190 Prof. Williamson on the Unit-volume of Gases. 



The great convenience of this hydrogen scale of densities, 

 which is already leading to its general adoption, arises from the 

 fact that it enables the chemist to calculate with facility the 

 density of any gas or vapour from a knowledge of its molecular 

 weight. For inasmuch as every molecule, with very few appa- 

 rent exceptions, is found to occupy two volumes in the state of 

 vapour, the vapour- density of every compound is equal to half 

 its molecular weight. 



For some years past I have been in the habit of using an ex- 

 tension of this natural system of volume-notation, which affords 

 the means of calculating with rapidity the absolute volume of a 

 given weight of a gas or vapour, or inversely the absolute weight 

 of a given volume. This extension consists simply in substitu- 

 ting the word " gramme " for " part by weight " in the definition 

 of volume. A volume of hydrogen is the bulk of one gramme of 

 hydrogen at the normal temperature and pressure, and a volume 

 of any gas or vapour is a bulk of that gas or vapour equal to that 

 of one gramme of hydrogen. 



According to the determinations of oxygen, which are less 

 affected by errors of manipulation than the determination of 

 hydrogen, a volume is 11*19 litres at 0° C. and 760 millims. 

 mercurial pressure. For most purposes the volume may be de- 

 fined as 11*2 litres. 



In order to show the advantages derivable from this absolute 

 volume, it will be best to take a few examples of calculations per- 

 formed by the aid of it. 



Thus, it is required to find the volume of oxygen obtainable 

 by the decomposition of one kilogramme of potassic chlorate. 

 The formula KC10 3 tells us that the molecule of chlorate weigh- 

 ing 122*5 contains 48 parts of oxygen, so that the proportion 

 122*5 : 48 = 1000 : x gives us 391*8, say 392 grammes, as the 

 weight of oxygen contained in our kilogramme of chlorate. In 

 order to reduce this to litres, we have from the density of oxygen 

 (0 = 16= 1 vol.) the proportion 16 : 11*2 = 392 : y, whence y 

 = 274*4 litres as the measure of oxygen obtainable by the com- 

 plete decomposition of a kilogramme of chlorate. 



2. Given 500 grammes of zinc, required the volume of hy- 

 drogen obtainable by its action on sulphuric acid. The equation 

 Zn + H 2 S0 4 = H 2 + Zn SO 4 tells us that sixty-five parts by weight 

 of zinc displace two parts by weight of hydrogen from its sul- 

 phate; and 65 : 2 = 500 : x gives us 15*384 grammes as the 

 weight of hydrogen ; and this number, multiplied by 11*2, gives 

 us 174*3 as the number of litres of hydrogen. 



3. Given 150 grammes of paraflSne, find the volume of air 

 required for its combustion, assuming for parafline the formula 



