182 Densities of " Atmospheric Nitrogen" &c. 



The data may be divided into two groups : those of Leduc and 

 Schloesing ; and those of Rayleigh, Ramsay, and Kellas. 



From the first group it is possible to calculate the density of argon,. 

 i.e., the crude mixture left after separating oxygen and nitrogen from air. 



From the second group, the density of argon may be calculated; or 

 if that be assumed, both groups give data for the calculation of that 

 of " atmospheric " nitrogen. It has been thought better to express 

 the results in the form of the weight of one litre of the gas in 

 question ; but if it is desired to state them with reference to the 

 density of oxygen =16, the conversion may be made by means of 

 the weight of a litre of oxygen according to both Lord Kayleigh and 

 M. Leduc. 



The data are as follows : 



Weight of 1 litre of Leduc. Rayleigli. Scliloesing. Kellas. Eamsaj. 



Air 1-29316 1-29327 



Oxygen 1-42920 1*42952 



Nitrogen 1*25070 1-25092 



(atmo.) 1-25700 1-25718 



Argon ., 1-78151 ... 1-7816 



in " atmo." 



nitrogen ... 0*01183 0-01186 



Weight of 1 litre argon calculated from Leduc's and Schloesing's 

 figures : 



0-01183 = 1-25700 - (1-25070 x 0*98817); hence a; = 1-7828 



The difference from the value found is 7 in 10,000. 



Weight of 1 litre argon calculated from Rayleigh's and Kellas's 

 figures : 



0*011862* = 1*25718 -(1*25092x0*98814); hence x = 1-7791. 



The difference from the value found is 13 in 10,000. 



Both of these results are quite satisfactory, considering that the- 

 nature of the calculation involves a ratio of small differences. The 

 agreement is more striking if the density of " atmospheric " nitrogen 

 is calculated from the figures ; for this calculation, the weight of 1 

 litre of argon is assumed to be 1*7815 granis. 



Weight of 1 litre of " atmospheric " nitrogen from Leduc's and 

 Schloesing's figures : 



x = (1*7815x0*01183) + (1*25070x0-98817); whence z = 1*25698. 



Here the difference is only 2 in 125,000. 



From Lord Eayleigh's and Dr. Kellas's figures, we have : 



x = (1*7815x0*01186) + (1*25092x0-98814); whence x = 1-25721. 

 The difference here is only 3 in 125,000. 



