146 Mr. Harold B. Dixon on 



the formula used failed to express the hypothesis with 

 exactness. 



Let us examine the mode in which Berthelot calculates 

 the theoretical velocity, i.e., the mean rate of translation of 

 the products of combustion at the temperature of the 

 explosion. In Clausius' formula 



/I 



• »^=29-354V d 



M. Berthelot calculates the absolute temperature of the 

 explosion by dividing the quantity of heat developed in the 

 complete reaction by the specific heat of the products of 

 combustion taken at constant pressure. He argues that 

 each layer of gas, in transmitting the explosion, is heated 

 under constant pressure. I cannot follow his reasoning. 



"The combustion," he says, "in propagating itself from 

 layer to layer, is preceded by the compression of the gaseous 

 layer which it is about to transform. . . . The combustion of 

 each layer produces both heat and at the same time the 

 work necessary to compress the following layer — that is to 

 say, the layer loses on this score just as much heat as it 

 gained by its own compression. The whole proceeds, as 

 far as the elevation of temperature is concerned, precisely 

 as if we had operated under constant pressure." On the 

 facts of the case there is no dispute. The gas is exploded 

 in a closed vessel. Each layer is compressed before 

 being fired ; after firing it compresses the layer beyond 

 it. Now, as regards this preliminary compression, each 

 layer, in turn, expends the same energy as was previously 

 communicated to it, and therefore, it does no work of its 

 own. But a gas heated under conditions where it does 

 no work is raised to the same temperature as it would 

 be had its volume remained constant. M. Berthelot admits 

 that it would appear at first sight as if the gases were 

 heated at constant volume ; he adds that the concordance 



