Q. 2.10 Use a computer to produce a tabl... [FREE SOLUTION]

91影视

An Introduction to Thermal Physics

Daniel V. Schroeder

Physics

1st Edition 路 356 pages

ISBN: 9780201380279

Chapter 2: Q. 2.10 (page 60)

Use a computer to produce a table and graph, like those in this section, for the case where one Einstein solid contains $200$ oscillators, the other contains $100$ oscillators, and there are $100$ units of energy in total. 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 $67$ energy units out of the total energy units are in solid $A, q_{A} = 67$ and $q_{B} = 33$ , with a probability of $0.07315$ ; the least likely macrostate is when all energy units are in partition ${Bq}_{A} = 0$ or when $q_{B} = 100$ , with a probability of $2.6926 脳 10^{- 37}$ .

Step by step solution

Calculation for overall multiplicity

Probability is,

P (q_{A}) = \frac{惟_{A} 惟_{B}}{惟_{overall}}

Over all multiplicity is,

$惟_{overall} (N_{overall}, q_{overall}) = (\begin{matrix} q_{overall} + N_{overall} - 1 \\ q_{overall} \end{matrix})$

$= \frac{(q_{overall} + N_{overall} - 1)!}{q_{overall}! (N_{overall} - 1)!}$

Where,

$q_{overall} = q_{A} + q_{B} = 100$

$N_{overall} = N_{A} + N_{B} = 300$

Then,

$惟_{overall} = (\begin{matrix} 100 + 300 - 1 \\ 100 \end{matrix})$

$= \frac{(100 + 300 - 1)!}{100! (300 - 1)!}$

$= 1.681 脳 10^{96}$

Calculation for total multiplicity and probability

Total multiplicity is,

$惟_{total} = 惟_{A} 惟_{B}$

$N_{A} = 200$

$惟_{A} = (\begin{matrix} q_{A} + 199 \\ q_{A} \end{matrix})$

$= \frac{(q_{A} + 199)!}{q_{A}! (199!)}$

$惟_{B} = (\begin{matrix} q_{B} + N_{B} - 1 \\ q_{B} \end{matrix})$

For $N_{B} = 100$ , $q_{B} = 100 - q_{A}$

$惟_{B} = (\begin{matrix} 100 - q_{A} + 100 - 1 \\ 100 - q_{A} \end{matrix})$

$= \frac{(199 - q_{A})!}{(100 - q_{A})! (99)!}$

probability is,

$P (q_{A}) = \frac{惟_{A} 惟_{B}}{惟_{overall}}$

$= \frac{1}{1.681 脳 10^{96}} \frac{(q_{A} + 199)!}{q_{A}! (199!)} \frac{(199 - q_{A})!}{(100 - q_{A})! (99)!}$

$P (q_{A}) = \frac{1}{1.681 脳 10^{96}} \frac{(q_{A} + 199)!}{q_{A}! (199!)} \frac{(199 - q_{A})!}{(100 - q_{A})! (99)!}$

Python program for creation of graph

Graph for probability versus energy

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91影视

Short Answer

Step by step solution

Calculation for overall multiplicity

Calculation for total multiplicity and probability

Python program for creation of graph

Graph for probability versus energy

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