28 inches barometrical pressure; that is to say, the same gas or air
which at 28.5 inches of the barometer occupies 100 cubical inches of
volume, will occupy 101.786 cubical inches when the barometer is at 28
inches. It is equally easy to calculate the weight of this gas,
occupying 100 cubical inches, under 28.5 inches of barometrical
pressure; for, as it corresponds to 101.786 cubical inches at the
pressure of 28, and as, at this pressure, and at 10 (54.5 ) of
temperature, each cubical inch of oxygen gas weighs half a grain, it
follows, that 100 cubical inches, under 28.5 barometrical pressure, must
weigh 50.893 grains. This conclusion might have been formed more
directly, as, since the volume of elastic fluids is in the inverse ratio
of their compression, their weights must be in the direct ratio of the
same compression: Hence, since 100 cubical inches weigh 50 grains, under
the pressure of 28 inches, we have the following statement to determine
the weight of 100 cubical inches of the same gas as 28.5 barometrical
pressure, 28 : 50 :: 28.5 : x, the unknown quantity, = 50.893.
The following case is more complicated: Suppose the jar A, Pl. XII. Fig.
18. to contain a quantity of gas in its upper part ACD, the rest of the
jar below CD being full of mercury, and the whole standing in the
mercurial bason or reservoir GHIK, filled with mercury up to EF, and
that the difference between the surface CD of the mercury in the jar,
and EF, that in the cistern, is six inches, while the barometer stands
at 27.5 inches. It is evident from these data, that the air contained in
ACD is pressed upon by the weight of the atmosphere, diminished by the
weight of the column of mercury CE, or by 27.5 - 6 = 21.5 inches of
barometrical pressure. This air is therefore less compressed than the
atmosphere at the mean height of the barometer, and consequently
occupies more space than it would occupy at the mean pressure, the
difference being exactly proportional to the difference between the
compressing weights. If, then, upon measuring the space ACD, it is found
to be 120 cubical inches, it must be reduced to the volume which it
would occupy under the mean pressure of 28 inches. This is done by the
following statement: 120 : x, the unknown volume, :: 21.5 : 28
inversely; this gives x = 120 21.5 / 28 = 92.143 cubical inches.
In these calculations we may either reduce the height of the mercury in
the barometer, and the difference of level in the jar and bason, into
lines or decimal fractions of the inch; but I prefer the latter, as it
is more readily calculated. As, in these operations, which frequently
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