618 
Let us introduce the function § (the thermodynamic potential) 
instead of the function w (the free energy); then we have to calculate: 
KE 
148) RT 
Ero, Fo = SV,4 +f - dv : p( c= v4) :: . ĳ (1°) 
and 
EN 
k *(14-8,) RT 
Gro Be = SV, Ao + | i = a dv or P ( v= Vo) 7 id 7 (2e) 
wv 
Tos (0 
For the simplicity of the calculation of (1%), where we have to 
integrate with variable 3, we suppose that also the state v,, 8, is a 
gas-state, to which the simple law of Boyue applies. 
- 
«. Calculation of (1¢). As according to a well-known property: 
0g 0s 
ard Pig ae 
we get: 
$= (1—Bs)u, + 2u, =u, + B(— w, + 2u), 
feos ds 
with aa =p, and 2 = u, (u, and a, are therefore the molecular 
potentials of the components), and with 2, =1-— p, n, = 23. 
Now on account of the equilibrium (in (19) we have namely 
always states of equilibrium) — u, + 2u, = 0 5), hence simply: 
Gi 
Le. the fota/ potential of the mixture is equal to the molecular 
potential of the first (the dissociating) component [or also equal to 
twice the potential of the second component. 
For u, we may now further write at the large gasvolume J’: 
u, = C,—RT log V + RT + RT log (1--8£), 
in which C, is the known temperature function. Hence we may 
write for (1°): 
C1 
3 
fc babe eee hee weal Uae 
Ga | C,—R1 loyV+RT+ RI log(1—8) | + ne | —— dv — p(V—,). 
v 
Now for perfect gases (this follows from the condition of equili- 
brium — u, + 2u,— 0): 
: 5 De ‘ : OS dn os dn, 
1) For — =O is identical with — —2 = 
0p On, dp 5 On, d3 
ul) + us (2) = 0, or with — TS 2u, ==); 
= 01, e. with 
