280 Mr. J. J. Waterston on the Deviation from 
in passing through the plug 0°:26 C.,—let C D, fig. 1, represent 
Fig. 1. 
a cylinder of air at the pressure of 
one atmosphere and temperature 
10° C. Let the elastic force of 
the air in this cylinder be wholly 
maintained by the counterbalan- 
cing weight of the piston W, so 
that we have to suppose a perfect 
vacuum to exist on the upper side 
of it. By placing a small addi- 
tional weight w on W, suppose it 
to descend to g, so that the conse- RZ 
quent rise of temperature in the 
air contained in the cylinder may 
be 0°26 C. This increment of 
heat being withdrawn, the piston 
W descends from gto f/ We may 
now suppose w withdrawn from W, 
which consequently rises from f to 
e by the elastic force of the air in 
the cylinder. There is a conse- 
quent loss of heat, and the tempe- 
rature of the contained air’ sinks 
0°26. This decrement of heat 
being restored, the piston rises to 
D, its original position. AL 
The ratio of fg, or its equal De, to D C is nearly the same as 
that of 0°:26 to 273°+10°, or of the increment of temperature 
to the G temperature, or temperature reckoned from the zero 
of gaseous tension. The ratio of De to Df is nearly 1 to 4, as 
determined by the experiments of MM. Clement and Desormes, 
also of MM. Gay-Lussac and Welter (Mécanique Céleste, book 12). 
The ratio of Df to DC is thus nearly 1:04 to 288, or 1 to 272; 
and this is the amount of deviation from the law of Mariotte in 
air at 10° C. between the pressure of two and one atmospheres, 
as derived from Messrs. Thomson and Joule’s experiments on 
thermal effect. 
The above result may be exhibited in diagram by A B, another 
cylinder resting on the same plane as C D, and having one-half 
the transverse area. The weight of the piston U being equal to 
W, and the weight of air in each cylinder being the same, as well 
as the temperature, we have AB=fC equal to — of CD. 
§ 4. This computation, expressed by symbols, is 
nO 
ae 973 40? 
