CALCULATIONS. 15 



the elements of nitroglycerin, C 3 + H 5 + N 3 + 9 = C 3 H 5 (N0 3 ) 3 

 liquid, amounts to + 98'0 Cal. 



On the other hand, the formation of the products 



3(0 + 2 ) = 3C0 2 disengages + 94 x 3 = 282 



i[5(H 2 + 0) = 5H 2 0] +34-5x5 = 172-5 



Total 454-5 



The heat disengaged by the explosion will therefore be 

 + 454-5 - 98-0 = + 356-5 Cal. 



This is the heat set free by the decomposition of one equivalent 

 of nitroglycerin under atmospheric pressure about the tempera- 

 ture of 15. 



6. If the decomposition take place in a closed vessel, under 

 constant volume, rather more heat will be set free; because 

 the gases developed by the nitroglycerin in the open air, effect 

 a certain amount of work in driving back the atmosphere, and 

 this work consumes a corresponding amount of heat. 



The excess of heat resulting from an explosion in a closed 

 vessel may be calculated by the aid of the following formula : 



(1) Qtr = Qtp + (N' - N)0-54 + 0'002. 



Q,tp expresses the heat disengaged at constant pressure, Qtr the 

 heat disengaged at constant volume, t the surrounding tempera- 

 ture. 



N and N' are defined as follows : Let I be the number of 

 litres occupied by the original gas in the closed vessel in which 

 the explosion has taken place, the gases assumed to be reduced 

 to and 760 mm., and I' the number of litres occupied by the 

 gases after explosion, reduced to and 760 mm. Eeplace I by 

 the expression 22*32N, and I' by 22-32N', in order to compare 

 the volume of the gases with that occupied by 2 grms. of 

 hydrogen H 2 , taken as unity, namely 22*32 litres. 



The formula (1) establishes a general relation between the 

 heat of the reactions taking place at constant pressure and 

 those in constant volume. The author has demonstrated this in 

 his " Essai de Mecanique Chimique," torn. i. p. 44. 



Let this formula be applied to the decomposition of nitro- 

 glycerin, this substance being taken at 15. In this case we 

 may put N = 0. 

 Hence 



N , = 3X4 + 5X2 + 3X2 + 1 = 29 = ^ 



4 4 



Qfir = typ + 7-25 X 0-54 + 7'25 X 0*002 X 15 = Qtp + 

 4-13 = 360-6 Cal. 



This quantity applies to the weight represented by the formula 

 C 3 H 5 (N0 3 ) 3 , namely 227 grms. One grm. will therefore give 

 1590 small calories. 



