91影视

The Gibbs free energy for a magnetic system at constant temperature T, which obeys thermodynamic identity is given by

$d{G}_m=-SdT-{渭}_{鈭�}Md螚$

At boundary$G_m=G_p$

where$G_m,G_p$is Gibbs free energy for magnetic and para magnetic phase respectively.

The two phases must be equally stable therefore the changes in the Gibbs free energy must remain equal. Therefore,

$d{G}_m=d{G}_p$

The thermo-dynamic identity for G is

$$\begin{gathered} -S_{m}dT-{渭}_{鈭�}{M}_{m}d螚=-S_{p}dT-{渭}_{鈭�}{M}_{p}d螚 \\ 鈬�-S_m-{渭}_{鈭�}{M}_m\frac{d螚}{dT}=-S_p-{渭}_{鈭�}{M}_p\frac{d螚}{dT} \\ 鈬�-(S_m-S_p)-{渭}_{鈭�}(M_m-M_p)\frac{d螚}{dT}=0 \\ 鈬�{渭}_{鈭�}(M_m-M_p)\frac{d螚}{dT}=(S_m-S_p) \\ 鈬�\frac{dH}{dT}=\frac{(S_m-S_p)}{{渭}_{鈭�}(M_m-M_p)} \end{gathered}$$

This is equivalent to Clausius-Clapeyron equation.

In ferromagnetic phase diagram

$\frac{dH}{dT}=0$up to the critical point.

Therefore,

at the boundary phase,$S_m=S_{p}and{M}_m>M_p$for a ferro magnet.

Let phase 1 be normal phase and phase 2 be superconductor phase.

For a superconductor B=0

$$\begin{gathered} 鈬�-H=\frac{M}{V} \\ 鈬�M=HV \\ 鈬�\frac{dH}{dT}=\frac{-(S_1-S_2)}{{渭}_{鈭�}(M_1-M_2)} \\ 鈬�\frac{dH}{dT}=\frac{S_1-S_2}{{渭}_{鈭�}VH} \end{gathered}$$

We see the slope $\frac{dH}{dT}$is negative everywhere except at T=0.

Here, slope is 0. At every other point the entropy at superconductor phase is smaller than entropy at normal phase. At T=0, both phases have 0 entropy. this satisfies third law of thermodynamics.

At this point

$$\begin{gathered} \frac{dH}{dT}=0 \\ 鈬�H=cons\tan t \\ 鈬�H鈭�S_2-S_1 \end{gathered}$$

When H = 0

$S_1=S_2$.