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Q. 3.9

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An Introduction to Thermal Physics
Found in: Page 95
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 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.

The residual entropy for a carbon monoxide is calculated to be 5.76 JK-1.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given

Carbon monoxide is the given solid. CO is its chemical formula. Each molecule can be arranged in two different ways: CO or OC.

Step 2: Calculation

Multiplicity for N molecules is given as:

Ω=2N ………..(1)

And Residual entropy is given as:

Sresidual =klnΩ ……… (2)

Where, k is Boltzmann constant.

In the given case, N=6.022×1023

By substituting this value in equation (1), we get,

Ω=26.022×1023

Now, by substituting this value of Ω and k=1.38×10-23 JK-1 in equation (2), we get,

Sresidual =1.38×10-23ln26.022×1023Sresidual =1.38×10-236.022×1023ln2Sresidual =5.76 JK-1

Step 3: Final answer

Hence, the residual entropy is calculated to be 5.76 JK-1.

Most popular questions for Physics Textbooks

Can a "miserly" system, with a concave-up entropy-energy graph, ever be in stable thermal equilibrium with another system? Explain.

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

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.

Consider an Einstein solid for which both N and q are much greater than 1. Think of each oscillator as a separate "particle."

(a) Show that the chemical potential is

role="math" localid="1646995468663" μ=-kTlnN+qN

(b) Discuss this result in the limits N≫q and N≪q, concentrating on the question of how much S increases when another particle carrying no energy is added to the system. Does the formula make intuitive sense?

Use Table 3.1 to compute the temperatures of solid A and solid B when qA=1. Then compute both temperatures when qA=60. Express your answers in terms of ε/k, and then in kelvins assuming that ε=0.1 eV.
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