( 384 ) 
being given; also the value of pz. We get from: 
AL = H2 
(4) 2 dp, = (GE ) dT, 
Op) oT. P2 
If 7, is to be a minimum and therefore the cooling a maximum, 
04 : 02 
then Ge) and therefore also GE), must be 0. 
7; 
Pi T, 
Therefore: 
de; En d 1e) __ 
Gale fe der | qT, ; 
or 
a (. 7 at A ey a 
erk laden If rn 
If a is thought to be constant, this equation becomes: 
a _(lH(l-D)(l Het) 
vp (er — 5) 
3 
If, however, a is taken as a ra CLAusius does for CO, we 
find: 
da __(l+a)(l—5) Ata)’ 
dn (x) — 6)? 
In order to avoid needless calculations, I shall in what follows 
only examine the consequences if a is put constant, 
Then we find: 
rr mn LE 
p= (lot bne La 
If we had sought the value of v, for which the value of pv is a 
minimum, we had obtained: 
ee )=5 hs 
Syke 
From this appears that the value vj, for which a maximum value 
is obtained, is the same as that for which pe has a minimum value’ 
at a temperature equal to half 7}. 
