done on an Ideal Gas in a Reversible Adiabatic Process Method 2: From the 1 st Law • For a reversible process, W = Qr – ΔE. • So, for a reversible adiabatic process: W = – ΔE. • For an ideal gas, ΔE = CV ΔT = nc. V (T 2 – T 1). So, for a reversible adiabatic process in an ideal gas: W = – nc. V (T 2 – T 1). 59

Work Done on an Ideal Gas in a Reversible Adiabatic Process • So, for a reversible adiabatic process in an ideal gas: W = – nc. V (T 2 – T 1). • For an ideal gas PV = n. RT, 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]. 60

Summary: Reversible Processes for an Ideal Gas Adiabatic Isothermal Isobaric Isochoric Process PVγ = K T = constant P = constant V = constant γ = CP/CV W= – [1/(γ - 1)] [P 2 V 2 – P 1 V 1] ΔE = CV ΔT W = n. RT W = P V ln(V 2 /V 1) ΔE = 0 W=0 ΔE = CP ΔT ΔE = CV ΔT PV = n. RT, E = nc. VT, c. P – c. V = R, γ = c. P/c. V Monatomic ideal gas c. V = (3/2)R, γ = 5/3 61