334 AGRICULTURAL ANALYSIS 



Making the proper substitutions in the formula the equation 

 is reduced to the form below : 



V'=35 [ I +0.003665 (0-22 ) ] 



or V'= 35 ( J 0.08063) =32.18. 



Thirty-five cubic centimeters of nitrogen, therefore, measured 

 at 22 becomes 32.18 cubic centimeters when measured at o. 



When gases are to be converted into weight, after having been 

 determined by volume, their volume at o must first be deter- 

 mined; but this volume must also be calculated to some definite 

 barometric pressure. By common consent, this pressure has been 

 taken as that exerted by a column of mercury 760 millimeters in 

 height. Since the volume of a gas is inversely proportional to 

 the pressure to which it is subjected, the calculation is made 

 according to that simple formula. Let the reading of the barom- 

 eter, at the time of taking the volume of gas, be H, and any other 

 pressure desired H'. Then we have the general formula: 



HV 



V:V = H':H : and V = 



H' 



Example : Let the corrected reading of the barometer at the 

 time of noting the volume of the gas be 740 millimeters, and the 

 volume of the gas reduced to o be 32.18 cubic centimeters. 

 What will this volume be at a pressure of 760 millimeters ? 



Substituting the proper values in the formula, we have: 



32.18 X 740 

 V ~ = 31 - 33 ' 



Therefore, a volume of nitrogen which occupies a space of 

 35 cubic centimeters at a temperature of 22, and at a baro- 

 metric pressure of 740 millimeters, becomes 31.33 cubic centime- 

 ters at a temperature of o and a pressure of 760 millimeters. 



One liter of nitrogen at o and 760 millimeters pressure weighs 

 1.25456 grams; and one cubic centimeter, therefore, 0.00125456 

 gram. To find the weight of gas obtained in the above supposed 

 analysis, it will only be necessary to multiply this number by 

 the volume of nitrogen expressed in cubic centimeters under the 

 standard conditions; viz., 0.0125456X31.33=0.039305 gram. 



