CHANGE IN WEIGHT DURING CHEMICAL REACTION. 
239 
and interesting points. Let us consider the following three cases and the possible 
attendant errors :—• 
(1) The two vessels have identical f(.)rms but decidedly different volumes ; 
( 2 ) The two vessels are identical in form but possess slightly different volumes ; 
( 3 ) The two vessels are precisely equal in volume, but whilst the form of tlie one 
is truly sphericcd that of the other is slightly elUpisoidal. 
Case —^In one of my own experiments {vide infra) the difference in the volumes 
of the two vessels was found to he = 8 c.c. It was decided to equalise the volumes by 
adding to the vessel possessing the smaller volume a spherical glass bulb. Accordingly 
a bulb was formed and its diameter repeatedly altered until it was found to be 2’47 
instead of the exact value 2‘48 cm. demanded by theory. Neglecting the attached 
thin stem and hook, we find the external area of an 8 c.c. sphere to be = 19‘2 sq. cm. 
The volumes of the two reaction vessels were respectively 145‘2 c.c. and 137’2 c.c. ; 
and assuming the vessels are truly spherical in form, we find that the corresponding 
external areas are 133'6 sq. cm. and 128‘7 sq. cm. On adding the volume-compensating 
sphere to the smaller vessel, the respective and opposed external areas of the two 
become 133‘6 sq. cm. and 128‘7-tl9‘2 = 147’9 sq. cm. The difference in the areas is 
therefore = 14’3 sq. cm. Had Landolt’s plan been followed the air displacement of 
the two vessels would have been equalised by using a cylinder of the required volume 
prepared from a glass tube. Let the cylinder having V = 8 c.c. be made from tubing 
{a) 2 cm., and {h) 1 cm. in diameter. In the former case the length of the cylinder 
will be 2‘55 cm. and in the latter 10'20 cm. The external area of the shorter cylinder 
will be = 22'3 sq. cm., whilst that of the longer one will be = 33'6 sq. cm. The 
superficial areas of the tAvo cylinders will therefore exceed that of a sphere of the same 
volume by approximately 3 sq. cm. and 14 sq. cm. respectively. Or, the ratios of the 
opposed areas of the two reaction vessels will be changed from 147’9/l33'6 to the 
respective values of 15l/l33'6 and 162'3/l33'6. 
Case 2.—This case also may best be illustrated by means of the actual data acquired 
in a particular experiment. For the sake of brevity we term the two reaction vessels 
X and Y. For X, = 145‘15 c.c., and for Y, V 2 = 143‘56 c.c. ; and therefore 
Vj —V 2 = 1’59 c.c. The volume of the compensating sphere used was found to be 
1'68 c.c. instead of the required exact value 1'59 c.c. ; its external area therefore 
equalled 6'8 sq. cm. Regarding X and Y as spheres Ave find their respective superficial 
areas to be 133'5 sq. cm. and 132'G sq. cm.; therefore the total external area of Y plus 
that of the volume-compensating sphere = 139'4 sq. cm. ; and the ultimate difference 
in the areas of X and Y is equal to 5'9 sq. cm. 
If the volume compensator be formed from tubing having a diameter of 1 cm., Ave 
find that its external area will be = 8'3 sq. cm. ; the final ratio of the surfaces of X 
and Y exposed to the influence of the air avIII therefore = 140'9/l33’5. 
Case 3.—In this case it is assumed that the tAvo reaction vessels have strictly equal 
