44 QUANTITATIVE ASPECTS OF CHEMISTRY
oxygen cylinder be the sum of the partial pressures of the oxygen
and the water vapor.
This deduction can be verified experimentally. Let dry oxygen
run into the dry cylinder until the gauge stands at 760 mm. at 20°.
Then run a thin layer of water into the bottom of the cylinder.
The gauge begins slowly to rise, and finally it stops at 777.4 mm.
if the temperature is still 20°. Thus the partial pressure of sat-
urated water vapor is 17.4 mm. and is not affected by the pres-
ence of the oxygen gas. It takes a much longer time to saturate
the space in the cylinder when oxygen is present because the water
vapor has to diffuse through the oxygen, but the final result is
exactly the same. The oxygen does not diminish the capacity
of the space for the water vapor.
Suppose now that we have to measure a quantity of oxygen
which has been collected in a measuring tube over water in a
trough. The oxygen has bubbled up through the water and we may
assume that it has in this way become fully saturated with water
vapor. We raise or lower the measuring tube until the level of
the water is the same inside and outside the tube. On the outside
surface of the water the atmosphere is exerting its pressure which
is transmitted through the liquid to the gas within the tube.
Let us say that the barometer reads 740 mm. Then the pressure
of the gas in the tube is 740 mm. This pressure, however, is the
total of two partial pressures, that of the oxygen and that of the
water vapor. Let us say that the temperature is 20°; then the
partial pressure of the water vapor is 17.4 mm.; and the pressure
of the oxygen is 740 — 17.4 = 722.6 mm. The volume read in
the measuring tube is, say, 60 cc. We want to calculate the volume
of the oxygen under standard conditions. Substituting in the
general gas equation we have:
722.6 X 60 _ 760 X Vsi
273 + 20 273
or
reox^273 x^722 -^6
V bU X 20 X
The general formula for reducing to standard conditions the vol-
ume of a gas measured over water is
(^273) v P ~ aq-tens"
X 760