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

Suppose you have a mixture of gases (such as air, a mixture of nitrogen and oxygen). The mole fraction xiof any species iis defined as the fraction of all the molecules that belong to that species: xi=Ni/Ntotal. The partial pressure Piof species iis then defined as the corresponding fraction of the total pressure: Pi=xiP. Assuming that the mixture of gases is ideal, argue that the chemical potential μiof species i in this system is the same as if the other gases were not present, at a fixed partial pressure Pi.

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).

In solid carbon monoxide, each CO molecule has two possible orientations: CO or OC. Assuming that these orientations are completely random (not quite true but close), calculate the residual entropy of a mole of carbon monoxide.

Use the result of Problem 2.42 to calculate the temperature of a black hole, in terms of its mass M. (The energy is Mc2. ) Evaluate the resulting expression for a one-solar-mass black hole. Also sketch the entropy as a function of energy, and discuss the implications of the shape of the graph.

The results of either of the two preceding problems can also be applied to the vibrational motions of gas molecules. Looking only at the vibrational contribution to the heat capacity graph for H2shown in Figure 1.13, estimate the value of εfor the vibrational motion of anH2 molecule.

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