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 in a 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=da; then II+d,IIda will ultimately be the pressure at Q. Now it is found that s - r is proportional to a, we may therefore T= cx or T=s-cx; therefore if g be the force of gravity, μ the ratio of the pressure of air to its density at 0o, 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 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 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 = log. 10. {log10h-log1ok-log10€.e(s− t)}, .. x=log. 10. {1+E(S+T)}. {log 10h-log1k — log1€. e (8 − t)} . μ g log 10. =60345+155.cos 2λ 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, x= = {60345+121(s+T)+155 cos2λ} {logioh-log1k-(0.00008) (8−t)} When MP is very small, x= {26207+52(s+T)} 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 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, = log. 10 (log1h — log1k—logie.e (s—t)+logie. 2); (logioh — logik — logie.e (§ − t) + logwe. 2 — ) ; log10€. e=0,000078, log10€. 0,0000000416; 2 .. a= {60158 + 120. ($ + T) + 155. cos 2λ + (0,0029) x} {log19h logik (0,000078) (s − t) + (0,0000000416) x}. An approximate value of x must be first obtained from the equation = {60345+121. (s+T)} {logioh-log10k (0,00008) (s—t)}, and this substituted for a in the small terms will give a nearer value of x. g The values of, and E are adapted to the mixture of air and watery vapour, constituting the atmosphere in its ordinary state. The vapour of water is lighter than air, under the same pressure, and the quantity of it contained in a given quantity μ of air increases with the temperature. Hence, and E arè g larger than if the atmosphere consisted of perfectly dry air. 37. The pressure of vapour not in contact with the fluid from which it was produced, is found to be inversely proportional to the space it occupies, and its expansion, on being heated, is the same as that of air. If however the temperature, or the volume of a given quantity of vapour, be diminished beyond a certain point, a portion of it will return to the state of a liquid; and then, if the temperature of the vapour be invariable, its volume may be diminished till the whole becomes liquid, without increasing its pressure. It appears probable from the experiments of Mr. Faraday, that every gas may be made to assume the form of a liquid by diminishing its volume. When the condensation of a gas is carried on nearly to the point at which it begins to liquefy, the ratio of its pressure to its density, at a given temperature is no longer constant. The value of this ratio for dry atmospheric air does not perceptibly change under the pressure of a column of mercury nearly ninety feet high. SECTION IV. ON THE EQUILIBRIUM OF FLUIDS ACTED ON BY ANY FORCES. ART. 38. To find the pressure at any point in a mass of fluid at rest acted on by any forces. Let PQ (fig. 21) be the edge of a very small prism of fluid in the interior of a mass of fluid at rest, R the accelerating force at P, S the resolved part of R in the direction PQ. Let the prism become solid; then (Art. 4) it will remain at rest; and since S. (mass prism), and the pressures on its ends are the only forces that act upon it in a direction parallel to PQ, they must be in equilibrium, ..press. on the end Q-press. on the end P=S. (mass prism). Let x, y, ; x + dx, y + dy, z+d be the co-ordinates of P, Q referred to rectangular axes Ox, Oy, Oz. Construct a parallelopiped LMN, of which PQ is the diagonal, having its edges PL, PM, PN parallel to Ox, Oy, Ox respectively. Let X, Y, Z be the components of R resolved parallel to Ox, Oy, Ox; the area of the base of the prism; p the density of the fluid; p the pressure at P, and therefore p+d.p.da + dyp.dy+d2p.dx, ultimately the pressure at Q. Then if the sides of the base of the prism be very small compared with its length, pressure on the end Q-pressure on the end P = K(d ̧p.Sx+dyp. Sy + d,p. S≈). S=X.cos QPL+ Y. cos QPM + Z. cos QPN, and the mass of the prism=pK. PQ, . S. (mass prism) =pK. PQ.(X.cos QPL+ Y. cos QPM + Z. cos QPN) = pK (X. Sa + Y. Sy + Z.S3), |