882 B. Silliman on determining the Photometric power, etc. 
diluant. We might make a thousand observations by the 
means now at command, and not obtain one with an exact 0 for 
hydrogen. A very trifling error in the admeasurement greatly 
influences the result. 
We may therefore safely conclude, as it appears to me— 
Ist. That in all illuminating gas we have a certain substratum 
of non-luminous gas, holding in solution a variable volume of 
luminous gas (olefines). 
2 at when a gas is too rich in illuminants to permit of 
accurate photometric admeasurement by the usual standards of 
intensity, it may be diluted with a poor gas of known value 
and volume to such a standard as is consistent with the accurate 
employment of the usual photometric apparatus, its true value 
being then calculated from the known values employed. 
P. 8. I find in the manuscript records of the Manhattan Gas 
Company, mention of four experiments made many years since 
by Mr. Schultz, chemist of that company, in which he mixed 
5 per cent and 10 per cent of hydrogen with gas of very high 
illuminating power. The results are less satisfactory than they 
would have been, had the volume of hydrogen employed been 
much larger and the intensity of the coal gas not over 15 
candles. Moreover at that time the means of admeasurement 
at the command of the observer in the laboratory of the Man- 
hattan Company of small volumes of gas were much less exact 
than they now are. The results obtained are as follows: 
lst Exeperiment. 
Taken 90 volumes of coalgas = =22°53 candles, observed. 
= 10 ‘ bydroger ==... 
eo re tre cee0-11 . 
Tf the hydrogen istakenasoO “ =20°27 “ calculated. 
Showing apparent error of soe D1. i 
2d Experiment. 
Taken 95 volumes of coalgas = =22-53 candles, observed. 
= 5 - hydrogen = ..... 
. 100 mixture <=20°36  * if 
If the hydrogenistakenasoO “ =21:39 “ calculated. 
Apparent error =—104 * 
’ 38d Experiment. 
‘Taken 90 volumes coal gas 20°83 candles, observed. 
* 10 “ hydrogen Pees 
“a 100 +=“ mixture i946 “a 
Ifhydrogen is taken—=0 “ =1829 “ calculated. 
Apparent error, — 16 % e 
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