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.
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

3. 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!

4. 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.

5. 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.

6. 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

7. 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

8. 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.

9. 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

10. 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

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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

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

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

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

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

22. 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

23. (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

24. 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

25. 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

26. 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

27. 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

28. 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

29. 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

30. 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

31. 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

32. 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

33. 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

34. 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