Lecture 11 Equilibrium Between Particles Agenda for today

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Lecture 11Equilibrium Between ParticlesAgenda
for today

• Free Energy and Chemical Potential
• Ideal gases
• Law of Atmospheres
• Carriers in semiconductors

Reference for this Lecture Elements Ch 11
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Free Energy F U - TS
Equilibrium corresponds to minimum free energy if
system is in contact with resevoir at temperature
T.
Next, consider exchange of gas between two
containers
Equilibrium condition
(dF/dN1 0)
Call
For these 2 subsystems exchanging particles, we
see that the condition for chemical equilibrium
is

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Volume exchange p1 p2
mechanical equilibrium Energy exchange T1
T2 thermal equilibrium Particle
exchange m1 m2 chemical
equilibrium
Particles flow toward low m, just as heat flows
toward low T. Why ?
Suppose m1 gt m2 , then
So total free energy can be lowered by making dN1
lt 0. Particles want to flow from V1 into V2 to
lower free energy.
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Physical significance of Chemical Potential
Consider the case of no external potential
m kT ln(n) - constant
high m low m
mu is kT log density
Diffusion results from a gradient in chemical
potential.
Equilibrium when m is constant.
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Well need the chemical potential of an ideal
gas, for many applications.
V and T held fixed
For the entropy, think of dividing volume V into
M cells, each of volume ?V. So M V/ ?V.
Distribute N indistinguishable particles into M
cells.
(Used Stirling approximation for ln N!)
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For ?V, need some quantum mechanics. Consider
deBroglie wavelength ? h/p for a particle with
average kinetic energy kT,
quantum volume
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nQ, (also called nT) is known as the quantum
density
Putting all this together, we obtain the chemical
potential of an ideal gas
Actual density
quantum density
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Chemical potential of particles with potential
energy)
The potential energy gets added to the
translational energy
PE potential energy per particle
The potential energy makes an additional
contribution (NPE) to the free energy F U -
TS.
So, the chemical potential (m dF/dN) gains an
additional contribution d(NPE)/dN PE.

So,
Examples



(For now we are ignoring the rotation and
vibration of molecules.)

Remember Chemical potential is the change
in free energy when we add one particle (at
constant V,T).
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The Law of Atmospheres Revisited

• Problem Find how the gas density n and pressure
  p of the gas vary with height h from the earths
  surface?
• T T1T2 (uniform temperature)

Pressure p2 p(h) p1 1 atm
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Model system 2 boxes in contact with a thermal
reservoir at temperature T
They each have volume V and are connected by a
narrow tube of negligible volume.
1. Equilibrium Condition
2. Chemical Potentials for the ideal gases
3. Solution
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Carriers in Semiconductors
Conduction band (at T 0 unpopulated with
electrons)
Energy gap, D
Valence band (at T 0 totally filled with
electrons)
At non-zero temperature, electrons are thermally
excited from the valence band to the conduction
band. For simplicity, assume every electron in
the conduction band has energy ? more than every
hole in the valence band.
The activated free electrons and the remaining
holes left behind act as two ideal gases!! We
can compute exactly the density of thermally
excited electrons (and holes) by minimizing
Felectron Fhole .
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Intrinsic Semiconductors Simplified
Ne electrons and Nh holes
Ne Nh by electro-neutrality
Total free energy, F Fe Fh, is a minimum.
For each electron, a hole is created
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Holes
Electrons
(Note Electron and holes masses are not exactly
free electron mass, so nQ depends upon the
specific semiconductor, Si, Ge, etc.)
For an intrinsic semiconductor,
So,
For silicon at room temperature,
D 1.14 eV , kT .026 eV
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Digital Thermometers

• Many modern digital thermometers use a
  thermistor, a semiconductor device whose
  conductance (and hence, whose resistance) depends
  on temperature. When the temperature increases,
  the thermal energy (kT) is enough to promote
  some of the electrons from the valence band into
  the conduction band. The resistance of the
  material drops markedly, and can be used to
  determine the temperature.

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Work from Free Energy due to non-equilibrium m

• Mechanical Work (isothermal expansion of N
  particles at temp T)

n1 n2
Quasi-static expansion (requiring an
external force on piston)
Thermal Bath, T
Change in internal energy of gas
Change in entropy of gas
Free energy
Change in chemical potential
Wby -DF -DU T DS -Dm ? N
NkT (ln(ni) - ln(nf)) NkT
ln(ni/nf) NkT ln(Vf/Vi) Look familiar? Same
old physics, different language.
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ACT 3 Work from Free Energydue to
non-equilibrium m
NG NP
How about this situation (equal pressures,
different molecule types)
If these are ideal gases each occupying half of
V, how much work, if any, can be extracted from
this situation? a) 0 b) (NG - NP)
kT c) (NGNP) kTln(2)
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ACT 3 Work from Free Energydue to
non-equilibrium m Solution
NG NP
How about this situation (equal pressures,
different molecule types)
If these are ideal gases each occupying half of
V, how much work, if any, can be extracted from
this situation? a) 0 b) (NG - NP)
kT c) (NGNP) kTln(2)
Each of the components individually loses free
energy in the expansion (after the expansion,
theres obviously no longer the possibility of
expanding and doing work!) therefore, each
component individually could do work up to DF N
kT ln2.
How to realize need a sort of single-species
permeable piston.
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Chemical reactions (more next lecture)
2H2 O2 ? 2 H2O
Free energy of components on left gt Free energy
of products
W - ?F
Use change in free energy to do electrical work
fuel cell.
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Fuel cells
Unlike typical heat engines which first burn
fuel to make heat, theres no conversion to heat
-gt no intrinsic inefficiency
Hydrogen fuel cells on the space shuttle
Directly extract electrical energy from
gradients in m. 2H2 O2 -gt 2H2O energy
high m -gtlow m energy
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Discussion Converting fuel to work

• The work output comes from the excess free energy
  (above the equilibrium value) of the starting
  chemicals (fuel).
• In an atmosphere containing O2, H2 is fuel.
• Theres lots of H2 in the H2O in the oceans.
• Do we therefore have plenty of fuel?
• a) yes b) no c) maybe

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Home Exercise

• Lets say that some gas (helium) is present at
  five parts per million in the atmosphere, at
  300K. You want a 99 pure sample of it at 1 atm.
  About how much work must be done per mole to get
  that? (ignore changes of F due to the other
  constituents.)
• Whats the difference in m between the inside and
  outside of the sample tank? Which has bigger
  m?
• How much DF per mole?

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Home Exercisesolution

• Lets say that some gas (helium) is present at
  five parts per million in the atmosphere, at
  300K. You want a 99 pure sample of it at 1 atm.
  About how much work must be done per mole to get
  that? (ignore changes of F due to the other
  constituents.)
• Whats the difference in m between the inside and
  outside of the sample tank? Which has bigger
  m?
• How much DF per mole?
• 5x10-20J (6 x 1023) 3 x 104J
• Thats about 0.01 kWhr, would cost about 0.001
• unfortunately real life extraction methods are
  much less efficient