zero, and absolute temperatures are measured from it as Centigrade temperatures are measured from the meltingpoint of ice. Any temperature expressed on the Centigrade scale is converted into the absolute scale by adding 273 to it. Thus 13° C. is - 13+ 273 260° on the absolute scale. = (16) EXPANSION BY HEAT. Solids and liquids usually increase in volume when heated and decrease in volume when cooled, but the change is so small, that in ordinary chemical experiments it may be neglected. Thus 5550 volumes of mercury at 0° C. increase by one volume for every 1°C. through which they are heated, becoming 5550+ 100 5650 volumes at 100° C. = The expansion of gases however is so considerable that a change of a very few degrees may affect an experimental result. Gay Lussac's Law. When 273 volumes of a gas at 0° C. are heated, they increase by one volume for every 1°C. through which they are heated. Thus 273 volumes of gas at 0° C. become at I° C. 273 + 1 volumes, where t and T express any number of degrees on the Centi grade scale. Hence 273+t volumes of gas at t° C. become at 7° C. 273+T and v volumes at t° C. become at 7° C. v × volumes; 273+t which, if I stand for the volume of the gas after a change of temperature from t° C. to 7" C., is usually written To find the new volume if 1000 c.c. of gas are heated from 17° C. to 27° C. If the pressure on the gas as well as its temperature be changed, the above formula must be combined with the one If 500 c.c. of gas are cooled from 39° C. to 13 C., the pressure being decreased from 800 m.m. to 300 m.m., find the new volume. *In certain cases the fraction, which is at least equally correct, is more convenient for calculation than 75. (3000+117) Then V = 3000+ 11t Thus, to find the new volume if 1000 c.c. of gas are heated from 0° C. to 300° C., In other cases, and especially when logarithms are used, it is more convenient to use formulæ containing decimal instead of vulgar fractions. Thus : -. 1 volume of gas at 0°C. when heated to 1° C. becomes 1 + 00367 × 1 vols. The fraction 00367 is called the coefficient of expansion of a gas, and is frequently expressed by a (17) THE DENSITY OF GASES. When the pressure on and temperature of a gas are not mentioned, it is supposed to be at 0° C. and 760 m.m. A gas under these conditions is said to be normal. The formulæ given in this and the succeeding section only apply to normal gases; hence, when necessary, the gas under consideration must be rendered normal by using the formula: And conversely the volume found by these formulæ is normal and must be reduced to the required temperature and pressure by It has already been stated in (9) that the density of a substance is the mass of unit-volume of it, but in the case of gases the cubic centimetre and gram, which are used in the case of solids and liquids, are not convenient units. In the case of gases, the litre = 1000 c.c. is taken as the unit of volume, and the mass of one litre of normal hydrogen, called a crith 0896 gram, is taken as the unit of mass. = The density of a gas then is the number of criths contained in one litre of it measured at 0° C. and 760 m.m.; or the number of times it is heavier than an equal volume of hydrogen. Hence the mass in grams of a litre of any normal gas can be found by multiplying its density by 0896. ̈ ́ Thus the density of carbon monoxide being 14, a litre of carbon monoxide weighs 14 × 0896 1.2544 grams. It is found that with few exceptions the density of any elementary gas is expressed by the same number as its atomic weight, and that of any compound gas is expressed by the same number as half its molecular weight. Thus : = Hydrogen H = 1. Density 1, or 1 litre weighs 1 crith. = 22 criths. If the density be doubled, or, in other words, if the unit of hydrogen be taken as 2 instead of 1, the double-density of almost every gas is expressed by the same number as its molecular weight. The densities of gases used to be referred to air, which is 14.436 times as heavy as hydrogen. In such a case, to find the density referred to hydrogen, multiply the density referred to air by 14-436. Thus sulphur dioxide is 2.22 times as heavy as air: find its density and molecular weight. 14·436 × 2·22 = 32.042 is the density referred to hydrogen, and 32.042 × 2 X = 64.084 is the molecular weight. = If the density of a gas referred to air is given, to find the mass of one litre of it, multiply the given density by 1.293, the mass in grams of one litre of normal air. Thus carbon dioxide is 1.529 times as heavy as air, to find the mass of one litre of it 1.529 × 1.293 = 1.977 grams. Conversely, if it be required to find the density referred to air from that referred to hydrogen, multiply the density referred to hydrogen by 06926, the density of hydrogen referred to air. |