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273 
*«* As the vapours of different liquids have at their respective 
boiling points the same elastic force, equal volumes of them 
will produce equal mechanical effects. In order, therefore, to 
the solution of the question under consideration, it will only 
be necessary to calculate the weights of the different liquids, 
water included, which give equal volumes of vapours, and to 
determine the quantities of caloric necessary for the conver- 
sion of them into vapour. 
‘‘ Now as the volume of a vapour, like that ofany other form 
of matter, is represented by its weight or mass, divided by its 
specific gravity, if we put 
zx being the weight of any vapour, whose specific gravity is s’‘, 
and s the specific gravity of the vapour of water, we will get 
/ 
s 
f= 
s 
that is, the weight of any liquid which, at its boiling point, gives 
a volume of vapour equal to that given by a weight of water 
represented by unity at its boiling point, is got by dividing the 
specific gravity of the vapour by that of steam. But the specific 
gravities to be used in this computation are not those usually 
given in books, each of which is referred to a different unit, 
viz., air at the same temperature, and under the same pressure 
as the vapour, but the specific gravities of the vapours at the 
respective boiling points of the several liquids, compared to 
the standard unit, viz., air at 60°, and under a pressure of 
30°. In the following Tables, the former specific gravities 
are found in the second, and the others in the third column, 
the latter being in each case got by multiplying the former by 
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vapour. Inthe fourth column we have the weights, which would 
t being the boiling point of the liquid which yields the 
