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Q. 2.19

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An Introduction to Thermal Physics
Found in: Page 64
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

Use Stirling's approximation to find an approximate formula for the multiplicity of a two-state paramagnet. Simplify this formula in the limit N↓≪Nto obtain Ω≈Ne/N↓N↓. This result should look very similar to your answer to Problem 2.17; explain why these two systems, in the limits considered, are essentially the same.

The two state of paramagnet and the system is formaly equivalent to an Einstein solid

Ω≈NeN↓N↓

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1 Two state of paramagnet 

Consider two-state paramagnet with N magnetic dipoles and N↓ energy quanta, since the system is formally equivalent to an Einstein solid (we're distributing the energy quanta among dipoles rather than oscillators). The multiplicity of the paramagnet is then:

ΩN',N↓=N↓+N-1N↓

Writing out the binomial coefficient:

ΩN,N↓=N↓+N-1N↓=N↓+N-1!N↓!(N-1)!

since N>>N↓, and they are large numbers, so:

ΩN,N↓=N↓+N!N↓!N!

take the natural logarithms for both sides:

ln(Ω)=lnN↓+N!N↓!N!

but, we have these logarithmic relations:

lnab=ln(a)-ln(b) ln(ab)=ln(a)+ln(b)

Step 2 Logarithm

ln(Ω)=lnN↓+N!N↓!N!=lnN↓+N!-ln(N!)-lnN↓!

use Stirling's approximation for the logarithm of a factorial:

ln(n!)≈nln(n)-n

so,we get:

ln(Ω)≈N↓+NlnN↓+N-N↓+N-Nln(N)+N-N↓lnN↓+N↓

ln(Ω)≈N↓+NlnN↓+N-Nln(N)-N↓lnN↓

If we now use the assumption that N>>N↓, we get:

ln(Ω)≈N↓+NlnNN↓N+1-Nln(N)-N↓lnN↓

we factored out N from the first logarithm:

ln(Ω)≈N↓+Nln(N)+lnN↓N+1-Nln(N)-N↓lnN↓

but, we have the logarithmic relation:

lnab+1≈ab

for b>>a, so:

Step 3 To neglect second term 

ln(Ω)≈N↓ln(N)+N↓N↓N+Nln(N)+NN↓N-Nln(N)-N↓lnN↓ln(Ω)≈N↓ln(N)+N↓2N+N↓-N↓lnN↓

ln(Ω)≈N↓+Nln(N)+N↓N-Nln(N)-N↓lnN↓

ln(Ω)≈N↓ln(N)-lnN↓+N↓2N+N↓

but, =ln(Ω)≈N↓lnNN↓+N↓, so:

ln(Ω)≈N↓lnNN↓+N↓2N+N↓

use the assumption thatN>>N↓, to neglect the second term:

ln(Ω)≈N↓lnNN↓+N↓

Step 4 Exponentiating both sides 

Exponentiating both sides this equation gives the approximate value for Ω, so:

e(ln(Ω))≈eN↓lnNN↓+N↓Ω≈elnNN↓eN↓Ω≈elnNN↓N↓eN↓\

but,e^{\ln (x)}=x, SO:

Ω≈NN↓N↓eN↓

→Ω≈NeN↓N↓

The corresponding result in N<<N↓ case is:

Ω≈NeN↓N↓

which could have been predicted easily, since NandN↓appear symmetrically in the following approximation:

ΩN,N↓=N↓+N!N↓!N!

Most popular questions for Physics Textbooks

Consider a system of two Einstein solids, $A$ and $B$, each containing 10 oscillators, sharing a total of 20 units of energy. Assume that the solids are weakly coupled, and that the total energy is fixed.

(a) How many different macrostates are available to this system?

(b) How many different microstates are available to this system?

(c) Assuming that this system is in thermal equilibrium, what is the probability of finding all the energy in solid $A$ ?

(d) What is the probability of finding exactly half of the energy in solid $A$ ?

(e) Under what circumstances would this system exhibit irreversible behavior?

Show that during the quasistatic isothermal expansion of a monatomic ideal gas, the change in entropy is related to the heat input Qby the simple formula

∆s=QT

In the following chapter I'll prove that this formula is valid for any quasistatic process. Show, however, that it is not valid for the free expansion process described above.

Fun with logarithms.
a Simplify the expression ealn⁡b. (That is, write it in a way that doesn't involve logarithms.)b Assuming that b <<a, prove that ln⁡(a+b)≈(ln⁡a)+(b/a). (Hint: Factor out the afrom the argument of the logarithm, so that you can apply the approximation of part d of the previous problem.)

Use a pocket calculator to check the accuracy of Stirling's approximation for N=50 . Also check the accuracy of equation 2.16 for ln N! .

For either a monatomic ideal gas or a high-temperature Einstein solid, the entropy is given by times some logarithm. The logarithm is never large, so if all you want is an order-of-magnitude estimate, you can neglect it and just say . That is, the entropy in fundamental units is of the order of the number of particles in the system. This conclusion turns out to be true for most systems (with some important exceptions at low temperatures where the particles are behaving in an orderly way). So just for fun, make a very rough estimate of the entropy of each of the following: this book (a kilogram of carbon compounds); a moose of water ; the sun of ionized hydrogen .
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