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Q. 3.25

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An Introduction to Thermal Physics
Found in: Page 108
An Introduction to Thermal Physics

An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

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Short Answer

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

Ω(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=qϵ, where ϵ 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, ex≈1+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/Nk vs. the dimensionless variable t=kT/ϵ, 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 ϵ, 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 (ϵ/kT)2 in the final answer. When the smoke clears, you should find C=Nk1-112(ϵ/kT)2.

(a) S=k[(q+N)ln(q+N)-qlnq-NlnN]

(b) ϵkT=lnU+NεU

(c) U=Nϵ(eϵkT−1), C=ϵ2NeϵkTkT2(eϵkT−1)2

(d) It is shown that in the limit T→∞, the heat capacity is C=Nk.

(e) The graph can be made as:

CNk=(ϵkT)2kNeεkT(eεkT−1)2εlead=6.875×10−3 eVεaluminum=2.587×10−2 eVεdiamond=0.19233 eV

(f) C≈Nk(1−(ϵkT)212)

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Part (a) Step 1: Given Information

The multiplicity of an Einstein solid is given as:

Ω(N,q)≈q+Nqqq+NNN

Part (a) Step 2: Calculation

The entropy is given as:

S=klnΩ

By substituting the given value in the above equation, we get,

S=kln{(q+Nq)q(q+NN)N}S=k[qln(q+Nq)+Nln(q+NN)]S=k[qln(q+N)−qlnq+Nln(q+N)−NlnN]S=k[(q+N)ln(q+N)−qlnq−NlnN]

The multiplicity for Einstein solid can be given as:

Ω=⌊q+N−1⌊q⌊N−1Ω=⌊q+N−1⌊q⌊N−1×N(q+N)N(q+N)Ω=⌊q+N⌊q⌊N×N(q+N)

But,

Stirling approximation of large n is given as:

localid="1647403887405" ⌊n=2πnnne-n

Hence,

Ω=N(q+N)×2π(q+N)(q+N)q+Ne−(q+N)2π(q)(q)qe−(q)2π(N)(N)Ne−(N)Ω=N(q+N)×q+NqN×(q+N)q+Ne−(q+N)qqNNe−(q+N)Ω=N2πq(q+N)(q+Nq)q(q+NN)N

For large q and N, when compared to the power terms, the square root terms are small. As they have a negligible effect, they can be neglected. Hence, the approximate value can be written as:

Ω=N2πq(q+N)(q+Nq)q(q+NN)NΩ≈(q+Nq)q(q+NN)N

Part (a) Step 3: Final answer

Hence, the entropy of an Einstein solid containing N oscillators and q energy units is given as:

S=k[(q+N)ln(q+N)−qlnq−NlnN]

For large values of q and N, the square root terms are small when compared to the power terms. Hence they are neglected as they have a very negligible effect on the entropy. Therefore the approximate value of multiplicity is given as:

Ω≈(q+Nq)q(q+NN)N

Part (b) Step 1: Given Information

The total energy of the system is given as:

U=qε

The entropy of the system is given as:

S=k[(q+N)ln(q+N)-qlnq-NlnN]

Part (b) Step 2: Calculation

The temperature can be given as:

1T=∂S∂U1T=∂[k[(q+N)ln(q+N)-qlnq-NlnN]]ε∂q1T=kϵ[ln(q+N)+1-lnq-1-0]1T=kϵlnq+Nq×ϵϵϵkT=lnU+NεU

Part (b) Step 3: Final answer

The temperature as a function of its energy can be given as:

ϵkT=lnU+NεU

Part (c) Step 1: Given Information

The temperature as a function of energy is given as:

ϵkT=lnU+NεU

Part (c) Step 2: Calculation

As given:

ϵkT=ln⁡U+NεU

On simplifying, we get,

eϵkT=U+NϵUUeϵkT−U=NεU=NϵeϵkT−1

Now, differentiating the energy with respect to temperature to get the heat capacity as:

C=∂U∂TC=∂∂TNεeϵkT-1C=Nε-eϵkT-1-2eϵkTϵkT2C=ϵ2NeϵkTkT2eϵkT-12

Part (c) Step 3: Final answer

The energy as a function of temperature can be given as:

U=NϵeϵkT−1

Expression of Heat capacity is:

C=ϵ2NeϵkTkT2eϵkT-12

Part (d) Step 1: Given Information

Heat capacity in terms of temperature is given as:

C=ϵ2NeϵkTkT2eϵkT-12

For very small values of x, ex≈1+x.

Part (d) Step 2: Calculation

By expanding the exponential for higher temperatures, we get

eϵkT≈1+εkT

The heat capacity for very large values of T can be given as:

C=ϵ2NeϵkTkT2(eϵkT−1)2C≈ϵ2N(1+ϵkT)kT2(ϵkT)2C≈Nk(1+ϵkT)C≈Nk

Part (d) Step 3: Final answer

The derived expression is analogous to the expression of heat capacity at high temperatures.

So, C≈Nk is approximately the expected result.

Part (e) Step 1: Given Information

The heat capacity is given as:

C=ϵ2NeϵkTkT2eϵkT-12

Part (e) Step 2: Calculation for heat capacity

The heat capacity for low temperature can be given as:

C=ϵ2NeϵkTkT2eεkT-12C=ϵ2kNeεkTk2T2eεkT-12CNk=εkT2eεkTeεkT-12

Part (e) Step 3: Graph

Based on the derived equation, the graph of CNk versus t=kTϵ can be made as follows:

Part (e) Step 4: Calculation of ε

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

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

For aluminum, heat capacity is maximum at T=300 K

εaluminium=kTεaluminium=1.38×10-23×300εaluminium=4.14×10-21 Jεaluminium=2.587×10-2 eV

For diamond, heat capacity is maximum at T=2230 K

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

Part (e) Step 5: Final answer

The heat capacity for low temperatures can be given as:

CNk=εkT2eεkTeεkT-12

The graph can be made as:

The values of ε are as follows:

εlead=6.875×10−3 eVεaluminum=2.587×10−2 eVεdiamond=0.19233 eV

The graph of heat capacity versus temperature for lead, aluminum, and diamond is approximately anomalous as in figure 1.14.

Part (f) Step 1: Given Information

The heat capacity is given as:

C=ϵ2NeϵkTkT2eϵkT-12

Part (f) Step 2: Calculation

Let,

x=εkT

The heat capacity can be modified as:

C=ε2NeεkTkT2(eϵkT−1)2CNk=(ϵkT)2eϵkT(eϵkT−1)2CNk=(x)2ex(ex−1)2

Now,

The Taylor expansion is given as:

ex=1+x+x22!+x33!+....

Hence, the above equation becomes:

CNk=(x)2(1+x+x22!+x33!)(x+x22!+x33!)2CNk=(1+x+x22!+x33!)(1+x2!+x23!)2CNk=(1+x+x22!+x33!)(1+x2!+x23!)−2

Using the Taylor series, the above equation becomes:

CNk=1+x+x22+x365x212-x+1

Solving further and neglecting the higher powers than 2, we get,

C≈Nk1-ϵkT212

Part (f) Step 3: Final answer

Hence, the required expression is:

C≈Nk1-ϵkT212

Most popular questions for Physics Textbooks

Show that the entropy of a two-state paramagnet, expressed as a function of temperature, is S=Nk[ln(2coshx)−xtanhx], where x=μB/kT. Check that this formula has the expected behavior as T→0 and T→∞.

In Problem 1.55 you used the virial theorem to estimate the heat capacity of a star. Starting with that result, calculate the entropy of a star, first in terms of its average temperature and then in terms of its total energy. Sketch the entropy as a function of energy, and comment on the shape of the graph.

As shown in Figure 1.14, the heat capacity of diamond near room temperature is approximately linear in T. Extrapolate this function up to 500 K, and estimate the change in entropy of a mole of diamond as its temperature is raised from 298 K to 500 K. Add on the tabulated value at 298 K (from the back of this book) to obtain S(500 K).

In Section 2.5 I quoted a theorem on the multiplicity of any system with only quadratic degrees of freedom: In the high-temperature limit where the number of units of energy is much larger than the number of degrees of freedom, the multiplicity of any such system is proportional to UNf/2, where N f is the total number of degrees of freedom. Find an expression for the energy of such a system in terms of its temperature, and comment on the result. How can you tell that this formula for Ω cannot be valid when the total energy is very small?

Suppose you have a mixture of gases (such as air, a mixture of nitrogen and oxygen). The mole fraction xi of any species i is defined as the fraction of all the molecules that belong to that species: xi=Ni/Ntotal . The partial pressure Pi of species i is 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 μi of species i in this system is the same as if the other gases were not present, at a fixed partial pressure Pi.
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