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Q. 8.13

Expert-verifiedFound in: Page 338

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

$U\approx \frac{3}{2}NkT+\frac{{N}^{2}}{V}\xb72\pi {\int}_{0}^{\infty}{r}^{2}u\left(r\right){e}^{-\beta u\left(r\right)}dr$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 $U=\frac{3}{2}NkT+2\pi \times \frac{{N}^{2}}{V}\times \int {r}^{2}\times u\left(r\right){e}^{-\beta \times u\left(r\right)}dr$.

To evaluate the integral $U\approx \frac{3}{2}NkT+\frac{{N}^{2}}{V}\xb72\pi {\int}_{0}^{\infty}{r}^{2}u\left(r\right){e}^{-\beta u\left(r\right)}dr$ 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.

The system's energy is provided by:

$dU=-\frac{1}{Z}\times \frac{dZ}{d\beta}=-\frac{d}{d\beta}\left(\frac{dZ}{Z}\right)$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:${d}^{3}r={r}^{2}dr{\int}_{0}^{2\pi}d\varphi \times {\int}_{0}^{\pi}\mathrm{sin}\left(\theta \right)d\theta =4\pi \times {r}^{2}dr$

To estimate volume, utilize spherical coordinate:

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

Let's put that into the equation above:

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

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

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

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

Then the energy term is:

$U={U}_{id}+{U}_{e}=-\frac{d}{d\beta}\mathrm{ln}\left({Z}_{id}\right)-\frac{d}{d\beta}\mathrm{ln}\left({Z}_{e}\right)$

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

The following factors contribute to Lennard Jones's potential:

$u\left(r\right)=4\u03f5\times \left({\left(\frac{\sigma}{r}\right)}^{12}-{\left(\frac{\sigma}{r}\right)}^{6}\right)$

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

$\frac{u\left(r\right)}{r}=-4\u03f5\times \left(\frac{12{\sigma}^{12}}{{r}^{13}}-\frac{6{\sigma}^{6}}{{r}^{7}}\right)=0$

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

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

At that point, the value of the Lennard-Jones potential is equal to:

$u\left(r={2}^{1/6}\sigma \right)=4\u03f5\times \left({\left(\frac{\sigma}{{2}^{1/6}\sigma}\right)}^{12}-{\left(\frac{\sigma}{{2}^{1/6}\sigma}\right)}^{6}\right)$

$=4\u03f5\left(\frac{1}{4}-\frac{1}{2}\right)$

$=-\u03f5$

The second coefficient as:

${U}_{e}=-2\pi \times \frac{{N}^{2}}{V}\times \int \left(\mathrm{exp}-4\beta \u03f5\times \left({\left(\frac{\sigma}{r}\right)}^{12}-{\left(\frac{\sigma}{r}\right)}^{6}\right)-1\right){r}^{2}dr$

Change of variables as:

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

${\beta}^{*}=\beta \times \u03f5$

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

The second coefficient as:

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

The temperature-dependent part of the correction term is provided as below:

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

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