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XIX
ADIABATIC CHANGES IN MOIST AIR
Now that in this way we have determined the whole of
the series of states which the mixture traverses, we may for
each individual state find as follows the remaining quantities
which interest us:-
1. The dotted line, on which we are, shows at once how
many grammes of water are present as vapour in the corre-
sponding state. If we subtract this from the total amount μ
of water present originally, we get the amount of water
already condensed.
2. The density 8 of the mixture with the approximations
introduced may for all states be calculated from the formula
8=p/RT or log 8 = log p - log T — log R. Graphically it
would be read off at once if the
system of lines of equal density.
-
diagram were covered by a
These lines are seen to be
a system of parallel straight lines. Only one of these lines,
8, is actually drawn on the diagram, so as not to overload it.
But we may by the help of this line alone compare the
densities in two states 1 and 2 according to the following
rule. From the points 1 and 2 draw two straight lines
parallel to 8 and cutting the isothermal for 0° C.; and read
off the pressures p₁ and p₂ at these intersections.
Pi P2
The densi-
ties at 1 and 2 are in the ratio P₁: P2. For the densities in
the states p₁, 0° and P2, 0° are by Boyle's law in the ratio
P₁: P₂, and they are equal to the densities in 1 and 2, as they
lie on lines of equal density.
3. The difference of height h, which corresponds to the
passage from the state po to the state p, on the assumption of
an adiabatic equilibrium state, is given by the equation
Po
Po
dp
h = 42.
p
vdp=RT
P
Ρ
Here we would now find T as a function of p from the
diagram and then evaluate the integral mechanically. In
actual practice the supposition of an adiabatic equilibrium
will never be satisfied so nearly as to make an exact develop-
ment of its consequences of any importance. And again we
shall only commit a comparatively slight error for moderate
heights if we give to T a mean value and then regard it as