( 518 ) 

 4. Let us examine the formula 



T=T„ 



0,0453 f I 



1— 0,396 Zor/(l—,t') 



more closely. 'With small values of .v, it passes into 



1 ^ 0,0453 x' 

 T— 1\ — -— z=: 2\ [1—0,396 X + 0,004 .^•n. 



" 1 -f- 0,396 (a- 4- 7, x"") n L . -r . J 



Because the coeflicient of .v^ is nccidcntaJhj nearly 0, the melting- 

 point-line in this case runs over a fairly large region (from 232° to 

 120") as an almost straight line. To ensure this, it is generally necessary, 

 that 0^ — h -{- a is very small or 0. 



As, for equilibrium between the solid tin and the tin in the amal- 

 gam, ƒ< = Ml or 



— M + Ml = 0, 

 we also have : 



Ö d - d.v 



d ' q 



Now according to a well-known theorem — (— ji -f f'l) = — 77,- 



The molecular potential n for the solid phase is moreover not depen- 

 dent on d'. Therefore: 



n du, d.v 

 3' ^ ö.^• dT ~ ' 

 and consecpiently 



dT_Tdii, 

 dx q d.i" 



We therefore see, that supposing the solid phase (as in this case) 



dT dftj 



contains no mercury, — - cannot become 0, unless ^-— = 0. But 



dx dx 



then the liquid phase will be unstable, and we iind ourselves on the 



spinodal line, so that the liquid amalgam would long ago have broken 



up into two phases of different composition. 



d|Ltj dT 



Now, -r— and therefore also — mav become zero in the case of 

 Ox dx 



two values for .t ; there are therefore in this case always two hori- 

 zontal tangents. A limiting case of this is of course a point of 

 inflection with only one horizontal tangent. 



As 



