Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Q. 2.11 (page 60)

Use a computer to produce a table and graph, like those in this section, for two interacting two-state paramagnets, each containing 100 elementary magnetic dipoles. Take a "unit" of energy to be the amount needed to flip a single dipole from the "up" state (parallel to the external field) to the "down" state (antiparallel). Suppose that the total number of units of energy, relative to the state with all dipoles pointing up, is80; this energy can be shared in any way between the two paramagnets. What is the most probable macrostate, and what is its probability? What is the least probable macrostate, and what is its probability?

Short Answer

Expert verified

The most likely macrostate is when the energy units are evenly distributed,qA=qB=40 , with a probability of 0.07513. The least likely state is when all the energy units are in partition BorA,qA=40 , or when qB=40, with a chance of 7.8726×10-20.

Step by step solution

01

Expression for overall multiplicity

The probability of PqAis,

PqA=ΩtotalΩoverall

The overall multiplicity is,

ΩoverallNoverall,qoverall=qoverall+Noverall-1qoverall

The total multiplicity is ,

Ωtotal=ΩAΩB

ΩA=qA+NA-1qA

ΩB=qB+NB-1qB

02

Calculation for total multiplicity

Multiplicity is,

Ωoverall=qoverall+Noverall-1!qoverall!Noverall-1!

qoverall=qA+qB=80

Noverall=NA+NB=200

So,

role="math" localid="1650306409339" Ωoverall=(80+200-1)!80!(200-1)!

=2.1225×1071

Multiplicity of Ais,

ΩA=qA+99qA=qA+99!qA!(99)!

Multiplicity of Bis,

ΩB=qB+99qB

Substitute qB=80-qA

so,

role="math" localid="1650306390399" ΩB=179-qA80-qA

=179-qA!80-qA!(99)!

Probability is,

PqA=ΩAΩBΩoverall

=12.1225×1071qA+99!qA!(99)!179-qA!80-qA!(99)!

03

Python program for creation of graph

04

Graph for probability and energy

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A black hole is a region of space where gravity is so strong that nothing, not even light, can escape. Throwing something into a black hole is therefore an irreversible process, at least in the everyday sense of the word. In fact, it is irreversible in the thermodynamic sense as well: Adding mass to a black hole increases the black hole's entropy. It turns out that there's no way to tell (at least from outside) what kind of matter has gone into making a black hole. Therefore, the entropy of a black hole must be greater than the entropy of any conceivable type of matter that could have been used to create it. Knowing this, it's not hard to estimate the entropy of a black hole.
aUse dimensional analysis to show that a black hole of mass Mshould have a radius of order GM/c2, where Gis Newton's gravitational constant and cis the speed of light. Calculate the approximate radius of a one-solar-mass black holeM=2×1030kg .
bIn the spirit of Problem 2.36, explain why the entropy of a black hole, in fundamental units, should be of the order of the maximum number of particles that could have been used to make it.

cTo make a black hole out of the maximum possible number of particles, you should use particles with the lowest possible energy: long-wavelength photons (or other massless particles). But the wavelength can't be any longer than the size of the black hole. By setting the total energy of the photons equal toMc2 , estimate the maximum number of photons that could be used to make a black hole of mass M. Aside from a factor of 8π2, your result should agree with the exact formula for the entropy of a black hole, obtained* through a much more difficult calculation:

Sb.h.=8π2GM2hck

d Calculate the entropy of a one-solar-mass black hole, and comment on the result.

Consider again the system of two large, identical Einstein solids treated in Problem 2.22.

(a) For the case N=1023, compute the entropy of this system (in terms of Boltzmann's constant), assuming that all of the microstates are allowed. (This is the system's entropy over long time scales.)

(b) Compute the entropy again, assuming that the system is in its most likely macro state. (This is the system's entropy over short time scales, except when there is a large and unlikely fluctuation away from the most likely macro state.)

(c) Is the issue of time scales really relevant to the entropy of this system?

(d) Suppose that, at a moment when the system is near its most likely macro state, you suddenly insert a partition between the solids so that they can no longer exchange energy. Now, even over long time scales, the entropy is given by your answer to part (b). Since this number is less than your answer to part (a), you have, in a sense, caused a violation of the second law of thermodynamics. Is this violation significant? Should we lose any sleep over it?

For a single large two-state paramagnet, the multiplicity function is very sharply peaked about N↑=N/2.

(a) Use Stirling's approximation to estimate the height of the peak in the multiplicity function.

(b) Use the methods of this section to derive a formula for the multiplicity function in the vicinity of the peak, in terms of x≡N↑−(N/2). Check that your formula agrees with your answer to part (a) when x=0.

(c) How wide is the peak in the multiplicity function?

(d) Suppose you flip 1,000,000coins. Would you be surprised to obtain heads and 499,000 tails? Would you be surprised to obtain 510,000 heads and 490,000 tails? Explain.

According to the Sackur-Tetrode equation, the entropy of a monatomic ideal gas can become negative when its temperature (and hence its energy) is sufficiently low. Of course this is absurd, so the Sackur-Tetrode equation must be invalid at very low temperatures. Suppose you start with a sample of helium at room temperature and atmospheric pressure, then lower the temperature holding the density fixed. Pretend that the helium remains a gas and does not liquefy. Below what temperature would the Sackur-Tetrode equation predict that Sis negative? (The behavior of gases at very low temperatures is the main subject of Chapter 7.)

For an Einstein solid with each of the following values of N and q , list all of the possible microstates, count them, and verify formula Ω(N,q)=q+N−1q=(q+N−1)!q!(N−1)!

(a) N=3,q=4

(b)N=3,q=5

(c) N=3,q=6

(d) N=4,q=2

(e) N=4,q=3

(f) N=1,q=anything

(g) N= anything, q=1
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Astrophysics

Read Explanation

Fluids

Read Explanation

Modern Physics

Read Explanation

Work Energy and Power

Read Explanation

Atoms and Radioactivity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free