If HK (fig. 17.)=h, t the temperature of the mercury in At the level of the sea in latitude 50° the mean value of h is 30,035 inches, the temperature of the air being 12o.2. 30. The pressure of air at a given temperature varies inversely as as the space it occupies.. (I.) Let ABD (fig. 18.) be a glass tube having an open capillary termination at A, and bent at B so that the branches AB, BD may be parallel, and PC the axis of AB vertical. Pour mercury into DB till it rises to P, cutting off the communication between AB and BD, and then seal the aperture at A. Let mercury be now poured in till the horizontal plane through the surface of the mercury in BD meets PC in any point C; and let the surface of the mercury in AB meet PC in M. Then if h be the altitude of the mercury in the barometer, and σ its density at the time of making the experiment; u, v the capacities of the portions AM, AP of the tube; M, II the pressures of the air in the tube when occupying the spaces u, v: we have (Arts. 29. 12.) II=pressure of the exterior air=goh, M=goh+go. CM. But if u, v be measured, it will be found that v u=h+MC: h, .. M: ПI=v u. (II.) Let AB (fig. 19.) be a glass tube having an open capillary termination at A; P, C any two points in its axis. Immerse AB vertically in mercury till the surface of the mercury meets PC in P, and seal the aperture at A. Elevate the tube till the plane of the surface of the mercury outside meets PC in C; and let the surface of the mercury within the tube meet PC in M. Then as before if u, v be the capacities of the portions AM, AP of the tube; M, II the pressures of the air in the tube when occupying the spaces u, v: we have II=pressure of the exterior air =goh, M=goh-go.MP. And if u, v be measured, it will be found that v : u=h− MP: h, ... M : II= = √ : Ղ.. Hence the pressure of air at a given temperature varies inversely as the space it occupies, when the pressure is less than that of the atmosphere, as well as when it is greater. Po 31. Let po be the density of atmospheric air under the pressure II, at the temperature of melting snow; then since the pressure of air at a given temperature varies inversely as the space it occupies, and therefore directly as its density, II=μ pos where μ is constant. If h be the altitude of the mercury in the barometer, σ its density, II=goh, therefore gohu po Το It is found that =10467, σ being the density of mercury Po at 0°, when h=0,76 metres = 29,9218 inches, and g=9,8088 metres = 386,18 inches, ..√(u)=279,33 metres=916,46 feet. 32. The expansion of air between the temperatures of melting snow and boiling water, under a constant pressure, is equal to 0,375 of its volume at the temperature of melting snow, and the increment of its volume is proportional to its temperature above 0o, as indicated by a mercurial thermometer. Hence if be the densities of air at the temperatures 0°, T° Po Pr under the same pressure, po= {1+(0,00375)T}p, therefore П= = μ. {1+ (0,00375)T}Pr 33. When a given mass of atmospheric air is suddenly compressed or dilated, its temperature is increased or diminished according to the following law. Let دم P19 To be the corresponding densities and tem peratures of a given mass of air: then, T, — T = (111,25). (1 — ・e). If, in making the first experiment described in Art. 30, the mercury be suddenly poured into the tube DB, the temperature C in a of the air in AB will be increased; and if the altitude CM (fig. 18.) be observed before the air in AB has cooled down to its original temperature, the pressure will appear to vary higher inverse ratio than that of the first power of the space occupied by the air. The same observation applies to the second experiment. 34. It appears from reasoning similar to that employed in Art. 11. that when an elastic fluid of uniform temperature, acted on by gravity, is at rest, its pressure, and therefore its density is the same at all points in the same horizontal plane. This is also true when the temperature at any point depends only on the distance of the point from a given horizontal plane. 35. To find the difference of the altitudes of two stations by means of the barometer. = Let M, II be the pressures of the air at the points M, P in the vertical QPM (fig. 20.); s the temperature at M, т the temperature at P; Q a point very near to P; MP=x, PQ-8; then II+d,IIda will ultimately be the pressure at Q. Now it is found that s T is proportional to a, we may therefore Tcx or Ts-cx; therefore if g be the force of gravity, μ the ratio of the pressure of air to its density at 0°, E the expansion of air for one degree of heat under a constant pressure; we shall have the density of the air at P assume s Π = μ 1+ET μ and the density between P and Q may be considered uniform, therefore ultimately (Art. 12.) pressure at P=g (density at P). PQ+ pressure at Q, or Let h be the altitude, s the temperature of the mercury in the barometer at M; k the altitude, t the temperature of the mercury at any point in a horizontal plane passing through P; e the expansion of mercury for one degree of heat: then μ k 1+ es = log. 10. {log10h-log1ok-log10.e (s—t)}, .. x=log, 10. {1+E(S+T)}. {log 10h-log10 — log10€.e (8−t)} . g log. 10. =60345+155.cos2 feet, λ being the latitude of the g place of observation; E = 0.002, logoe.e=0.00008, the temperatures being expressed in degrees of the centigrade thermometer; {60345 + 155. cos2λ} {1 + (0.002) (s + T)} = 60345 +121(s+T)+155.cos2λ nearly; therefore if a be the number of feet contained in MP, · {60345+121(s+T)+155 cos2)} {logioh-log1k-(0.00008) (s-t)} When MP is very small, 36. When the difference of the altitudes of the stations is large, it becomes necessary to take into account the variation of gravity in the same vertical. Let r be the distance of M (fig. 20.) from the centre of the earth, g the force of gravity at P, g' the force of gravity at M; then, retaining the notation of the preceding Article, |