Problems and Solutions on Thermodynamics and Statistical Mechanics

142 Problems €4 Solutions on Thermodynamica €4 Statistical Mechanics

S phase: B = 0, M = -H/4a.
Integrating dG = -MdH, we obtain

Gs = H2/87r + const

Noting that Gs (H,, 7') = Go(T) at the transition point, we have

1

871.

Gs = Go(T) + -(H2 - H:).

It follows that

Gs - GN = -(H^12 2

a71. - H,).

(b) Since

s=-(g) H ,

we have

= 4a [ ( s)2 + H, ( a2Hc aii)]

• 
  
  
  
  • -- H; T [3 (f)2 - 11
    2a T,"
  
  
  
  

When H = 0, Cs - CN = H;/aT,

(d) At H = 0, L = 0, Cs - CN # 0, therefore the phase transition is

second order.

1145

The phase boundary between the superconducting and normal phases

of a metal in the He - T plane (He = magnitude of applied external field)

is given by Fig. 1.43.