Use a computer to study the entropy, temperature, and heat capacity of an Einstein solid, as follows. Let the solid contain 50 oscillators (initially), and from 0 to 100 units of energy. Make a table, analogous to Table 3.2, in which each row represents a different value for the energy. Use separate columns for the energy, multiplicity, entropy, temperature, and heat capacity. To calculate the temperature, evaluate ΔU/ΔSfor two nearby rows in the table. (Recall that U=qϵfor some constant ϵ.) The heat capacity (ΔU/ΔT)can be computed in a similar way. The first few rows of the table should look something like this:

(In this table I have computed derivatives using a "centered-difference" approximation. For example, the temperature .28is computed as 2/(7.15-0).) Make a graph of entropy vs. energy and a graph of heat capacity vs. temperature. Then change the number of oscillators to 5000 (to "dilute" the system and look at lower temperatures), and again make a graph of heat capacity vs. temperature. Discuss your prediction for the heat capacity, and compare it to the data for lead, aluminum, and diamond shown in Figure 1.14. Estimate the numerical value of εin electron-volts, for each of those real solids.

Short Answer

Expert verified

The table can be prepared as:

The graphs can be made as:

For 50 oscillations

For 5000 oscillations

The values of εare:

εlead=6.875×103eVεaluminium=2.587×102eVεdiamond=0.19233eV

Step by step solution

01

Given Information

The given Einstein solid contains 50 oscillators from 0 to 100 units of energy.

That is,

N=50q=1to100

U=qϵ

02

Calculation for table values

The entropy of the system is given as:

Sk=lnΩSk=lnq+N-1qN-1

The temperature is given as:

1T=SUT=ΔUΔST=Δ(εq)Δ(klnΩ)kTϵ=Δ(q)Δ(lnΩ)

Heat capacity per oscillation is given as:

CN=ΔUNΔTCN=Δ(εq)NΔΔ(εq)Δ(klnΩ)CNk=Δ(q)NΔΔ(q)Δ(lnΩ)

The table consisting of entropy, temperature, and heat capacity of an Einstein solid for 50 oscillations with 0 to 100 units of energy can be prepared as follows:

03

Graph

The graphs of entropy vs energy and heat capacity vs temperature can be sketched as follows:

For 50 oscillations:

The graph of entropy vs energy can be made as:

The graph of heat capacity vs temperature can be made as:

For 5000 oscillations:

The graph of entropy vs energy can be made as:

The graph of heat capacity vs temperature can be made as:

The heat capacity is substantially smaller than the values in figure 1.14, since, for 5000 oscillations with the same units of energy, the energy is lowered, lowering the system's temperature.

04

Calculation of ε

For q=1,Ω=1

Solving for ε,we get,

U=ϵqε=Uqϵ=Tk(lnΩ)qε=kT

For lead, heat capacity is maximum at T=80K

εlead=kTεlead=1.38×10-23×80εlead=1.1×10-21Jεlead=6.875×10-3eV

For aluminum heat capacity is maximum at T=300K

εaluminum=kTεaluminum=1.38×10-23×300εaluminum=4.14×10-21Jεaluminum=2.587×10-2eV

For diamond heat capacity is maximum at T=2230K

εdiamond=kTεdiamond=1.38×10-23×2230εdiamond=3.0774×10-20Jεdiamond=0.19233eV

05

Final answer

The table of the values can be prepared as:

The graphs can be prepared as:

For 50 oscillations:

For 5000 oscillations:

The value of εare:

ϵlead=6.875×103eVϵaluminium=2.587×102eVϵdiamond=0.19233eV

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Most popular questions from this chapter

Experimental measurements of the heat capacity of aluminum at low temperatures (below about 50K) can be fit to the formula

CV=aT+bT3

where CVis the heat capacity of one mole of aluminum, and the constants aand bare approximately a=0.00135J/K2and b=2.48×10-5J/K4. From this data, find a formula for the entropy of a mole of aluminum as a function of temperature. Evaluate your formula at T=1Kand at T=10K, expressing your answers both in conventional units (J/K)and as unitless numbers (dividing by Boltzmann's constant).

A cylinder contains one liter of air at room temperature ( 300K) and atmospheric pressure 105N/m2. At one end of the cylinder is a massless piston, whose surface area is 0.01m2. Suppose that you push the piston in very suddenly, exerting a force of 2000N. The piston moves only one millimeter, before it is stopped by an immovable barrier of some sort.

(a) How much work have you done on this system?

(b) How much heat has been added to the gas?

(c) Assuming that all the energy added goes into the gas (not the piston or cylinder walls), by how much does the internal energy of the gas increase?

(d) Use the thermodynamic identity to calculate the change in the entropy of the gas (once it has again reached equilibrium).

In the text I showed that for an Einstein solid with three oscillators and three units of energy, the chemical potential is μ=-ϵ(where ϵis the size of an energy unit and we treat each oscillator as a "particle"). Suppose instead that the solid has three oscillators and four units of energy. How does the chemical potential then compare to -ϵ ? (Don't try to get an actual value for the chemical potential; just explain whether it is more or less than -ϵ.)

Fill in the missing algebraic steps to derive equations 3.30, 3.31, and 3.33.

In the experiment of Purcell and Pound, the maximum magnetic field strength was 0.63Tand the initial temperature was 300K. Pretending that the lithium nuclei have only two possible spin states (in fact they have four), calculate the magnetization per particle, M/N, for this system. Take the constant μto be 5×10-8eV/T. To detect such a tiny magnetization, the experimenters used resonant absorption and emission of radio waves. Calculate the energy that a radio wave photon should have, in order to flip a single nucleus from one magnetic state to the other. What is the wavelength of such a photon?

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