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Starting with the result of Problem 3.5, calculate the heat capacity of an Einstein solid in the low-temperature limit. Sketch the predicted heat capacity as a function of temperature.

Short Answer

Expert verified

The required expression is Cv=N2kT2e-kT.

The graph of the predicted heat capacity as a function of temperature can be sketched as follows:

Step by step solution

01

Given

The equation for Einstein solid at low temperature is calculated as:

U=Ne-kT............(1)

Here, Nis number of oscillator, is the amount of energy quanta, kis Boltzmann constant and Tis temperature.

02

Calculation of heat capacity

Heat capacity at constant volume is given as:

Cv=UTN,V

Where, Uis internal energy.

By substututing the value of Uin the above equation, we get,

Cv=TNe-kTCv=N2kT2e-kT

03

Graph of the heat capacity as a function of temperature

Consider the equation which gives the relation of the heat capacity as a function of temperature:

Cv=N2kT2e-kT

Now, by considering the rest other factors as a constant, heat capacity as a function of temperature can be given as:

Cv1T2e-1T

Based on the above relation, the graph can be plotted as below:

04

Final answer

The required expression is derived as Cv=N2kT2e-kTand the graph showing the heat capacity as a function of temperature can be made as follows:

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Most popular questions from this chapter

In Problem 2.18 you showed that the multiplicity of an Einstein solid containing N oscillators and q energy units is approximately

(N,q)q+Nqqq+NNN

(a) Starting with this formula, find an expression for the entropy of an Einstein solid as a function of N and q. Explain why the factors omitted from the formula have no effect on the entropy, when N and q are large.

(b) Use the result of part (a) to calculate the temperature of an Einstein solid as a function of its energy. (The energy is U=q, where is a constant.) Be sure to simplify your result as much as possible.

(c) Invert the relation you found in part (b) to find the energy as a function of temperature, then differentiate to find a formula for the heat capacity.

(d) Show that, in the limit T, the heat capacity is C=Nk. (Hint: When x is very small, ex1+x.) Is this the result you would expect? Explain.

(e) Make a graph (possibly using a computer) of the result of part (c). To avoid awkward numerical factors, plot C/Nkvs. the dimensionless variable t=kT/, for t in the range from 0 to about 2. Discuss your prediction for the heat capacity at low temperature, comparing to the data for lead, aluminum, and diamond shown in Figure 1.14. Estimate the value of , in electron-volts, for each of those real solids.

(f) Derive a more accurate approximation for the heat capacity at high temperatures, by keeping terms through x3 in the expansions of the exponentials and then carefully expanding the denominator and multiplying everything out. Throw away terms that will be smaller than(/kT)2 in the final answer. When the smoke clears, you should find C=Nk1-112(/kT)2.

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=2.4810-5J/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).

Starting with the result of Problem 2.17, find a formula for the temperature of an Einstein solid in the limit qN. Solve for the energy as a function of temperature to obtain U=Ne-/kT (where is the size of an energy unit).

When the sun is high in the sky, it delivers approximately 1000 watts of power to each square meter of earth's surface. The temperature of the surface of the sun is about 6000K, while that of the earth is about 300K.

(a) Estimate the entropy created in one year by the flow of solar heat onto a square meter of the earth.

(b) Suppose you plant grass on this square meter of earth. Some people might argue that the growth of the grass (or of any other living thing) violates the second law of thermodynamics, because disorderly nutrients are converted into an orderly life form. How would you respond?

Use the thermodynamic identity to derive the heat capacity formula

CV=TSTV

which is occasionally more convenient than the more familiar expression in terms of U. Then derive a similar formula for CP, by first writing dHin terms of dSand dP.

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