30 ELECTROLYSIS. 



AMOUNT OF WORK NECESSARY FOR THE DECOMPOSITION OF 

 WATER. We will propose to determine the work corresponding 

 to the decomposition of one kilogramme of water in one hour, 

 for example. 



We have seen that the said work was represented by the 

 two expressions 



CE . , C 2 E 



w = and w = 



9 9 



P* Tf 



In this instance we have only to deal with , that is to 



9 

 say, the work absorbed by the decomposition of the water 



exclusive of the calorific energy expended by the conductors. 

 One ampere current decomposing 0-00009328 gramme, or 

 0' 00000009328 kilogramme of water, to decompose one kilo- 

 gramme, Q. QQQQ()0()9328 am P eres wil1 be required ; and as the 



electromotive force is 1*4856 volts, the corresponding work will 

 be 



w = _-., * _r ^ = 1623430 ki 



In order to verify this number, it is sufficient to calculate 

 the work which the hydrogen and the oxygen contained in one 

 kilogramme of water can develop in combining together. 

 One equivalent of water or 9 grammes corresponds to 

 34-462 calories, 1 kilogramme will therefore correspond to 



344R9 



- = 3829-111 calories, or to 3829-11 x 424 = 1633542 



y 



kilogrammetres (the slight discrepancy existing between these 

 two results arising solely from the neglect, in both the calcula- 

 tions, of some decimals).* 



* Edouard Japing, in his recent work on Electrolysis (Vienna, 1883), has trans- 

 lated a paragraph published in the journal ' La Lumiere Electrique,' in January 

 1882, and in which the work required for the electrolysation of 1 kilogramme of 

 water is valued at 1,925,000 kilogrammetres, and the work resulting from the 



1 8 



combination of - kilogramme of hydrogen with - kilogrammes of oxygen is given 



as 3,636,000 kilogrammetres. In a journal such an error can be understood to 

 occur, owing to the hurried manner in which the work of publication is gone 

 through ; but it ought not to be so easily excused in the case of the publication 



