( 683 ) 
have a phasis, say a liquid phasis, at constant temperature. This 
phasis fills the whole volume, except a very small part, which is 
filled with the coexisting vapour phasis. Now we make this gaseous 
part condense. For a simple substance we may imagine the con- 
densed part to be in exactly the same state as the liquid phasis. 
The volume now ceases to be totally filled. As second operation 
we imagine the substance to extend till the whole volume is filled 
homogeneously. In the first part of the process the system loses 
a certain positive quantity of energy, namely m times the amount 
of the internal latent heat of evaporation, if m represents the number 
of molecules in the vapour phasis. But in the second part the 
system gains energy. The loss is equal to m(é.—e) and the gain 
may be represented by: 
So the total loss is: 
de). 
mee) =a? ae En ) 
which agrees with the general expression for (&}), 
ams | —o)(5-) oa (-s—)(5~) — (Yye—y1) (=), 
if we put the quantities wj, #2, yo and y, equal to zero. 
For substances, for which the cohesion may be represented by 
a molecular pressure, we have: 
‘1 =) a( ees YQ—?} 
9—é,; == FS | Vg —— == CF, ° 
nd ries vy va Bet ze ae 
de 
dv; 
gain of energy in the second part is much greater than the loss 
in the first part. 
In the supposition, that the substance associates, &—é, would be 
From this we deduce, that sel )= Lene), so that the 
Uv} 
Be od Vie nee 
greater than a (- a -), but also in this case this quantity is equal 
Pio ee 
to the internal latent heat. And even for water the amount of 
this latent heat does not differ so much from the value, which it 
would have for a normal substance, that this might account for the 
reversal of the sign of (&).. So the positive amount of (&)y for 
water below 4° is not to be ascribed to the value of &—é, but to 
de de Vi", 
that of the second part, namely to (5 )tes- vi). As laa is equal 
dvi dv, 
to F ea —p, and 3 is in this case negative, (&aj)v consists 
Vv v 
or 
