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Expert-verified Found in: Page 224 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # Consider a system of two Einstein solids, where the first "solid" contains just a single oscillator, while the second solid contains 100 oscillators. The total number of energy units in the combined system is fixed at 500. Use a computer to make a table of the multiplicity of the combined system, for each possible value of the energy of the first solid from 0 units to 20. Make a graph of the total multiplicity vs. the energy of the first solid, and discuss, in some detail, whether the shape of the graph is what you would expect. Also plot the logarithm of the total multiplicity, and discuss the shape of this graph.

Therefore, the first graph is exponential decaying of -qA. In second graph the slope is $-ϵ/kT$ where $ϵ$ is the size of energy unit and T is the temperature.

See the step by step solution

## Step 1: Given information

A system of two Einstein solids, where the first "solid" contains just a single oscillator, while the second solid contains 100 oscillators. The total number of energy units in the combined system is fixed at 500.

## Step 2: Explanation

Consider two Einstein solids, one of which has NA=1 oscillators and the other of which includes NB=100 oscillators. Because there are 500 energy units, the following formula is used:

${q}_{A}+{q}_{B}=500\left(1\right)$

We'll need to make a table of the system's overall multiplicity. The total multiplicity is calculated as follows:

${\Omega }_{\text{tot.}}=\Omega \left({N}_{A},{q}_{A}\right)\Omega \left({N}_{B},{q}_{B}\right)$

Where,

$\Omega \left({N}_{A},{q}_{A}\right)=\frac{\left({q}_{A}+{N}_{A}-1\right)!}{{q}_{A}!\left({N}_{A}-1\right)!}\Omega \left({N}_{B},{q}_{B}\right)=\frac{\left({q}_{B}+{N}_{B}-1\right)!}{{q}_{B}!\left({N}_{B}-1\right)!}$

Thus,

${\Omega }_{tot.}=\frac{\left({q}_{A}+{N}_{A}-1\right)!}{{q}_{A}!\left({N}_{A}-1\right)!}\frac{\left({q}_{B}+{N}_{B}-1\right)!}{{q}_{B}!\left({N}_{B}-1\right)!}$

Substitute from (1) with qB

${\Omega }_{tot.}=\frac{\left({q}_{A}+{N}_{A}-1\right)!}{{q}_{A}!\left({N}_{A}-1\right)!}\frac{\left(499-{q}_{A}+{N}_{B}\right)!}{\left(500-{q}_{A}\right)!\left({N}_{B}-1\right)!}$

Substitute with NA and NB

${\Omega }_{\text{tot.}}=\frac{\left(599-{q}_{A}\right)!}{\left(500-{q}_{A}\right)!\left(99\right)!}$

## Step 3: Calculations

Using python, table is generated, the code used is: The table is:

$\begin{array}{cc}{q}_{A}& {\Omega }_{\text{tot}}\\ 1& 1.852×{10}^{115}\\ 2& 1.546×{10}^{115}\\ 3& 1.290×{10}^{115}\\ 4& 1.076×{10}^{115}\\ 5& 8.975×{10}^{114}\\ 6& 7.481×{10}^{114}\\ 7& 6.234×{10}^{114}\\ 8& 5.193×{10}^{114}\\ 9& 4.325×{10}^{114}\\ 10& 3.600×{10}^{114}\\ 11& 2.996×{10}^{114}\\ 12& 2.492×{10}^{114}\\ 13& 2.073×{10}^{114}\\ 14& 1.723×{10}^{114}\\ 15& 1.432×{10}^{114}\\ 16& 1.190×{10}^{114}\\ 17& 9.882×{10}^{113}\\ 18& 8.204×{10}^{113}\\ 19& 6.808×{10}^{113}\\ 20& 5.648×{10}^{113}\end{array}$

To plot the graph between ${q}_{A}\text{and}{\Omega }_{tot}$, the code is The graph plotted is: ## Step 4: Explanation

Now we need to plot a graph between ${q}_{A}\text{and}\mathrm{ln}\left(\Omega \right)$, the code used is: The graph plotted is: Therefore, the first graph is exponential decaying of -qA. In second graph the slope is $-ϵ/kT$ where $ϵ$ is the size of energy unit and T is the temperature. ### Want to see more solutions like these? 