( 40 ) 



a CL 

 R2\n =r tut) 



(1-z y( 1 - S*)(l - %) L Jl - --) '( 1 - 8«)'(l-y) "I 



|_ i(y s )»(6z'-8z+3-y) % 

 as iuo (<p + ■'') = s and (1 -(- »,r) to = y. Reduction yields 

 (1_*)«(1_8«) 



2« s 

 R'J m z=. — nt» 



$ 4( y - g y(6 g *-8z+3-y) 



l_ 3 *),l_y)_(l_*)(l_3y) . 



1 



The expression between [ ] is = 2(jy — z), hence, - being — , we get: 



P "\ 



RT„ 



a ! to (1 - z y(l — 3z) 



b x (y-*)(6**-8*+3-y) 

 Let ns express this in 7',, the critical temperature of one com- 

 ponent. (?; < r s ). We find: 



\/a x 



Now 



_ 8 a, _ 8 «V 

 1 ~~ 27 AT 27T - 



= (f was put. At last we get 



3^» = 27 w_ (1-^(1-3.-) 



(4) 



^ = «co (y> -)- ,r) ; y = (1 -|- nx)a>, 

 from which we solve : 



hence : 



nm =z — ; y = to -| — ■ , 



«« 1 y 



u> = y ; — = — tv+' r ) — x 



<p -\- x n z 



(5) 



Now to and n have been expressed in y and z, as (</> -\- .r)«, and .?,, 

 had already been expressed in // and z by (2) and (3). 

 As further : 



6. 



l + « = l + 



i "i 



b, Jt 



and 



1 _ « _ l/a, _ 8 



<p j/ttj y/a i y/jt 



when 0= T„. T.= T : — and .t = » a : />, = — : — , we have also: 



1 + n (1 + nf ' 



so that also 6 and .t can be expressed in y and c. 



(6) 



