If v be the measured volume of gas and moisture in c.c. at to C. and H m.m., d being the density of the gas and F the tension of aqueous vapour at to C., Thus, to find the mass of nitrogen and of aqueous vapour in 200 c.c. of moist nitrogen measured at 15° C. and 730 m.m. (F = 12·7), *(21) THE DIFFUSION OF GASES. Graham's Law. The volumes of two gases, which diffuse in equal times under the same circumstances, are inversely proportional to the square roots of their densities. Or, when two gases diffuse through the same apparatus for equal times under similar conditions, the volume of the one gas diffused multiplied by the square root of its density is equal to the volume of the other gas diffused multiplied by the square root of its density. V√D=v√d. 4 litres of hydrogen diffuse through an apparatus in 10 minutes, and 1 litre of oxygen diffuses in an equal time under similar conditions: find the density of oxygen. 1 √D = 4√I; .. D= 16. 10 litres of oxygen (016) diffuse through a certain apparatus in a certain time. What volume of hydrogen iodide (HI = 128) will diffuse in an equal time under similar conditions? 64. The density of oxygen is 16, and that of hydrogen iodide And how much ammonia (NH ̧ = 17) would diffuse under similar conditions? The law of effusion, or of the passage of gases through a minute hole in a thin plate, is identical with that of diffusion. *(22) THE DETERMINATION OF VAPOUR DENSITIES. Since the double-density of a vapour referred to hydrogen is expressed by the same number as its molecular weight, the approximate determination of vapour densities is a subject of considerable importance to Chemists. The density of a vapour is usually determined by finding how many times a certain volume of it is heavier than an equal volume of hydrogen, under the same conditions of temperature and pressure. Numerous methods for accomplishing this have been proposed, but they may almost all be classified under two heads: (i) In the methods of Gay Lussac, Hofmann, Victor Meyer, and others, a known mass (M grams) of the substance is heated in an enclosed space, and the volume (V c.c.) of its vapour at T° C. under the barometric pressure H m.m., in some cases increased or decreased by a column of mercury &c. equivalent to h m.m. of mercury, is measured. If the vapour is measured over mercury at a high temperature the tension ƒ due to the mercury-vapour must be subtracted from the pressure on the vapour. 120° C. 140° C. 160° C. 180° C. 200 C. f=746 m.m. 1534 3.06 5.9 11 19.9. The pressure of h m.m. of mercury at 7° C. is equivalent to the pressure of 0° C. h m.m. at the temperature of (At 448° C. the density of fusible alloy is about of that of mercury at 0° C.) The vessel is usually of glass and graduated at 0°C., hence at 7 C. the volume of the vessel becomes (1 + ·000037′). The mass of an equal volume of hydrogen under the same conditions as the vapour would be : V h √ (1+·00003T) (H=1+·000197", (1+·003677) × 760 × 0000896 grams. Hence the density of the vapour is found from the The following numbers were obtained in the determination of the vapour-density of ethyl-propinyl (C,H,0 = 84) by Hofmann's method: M=0518 grams, y = 52.5 c.c., T = 100° C., H=752.5 m.m., h = 484 m.m. ·0518 (1 + ·367) × 760 52·5 (1+ ·003) (752·5 −475) × ·0000896 0518 x 1.367 × 760 52.5 x 1·003 × 277·5 × ·0000896 = 41.1. (ii) In the methods of Dumas, Deville and Troost, and others, a vessel, usually a glass globe holding V c.c. at 0° C., is weighed full of dry air at t° C. and H, m.m. Suppose that it weighs w grams, A quantity of the substance is placed in the globe, which is heated to 7° C., and closed when no more vapour issues from it under the pressure H ̧. It is cooled, cleaned, and found to weigh W grams. The vessel is opened under mercury and its volume V c.c. at 0° C. determined by measurement and calculation; if all the air has not been expelled, the volume v c.c. of the residual air at t° C. and Hmm. is determined. If the vessel is of glass its volume V c.c. at 0° C. becomes at t° C. (1 +00003t,) and at 7° C., V (1 +000037′). The mass of the air which filled the vessel during the first weighing is V(1+00003,) × H1·0012932 grams. Hence the true mass of the vessel alone is V(1+00003t,) × H1×·0012932 and the true mass of the vapour is W-w+ grams, But the mass of a volume of hydrogen equal to and under the same conditions as the vapour at the moment of sealing the vessel is V (1 + ·000037) × H ̧ × ·0000896 2 (1 + 003677) × 760 grams. And hence the density of the vapour is found to be W-w+ D= V (1+00003t,)x H, x*0012932 (1+00367) x 760 V (1+·000037) × H ̧ × ·0000896 (1 + .003677) × 760 The presence of residual air must be avoided as far as possible, since it is apt to render the results untrustworthy. The mass of the vapour found above is too great by the mass of the residual air, which must therefore be subtracted from the numerator of the fraction. v × 3 grams. The volume of the vapour and therefore the volume of the hydrogen has been taken too large by a volume equal to the volume, which the residual air occupied at the moment |