273 



" As the vapours of different liquids have at their respective 

 boding 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 of any other form 

 of matter, is represented by its weight or mass, divided by its 

 specific gravity, if we put 



x being the weight of any vapour, whose specific gravity is s\ 

 and s the specific gravity of the vapour of water, we will get 



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 



518 

 j^r , t beiug the boiling point of the liquid which yields the 



vapour. In the fourth column we have the weights, which would 



