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.”
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.”
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 2nd Law of Thermodynamics
- I can offer you no hope; there is
nothing for it but to collapse in deepest
humiliation!”
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
• 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.
• 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) = CTn (C = constant, n > 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)
• Some useful manipulation:
αJ  (∂T/∂V)E = – (∂T/∂E)V(∂E/∂V)T
 – (∂E/∂V)T/CV
Combined 1st & 2nd Laws:
dE = T dS – pdV.
• 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!
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.
Adiabatic Wall
Thermometers
p1,V1,T1
p2,V2,T2
Porous Plug
Throttling Expansion p1 > p2
• 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 = 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.
• 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/CP.
1st & 2nd Laws: dH = TdS + V dp.
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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.
The 1st 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]/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 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 expressed as
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)
 0
 0
• 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
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/v2 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
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equation approaches ideal gas equation Pv = RT.
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 & cV are specific heats.
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• 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 + 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
γ
PV = constant.
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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].
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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 = CV ΔT = 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].
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Summary: Reversible Processes for an Ideal Gas
Adiabatic
Process
PVγ = K
γ = CP/CV
W=
Isothermal
Process
Isobaric
Process
Isochoric
Process
T = constant P = constant V = constant
W=
– [1/(γ –1)]
[P2V2 – P1V1]
nRT ln(V2 /V1)
ΔE = CV ΔT
ΔE = 0
W = P V
W=0
ΔE = CV ΔT ΔE = CV ΔT
PV = nRT, E = ncVT, cP – cV = R, γ = cP/cV
Monatomic ideal gas cV = (3/2)R, γ = 5/3
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