One litre of CO_{2} = 22 x .0896 = 1.9712 grm.
" H_{2}O = 9 x " = 0.8064 "
" N_{2} = 14 x " = 1.2544 "
" O_{2} = 16 x " = 1.4336 "
The volume of permanent gases at 0° and 760 mm. is constant, and assuming the gramme as the unit of mass, is found to be 22.32 litres. Thus:—
Volume of 44 of CO_{2}, at 0° and 760 mm. = 44/1.9712 = 22.32 litres. 18 " H_{2}O " " = 18/0.8044 = 22.32 " 28 " N_{2} " " = 28/1.2544 = 22.32 " 32 " O_{2} " " = 32/1.4366 = 22.32 "
Therefore
132 grms. of CO_{2} at 0° C and 760 mm. = 22.32 x 3 = 66.96 litres.
45 " H_{2}O " " = 22.32 x 2-1/2 = 55.80 "
42 " N_{2} " " = 22.32 x 1-1/2 = 33.48 "
8 " O_{2} " " = 22.32 x 1/4 = 5.58 "
____________
161.82 " Therefore 1 gram-molecule or 227 grms. of nitro-glycerine when exploded, produces 161.82 litres of gas at 0° C and 760 mm.
To determine the volume of gas at the temperature of explosion, we simply apply the law of Charles.[A] Thus—
V : V' :: T : T' or V' = VT'/T
in which V represents the original volume.
V' " new volume.
T " original temperature on the absolute scale.
T' " new temperature of the same scale
In the present case T' = 6001°.
Therefore substituting, we have