Thermodynamics/ Statistical Mechanics Resource Packet (comments to: moloney@nextwork.rose-hulman.edu)
Suggestions for 'Active' Classroom Learning
You may want students verify during class time the number of ways of distributing n units of energy among N distinguishable harmonic oscillators for small values of n and N . This is a prelude to the problem of entropy, energy and temperature. (Discussion of method).
When students have some ideas about the first law of thermodynamics, and some sense of dQ = TdS, they should be ready to work out for themselves in classroom groups how to come up with the temperature of a group of N distinguishable oscillators sharing n units of energy. (more details). A number of hints may be in order, depending on how the groups are progressing. The volume is constant so p dV =0. For S to change, either n or N will have to change. The problem can perhaps be stated in terms of a fixed amount of substance, so students realize that N must be constant. If studetnts are led to focus on the need for S to change, they should come to realize than n must change. There may still be some problem with the 'reduced' energy units being used. The dimensionless 'temperature' is in fact kT/e, where e is the size of oscillator energy steps. A change of 1 unit in n will lead to a change in S, and also a change in U. Then the way should be open to find the reduced temperature as (ds/dn)^(-1), with s = ln(number of ways).
The next level of difficulty is to connect two groups of identical oscillators (N1 and N2) which can initially have different amounts of energy (n1 and n2). Students can then be confronted in class with the question of where these two groups will come to equilibrium when energy units can be exchanged among them. Student ideas ought to be jelled by classroom discussion before they tackle the solution on a spreadsheet or other means (more details)
Exercises involving the boltzmann factor
In a simplified view of liquid water,
the liquid state could be thought of as the 'ground state' of a 2-level
system, where the gas is the 'excited state', separated from the ground
state by the heat of vaporization (about 540 cal/g, or about 2200 j/g.)
In this view, the vapor pressure of water would be proportional to the
fraction of molecules in the excited state. a) Use the boltzmann factor
( exp(-E/(kT)) to calculate the fraction of particles in the gas at the
boiling point of water, 373 K. b) Determine the vapor pressure of water
at 50 degrees C, assuming the heat of vaporization is constant. c) Find
the vapor pressure of water just before it freezes, in this model.
Assuming
the Earth's atmosphere is at constant temperature of 300 K (it is not),
determine the ratio of oxygen to nitrogen at a height of 5 km above the
Earth's surface.
Electrons
behave according to Fermi-Dirac statistics, but in many circumstances their
behavior is well described by a boltzmann distribution. This is true of
electrons in the conduction band of silicon, when it comes to describing
concentrations of electrons with a given energy above the bottom of the
conduction band. In silicon whose doping gives a total electron concentration
in the conduction band of 5 * 10^(17) /cm^3, determine the concentration
of electrons whose energy at 300 K is 0.6 eV above the bottom of the conduction
band. [These electrons will be able to flow from the n region of a diode
into the p-region when the barrier between regions is reduced below 0.6v.]
J. Perrin (1908) mixed small particles of volume V =9.8*10^(-15) cm^3 and density D=1.35 g/cm^3 in a solution of water at 300 K. When the mixture reached equilibrium, the number of particles N at a height y above the bottom was described by the Boltzmann factor N = No exp(-mgy/(kT)), where m=DV is the mass of a particle and k is Boltzmann's constant. To account for the buoyant force of the water, one must use the difference in densities between the particles and the water. For the following (y,N) data, determine the 'best' value of Boltzmann's constant. (0 cm,200), (0.0025 cm,170), (0.0050 cm,146), (0.0075 cm,116), (0.0100 cm, 100). Since k is the gas constant per particle, it can be multiplied by Avogadro's number and compared to the gas constant R = 8.314 j/(mole-K). {Have class discuss methods of approaching this problem on a spreadsheet, or with some other package.}
Given the distance between H and Br in HBr is 2.6 A = 0.26 nm, (m(H) = 1 amu, and m(Br) = 80 amu, (1 amu = 1.66 x 10^-27 kg)
a) calculate the moment of inertia of HBr with respect to its center of mass.
b) Since for a rotating body the kinetic energy is 1/2 I w^2 = 1/2 L^2/I , and angular momentum L is quantized in steps of h/(2 pi), calculate the size of each rotational 'energy step'.
c) Find the ratio of HBr molecules in the first rotational excited state (one unit of angular momentum) to the number in the ground state (zero units of angular momentum) at 500K.
d) Plot the ratio in part c) from 100K to 600K
e) Plot the ratio of population in the 10th excited state to that in the ground state from 100 to 600 K.
A possible interpretation of thermodynamic potentials (after a Dan Schroeder post on phys-L).
(Students might be given one or two
and asked to try and come up with one of the others.)
u (internal energy)
kinetic and potential energy of the system
u+pV = h (enthalpy)
work needed to give the system its energy plus the work to push the surrounddings apart to make room for the system's volume. When system pressure and volume change (as in a throttling process), h is of interest, since creating the system from scratch on either side of the throttle valve should involve the same amount of energy or work.
u -Ts = f (helmholtz free energy)
'free' energy available from a system
where volume changes are not of interest. We could reclaim internal energy
if we were to dispose of the system, but the system entropy must be disposed
of .
We might do this by sending heat Q = Ts to a reservoir at temperature T,
and that energy would not be available to us.
u+pV -Ts = g (gibbs free energy)
'free' energy available from the system. In disposing of this system, we could reclaim internal energy, and get back work done by the surroundings as the volume disappears, but the entropy must be disposed of (we imagine) by sending heat to a reservoir at temperature T.
Equipartition of energy questions