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Expert-verified Found in: Page 108 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # In Problem 2.18 you showed that the multiplicity of an Einstein solid containing N oscillators and q energy units is approximately$\Omega \left(N,q\right)\approx {\left(\frac{q+N}{q}\right)}^{q}{\left(\frac{q+N}{N}\right)}^{N}$(a) Starting with this formula, find an expression for the entropy of an Einstein solid as a function of N and q. Explain why the factors omitted from the formula have no effect on the entropy, when N and q are large.(b) Use the result of part (a) to calculate the temperature of an Einstein solid as a function of its energy. (The energy is $U=qϵ$, where $ϵ$ is a constant.) Be sure to simplify your result as much as possible.(c) Invert the relation you found in part (b) to find the energy as a function of temperature, then differentiate to find a formula for the heat capacity.(d) Show that, in the limit $T\to \infty$, the heat capacity is $C=Nk$. (Hint: When x is very small, ${e}^{x}\approx 1+x$.) Is this the result you would expect? Explain.(e) Make a graph (possibly using a computer) of the result of part (c). To avoid awkward numerical factors, plot $C/Nk$ vs. the dimensionless variable $t=kT/ϵ$, for t in the range from 0 to about 2. Discuss your prediction for the heat capacity at low temperature, comparing to the data for lead, aluminum, and diamond shown in Figure 1.14. Estimate the value of $ϵ$, in electron-volts, for each of those real solids.(f) Derive a more accurate approximation for the heat capacity at high temperatures, by keeping terms through ${x}^{3}$ in the expansions of the exponentials and then carefully expanding the denominator and multiplying everything out. Throw away terms that will be smaller than $\left(ϵ/kT{\right)}^{2}$ in the final answer. When the smoke clears, you should find $C=Nk\left[1-\frac{1}{12}\left(ϵ/kT{\right)}^{2}\right]$.

(a) $S=k\left[\left(q+N\right)\mathrm{ln}\left(q+N\right)-q\mathrm{ln}q-N\mathrm{ln}N\right]$

(b) $\frac{ϵ}{kT}=\mathrm{ln}\left(\frac{U+N\epsilon }{U}\right)$

(c) $U=\frac{Nϵ}{\left({e}^{\frac{ϵ}{kT}}-1\right)}$, $C=\frac{{ϵ}^{2}N{e}^{\frac{ϵ}{kT}}}{k{T}^{2}{\left({e}^{\frac{ϵ}{kT}}-1\right)}^{2}}$

(d) It is shown that in the limit $T\to \infty$, the heat capacity is $C=Nk$.

(e) The graph can be made as: $\begin{array}{rl}& \frac{C}{Nk}=\frac{{\left(\frac{ϵ}{kT}\right)}^{2}kN{e}^{\frac{\epsilon }{kT}}}{{\left({e}^{\frac{\epsilon }{kT}}-1\right)}^{2}}\\ & {\epsilon }_{\text{lead}}=6.875×{10}^{-3}\mathrm{eV}\\ & {\epsilon }_{\text{aluminum}}=2.587×{10}^{-2}\mathrm{eV}\\ & {\epsilon }_{\text{diamond}}=0.19233\mathrm{eV}\end{array}$

(f) $C\approx Nk\left(1-\frac{{\left(\frac{ϵ}{kT}\right)}^{2}}{12}\right)$

See the step by step solution

## Part (a) Step 1: Given Information

The multiplicity of an Einstein solid is given as:

$\Omega \left(N,q\right)\approx {\left(\frac{q+N}{q}\right)}^{q}{\left(\frac{q+N}{N}\right)}^{N}$

## Part (a) Step 2: Calculation

The entropy is given as:

$S=k\mathrm{ln\Omega }$

By substituting the given value in the above equation, we get,

$S=k\mathrm{ln}\left\{{\left(\frac{q+N}{q}\right)}^{q}{\left(\frac{q+N}{N}\right)}^{N}\right\}\phantom{\rule{0ex}{0ex}}S=k\left[q\mathrm{ln}\left(\frac{q+N}{q}\right)+N\mathrm{ln}\left(\frac{q+N}{N}\right)\right]\phantom{\rule{0ex}{0ex}}S=k\left[q\mathrm{ln}\left(q+N\right)-q\mathrm{ln}q+N\mathrm{ln}\left(q+N\right)-N\mathrm{ln}N\right]\phantom{\rule{0ex}{0ex}}S=k\left[\left(q+N\right)\mathrm{ln}\left(q+N\right)-q\mathrm{ln}q-N\mathrm{ln}N\right]$

The multiplicity for Einstein solid can be given as:

$\mathrm{\Omega }=\frac{⌊q+N-1}{⌊q⌊N-1}\phantom{\rule{0ex}{0ex}}\mathrm{\Omega }=\frac{⌊q+N-1}{⌊q⌊N-1}×\frac{N\left(q+N\right)}{N\left(q+N\right)}\phantom{\rule{0ex}{0ex}}\mathrm{\Omega }=\frac{⌊q+N}{⌊q⌊N}×\frac{N}{\left(q+N\right)}$

But,

Stirling approximation of large n is given as:

localid="1647403887405" $⌊n=\sqrt{2\pi n}{n}^{n}{e}^{-n}$

Hence,

$\mathrm{\Omega }=\frac{N}{\left(q+N\right)}×\frac{\sqrt{2\pi \left(q+N\right)}\left(q+N{\right)}^{q+N}{e}^{-\left(q+N\right)}}{\sqrt{2\pi \left(q\right)}\left(q{\right)}^{q}{e}^{-\left(q\right)}\sqrt{2\pi \left(N\right)}\left(N{\right)}^{N}{e}^{-\left(N\right)}}\phantom{\rule{0ex}{0ex}}\mathrm{\Omega }=\frac{N}{\left(q+N\right)}×\sqrt{\frac{q+N}{qN}}×\frac{\left(q+N{\right)}^{q+N}{e}^{-\left(q+N\right)}}{{q}^{q}{N}^{N}{e}^{-\left(q+N\right)}}\phantom{\rule{0ex}{0ex}}\mathrm{\Omega }=\sqrt{\frac{N}{2\pi q\left(q+N\right)}{\left(\frac{q+N}{q}\right)}^{q}{\left(\frac{q+N}{N}\right)}^{N}}$

For large $q$ and $N$, when compared to the power terms, the square root terms are small. As they have a negligible effect, they can be neglected. Hence, the approximate value can be written as:

$\mathrm{\Omega }=\sqrt{\frac{N}{2\pi q\left(q+N\right)}}{\left(\frac{q+N}{q}\right)}^{q}{\left(\frac{q+N}{N}\right)}^{N}\phantom{\rule{0ex}{0ex}}\mathrm{\Omega }\approx {\left(\frac{q+N}{q}\right)}^{q}{\left(\frac{q+N}{N}\right)}^{N}$

## Part (a) Step 3: Final answer

Hence, the entropy of an Einstein solid containing $N$ oscillators and $q$ energy units is given as:

$S=k\left[\left(q+N\right)\mathrm{ln}\left(q+N\right)-q\mathrm{ln}q-N\mathrm{ln}N\right]$

For large values of $q$ and $N$, the square root terms are small when compared to the power terms. Hence they are neglected as they have a very negligible effect on the entropy. Therefore the approximate value of multiplicity is given as:

$\mathrm{\Omega }\approx {\left(\frac{q+N}{q}\right)}^{q}{\left(\frac{q+N}{N}\right)}^{N}$

## Part (b) Step 1: Given Information

The total energy of the system is given as:

$U=q\epsilon$

The entropy of the system is given as:

$S=k\left[\left(q+N\right)\mathrm{ln}\left(q+N\right)-q\mathrm{ln}q-N\mathrm{ln}N\right]$

## Part (b) Step 2: Calculation

The temperature can be given as:

## Part (b) Step 3: Final answer

The temperature as a function of its energy can be given as:

$\frac{ϵ}{kT}=\mathrm{ln}\left(\frac{U+N\epsilon }{U}\right)$

## Part (c) Step 1: Given Information

The temperature as a function of energy is given as:

$\frac{ϵ}{kT}=\mathrm{ln}\left(\frac{U+N\epsilon }{U}\right)$

## Part (c) Step 2: Calculation

As given:

$\begin{array}{r}\frac{ϵ}{kT}=\mathrm{ln}\left(\frac{U+N\epsilon }{U}\right)\end{array}$

On simplifying, we get,

${e}^{\frac{ϵ}{kT}}=\frac{U+Nϵ}{U}\phantom{\rule{0ex}{0ex}}U{e}^{\frac{ϵ}{kT}}-U=N\epsilon \phantom{\rule{0ex}{0ex}}U=\frac{Nϵ}{\left({e}^{\frac{ϵ}{kT}}-1\right)}$

Now, differentiating the energy with respect to temperature to get the heat capacity as:

$C=\frac{\partial U}{\partial T}\phantom{\rule{0ex}{0ex}}C=\frac{\partial }{\partial T}\left[\frac{N\epsilon }{\left({e}^{\frac{ϵ}{kT}}-1\right)}\right]\phantom{\rule{0ex}{0ex}}C=N\epsilon \left(-{\left({e}^{\frac{ϵ}{kT}}-1\right)}^{-2}\right){e}^{\frac{ϵ}{kT}}\left(\frac{ϵ}{k{T}^{2}}\right)\phantom{\rule{0ex}{0ex}}C=\frac{{ϵ}^{2}N{e}^{\frac{ϵ}{kT}}}{k{T}^{2}{\left({e}^{\frac{ϵ}{kT}}-1\right)}^{2}}$

## Part (c) Step 3: Final answer

The energy as a function of temperature can be given as:

$U=\frac{Nϵ}{\left({e}^{\frac{ϵ}{kT}}-1\right)}$

Expression of Heat capacity is:

$C=\frac{{ϵ}^{2}N{e}^{\frac{ϵ}{kT}}}{k{T}^{2}{\left({e}^{\frac{ϵ}{kT}}-1\right)}^{2}}$

## Part (d) Step 1: Given Information

Heat capacity in terms of temperature is given as:

$C=\frac{{ϵ}^{2}N{e}^{\frac{ϵ}{kT}}}{k{T}^{2}{\left({e}^{\frac{ϵ}{kT}}-1\right)}^{2}}$

For very small values of $x,{e}^{x}\approx 1+x$.

## Part (d) Step 2: Calculation

By expanding the exponential for higher temperatures, we get

${e}^{\frac{ϵ}{kT}}\approx 1+\frac{\epsilon }{kT}$

The heat capacity for very large values of $\mathrm{T}$ can be given as:

$C=\frac{{ϵ}^{2}N{e}^{\frac{ϵ}{kT}}}{k{T}^{2}{\left({e}^{\frac{ϵ}{kT}}-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}C\approx \frac{{ϵ}^{2}N\left(1+\frac{ϵ}{kT}\right)}{k{T}^{2}{\left(\frac{ϵ}{kT}\right)}^{2}}\phantom{\rule{0ex}{0ex}}C\approx Nk\left(1+\frac{ϵ}{kT}\right)\phantom{\rule{0ex}{0ex}}C\approx Nk$

## Part (d) Step 3: Final answer

The derived expression is analogous to the expression of heat capacity at high temperatures.

So, $C\approx Nk$ is approximately the expected result.

## Part (e) Step 1: Given Information

The heat capacity is given as:

$C=\frac{{ϵ}^{2}N{e}^{\frac{ϵ}{kT}}}{k{T}^{2}{\left({e}^{\frac{ϵ}{kT}}-1\right)}^{2}}$

## Part (e) Step 2: Calculation for heat capacity

The heat capacity for low temperature can be given as:

$C=\frac{{ϵ}^{2}N{e}^{\frac{ϵ}{kT}}}{k{T}^{2}{\left({e}^{\frac{\epsilon }{kT}}-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}C=\frac{{ϵ}^{2}kN{e}^{\frac{\epsilon }{kT}}}{{k}^{2}{T}^{2}{\left({e}^{\frac{\epsilon }{kT}}-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}\frac{C}{Nk}=\frac{{\left(\frac{\epsilon }{kT}\right)}^{2}{e}^{\frac{\epsilon }{kT}}}{{\left({e}^{\frac{\epsilon }{kT}}-1\right)}^{2}}$

## Part (e) Step 3: Graph

Based on the derived equation, the graph of $\frac{C}{Nk}$ versus $t=\frac{kT}{ϵ}$ can be made as follows: ## Part (e) Step 4: Calculation of ε

For lead, heat capacity is maximum at $T=80\mathrm{K}$

${\epsilon }_{\text{lead}}=kT\phantom{\rule{0ex}{0ex}}{\epsilon }_{\text{lead}}=\left(1.38×{10}^{-23}\right)×80\phantom{\rule{0ex}{0ex}}{\epsilon }_{\text{lead}}=1.1×{10}^{-21}\mathrm{J}\phantom{\rule{0ex}{0ex}}{\epsilon }_{\text{lead}}=6.875×{10}^{-3}\mathrm{eV}$

For aluminum, heat capacity is maximum at $T=300\mathrm{K}$

${\epsilon }_{\text{aluminium}}=kT\phantom{\rule{0ex}{0ex}}{\epsilon }_{\text{aluminium}}=\left(1.38×{10}^{-23}\right)×300\phantom{\rule{0ex}{0ex}}{\epsilon }_{\text{aluminium}}=4.14×{10}^{-21}\mathrm{J}\phantom{\rule{0ex}{0ex}}{\epsilon }_{\text{aluminium}}=2.587×{10}^{-2}\mathrm{eV}$

For diamond, heat capacity is maximum at $T=2230\mathrm{K}$

${\epsilon }_{\text{diamond}}=kT\phantom{\rule{0ex}{0ex}}{\epsilon }_{\text{diamond}}=\left(1.38×{10}^{-23}\right)×2230\phantom{\rule{0ex}{0ex}}{\epsilon }_{\text{diamond}}=3.0774×{10}^{-20}\mathrm{J}\phantom{\rule{0ex}{0ex}}{\epsilon }_{\text{diamond}}=0.19233\mathrm{eV}$

## Part (e) Step 5: Final answer

The heat capacity for low temperatures can be given as:

$\frac{C}{Nk}=\frac{{\left(\frac{\epsilon }{kT}\right)}^{2}{e}^{\frac{\epsilon }{kT}}}{{\left({e}^{\frac{\epsilon }{kT}}-1\right)}^{2}}$

The graph can be made as: The values of $\epsilon$ are as follows:

$\begin{array}{l}{\epsilon }_{\text{lead}}=6.875×{10}^{-3}\mathrm{eV}\\ {\epsilon }_{\text{aluminum}}=2.587×{10}^{-2}\mathrm{eV}\\ {\epsilon }_{\text{diamond}}=0.19233\mathrm{eV}\end{array}$

The graph of heat capacity versus temperature for lead, aluminum, and diamond is approximately anomalous as in figure 1.14.

## Part (f) Step 1: Given Information

The heat capacity is given as:

$C=\frac{{ϵ}^{2}N{e}^{\frac{ϵ}{kT}}}{k{T}^{2}{\left({e}^{\frac{ϵ}{kT}}-1\right)}^{2}}$

## Part (f) Step 2: Calculation

Let,

$x=\frac{\epsilon }{kT}$

The heat capacity can be modified as:

$\begin{array}{l}C=\frac{{\epsilon }^{2}N{e}^{\frac{\epsilon }{kT}}}{k{T}^{2}{\left({e}^{\frac{ϵ}{kT}}-1\right)}^{2}}\\ \frac{C}{Nk}=\frac{{\left(\frac{ϵ}{kT}\right)}^{2}{e}^{\frac{ϵ}{kT}}}{{\left({e}^{\frac{ϵ}{kT}}-1\right)}^{2}}\\ \frac{C}{Nk}=\frac{\left(x{\right)}^{2}{e}^{x}}{{\left({e}^{x}-1\right)}^{2}}\end{array}$

Now,

The Taylor expansion is given as:

${e}^{x}=1+x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!}+....$

Hence, the above equation becomes:

$\begin{array}{l}\frac{C}{Nk}=\frac{\left(x{\right)}^{2}\left(1+x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!}\right)}{{\left(x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!}\right)}^{2}}\\ \frac{C}{Nk}=\frac{\left(1+x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!}\right)}{{\left(1+\frac{x}{2!}+\frac{{x}^{2}}{3!}\right)}^{2}}\\ \frac{C}{Nk}=\left(1+x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!}\right){\left(1+\frac{x}{2!}+\frac{{x}^{2}}{3!}\right)}^{-2}\end{array}$

Using the Taylor series, the above equation becomes:

$\frac{C}{Nk}=\left(1+x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{6}\right)\left(\frac{5{x}^{2}}{12}-x+1\right)$

Solving further and neglecting the higher powers than 2, we get,

$C\approx Nk\left(1-\frac{{\left(\frac{ϵ}{kT}\right)}^{2}}{12}\right)$

## Part (f) Step 3: Final answer

Hence, the required expression is:

$C\approx Nk\left(1-\frac{{\left(\frac{ϵ}{kT}\right)}^{2}}{12}\right)$ ### Want to see more solutions like these? 