Chapter 14, Elastic bands, rods, bubbles, magnets Video Solutions, Thermodynamics: A complete undergraduate course

Problem 1

Show that, for an elastic rod,
$$
\left.\frac{\partial C_L}{\partial L}\right|_T=-\left.T \frac{\partial^2 f}{\partial T^2}\right|_L,
$$
where $C_L$ is the heat capacity at constant length.

Nick Johnson

Numerade Educator

04:04

Consider the ideal elastic, whose equation or state is given by $(0.20)$.
(i) Show that the internal energy is a function of temperature alone (in this respect the ideal elastic substance is like an ideal gas. It is a reflection of the fact that the energy is associated with kinetic not potential energy of the molecules).
(ii) If the heat capacity $C_L$ is independent of temperature at $L=L_0$, show that the Helmholtz function is given by
$$
F=F_0-T C_L \ln \frac{T}{T_0}+K T\left[\frac{L^2}{2 L_0}+\frac{L_0^2}{L}\right],
$$
where $F_0(T)=U_0-T S_0+\left(T-T_0\right) C_L-(3 / 2) L_0 K T$. [Use equation (13.53), suitably modified.]
(iii) Show that the difference between the primary heat capacities is
$$
C_L-C_f=-\frac{K\left(L^3-L_0^3\right)^2}{L_0 L\left(L^3+2 L_0^3\right)}
$$

Vipender Rao

Numerade Educator

01:30

Problem 3

The surface tension of liquid argon is given by $\sigma=\sigma_0\left(1-T / T_c\right)^{1.28}$, where $\sigma_0=0.038 \mathrm{~N} / \mathrm{m}$ and the critical temperature $T_c=151 \mathrm{~K}$. Find the surface entropy per unit area at the triple point, $T=83 \mathrm{~K}$.

Manik Pulyani

Numerade Educator

02:19

Problem 4

Obtain equations (14.33)-(14.38) from (14.32).

Adriano Chikande

Numerade Educator

01:09

Problem 5

Calculate $(\partial S / \partial U)_B$ from equations (14.34) and (14.35), and comment

Adriano Chikande

Numerade Educator

Problem 6

Obtain equation (14.40) from the Curie-Weiss law, as follows. Consider the magnetic Helmholtz function $F=U-S T$. We have $\mathrm{d} F=-S \mathrm{~d} T-$ $m d B$ and it is useful to consider $F(T, B)$. This can be written
$$
F(T, B)=F(T, 0)+\left.\int \frac{\partial F}{\partial B}\right|_T \mathrm{~d} B
$$
Use this to obtain
$$
F(T, B)=F(T, 0)-\frac{a V B^2}{2 \mu_0\left(T-T_c\right)}
$$
and hence
$$
S(T, B)=S(T, 0)-\frac{a V}{2 \mu_0} \frac{B^2}{\left(T-T_c\right)^2}
$$
The first term is the entropy at $B=0$ for any given temperature. The second term is the magnetic contribution. If we assume that the change in entropy is dominated by the latter contribution, then for an adiabatic process, explain why equation (14.40) is obtained. (This derivation illustrates the concept introduced in equation (13.30): an equation of state does not give the heat capacity directly, but it can be used to extend partial knowledge of the heat capacities.)

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01:59

Problem 7

Alternative route for the ideal paramagnet. Introduce $y=m / N \mu$ in equation (14.87) and show that
$$
S=\frac{N k_{\mathrm{B}}}{2}\left(y \ln \frac{1-y}{1+y}-\ln \frac{1}{4}\left(1-y^2\right)\right) .
$$
Then introduce $x$ defined by $y \equiv \tanh x$, and obtain equation (14.34). Then find an expression for $T$ in terms of $x$, using $(1 / T)=(\partial S / \partial U)_B$, and hence obtain (14.31). Finally, use (14.85) to obtain (14.35), and hence (14.32).

Susan Hallstrom

Numerade Educator

05:03

Problem 8

Geometric factor. A long straight solenoid has cross-section $A$ and contains a long straight paramagnetic sample with cross-section $A_m$. Show that, in the case $\chi \ll 1$, the total magnetic flux through the system is $\Phi=\mu_0 n I\left(A+A_m \chi\right)$, and hence
$$
B_0=B_{0 I}\left(1+\chi A_m / A\right)
$$
Thus $B_0=B_{O I}$ when the sample is thin and does not fill much of the solenoid, and $B_0=(1+\chi) B_{0 I}$ when the sample fills the solenoid. In general we define a geometric factor $\gamma_0$ such that $B_0=\left(1+\gamma_0 \chi\right) B_{0 I}$ for $\chi \ll 1$. These results are modified when $\chi$ is not small (see question 14.9).

Keshav Singh

Numerade Educator

05:05

Problem 9

Demagnetizing factor. For a strongly magnetic material, the internal $B_i$ field may differ considerably from $\mu_0 H_1$. One has $M=\chi H_i$ and
$$
H_i=H_{0 I}-\alpha M
$$
where $\alpha$ is called the demagnetizing factor, its value depends on the shape of the sample. For a long cylindrical sample, $\alpha=0$; for a sphere, $\alpha=$ $1 / 3$.
(i) Show that $H_i=H_{0 I} /(1+\alpha \chi)$ and hence
$$
M=\frac{\chi}{1+\alpha \chi} H_{0 I} .
$$
This means that the influence of the magnetization on the field can be accounted for in many expressions by replacing the susceptibility by $\chi /(1+\alpha \chi)$, but this simple rule is not universally true because we still have $B_i=\mu_0(1+\chi) H_i$.
(ii) Show that the generalization of $(14.20)$ is
$$
B_0=B_{0 I}\left(1+\gamma_0 \frac{(1-\alpha) \chi}{1+\alpha \chi}\right),
$$
where $\gamma_0$ is the geometric factor defined in question 14.9. Note, since $\alpha$ also depends on the geometry, it is common to absorb the factor $(1-\alpha)$ into the geometric factor, such that one has
$$
B_0=B_{0 I}\left(1+\gamma \frac{\chi}{1+\alpha \chi}\right),
$$
where $\gamma=(1-\alpha) \gamma_0$.

Zhaojie Xu

Numerade Educator

Problem 10

Show that for strong as well as weak magnetization,
$$
\mu_0 M=\frac{\chi}{1+\alpha \chi} B_{0 I}=\frac{\chi}{1+(\gamma+\alpha) \chi} B_0 .
$$
Hence, show that, for a sample described by the Curie-Weiss law, the equation of state can be written
$$
\frac{\mu_0 M}{B}=\frac{a}{T-T_c^v}
$$
where $T_c^{\prime}=T_c-\alpha a$ for $B=B_{0 I}$ and $T_c^{\prime}=T_c-(\gamma+\alpha) a$ for $B=B_0$. Hence for strong magnetization the Curie-Weiss law is modified simply by a shift in the parameter $T_c$. Note, this is not strictly the transition temperature, but a parameter in a model that is valid for $T \gg T_c$ (the actual critical point is reached at a temperature slightly below this $T_c$ ).

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01:17

Problem 11

Compare the elastic band experiment described at the end of Section 14.2 with the process of cooling by adiabatic demagnetization.

Manik Pulyani

Numerade Educator

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11