XIX
297
ADIABATIC CHANGES IN MOIST AIR
with the diagram in hand, we shall be able to judge of its
utility and to see how it is used, without needing to wade
through the calculations in A and B.
A. Suppose that 1 kilogramme of a mixture of air and
water-vapour contains λ parts by weight of dry air, and μ
parts by weight of unsaturated water-vapour. Let the pressure
of the mixture be p, its absolute temperature T.
The question
is: What changes will the mixture undergo as the pressure
gradually diminishes to zero without heat being supplied?
We must distinguish several stages.
Stage 1. The vapour is unsaturated, and no liquid water
is present. We assume that the unsaturated mixture obeys
the laws of Gay-Lussac and Boyle. If then e be the partial
pressure of the water-vapour, p-e that of the dry air, the
volume of 1 kilogramme of the mixture, we have
ᎡᎢ
pe = λ =,
R₂T
c = μ
=
v
where R, R, are constants of known meaning and magnitude.
Since the total pressure is the sum of these two values, we
get pv (XR+μR)T, and this is the so-called characteristic
equation of the mixture. Further, let c, denote the specific
heat at constant volume of the air, c, that of water-vapour;
then in order to produce the changes dv and dT, we must
supply the dry air with an amount of heat
dv
¿Q₁ = x ( </T + ART!).
dQ1
and the water-vapour with
dv
= (c,'dT + AR,T");
dQ₂ =μ
v
or both together with the amount of heat
dQ = (λc¸ + µc,')dT + A(\R+µR₁)Tdv
v
But this amount of heat vanishes for the change we are
considering. In order to integrate the differential equation
¹ Clausius, Mechanische Wärmetheorie, vol. i. p. 51, 1876.