298
XIX
ADIABATIC CHANGES IN MOIST AIR
resulting from putting dQ= 0, we divide it by T. We know
a priori from the theory of heat that this operation renders
the equation integrable, and we find it confirmed a posteriori.
Performing the integration, eliminating v by the characteristic
equation, and remembering that c,+AR equals c, the specific
heat at constant pressure, we obtain
T
P
P
0 = (Xc+μc₁₂) log
- A(λR+µR₁) log
(I).
To
Po
The right-hand member of this equation has a physical mean-
ing; it is the difference of the entropy of the mixture in the
two states defined by the quantities p, T and Po, To Clearly
the mixture behaves like a gas, whose density and specific
heat have values which are the means of those of the water-
vapour and of the air.
We must now calculate the limiting value of p down to
which equation (I) may be used. Now and in what follows
let e denote the pressure of saturated aqueous vapour at the
temperature T. e is a function of T, but of T alone. Then
the quantity of saturated aqueous vapour contained in the
volume v at temperature T, is
ve
V =
R₂T
(a).
and this quantity must exceed μ, as long as the vapour remains
unsaturated. Thus the limit is reached as soon as µ = v. If
we introduce the value of v from the characteristic equation,
this condition takes the form
p =
λR+μR₁
μR₁
e
(b).
As soon as p and T reach values satisfying this condition, we
must leave equation 1, and enter on-
Stage 2. The air is saturated with aqueous vapour, and
contains liquid water in addition. We neglect the volume of
the latter. Then we may here also consider the air by itself,
and the water with its vapour by themselves, in each case as
if the other were absent. Both have the same volume v and
¹ Clausius, Mechanische Wärmetheorie, vol. i. p. 51, 1876.