Question

For a magnetic system held at constant TT and HH (see Problem 5.17 ), the quantity that is minimized is the magnetic analogue of the Gibbs free energy, which obeys the thermodynamic identity

Phase diagrams for two magnetic systems are shown in Figure 5.14 ; the vertical axis on each of these figures is μ0Hμ0H (a) Derive an analogue of the Clausius-Clapeyron relation for the slope of a phase boundary in the HH - TT plane. Write your equation in terms of the difference in entropy between the two phases. (b) Discuss the application of your equation to the ferromagnet phase diagram in Figure 5.14. (c) In a type-I superconductor, surface currents flow in such a way as to completely cancel the magnetic field (B, not H)(B, not H) inside. Assuming that MM is negligible when the material is in its normal (non-superconducting) state, discuss the application of your equation to the superconductor phase diagram in Figure 5.14.5.14. Which phase has the greater entropy? What happens to the difference in entropy between the phases at each end of the phase boundary?

Short answer

The Clausius- Clapeyron relation is achieved; both the phases have equal entropy; superconducting phase has less entropy than normal phase except at the T=0

Step-by-step solution

given information

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

$d{G}_m=-SdT-{\mu }_{\circ }Md{\rm H}$

Part (a)

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

$$\begin{gathered} -S_{m}dT-{\mu }_{\circ }{M}_{m}d{\rm H}=-S_{p}dT-{\mu }_{\circ }{M}_{p}d{\rm H} \\ \Rightarrow -S_m-{\mu }_{\circ }{M}_m\frac{d{\rm H}}{dT}=-S_p-{\mu }_{\circ }{M}_p\frac{d{\rm H}}{dT} \\ \Rightarrow -(S_m-S_p)-{\mu }_{\circ }(M_m-M_p)\frac{d{\rm H}}{dT}=0 \\ \Rightarrow {\mu }_{\circ }(M_m-M_p)\frac{d{\rm H}}{dT}=(S_m-S_p) \\ \Rightarrow \frac{dH}{dT}=\frac{(S_m-S_p)}{{\mu }_{\circ }(M_m-M_p)} \end{gathered}$$

This is equivalent to Clausius-Clapeyron equation.

Part(b)

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.

Part (C)

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

For a superconductor B=0

$$\begin{gathered} \Rightarrow -H=\frac{M}{V} \\ \Rightarrow M=HV \\ \Rightarrow \frac{dH}{dT}=\frac{-(S_1-S_2)}{{\mu }_{\circ }(M_1-M_2)} \\ \Rightarrow \frac{dH}{dT}=\frac{S_1-S_2}{{\mu }_{\circ }VH} \end{gathered}$$

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

$$\begin{gathered} \frac{dH}{dT}=0 \\ \Rightarrow H=cons\tan t \\ \Rightarrow H\propto S_2-S_1 \end{gathered}$$

When H = 0

$S_1=S_2$.