From (combin.ms). This calculation goes toward showing that the number of ways of sharing n indistinguishable energy units among N distinguishable oscillators is

Case of N oscillators sharing 1 energy unit:

Looking toward the general method, we could separate off one oscillator (call it A) and divide the problem into subproblems where A has 0 or 1 energy units. When A has 1 unit of energy, there is only one way the rest can share 0 units. When A has 0 units there are N-1 ways the remaining oscillators can share the 1 unit of energy. Adding these sub-problems together we get 1 + N-1 = N = ways of sharing 1 energy unit among N oscillators.

Case of N oscillators sharing 2 energy units:

For 2 energy units, we have 3 sub-problems; the 'A' oscillator has 2, 1, or 0 units of energy. When A has 2 units of energy, there is only 1 way for the rest to share 0 units. When A has 1 unit of energy, the other N-1 have to share 1 unit. When A has 0 units, the other N-1 have to share 2 units.

We can write an equation for N oscillators sharing 2 energy units. It will contain 3 terms, one for A having 0, 1, or 2 energy units. We can also write an equation also for N-1 oscillators sharing 2 units of energy, also containing 3 terms. We can keep on writing 3-term equations until we reach the case of 1 oscillator 'sharing' 2 units of energy. This 'last' equation will say there is only one way for 1 oscillator to share 2 units (or any other number of units) of energy.

Adding these equations together gives the number of ways of N oscillators to share 2 energy units. This number of ways equals the sum from 1 to N-1 of sharing 0 energy units, plus the sum from 1 to N-1 of sharing 1 energy unit, plus the number of ways 1 oscillator shares 2 energy units (only 1 way).

Case of N oscillators sharing 3, 4, etc. energy units.

Statistics of a group of Harmonic Oscillators.

Consider a group of harmonic oscillators, equally spaced levels, ground state energy of zero. The number of ways of sharing n energy units among N such oscillators is

The entropy is S = k ln W . We will be considering two kinds of oscillators, 'large' and 'small' . The unit of energy for 'small' oscillators will be 'e', and for large oscillators, it will be L times larger, or L e. A small oscillator having n units of energy will have U = internal energy = n e, while a large oscillator having n 'large' energy units will have internal energy U = n L e .

If particle number doesn't change and energy units n can change, then at constant volume and particle number we have TdS = dU. For small oscillators dU = e dn, and for large oscillators dU = L e dn. The temperature is then T = 1/[ dS/dU ] . For small oscillators, T = 1/ [ (k/e) d(ln W )/dn ]. [N is held constant.].

When a unit of heat is added, the internal energy changes at constant volume: dQ = TdS = dU =Cv dT. The heat capacity Cv is then dU/dT. dU will be one unit of energy, and dT will be the difference in temperatures.

When particle number changes, we have TdS = dU - mu dN. The idea is to set dS=0 and find an expression for dN. We write dS in terms of partials (shown here as ordinary derivatives)

(1)    dS = (dS/dn) dn + (dS/dN) dN = 0.

From TdS = dU - mu dN,

(2)    dS = 1/T [ e dn - mu dN ] = 0 . 

Solving (2) for mu gives e dn/dN, and from (1) this becomes

(3)    mu = -e dS/dN / dS/dn.

We deal with 'reduced' parameters in this program. S is written without the factor of k, the temperatures are written without the factor of e/k, and mu is written without the factor of e. In effect, the reduced temperature is kT/e, and the reduced mu is mu/e.

Bose - Einstein statistics deals with the number of ways w of putting N indistinguishable "things" into g distinguishable "boxes". One possible application of these statistics would be the case of an "Einstein solid", a system of N identical quantum harmonic oscillators (N/3 atoms each oscillating in three directions). If one takes the viewpoint that the system contains n indistinguish- able energy quanta to be given to N distinguishable oscillators, then B-E statistics may be applied to this case. The number of ways of giving n (indistinguishable) quanta to N (distinguishable) oscillators is: w = [(n + N - 1)!] / [n!(N - 1)!]

For example, if there were only two energy quanta to give to three oscillators, then n = 2, N = 3, and w = 6 as shown below.

The total number of  ways (multiplicity) is six .

Suppose we have two Einstein solids in thermal contact and suppose system #1 has 200 oscillators, #2 has 300 oscillators, and 100 energy quanta are shared between them. Excel or QuattroPro may be used to set up a table in which columns of data include n1 (the number of quanta assigned to system #1), w1 (the number of ways of giving n1 quanta to N1 oscillators), n2 (the number of quanta assigned to system #2), w2 (the number of ways of giving n2 quanta to N2 oscillators), wtotal = w1*w2 (the total multiplicity of the two systems), and S = ln(wtotal) (a number proportional to the entropy of this particular macrostate). [The most probable arrangement is of course the "equilibrium" distribution (the one with the highest entropy).]

[In the following table w1 and w2 are calculated using the Excel function COMBIN(n+N-1,n).] ...

more from statme1.xls

Suppose we now look at one of these Einstein solids, the one that has 200 oscillators (n=100 still). We make a table showing n1, w1, and S1=ln(w1), but note that in changing from one macrostate to the next the total energy of the system has been changed by one quanta (heat in or out). Since the heat input or output dQ (=dU) is temperature T times the change in entropy dS, the temperature associated with a particular macrostate may be taken as (approximately) the change dU in energy between that macrostate and the next, divided by the change dS in entropy. Also, the heat capacity can be caculated by noting that it is the energy change dU between adjacent macrostates divided by the temperature change dT . In the following table, the energy change between one row and the next is one energy unit so T = 1/dS and heat capacity is calculated per oscillator [C = (dU/dT)/N] . The last column calculates, for comparison purposes, the Einstein heat capacity function T^(-2)*exp(1/T)/(exp(1/T)-1)^(-2)

The first graph shows how the temperature of system #1 changes as additional energy units are added. The second graph shows how the heat capacity changes with temperature. Note in the latter graph how the heat capacity falls off as the temperature is lowered, as it should in the Einstein model.

(from comb2osc.ms)

We want to model different degrees of freedom, and will do this by having two types of oscillators, N of each, but with different step sizes. Each type of oscillator is supposed to represent a different 'degree of freedom' in a set of atoms. The intent is to see where the entropy is a maximum when the two groups of oscillators have to share a fixed amount of energy. This is supposed to get at the idea of 'freezing out' degrees of freedom at low temperatures. The temperature obtained is 'reduced' in that it represents kT/step , where step represents the size of a step for a given oscillator.

The 'large' step oscillators will have an energy step L times that of the smaller- step oscillators. Nbig will be the number of larger energy units to be shared between the two groups of oscillators.

Given Nbig and N, we will find a maximum in the total system entropy. We will see what fraction of the Nbig energy units are in each group of oscillators.

VAN DER WAALS EQUATION OF STATE (from vander.ms)

The equation of state for an ideal gas is Pv = RT, where v is the specific volume (volume per mole). The model on which this equation is based assumes negligible forces between the molecules and a negligible amount of space taken up by the molecules themselves. Both of assumptions are reasonable except when the volume and temperature are low enough that the molecules spend a significant amount of time in the neighborhood of other molecules, where they might experience forces exerted by them. If these forces are attractive it is reasonable to assume that the pressure on the walls of a container would be reduced, since a molecule that is about to collide with a wall will be attracted by its nearest neighbors in the interior and momentarily slowed down. In the van der Waals model the pressure is reduced by an amount a/(v)^2 . Also the van der Waals model assumes that the volume in which molecules can move around is less than the volume of the container because of the space taken up by the molecules themselves, so that the v in the equation of state should be replaced by a smaller volume v-b. The van der Waals equation of state for a (non-ideal) gas then becomes:

(P+a/(v)^2)*(v-b) = R*T

(a) Write down the van der Waals equation and solve for P. Substitute the value of the ideal gas constant R = 8314 Joules per kilomole, and the appropriate constants a and b for the gas. (b) Choose three temperatures (about 40 degree intervals) for which you think that graphs of P vs v might look non-ideal in appropriate ranges of specific volume, and substitute them into the expression for P. (c) Plot on the same graph three P vs. v curves for the three temperatures and specific volume ranges from roughly 0.06 to about 0.4 cubic meters per kilomole . Your lowest temperature line should have a temperature such that there is a region of slope zero, but not so low that there is a region of positive slope (positive slope regions occur in the liquid region where the van der Waals equation is not valid). (d) Repeat for an ideal gas (P=RT/v) and compare the graph with that obtained using van der Waals' equation of state.

JOULE-THOMSON COOLING (rev. 1) (from jthom.ms)

When a stream of gas at pressure p1 and temperature T1 is forced through a throttling valve, it emerges at a lower pressure and a temperature which may be lower or higher, depending on the gas and the initial conditions. For this effect to be observed the gas must be nonideal; there must be forces between the molecules. The process increases the time average distance between molecules. If conditions are such that the average intermolecular force are attractive (rather than repulsive) as the average intermolecular distances increase, then the process increases the potential energy of the system; we might expect at least some of this increase to come at the expense of kinetic energy. A drop in kinetic energy means a lower temperature. However, if conditions are such that average intermolecular forces are repulsive as intermolecular distances increase, potential energy will decrease and we might expect an increase in kinetic energy and temperature. Whether the temperature increases or decreases depends on the intial state of the gas (T1, p1), although in all cases the enthalpy h=u+pv is constant. Generally, if the initial pressure is high (pushing the molecules close together) repulsive forces will dominate and the process will produce a heating effect, but if the initial pressure is lower and the temperature is in the right range there can be a cooling effect. This cooling effect has been used in the process of liquefying gases.

The Joule-Thomson coefficient is the indicator of whether the throttling process produces cooling or heating and how effectively it does so. It is defined as the rate of change of temperature with respect to pressure as the enthalpy is held constant (partial derivative of T w.r.t. p, constant h). From the first and second laws of thermodynamics, the Joule-Thomson coefficient (dT/dp) = - (1/cp)*(dh/dp) = (1/cp)[T(dv/dT)-v]. If this coefficient is positive, reductions in pressure will cause reductions in temperature; this is what we would expect to happen at lower initial pressures (in the proper temperature range). If the coefficient is negative, then reductions in pressure cause increases in temperature; we would expect this to happen at higher initial pressures. Therefore, for a gas which has some particular enthalpy h, decreasing the initial pressure (in the proper temperature range) causes the effect to change from one that causes heating to one that causes cooling at some "inversion" initial pressure and temperature. The "inversion curve" is the locus of these inversion points on a p-T graph..

The following exercise uses the van der Waals model of a gas, which takes account of intermolecular forces by adding a/(v)^2 to the pressure term in the equation of state (see the van der Waals exercise):

(p+a/(v)^2)*(v-b)=R*T

The calculations use several relationships that come from the first and second laws of thermodynamics.

{This problem follows Sommerfeld's Vol V, Thermodynamics and Statistical Mechanics}

(a) Solve van der Waals' equation for pressure and calculate partial derivatives dp/dT, dp/dv, and dv/dT=-(dp/dT)/(dp/dv). Also find the volume expansion coefficient beta=(1/v)*(dv/dT) and the isothermal compressibility kappa=-(1/v)*(dv/dp). Simplify these expressions as much as possible.

(b) Find an expression for the rate of change of internal energy with respect to changes in volume when the temperature is held constant:

du/dv=T*(dp/dT)-p Simplify this expression.

(c) Find an expression for the Joule-Thomson coefficient (see the discussion above) and simplify it. Set the coefficient equal to zero and solve for T to find an expression for the inversion temperature in terms of volume v. Solve this equation for v and substitute into the van der Waals equation to get an expression relating the inversion temperature to pressure. Solve for the inversion pressure and simplify.

(d) Substitute the value for R (8314 J/kilomole K) and the appropriate constants a (3440 J m^3/kilomole^2) and b (0.0234 m^3/kilomole) for Helium. Plot the inversion curve on a p-T graph for the temperature range from 4 to 35K. For what part of the graph would the throttling process produce a cooling effect. In order to use the throttling process to cool Helium for the purpose of liquefying it, to what temperature must you get it first by other methods?

(e) Repeat part (d) for Oxygen (a = 138000 Jm^3/kilomole^2, b = 0.0318 m^3/kilomole) and find the temperature-pressure range for which the Joule-Thomson effect results in cooling. In the graph, eliminate negative pressures. [There are temperatues and pressures that fit van der Waals equation that are completely "unphysical"; the van der Waals model doesn't fit reality everywhere.]

MAXWELL (from maxwell.ms)

In the Maxwell theory of an ideal gas the fraction of molecules with speeds between v and v+dv is given by f(v)dv = 4*Pi*(m/(2*Pi* k*T))^(3/2)*v^2*exp(-m*v^2/(2*k*T)), where m is the mass of the gas molecule, k is the Boltzmann constant, and T is the absolute (Kelvin) temperature. The most probable speed is the one that that gives a maximum value for f(v). The average speed is obtained by integrating the product vf(v) over the entire range of speeds. Similarly the mean square speed is found by integrating the v^2*f(v) over the entire range; the root mean square is the square root of the mean square speed. The average molecular kinetic energy is (1/2) m (vrms)^2 .

(a) For oxygen molecules at 300K, calculate kT/m. Then make a plot of the speed distribution function f(v) versus the speed v for a speed range from 0 to 1200 m/s.

(b) In order to find the most probable speed v, differentiate f(v) with respect to v and set it equal to zero (look for a solution in the range v=1..1200). Compare your result with your graph of part (a) and with (2*k*T/m)^(1/2).

(c) Find the average molecular speed and the rms speed and locate them on your graph. Compare your average v with (8*k*T/(Pi*m))^(1/2) the rms speed with (3*k*T/m)^(1/2) .

(d) Calculate the fraction of the molecules having speeds which are less than the average speed .

(e) Calculate the average kinetic energy and compare with 3kT/2 .