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.
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
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
The 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.
Thermometers
Adiabatic Wall
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.
• 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.
• 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  E+ 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.
st
•1
&
nd
2
Laws:
dH = TdS + V dp.
16
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: dH = TdS + 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: 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
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 written 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, 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
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
on the next slide.
24
Van der Waals Equation of State:
2
(P + a/v )(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 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.
25
Van der Waals Equation of State:
Some Typical P vs V curves for Different T (isotherms)
• Below a critical
temperature Tc, the curves
show maxima and
minima.
P
Isotherms for
different T
• C is a critical point.
• A vapor, which occurs
C
Tc
V
below Tc, differs from a
gas in that it may be
liquefied by applying
pressure at constant
temperature.
26
Van der Waals Equation of State:
More Typical P vs V curves for various T (isotherms)
P
Isotherms
vapor
Tc
C
gas
• An inflection point,
which occurs on the
curve at the critical
temperature Tc, gives
the critical point
V
(Tc,Pc).
27
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).
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.
29
• 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
30
• 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
γ
PV = constant.
31
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].
32
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).
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].
34
Summary: Reversible Processes for an Ideal Gas
Adiabatic Isothermal Isobaric Isochoric
Process
Process
Process
Process
PVγ = K T = constant P = constant V = constant
γ = CP/CV
W=
– [1/(γ - 1)]
[P2V2 – P1V1]
ΔE = CV ΔT
W = nRT W = P V
 ln(V2 /V1)
ΔE = 0
W=0
ΔE = CP ΔT ΔE = CV ΔT
PV = nRT, E = ncVT, cP – cV = R, γ = cP/cV
Monatomic ideal gas cV = (3/2)R, γ = 5/3
35