Q. 3.39 In Problem 2.32聽you computed th... [FREE SOLUTION]

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An Introduction to Thermal Physics

Daniel V. Schroeder

Physics

1st Edition 路 356 pages

ISBN: 9780201380279

Chapter 3: Q. 3.39 (page 121)

In Problem $2.32$ you computed the entropy of an ideal monatomic gas that lives in a two-dimensional universe. Take partial derivatives with respect to $U, A$ , and N to determine the temperature, pressure, and chemical potential of this gas. (In two dimensions, pressure is defined as force per unit length.) Simplify your results as much as possible, and explain whether they make sense.

Short Answer

Expert verified

$\begin{aligned} T = \frac{U}{N k} \\ P = \frac{N k T}{A} \\ 渭 = 鈭� k T \ln 鈦� (\frac{V}{N} {(\frac{2 m m k T}{h^{2}})}^{\frac{3 / 2}{2}}) \end{aligned}$

Step by step solution

Given Information

The entropy of an ideal gas is,

$S = N k [\ln (\frac{2 蟺 m A U}{(N h)^{2}}) + 2]$

Calculation

To get the temperature, we partial differentiate with respect to U,

$\begin{matrix} \frac{1}{T} = {(\frac{鈭� S}{鈭� U})}_{A, N} \\ \frac{1}{T} = \frac{鈭�}{鈭� U} [N k [\ln 鈦� (\frac{2 m m A U}{(N h)^{2}}) + 2]] \\ = N k [\frac{(N h)^{2}}{2 x m A U} \frac{2 nmA}{(N h)^{2}}] \\ = \frac{N k}{U} \\ T = \frac{U}{N k} \end{matrix}$

Partial differentiating with respect to $A$ , we get the pressure as,

$\begin{matrix} P = T (\frac{鈭� S}{鈭� A}) U, N \\ = T \frac{鈭�}{鈭� A} [N k [\ln 鈦� (\frac{2 m n U U}{(N h)^{2}}) + 2]] \\ = N k T [\frac{(N h)^{2}}{2 蟺 n A U} 脳 \frac{2 m m U}{(N h)^{2}}] \\ = \frac{N k T}{A} \end{matrix}$

Chemical Potential

Partial differentiating the entropy with respect to N, we get the chemical potential as,

$\begin{matrix} 渭 = 鈭� T (\frac{鈭� S}{d N}) U A \\ = 鈭� T \frac{鈭�}{d N} [N k [\ln 鈦� (\frac{2 蟺 m A U}{(N h)^{2}}) + 2]] \\ = 鈭� k T [\ln 鈦� (\frac{2 蟺 m A U}{(N h)^{2}}) + 2] 鈭� k T N \frac{(N h)^{2}}{2 蟺 m A U} 脳 \frac{2 蟺 n A U}{h^{2}} 脳 \frac{鈭� 2}{N^{3}} \\ = 鈭� k T [\ln 鈦� (\frac{2 m m A U}{(N h)^{2}}) + 2] + 2 k T \\ = 鈭� k T [\ln 鈦� (\frac{2 m m A U}{(N h)^{2}}) + 2 鈭� 2] \\ = 鈭� k T [\ln 鈦� (\frac{2 m m A U}{(N h)^{2}})] \\ = 鈭� k T [\ln 鈦� (\frac{2 m m A (N k T)}{(N h)^{2}})] \\ = 鈭� k T \ln 鈦� {(\frac{V}{N} (\frac{2 蟺 m L T}{N^{2}}))}^{3 / 2}) \end{matrix}$

Conclusion

The temperature, pressure and entropy are given by:

$\begin{aligned} T = \frac{U}{N k} \\ P = \frac{N k T}{A} \\ 渭 = 鈭� k T \ln 鈦� (\frac{V}{N} {(\frac{2 蟺 m k T}{h^{2}})}^{\frac{3}{2}}) \end{aligned}$

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Short Answer

Step by step solution

Given Information

Calculation

Chemical Potential

Conclusion

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