This pressure varies from time to time, but is generally between 14 and 15 lbs.

82. The height of the homogeneous atmosphere.

If the density of the atmosphere were the same throughout the whole vertical column as it is at the sea level, its height would be less than 5 miles.

σ

To prove this, let σ, p be the densities of mercury and of air, each referred to water; then, if h be the height of the barometer, the atmospheric pressure = goh. Hence the height of the atmospheric column would be h. Now, it has been found that the ratio o p is about 10462: 1, and if we take h to be 30 inches, we shall find thath is a little less than 5 miles.

Ρ

83. The pressure of a given quantity of air, at a given temperature, varies inversely as the space it occupies.

The experimental proof of this law, due to Boyle and Marriotte, is as follows.

B

A bent glass tube, the shorter branch of which can have its end closed, is fixed to a graduated stand. Both ends being open, a little mercury is poured in, which rests with its surfaces P, P in the same horizontal plane. The end A is now closed and more mercury is poured in at B; the effect is a compression of the air in AP, the mercury rising to a height Q, which is however below the surface R of the mercury in BP.

After closing the end A the pressure of the air is equal to the atmospheric pressure, and when more mercury has been poured in, the pressure of the air in AQ is equal to that of the mercury at Q, the same level in the longer branch. This latter pres

P

R

Q

sure is due to the atmospheric pressure on the surface R, and the weight of the column RQ.

If now the spaces AQ, AP be compared, which may be effected by comparing the weights of the mercury they would contain, and if the height h of the barometer be observed, it will be found that

But, taking II as the original pressure of the air in AP, and II' as its pressure when compressed,

П=goh, and I'=II+goRQ=go(h+RQ);

.. II II: space AP: space AQ,

and this proves the law for a compression of air.

For a dilatation, employ a bent glass tube, of which both branches are long, and pour in mercury to a height P; then close the end A and withdraw some of the mercury from the branch B; let Q and R be the new surfaces.

It will now be found that

P

And .. II" II: space AP: space AQ.

In each case care must be taken to have the temperatures the same at the beginning and at the conclusion of the experiment.

Hence it follows that, since the density of a given quantity of air varies inversely as its volume, the pressure varies directly as the density. If p be the pressure, and the density, this is expressed by the equation

Ρ

p=kp,

where k is a quantity to be determined by experiment.

84. Effect of a change of temperature.

If the pressure remain constant, an increase of temperature of 10 centigrade, produces in a given mass of air an expansion .003665 of its volume.

This experimental law, combined with the preceding, enables us to express the relation between the pressure, density, and temperature of a given mass of air or gas.

Imagine a quantity of air confined in a cylinder by a piston to which a given force is applied, and let the

temperature be 0° C. Raise the temperature to to: the piston will then be forced out until the original volume (V) is increased by .003665 t. V or a tV, designating the decimal by a. If V be the new volume, we have

V=V1(1+at),

and therefore, if p, po be the densities at the temperatures t°, 0o,

85. The value of a is very nearly the sante for all gases, and moreover remains nearly the same for different pressures. M. Regnault has investigated the values of a for different substances; for instance, between 0° and 100° he finds the value of a for carbonic acid gas to be .003689. It has also been observed that the coefficients for two gases separate more from each other when the pressure is very much increased.

86. Illustration. The effect of heat in the expansion of air may be illustrated by a simple experiment.

Take a glass tube, open at one end, and ending in a bulb at the other; immerse the open end in water, and then apply the heat of a lamp to the bulb. The air in the bulb will expand, and will drive out a portion of the water in the tube.

If the lamp be removed, the air within will be cooled, and the water will then rise in the tube.

87. Determination of heights by the barometer.

It is found both from theory and from observation, that the height of the barometric column depends on its altitude above the sea level, and we are thus provided with a means of directly inferring from observation the height of any given station above the level of the sea.

For this purpose it is necessary to construct a formula which shall connect the height of the barometer with the height of its position above a given level, such as the sea level.

A general formula would be somewhat complicated, and difficult to obtain without the aid of the Integral Calculus, since the atmospheric pressure depends on the temperature and density of the air, which both vary with the height, and also on the intensity of gravity, which diminishes with an increase of height.

We shall however construct a formula on the supposition that the temperature and the force of gravity re

main constant: this will be practically useful for the determination of comparatively small differences of altitude.

88. If a series of heights be taken in arithmetic progression, the densities of the air decrease in geometric progression.

Take a vertical column of the atmosphere of a given height, and let it be divided into n horizontal layers of

represent the densities of the successive layers, measuring upwards.

These layers may be supposed each of the same density throughout, and, if we take the temperature the same in all, the pressures on the upper sides of the layers will be

kp1, kp2,...kpm,

k being the constant of variation for the particular temperature.

The difference between any two consecutive pressures must be equal to the weight of the air between them, and therefore, taking ther-1th and 7th

pressures,

that is, the densities diminish in geometric progression.

89. To find an expression for the difference of the altitudes of two stations.

If z be this difference, we have from the preceding article, putting y for