Problems and Solutions on Thermodynamics and Statistical Mechanics

Thermodynamics 145

Solution:
According to the problem F denotes the Gibbs function.
(a) F = minimum is the condition to determine the most probable
value of M in equilibrium. Thus M is determined from (~F/~M)T,H = 0.

(b) (~F/~M)T,H = -H + 2A(T - Tc)M + 4BM3 = 0. (*)
If 2A(T - T,)M >> 4BM3, that is, if T is far from T,, we have

H

2A(T - T,) '

M=

This is the Curie-Weiss law. The change of M with T is shown in

Fig. 1.44.

(c) If H = 0, the equation (*) has solutions

M = 0 , M = iJA(T, - T)/2B.

For stability consider

= ZA(T - T,) + 12BM2.

(%) T,H

When T > T,, the only real solution, M = 0, is stable;

'C

Fig. 1.44

when T < T,, the M = 0 solution is unstable, while the

when T = T,,M = 0, T, is the point of phase transition of the second

order. (If T > T,, the substance is paramagnetic; if T < T,, the substance

is ferromagnetic.)

If H # 0, (*) requires M # 0. Then as long as M2 > A(Tc - T.)/6B,

the system is stable. When T + TC,2A(T - T,)M << 4BM2, and (*)

has the solution M = (H/4B):. Thus T, is the point of first-order phase

transition.

M = iJA(Tc - T)/ZB solution is stable;