328 Dr, Thomson on [May, 



Volumes, 



138 

 140 

 144 

 142 

 144 

 142 



Tlie mean of these trials gives a residue of 141*6 volumes. The 

 smallest residue was 138, and. the greatest 144. 



The mean volume of gas which disappeared in this case was 

 58*4. Now of this, 21 volumes were oxygen gas, the remaining 

 37*4 volumes must have been deutoxide of azote ; so that the 

 mean of these experiments gives us 21 volumes of oxygen unit- 

 ing with 37*4 volumes of deutoxide of azote. This is only a very 

 little greater than 36 volumes, the quantity assigned by Mr. 

 Dalton. The extremes in the experiments are : 



Volumes. Volumes. 



21 oxygen + 35 deutoxide of azote 

 21 +41 



These variations are so great that I was induced to abandon 

 Dalton's method altogether. I find that a tube 0*9 inch in 

 diameter gives much more correct results. When we employ it, 

 the 21 volumes of oxygen just unite with 42 volumes of deutox- 

 ide of azote ; so that the oxygen is obtained by dividing the 

 diminution of bulk by 3. It is obvious that 36 volumes of deut- 

 oxide of azote is not the minimum quantity with which 21 

 volumes of oxygen gas are capable of uniting. The minimum, 

 instead of 36, is in reality 28 volumes. I have obtained a dimi- 

 nution not exceeding 51 volumes, when I employed very narrow 

 tubes ; but the process is disagreeable, and not nearly so accu- 

 rate as when we use tubes with a diameter of 0*9 inch. 



When we mix common air and deutoxide of azote in a com- 

 mon tumbler over water, the results are pretty uniform. The 

 following table exhibits the volume of residual gas when 100 

 volumes of deutoxide of azote were let up into 100 volumes of 

 common air in a tumbler about three inches in diameter : 



119 

 119 

 118 

 118 

 118 

 118 



Mean 118*3 



In these experiments, the mean diminution of bulk was 81 



