Question

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.)

   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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Thermodynamics: A complete undergraduate course
Thermodynamics: A complete undergraduate course
Andrew M. Steane 1st Edition
Chapter 14, Problem 6 ↓

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Start with the equation $F(T, B) = F(T, 0) + \left.\int \frac{\partial F}{\partial B}\right|_T \mathrm{~d} B$.  Show more…

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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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