6.A. Partial pressures for an ideal gas. Several questions came up last week
with the hemoglobin problem about partial pressure. In this problem, you get to
derive the relation between partial pressure and chemical potential, and then
show that each gas can independently act as an ideal gas. (from Sturge, 7.2)
a) By combining the relationship G = µ i N i and the standard differential
i
dG = SdT + Vdp + µ i dN i , derive the Gibbs-Dunham relation:
i
µ 1
+ SdT Vdp = 0 , and use this to show that i = , where i
ni
p T
i
denotes each distinguishable species in a mix, Ni is the number of that species
and ni is its number density.
N dµ
i
i
b) The equation of state of an ideal gas is p= nkBT, where n is the particle density
(N/V). Use the result in (a) to re-derive the form of the ideal gas chemical
potential, µ i = kbT ln ( Ani ) where A in this derivation can only be specified as a
parameter that depends only on temperature (we know it to be the quantum
volume, 1/nQ).
c) In a mixture of gases, not necessarily ideal, show that the total pressure is
given by ptot = pi where we define pi i
µ (ni )
ni dµ i and the integral is at constant
0
temperature (but variable ni). pi is called the partial pressure of the Ith gas. You
will need to make the reasonable assumption that p = 0 when all the ni are zero.
d) Show that if the gases are ideal (with the chemical potential relationship to
density derived in (b), show that the partial pressures are the actual pressures
that the gases would exert had they been present alone: p = kbT ni .
i
Extra 6.B. Biopolymer equilibrium (From Kittel and Kroemer, 9.4)
Consider chemical equilibrium of a solution (which you should consider as ideal)
of linear polymers made up of identical units. The basic reaction step is
monomer + Nmer = (N + 1)mer . Let KN denote the equilibrium constant for adding
this Nth monomer to the chain.
a) Show from the law of mass action that the concentrations [1] of monomer and
[N+1] of (N+1)mer satisfy the relationship:
[1]N +1
[N + 1] =
K1K 2 K 3 K N
b) Following from the derivation of K in class, show that
3/2
nQ (N)nQ (1) (FN +1 FN F1 )
MN kbT KN =
e
, where nQ (N) = , MN = mass of one
2 2 n (N + 1)
Q
Nmer molecule and FN is its free energy.
c) Assume a large enough N that nQ (N) nQ (N + 1) and a negligible free energy
change for the basic reaction step ( F = FN +1 FN F1 = 0 ) and find the
[N + 1]
. For concreteness, use
[1]
parameters for amino acids in a bacterial cell, [1] ~ 1020 cm-3 and M1 ~ 200 MW.
You should find that the phase-space considerations alone (more available states
for the monomer than the polymer) bias the system towards free monomer.
concentration ratio of polymer to monomer: R =
d) Show that for the long molecules to be stable you need the change in free
energy to be negative, with F <~ 0.4eV . In reality, F ~ + 0.2 eV for this
reaction with an amino acid chain, so you wouldn’t expect to see polymers.
However, cells accomplish this by combining the process of building a
biopolymer with breaking apart ATP, which is a very exothermic reaction, and
by combining both of these at the same time, the net Fis sufficiently negative to
build the polymer, and there are kinetic barriers to its dissolution (see C. Kittel,
Am. J. Phys. 40 (1972) 60).
Extra 6.C. Thermal Ionization of Hydrogen (From Kittel 9.2)
Consider the reaction e + H + H , where a hydrogen atom splits into an
electron e plus a proton H+.
a) Show the equilibrium concentrations satisfy the relation
[e][H + ]
nQe e Eo /kbT
[H]
3/2
m k T
where nQe = e b 2 , the electron quantum concentration (without spin) and
2 Eo is the ionization energy (13.6 eV). This is known as the Saha equation. Use the
approach outlined in class rather than that in your text.
b) If all the electrons and protons come from the ionization of hydrogen atoms,
1/2 Eo /2
e
show [e] = [H + ] = [H] nQe
. If additional electrons are added to the system,
what happens to the proton concentration [H+]?
c) Let [H*] be the concentration of excited (neutral) hydrogen atoms in their first
excited state, which is 0.75 Eo above the ground state. At the surface of the sun,
where [H] ~ 1023 cm-3 and T ~ 5000 K, compare the concentrations of excited
hydrogen [H*] to ionized hydrogen, [H+]. Note the first excited state is
effectively 8-fold degenerate (2s + 2p), and the ground state 2-fold (1s).
Extra 6.D. Neutron Star Collapse to Black Hole (from Sturge 12.10)
We (will) show in class (and your text outlines in problem 7.23) that balancing
the gravitational and kinetic energy in a white dwarf star leads to a situation
where if the mass is large enough (about 1.4 solar masses), the gravitational pull
to shrink the radius increases the electron density to the point where the
electrons are relativistic, and eventually the Fermi pressure of the electrons can
no longer counteract gravity. The white dwarf star then collapses until the
density approaches that of the nucleus. The electrons then combine with the
protons in the helium nuclei (the hydrogen have all fused to become He or larger
by this point) to form neutrons, and a new equilibrium ensues where the Fermi
pressure of the neutrons counteracts that of gravity.
In a neutron star, the process can repeat, with a heavy enough star collapsing to
form a black hole.
a) Using similar assumptions to the white-dwarf case, estimate the equilibrium
radius and mean density of a neutron star with a mass of 2 solar masses, where
the neutrons may be assumed to be non-relativistic. Compare the density to the
density of the sun (MS ~ 2 1030 kg, RS~ 7108 m) and to that of nuclear matter
( Rn ~ 1.2A1/3 fm , where A = mass in atomic mass units).
b) Estimate the limiting mass of a neutron star, beyond which the neutrons (now
relativistic) no longer have enough Fermi pressure to prevent collapse into a
black hole. Note that there are a lot of interactions that we are ignoring (like the
strong force and general relativity), so this is not as good an estimate as for the
Chandrasekhar mass of a white dwarf).