formula

p1 - p2
T1 - T2 = [gamma]-------.
T^2

In their experiments p2 was always 1 atmosphere, and the amount of p1
was not large. It would, therefore, be certainly wrong, even though for
a small difference in pressure the empiric formula might be
approximately correct, without closer investigation to make use of it
for the differences of pressure used in Linde's apparatus, where p1 =
200 and p2 = 18 atmospheres. For the existence of a most favourable
value of p1 is in contradiction with the formula, since it would follow
from it that T1 - T2 would always increase with the increase of p1. Nor
would it be right to regard as the cause for the existence of this most
favourable value of p1 the fact that the heat produced in the
compression of the expanded gas, and therefore p1/p2, must be kept as
small as possible, for the simple reason that the heat is produced in
quite another part of the apparatus, and might be neutralized in
different ways.

Closer examination of the process shows that if p2 is given, a most
favourable value of p1 must exist for the cooling itself. If p1 is taken
still higher, the cooling decreases again; and we might take a value for
p1 for which the cooling would be zero, or even negative.

If we call the energy per unit of weight [epsilon] and the specific
volume v, the following equation holds:--

[epsilon]1 + p1v1 - p2v2 = [epsilon]2,

or [epsilon]1 + p1v1 = [epsilon]2 + p2v2.

According to the symbols chosen by Gibbs, [chi]1 = [chi]2.

As [chi]1 is determined by T1 and p1, and [chi]2 by T2 and p2, we
obtain, if we take T1 and p2 as being constant,

/[delta][chi]1\ /[delta][chi]2\
(---------------) dp1 = ( ------------- ) dT2.
\ [delta]p1 /_T1 \ [delta]T2 /_p2

If T_2 is to have a minimum value, we have

/[delta][chi]1\ /[delta][chi]1)\
(---------------) = 0, or ( -------------- ) = 0.
\ [delta]p1 /_T1 \ [delta]v1 /_T1

From this follows

/[delta][epsilon]1\ /[delta](p1v1)\
( ----------------- ) + ( ------------- ) = 0.
\ [delta]v1 /_T1 \ [delta]v1 /_T1

As ([delta][epsilon]1/[delta]v1)T is positive, we shall have to take
for the maximum cooling such a pressure that the product p_v decreases
with v, viz. a pressure larger than that at which p_v has the minimum
value. By means of the equation of stat