285 
i.e. 
Frog et) 
1 en 
Le 
which properly becomes =O for }=1 (ideal substances, where 6 
is independent of v). 
For this may also be written, because (see above) pv + (1—8)b, = 
pe Oa 
=o 
0 log — = log v + tog (eb) ek FP lg — bg = 
Now v= aie hence v—b, =r and @ becomes: 
=> log = Plog at log (b—b,) slab +logb, log(1—8).") 
Thus we find for & =}: 
oo is Ob 11 /e6\ 1 708, 
Gar (EE) EE); 
or also: 
00 bog — db bo/a—b 0b, 
== b (b— nll b, (b—b,) (5 é 
Now v—b = 6 (b— = (°e/2—b) (see the note), hence finally: 
(5. ihe 0b ies 0b 
J=s3(x)- b,v—b (re): 
For . (log (v—b) + 6) may therefore simply be written 
Me be 1 0) 1 deh ae A 
a b, v—b (=): ob da neo 
0b 
= 20), HI) )y we have (S*) = (bo). — 1 (2) 
which quantity we shall represent by Ad,. This is accord- 
ingly the increase of volume, when */, double molecule passes 
into 1 single molecule. From the above given expression 
For as 0, 
db : : Belo) 
1 = lee, zn == Me 
ln 4 Je 5 we might at once have substituted v — b=b 1-6 
Be: from the above given expression for v) for v — b, and 
— Bb lj, |p uy pene 
im Bae =e +) db = i —1) log Ob) ~ 3 log b 
might have been written, but then the constant term (i.e. constant with regard 
to v) logby, essential for the differentiation with respect to x, would have been 
wanting, and 4 would not have become =O for @ =1 (the integral is indefinite). 
19 
