Chapter 3: Interactions and Implications. Start with Thermodynamic Identities

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Chapter 3 Interactions and Implications.Start
with Thermodynamic Identities
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Diffusive Equilibrium and Chemical Potential
Lets fix VA and VB (the membranes position is
fixed), but assume that the membrane becomes
permeable for gas molecules (exchange of both U
and N between the sub-systems, the molecules in A
and B are the same ).
For sub-systems in diffusive equilibrium
UA, VA, NA
UB, VB, NB
In equilibrium,
- the chemical potential
Sign - out of equilibrium, the system with the
larger ?S/?N will get more particles. In other
words, particles will flow from from a high ?/T
to a low ?/T.
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Chemical Potential examples
Einstein solid consider a small one, with N
3 and q 3.
lets add one more oscillator
To keep dS 0, we need to decrease the energy,
by subtracting one energy quantum.
?
Thus, for this system
Monatomic ideal gas
At normal T and P, ln(...) gt 0, and ? lt 0 (e.g.,
for He, ? - 510-20 J - 0.3 eV.
Sign - usually, by adding particles to the
system, we increase its entropy. To keep dS 0,
we need to subtract some energy, thus ?U is
negative.
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The Quantum Concentration
?
nN/V the concentration of molecules
0
when n increases
The chemical potential increases with the density
of the gas or with its pressure. Thus, the
molecules will flow from regions of high density
to regions of lower density or from regions of
high pressure to those of low pressure .
?
when n ? nQ, ? ? 0
- the so-called quantum concentration (one
particle per cube of side equal to the thermal de
Broglie wavelength). When nQ gtgt n, the gas is in
the classical regime.
At T300K, P105 Pa , n ltlt nQ. When n ? nQ, the
quantum statistics comes into play.
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Entropy Change for Different Processes
The partial derivatives of S play very important
roles because they determine how much the entropy
is affected when U, V and N change
Type of interaction Exchanged quantity Governing variable Formula
thermal energy temperature
mechanical volume pressure
diffusive particles chemical potential
The last column provides the connection between
statistical physics and thermodynamics.
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Thermodynamic Identity for dU(S,V,N)
? if monotonic as a function of U
(quadratic degrees of freedom!), may be
inverted to give
pressure
chemical potential
compare with
? shows how much the systems energy changes
when one particle is added to the system at fixed
S and V. The chemical potential units J.
- the so-called thermodynamic identity for U
This holds for quasi-static processes (T, P, ?
are well-define throughout the system).
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Thermodynamic Identities
- the so-called thermodynamic identity
With these abbreviations
? shows how much the systems energy changes
when one particle is added to the system at fixed
S and V. The chemical potential units J.
? is an intensive variable, independent of the
size of the system (like P, T, density).
Extensive variables (U, N, S, V ...) have a
magnitude proportional to the size of the system.
If two identical systems are combined into one,
each extensive variable is doubled in value.
The thermodynamic identity holds for the
quasi-static processes (T, P, ? are well-define
throughout the system)
The 1st Law for quasi-static processes (N
const)
This identity holds for small changes ?S provided
T and P are well defined.
The coefficients may be identified as
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The Equation(s) of State for an Ideal Gas
Ideal gas (fN degrees of freedom)
The energy equation of state (U ? T)
The pressure equation of state (P ? T)
- we have finally derived the equation of state
of an ideal gas from first principles! In other
words, we can calculate the thermodynamic
information for an isolated system by counting
all the accessible microstates as a function of
N, V, and U.
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Ideal Gas in a Gravitational Field
Pr. 3.37. Consider a monatomic ideal gas at a
height z above sea level, so each molecule has
potential energy mgz in addition to its kinetic
energy. Assuming that the atmosphere is
isothermal (not quite right), find ? and
re-derive the barometric equation.
note that the U that appears in the
Sackur-Tetrode equation represents only the
kinetic energy
In equilibrium, the chemical potentials between
any two heights must be equal
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An example of a non-quasistatic adiabatic process
Caution for non-quasistatic adiabatic processes,
?S might be non-zero!!!
Pr. 3.32. A non-quasistatic compression. A
cylinder with air (V 10-3 m3, T 300K, P 105
Pa) is compressed (very fast, non-quasistatic) by
a piston (A 0.01 m2, F 2000N, ?x 10-3m).
Calculate ?W, ?Q, ?U, and ?S.
holds for all processes, energy conservation
quasistatic, T and P are well-defined for any
intermediate state
quasistatic adiabatic ? isentropic
non-quasistatic adiabatic
?Q 0 for both
The non-quasistatic process results in a higher T
and a greater entropy of the final state.
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Direct approach
adiabatic quasistatic ? isentropic
adiabatic non-quasistatic
12
2
To calculate ?S, we can consider any quasistatic
process that would bring the gas into the final
state (S is a state function). For example, along
the red line that coincides with the adiabata and
then shoots straight up. Lets neglect small
variations of T along this path (? U ltlt U, so it
wont be a big mistake to assume T ? const)
P
?U Q 1J
1
Vi
Vf
V
The entropy is created because it is an
irreversible, non-quasistatic compression.
2
P
For any quasi-static path from 1 to 2, we must
have the same ?S. Lets take another path along
the isotherm and then straight up
?U Q 2J
isotherm
1
Vi
Vf
V
straight up
Total gain of entropy
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The inverse process, sudden expansion of an
ideal gas (2 3) also generates entropy
(adiabatic but not quasistatic). Neither heat nor
work is transferred W Q 0 (we assume the
whole process occurs rapidly enough so that no
heat flows in through the walls).
2
P
Because U is unchanged, T of the ideal gas is
unchanged. The final state is identical with the
state that results from a reversible isothermal
expansion with the gas in thermal equilibrium
with a reservoir. The work done on the gas in the
reversible expansion from volume Vf to Vi
3
1
Vi
Vf
V
The work done on the gas is negative, the gas
does positive work on the piston in an amount
equal to the heat transfer into the system
Thus, by going 1 ? 2 ? 3 , we will increase the
gas entropy by
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Systems with a Limited Energy Spectrum
The definition of T in statistical mechanics is
consistent with our intuitive idea of the
temperature  (the more energy we deliver to a
system, the higher its temperature) for many, but
not all systems.
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Unlimited Energy Spectrum
the multiplicity increase monotonically with U
? ? U f N/2
self-gravitating ideal gas (not in thermal
equilibrium)
Pr. 3.29. Sketch a graph of the entropy of H20
as a function of T at P const, assuming that CP
is almost const at high T.
ideal gas in thermal equilibrium
S
Pr. 1.55
?
U
At T ?0, the graph goes to 0 with zero slope. At
high T, the rate of the S increase slows down (CP
? const). When solid melts, there is a large ?S
at T const, another jump at liquidgas phase
transformation.
U
T
T gt 0
T gt 0
U
C
C
U
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Limited Energy Spectrum two-level systems
E
e.g., a system of non-interacting spin-1/2
particles in external magnetic field. No
quadratic degrees of freedom (unlike in an
ideal gas, where the kinetic energies of
molecules are unlimited), the energy spectrum of
the particles is confined within a finite
interval of E (just two allowed energy levels).
2N?B
the multiplicity and entropy decrease for some
range of U
S
U
in this regime, the system is described by a
negative T
T
U
Systems with T lt 0 are hotter than the systems
at any positive temperature - when such systems
interact, the energy flows from a system with T lt
0 to the one with T gt 0 .
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½ Spins in Magnetic Field
N? - the number of up spins N? - the number of
down spins
????????????????????
The magnetization
E
E2 ?B
an arbitrary choice of zero energy
0
The total energy of the system
E1 - ?B
? - the magnetic moment of an individual spin
Our plan to arrive at the equation of state for
a two-state paramagnet UU (N,T,B) using the
multiplicity as our starting point.
? (N,N?) ? S (N,N?) kB ln ? (N,N?) ?
? U U (N,T,B)
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From Multiplicity to S(N?) and S(U)
The multiplicity of any macrostate with a
specified N?
Max. S at N? N ? (N? N/2) SNkBln2
ln2 ? 0.693
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From S(U,N) to T(U,N)
The same in terms of N? and N?
Boltzmann factor!
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The Temperature of a Two-State Paramagnet
T ? and T - ? are (physically) identical
they correspond to the same values of
thermodynamic parameters.
E2
T
E1
E2
E1
N? B
0
U
- N? B
Systems with neg. T are warmer than the systems
with pos. T in a thermal contact, the energy
will flow from the system with neg. T to the
systems with pos. T.
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The Temperature of a Spin Gas
The system of spins in an external magnetic
field. The internal energy in this model is
entirely potential, in contrast to the ideal gas
model, where the energy is entirely kinetic.
Boltzmann distribution
B
At fixed T, the number of spins ni of energy Ei
decreases exponentially as energy increases.
spin 5/2 (six levels)
Ei
Ei
Ei
Ei
the slope ? T
T ?
no T
- lnni
- lnni
- lnni
- lnni
For a two-state system, one can always introduce
T - one can always fit an exponential to two
points. For a multi-state system with random
population, the system is out of equilibrium, and
we cannot introduce T.
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The Energy of a Two-State Paramagnet
? (N,N?) ? S (N,N?) kB ln ? (N,N?) ?
? U U (N,T,B)
The equation of state of a two-state paramagnet
U
U
N? B
? B/kBT
T
1
- N? B
- N? B
U approaches the lower limit (-N?B) as T
decreases or, alternatively, B increases (the
effective gap gets bigger).
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S(B/T) for a Two-State Paramagnet
S
NkBln2
U
Problem 3.23 Express the entropy of a two-state
paramagnet as a function of B/T .
0
N? B
- N? B
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S(B/T) for a Two-State Paramagnet (cont.)
B/T ? 0, S NkB ln2
B/T ? ?, S 0
high-T (low-B) limit
S/NkB
ln2 ? 0.693
ln2 ? 0.693
low-T (high-B) limit
kBT/?B x-1
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Low-T limit
Which x can be considered large (small)?
ln2 ? 0.693
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The Heat Capacity of a Paramagnet
The low -T behavior the heat capacity is small
because when kBT ltlt 2? B, the thermal
fluctuations which flip spins are rare and it is
hard for the system to absorb thermal energy (the
degrees of freedom are frozen out). This behavior
is universal for systems with energy
quantization. The high -T behavior N? N ? and
again, it is hard for the system to absorb
thermal energy. This behavior is not universal,
it occurs if the systems energy spectrum
occupies a finite interval of energies.
2? B
kBT
2? B
kBT
C
NkB/2
compare with Einstein solid
equipartition theorem (works for quadratic
degrees of freedom only!)
E per particle
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The Magnetization, Curies Law
The magnetization
M
The high-T behavior for all paramagnets (Curies
Law)
N?
1
? B/kBT
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Negative T in a nuclear spin system ? NMR ? MRI
Fist observation E. Purcell and R. Pound (1951)
An animated gif of MRI images of a human head. -
Dwayne Reed
Pacific Northwest National Laboratory
By doing some tricks, sometimes it is possible to
create a metastable non-equilibrium state with
the population of the top (excited) level greater
than that for the bottom (ground) level -
population inversion. Note that one cannot
produce a population inversion by just increasing
the systems temperature. The state of population
inversion in a two-level system can be
characterized with negative temperatures - as
more energy is added to the system, ? and S
actually decrease.
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Metastable Systems without Temperature (Lasers)
For a system with more than two energy levels,
for an arbitrary population of the levels we
cannot introduce T at all - that's because you
can't curve-fit an exponential to three arbitrary
values of , e.g. if occ. f (?E) is not
monotonic (population inversion). The latter
case an optically active medium in lasers.
E4
Population inversion between E2 and E1
E3
E2
E1
Sometimes, different temperatures can be
introduced for different parts of the spectrum.
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Problem

• A two-state paramagnet consists of 1x1022
  spin-1/2 electrons. The component of the
  electrons magnetic moment along B is ? ?B ?
  9.3x10-24 J/T.  The paramagnet is placed in an
  external magnetic field B 1T which points up.
• Using Boltzmann distribution, calculate the
  temperature at which N? N?/e.
• Calculate the entropy of the paramagnet at this
  temperature.
• What is the maximum entropy possible for the
  paramagnet? Explain your reasoning.

spin 1/2 (two levels)
(a)
?B
B
- ?B
E2 ?BB
kBT
E1 - ?BB
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Problem (cont.)
If your calculator cannot handle coshs and
sinhs
S/NkB
0.09
kBT/ ?B
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Problem (cont.)
(b)
the maximum entropy corresponds to the limit of T
? ? (N?N?) S/NkB ? ln2
For example, at T300K
E2
kBT
E1
ln2
S/NkB
T ? ? S/NkB ? ln2
T ? 0 S/NkB ? 0
kBT/ ?B