XIX
301
ADIABATIC CHANGES IN MOIST AIR
-
temperature just begins to fall below 0°. Here clearly σ = µ — v,
since ice and vapour alone are present. If we insert these
values, after introducing the pressure, we get
0λAR log
Po
- e
R
e
r+ q
R
e
+ λ
-
λ
Pi-
-e
R₁ Pi
-e
T
一​嶋
​R Po -e T
(III).
This equation connects the pressures P, and P₁, at which re-
spectively the third stage is entered and quitted.
It is not necessary to furnish e and T with an index, for
they have the same values at the beginning and end of the
stage.
Stage 4. When the temperature falls still lower, we have
only steam and ice. The investigation is the same as for
Stage 2, and the final formula also is the same. But here the
latent heat of vaporisation has a different value. Here it is
rq, for the heat necessary to evaporate the ice directly
must clearly be the same as that required to first melt it, and
then evaporate the water produced. If we wish to be quite
rigorous, we must not take q to be constant, but must suppose
it to vary slightly with the temperature; but the differences
are so small that they may be neglected here. Thus in this
fourth stage we shall reach those temperatures at which the
air itself can no longer be regarded as a permanent gas.
The four stages which we have distinguished might very
properly be called the dry, the rain, the hail, and the snow
stages.
If now we are compelled to follow exactly the changes in
a mixture containing a considerable percentage of water, there
is no choice but to make use of these complicated formulæ.
In that case we proceed as follows. We first substitute the
values of and μ in all the equations. Then we substitute
the values p, and T, for the initial state in equation I. The
Po
resulting equation and equation (b) we regard as two equations
to determine the unknown quantities p and T. Solving for
these we get the transition state between the first and the
second stage. The values obtained are then substituted for
Po and To in equation II. By putting T = 273 in the resulting