Chapter 5 Continued: More Topics in Classical
Thermodynamics
Slide 2
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.
Slide 3
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.
Slide 4
Eddington on Thermodynamics (1929) If someone points out to you
that your pet theory of the universe is in disagreement with
Maxwells Equations then so much the worse for Maxwells equations.
But if your theory is found to be against the 2 nd Law of
Thermodynamics - I can offer you no hope; there is nothing for it
but to collapse in deepest humiliation!
Slide 5
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 1 st
Law is: Q = E + W So, since Q = W = 0, the 1 st Law says that E =
0. Thus this is a very peculiar type of expansion and In a Free
Expansion, The Internal Energy of a Gas Does Not Change!
Slide 6
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 1
st Law says that E = 0. For an Ideal Gas it is easily shown that E
is independent of volume V, so that E = E(T) = CT n (C = constant,
n > 0)
Slide 7
Free Expansion For an Ideal Gas E = E(T) = CT n (C = constant,
n > 0) So, for Adiabatic Free Expansion of an Ideal Gas since E
= 0, T = 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.
Slide 8
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 J (T/V) E (= 0 for an ideal gas) Some useful
manipulation: J (T/V) E = (T/E) V (E/V) T (E/V) T /C V Combined 1
st & 2 nd Laws: dE = T dS pdV.
Slide 9
Joule Coefficient: J = (E/V) T /C V 1 st & 2 nd 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: J = (T/V) E
= [T(P/T) T p]/C V Obtained from the gas Equation of State J is a
measure of how close to ideal a real gas is!
Slide 10
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. p 1,V 1,T 1 p 2,V 2,T 2 Porous Plug Adiabatic Wall
Thermometers Throttling Expansion p 1 > p 2
Slide 11
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 = p 2 V 2 p 1
V 1. 1 st Law: E = 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. Recall the definition of Enthalpy: H pV. So in the
Joule-Thompson process, the Enthalpy H stays constant: H 2 = H 1 or
H = 0.
Slide 12
12 To analyze the Joule-Thompson Effect its convenient to
Define: The Joule-Thompson Coefficient (T/p) H ( > 0 for
cooling. < 0 for heating) Some useful manipulation: = (T/p) H =
(T/H) P (H/p) T = (H/p) T /C P. 1 st & 2 nd Laws: dH = TdS + V
dp.
Slide 13
Joule-Thompson Coefficient (T/p) H ( > 0 for cooling. < 0
for heating) Some useful manipulation: = (T/p) H = (T/H) P (H/p) T
= (H/p) T /C P. The 1 st 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: = (T/p) H = [T(V/T) T
V]/C P Obtained from the gas Equation of State
Slide 14
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 More on the Joule-Thompson Coefficient
Slide 15
Throttling A Constant Enthalpy Process H = E + PV = Constant
Characterized by the Joule-Thomson Coefficient, which can be
expressed as
Slide 16
Another Kind of Throttling Process! (From American slang!)
Slide 17
Throttling Processes: Typical T vs. p Curves Family of Curves
of Constant H (Reifs Fig. 5.10.3)
Slide 18
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
Slide 19
19 The Van der Waals Equation of State has the form: (P + a/v 2
)(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/v 2 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.
Slide 20
20 Adiabatic Processes in an Ideal Gas Ratio of Specific Heats:
c P /c V = C P /C V. 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 C V = (dE/dT). Also, PV = nRT and H =
E + PV, so that H =H(T). So, H = H(T) and C P = (dH/dT). Thus, C P
C V = (dH/dT) (dE/dT) = d(PV)/dT = nR. So, C P C V = nR. Mayers
EquationThis is sometimes known as Mayers Equation, & it holds
for ideal gases only. For 1 kmole, c P c V = R, where c P & c V
are specific heats.
Slide 21
21 Since dQ = 0 for an adiabatic process: dE = P dV & dE =
C V dT, so dT = (P/C V ) dV. For an ideal gas, PV = nRT so that P
dV +V dP = nR dT = (nRP/C V ) dV. & V dP + P (1 +nR/C V ) dV =
0. This gives, C V dP/P + (C V + nR) dV/V = 0. For an ideal gas
ONLY, C P C V = nR. so that C V dP/P + C P dV/V = 0, or dP/P + dV/V
= 0. Simple Kinetic Theory for a monatomic ideal gas (Ch. 6) gets E
= ( 3 / 2 )nRT so C v = ( 3 / 2 )nR & = (C p /C v ) = ( 5 / 3 )
Integration of the last equation in green gives: ln P + ln V =
constant, so that PV = constant.
Slide 22
22 Work Done on an Ideal Gas in a Reversible Adiabatic Process
Method 1: Direct Integration For a reversible adiabatic process, PV
= K. Since the process is reversible, W = PdV, so that W = K V dV =
[K/( 1)] V (1) = [1/( 1)] PV | (limits: P 1 V 1 P 2 V 2 ) So, W =
[1/( 1)] [P 2 V 2 P 1 V 1 ]. For an ideal monatomic gas, = 5/3, so
that W = (3/2)] [P 2 V 2 P 1 V 1 ].
Slide 23
23 Work Done on an Ideal Gas in a Reversible Adiabatic Process
Method 3: From the 1 st Law For a reversible process, W = Q r E.
So, for a reversible adiabatic process: W = E. For an ideal gas, E
= C V T = nc V T = nc V (T 2 T 1 ). So, for a reversible adiabatic
process in an ideal gas: W = nc V (T 2 T 1 ). For an ideal gas PV =
nRT, so that W = (c V /R)[P 2 V 2 P 1 V 1 ]. But, Mayers
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/( 1)] [P 2 V 2 P 1 V 1
].
Slide 24
24 Summary: Reversible Processes for an Ideal Gas Adiabatic
Process Isothermal Process Isobaric Process Isochoric Process PV =
K = C P /C V T = constantP = constantV = constant W = [1/( 1)] [P 2
V 2 P 1 V 1 ] W = nRT ln(V 2 /V 1 ) W = P V W = 0 E = C V TE = 0E =
C V T PV = nRT, E = nc V T, c P c V = R, = c P /c V Monatomic ideal
gas c V = (3/2)R, = 5/3