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Answers without the blur. Sign up and see all textbooks for free! Q. 8.13

Expert-verified Found in: Page 338 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # Problem 8.13. Use the cluster expansion to write the total energy of a monatomic nonideal gas in terms of a sum of diagrams. Keeping only the first diagram, show that the energy is approximately$U\approx \frac{3}{2}NkT+\frac{{N}^{2}}{V}·2\pi {\int }_{0}^{\infty }{r}^{2}u\left(r\right){e}^{-\beta u\left(r\right)}dr$Use a computer to evaluate this integral numerically, as a function of $T$, for the Lennard-Jones potential. Plot the temperature-dependent part of the correction term, and explain the shape of the graph physically. Discuss the correction to the heat capacity at constant volume, and compute this correction numerically for argon at room temperature and atmospheric pressure.

The correction numerically for argon at room temperature and atmospheric pressure is $U=\frac{3}{2}NkT+2\pi ×\frac{{N}^{2}}{V}×\int {r}^{2}×u\left(r\right){e}^{-\beta ×u\left(r\right)}dr$.

See the step by step solution

## Step 1: Given information

To evaluate the integral $U\approx \frac{3}{2}NkT+\frac{{N}^{2}}{V}·2\pi {\int }_{0}^{\infty }{r}^{2}u\left(r\right){e}^{-\beta u\left(r\right)}dr$ numerically, as a function of $T$, for the Lennard-Jones potential. To Plot the temperature-dependent part of the correction term, and to explain the shape of the graph physically.

## Step 2: Explanation

The system's energy is provided by:
$dU=-\frac{1}{Z}×\frac{dZ}{d\beta }=-\frac{d}{d\beta }\left(\frac{dZ}{Z}\right)$Equation of integration:
$U=-\frac{d}{d\beta }\mathrm{ln}\left(Z\right)$In the energy equation, having two terms:
$U={U}_{id}+{U}_{e}=-\frac{d}{d\beta }\mathrm{ln}\left({Z}_{id}\right)-\frac{d}{d\beta }\mathrm{ln}\left({Z}_{e}\right)$The first term is:
${U}_{id}=-\frac{d}{d\beta }\mathrm{ln}\left({Z}_{id}\right)=\frac{3}{2}NkT$

The second term is more difficult:
${U}_{e}=-\frac{d}{d\beta }\left(\frac{1}{2}\frac{{N}^{2}}{V}\right)\left(\int u\left(r\right){d}^{3}r\right)$To estimate volume, utilize spherical coordinate:${d}^{3}r={r}^{2}dr{\int }_{0}^{2\pi }d\varphi ×{\int }_{0}^{\pi }\mathrm{sin}\left(\theta \right)d\theta =4\pi ×{r}^{2}dr$

## Step 3: Explanation

To estimate volume, utilize spherical coordinate:

${d}^{3}r={r}^{2}dr{\int }_{0}^{2\pi }d\varphi ×{\int }_{0}^{\pi }\mathrm{sin}\left(\theta \right)d\theta =4\pi ×{r}^{2}dr$

Let's put that into the equation above:

${U}_{e}=-\frac{d}{d\beta }\left(\frac{1}{2}\frac{{N}^{2}}{V}\right)\left(\int u\left(r\right){d}^{3}r\right)$

$=-\frac{d}{d\beta }\left(\frac{1}{2}\frac{{N}^{2}}{V}\right)\left(4\pi ×\int {r}^{2}×u\left(r\right)dr\right)$

$=-2\pi ×\frac{{N}^{2}}{V}\left(\int {r}^{2}×u\left(r\right)\frac{d}{d\beta }\left({e}^{-\beta ×u\left(r\right)}\right)dr\right)$

$=-2\pi ×\frac{{N}^{2}}{V}×\int {r}^{2}×u\left(r\right){e}^{-\beta ×u\left(r\right)}dr$

## Step 4: Explanation

Then the energy term is:
$U={U}_{id}+{U}_{e}=-\frac{d}{d\beta }\mathrm{ln}\left({Z}_{id}\right)-\frac{d}{d\beta }\mathrm{ln}\left({Z}_{e}\right)$

$=\frac{3}{2}NkT+2\pi ×\frac{{N}^{2}}{V}×\int {r}^{2}×u\left(r\right){e}^{-\beta ×u\left(r\right)}dr$

The following factors contribute to Lennard Jones's potential:

$u\left(r\right)=4ϵ×\left({\left(\frac{\sigma }{r}\right)}^{12}-{\left(\frac{\sigma }{r}\right)}^{6}\right)$

Find the derivative of $u\left(r\right)$, setting it to $0$, then determine well-defined minimum:

$\frac{u\left(r\right)}{r}=-4ϵ×\left(\frac{12{\sigma }^{12}}{{r}^{13}}-\frac{6{\sigma }^{6}}{{r}^{7}}\right)=0$

$\frac{12{\sigma }^{12}}{{r}^{13}}=\frac{6{\sigma }^{6}}{{r}^{7}}$

$r={2}^{1/6}\sigma$

## Step 5: Explanation

At that point, the value of the Lennard-Jones potential is equal to:
$u\left(r={2}^{1/6}\sigma \right)=4ϵ×\left({\left(\frac{\sigma }{{2}^{1/6}\sigma }\right)}^{12}-{\left(\frac{\sigma }{{2}^{1/6}\sigma }\right)}^{6}\right)$

$=4ϵ\left(\frac{1}{4}-\frac{1}{2}\right)$

$=-ϵ$

The second coefficient as:

${U}_{e}=-2\pi ×\frac{{N}^{2}}{V}×\int \left(\mathrm{exp}-4\beta ϵ×\left({\left(\frac{\sigma }{r}\right)}^{12}-{\left(\frac{\sigma }{r}\right)}^{6}\right)-1\right){r}^{2}dr$

Change of variables as:

$x=\frac{r}{\sigma }$

${\beta }^{*}=\beta ×ϵ$

${N}_{0}=\frac{{N}^{2}}{V}$

## Step 6: Explanation

The second coefficient as:
${U}_{e}=-2\pi ×{N}_{0}×\int \left(\mathrm{exp}-4{\beta }^{*}×\left({x}^{-12}-{x}^{-6}\right)-1\right){x}^{2}dx$For the integral of a function $f\left(x\right)$, use Simpson's rule :${\int }_{a}^{b}f\left(x\right)dx=\frac{h}{3}×\left(f\left({x}_{0}\right)+4f\left({x}_{1}\right)+2f\left({x}_{2}\right)+\dots +2f\left({x}_{n-2}\right)+4f\left({x}_{n-1}\right)+f\left({x}_{n}\right)\right)$Also,$h=\frac{\left(b-a\right)}{n}$ and the points ${x}_{0},\dots ,{x}_{n}$ as${x}_{j}=a+jh,j=0,\dots ,n$.

## Step 7: Explanation

The temperature-dependent part of the correction term is provided as below: The correction numerically for argon at room temperature and atmospheric pressure is $U=\frac{3}{2}NkT+2\pi ×\frac{{N}^{2}}{V}×\int {r}^{2}×u\left(r\right){e}^{-\beta ×u\left(r\right)}dr$.

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