In order to find the most probable distribution of the molecules 
between gas and solid substance for given values of 7’ and V, 
which is the only one that need be considered for the thermo- 
dynamical equilibrium, we must see for what value of 7 the function 
Wn) is a maximum, so that JW ==0, or what comes to the same 
thing dlogW=0. 
As dn’ = — dn, x Eg dhe 4 = ==!) — cd : ib sl hj . ee, and / is 
u v v v Nn v n 
independent of n, we get the equation : 
: + 8 logy — log (kT) (11) 
= 
, C 4 poet 
0 = — logn + iT + $log (2am) + log v + 
7 
L 
We know from the ordinary kinetic gas theory that the pressure 
nkT 
p of the gas is ——, on account of which we may write for (11): 
ss 
| Reser ae En 3 Low (2m) —logp + ne —- 3 log p — + lor (/7’) (12) 
GEERT 2 YY Tr Y | nkT ‘ 2 iW = : 
The entropy S’ of a gramme molecule of solid substance is now 
(see § 2) = 3u (1 — log =) 
kT 
Equation (12) then becomes: 
Ob ets POS ee 4 Oe gee 
== iT +. 3 log (2am) — loy p + ET EN + log (kT) — dlogh +5 (15) 
It is further clear that the increase of the internal energy at the 
evaporation amounts to *): 
U=— Ne +&kNT — SINT =— Ne —4kNT. . . (14) 
If S is now the entropy of a gramme molecule of gas, the increase 
of entropy at the evaporation is: 
[ Bs N N pi 
| al a Ne N pv 
Ei eee ee (15) 
n 
Pr EAI 
We finally find for S from (18) and (15): 
S= kN |E log (kT) — loy p + $ log (2am) — 8 logh + 3} ~. (16) 
Konner and Wintrernirz*) have calculated the chemical constant 
of hydrogen for low temperatures, in which this behaves as a mona- 
! 
Dd 
1!) On account of the constant density mt 
[À 
2) A possible zero point energy had to be simply taken into account in c here 
and in the following 38. 
3) vy. KoHNER und P. WiINTERNITZ, Phys. 4.5. 15, 393 and 645 (1914). 
