3 7 2 D O C . 3 8 5 Q U A N T U M T H E O R Y O F I D E A L G A S I I
(24)
What happens, though, if at this temperature I now let the density of the sub-
stance increase even more (e.g., by isothermal compression)?
I assert that in this case a steadily growing number of molecules compared to the
total density will go over into the 1st quantum state (state without kinetic energy)
while the remainder of the molecules will distribute themselves according to the
parameter value λ = 1. This assertion thus means to say that something similar oc-
curs to isothermal compression of a vapor above the saturation volume. A separa-
tion sets in; one part “condenses,” the rest remains a “saturated ideal gas” (A = 0,
λ = 1).
One realizes that both parts do indeed form a thermodynamic equilibrium by
showing that per mole the “condensed” substance and the saturated ideal gas have
the same Planck function
.[4]
For the “condensed” substance, Φ
vanishes because S, E, and V individually
vanish.*)
For the “saturated
gas”
one ini-
tially has for A = 0, according to (12) and (13),
(25)
This summation can be written as an integral and transformed into a partial integra-
tion. One thus initially obtains
or, according to (8) and (11) and (15),
(26)
Hence it follows from (25) and (26) for the “saturated ideal gas”
*)
The “condensed” part of the substance does not claim any special volume because it
does not contribute anything to the
pressure.[5]
n
2πmκT)3 ( /2
h3
-----------------------------
-¦τ
3 2 /–
s
∞
=
[p. 4]
n
V
---
Φ S
E pV +
T
---------------- -–=
S
κ¦lg
1
e–α2)
– (
E
T
---. + –=
s
¦
s
e
cs2 /3
κT
–------------
1 e
cs2 /3
κT
–------------
–
-----------------------
2cs
3
----------------ds, -
1 3 /–
κT
⋅ ⋅
0
∞
³
–=
s
¦
2
3
- nsEs sd
0
∞
³
–--
2
3κT
-
E
–--------
pV
T
- –------. = = =