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Answers without the blur. Sign up and see all textbooks for free! Q. 2.11

Expert-verified Found in: Page 60 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # 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, is$80$; 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?

The most likely macrostate is when the energy units are evenly distributed,${\mathrm{q}}_{\mathrm{A}}={\mathrm{q}}_{\mathrm{B}}=40$ , with a probability of $0.07513$. The least likely state is when all the energy units are in partition $\mathrm{B}$or$\mathrm{A},{\mathrm{q}}_{\mathrm{A}}=40$ , or when ${\mathrm{q}}_{\mathrm{B}}=40$, with a chance of $7.8726×{10}^{-20}$.

See the step by step solution

## Step 1: Expression for overall multiplicity

The probability of $\mathrm{P}\left({\mathrm{q}}_{\mathrm{A}}\right)$ is,

$\mathrm{P}\left({\mathrm{q}}_{\mathrm{A}}\right)=\frac{{\mathrm{\Omega }}_{\text{total}}}{{\mathrm{\Omega }}_{\text{overall}}}$

The overall multiplicity is,

${\mathrm{\Omega }}_{\text{overall}}\left({\mathrm{N}}_{\text{overall}},{\mathrm{q}}_{\text{overall}}\right)=\left(\begin{array}{c}{\mathrm{q}}_{\text{overall}}+{\mathrm{N}}_{\text{overall}}-1\\ {\mathrm{q}}_{\text{overall}}\end{array}\right)$

The total multiplicity is ,

${\mathrm{\Omega }}_{\text{total}}={\mathrm{\Omega }}_{\mathrm{A}}{\mathrm{\Omega }}_{\mathrm{B}}$

${\mathrm{\Omega }}_{\mathrm{A}}=\left(\begin{array}{c}{\mathrm{q}}_{\mathrm{A}}+{\mathrm{N}}_{\mathrm{A}}-1\\ {\mathrm{q}}_{\mathrm{A}}\end{array}\right)$

${\mathrm{\Omega }}_{\mathrm{B}}=\left(\begin{array}{c}{\mathrm{q}}_{\mathrm{B}}+{\mathrm{N}}_{\mathrm{B}}-1\\ {\mathrm{q}}_{\mathrm{B}}\end{array}\right)$

## Step 2: Calculation for total multiplicity

Multiplicity is,

${\mathrm{\Omega }}_{\text{overall}}=\frac{\left({\mathrm{q}}_{\text{overall}}+{\mathrm{N}}_{\text{overall}}-1\right)!}{{\mathrm{q}}_{\text{overall}}!\left({\mathrm{N}}_{\text{overall}}-1\right)!}$

${\mathrm{q}}_{\text{overall}}={\mathrm{q}}_{\mathrm{A}}+{\mathrm{q}}_{\mathrm{B}}=80$

${\mathrm{N}}_{\text{overall}}={\mathrm{N}}_{\mathrm{A}}+{\mathrm{N}}_{\mathrm{B}}=200$

So,

role="math" localid="1650306409339" ${\mathrm{\Omega }}_{\text{overall}}=\frac{\left(80+200-1\right)!}{80!\left(200-1\right)!}$

$=2.1225×{10}^{71}$

Multiplicity of $\mathrm{A}$ is,

${\mathrm{\Omega }}_{\mathrm{A}}=\left(\begin{array}{c}{\mathrm{q}}_{\mathrm{A}}+99\\ {\mathrm{q}}_{\mathrm{A}}\end{array}\right)=\frac{\left({\mathrm{q}}_{\mathrm{A}}+99\right)!}{{\mathrm{q}}_{\mathrm{A}}!\left(99\right)!}$

Multiplicity of $\mathrm{B}$is,

${\mathrm{\Omega }}_{\mathrm{B}}=\left(\begin{array}{c}{\mathrm{q}}_{\mathrm{B}}+99\\ {\mathrm{q}}_{\mathrm{B}}\end{array}\right)$

Substitute ${\mathrm{q}}_{\mathrm{B}}=80-\mathrm{qA}$

so,

role="math" localid="1650306390399" ${\mathrm{\Omega }}_{\mathrm{B}}=\left(\begin{array}{c}179-{\mathrm{q}}_{\mathrm{A}}\\ 80-{\mathrm{q}}_{\mathrm{A}}\end{array}\right)$

$=\frac{\left(179-{\mathrm{q}}_{\mathrm{A}}\right)!}{\left(80-{\mathrm{q}}_{\mathrm{A}}\right)!\left(99\right)!}$

Probability is,

$\mathrm{P}\left({\mathrm{q}}_{\mathrm{A}}\right)=\frac{{\mathrm{\Omega }}_{\mathrm{A}}{\mathrm{\Omega }}_{\mathrm{B}}}{{\mathrm{\Omega }}_{\text{overall}}}$

$=\frac{1}{2.1225×{10}^{71}}\frac{\left({\mathrm{q}}_{\mathrm{A}}+99\right)!}{{\mathrm{q}}_{\mathrm{A}}!\left(99\right)!}\frac{\left(179-{\mathrm{q}}_{\mathrm{A}}\right)!}{\left(80-{\mathrm{q}}_{\mathrm{A}}\right)!\left(99\right)!}$

## Step 3: Python program for creation of graph ## Step 4: Graph for probability and energy  ### Want to see more solutions like these? 