V. MASS. 73 



Example 1. 



T=the reservoir filled with mercury weighs at 19 Cels.= 109r6 

 R=the empty reservoir - =- 329*6 



Q=T R. Consequently mercury 762*0 grammes. 

 Consequently contents of the reservoir at 19 Gels. 



T-R 762*0 



= 13-59(1-* 0-0018) = lF55 = 56 ' 236 Cub " Centlm ' 



If now 60 grammes ( = P) gunpowder is filled into the reservoir, and the 

 remaining space of the same with mercury, we obtain 

 611*6 grammes ( = TJ 



Reservoir empty =329-6 

 60 grammes powder = 60-0 



R + P= 389 -6 deducted, 



remains Q = 222.0 grammes weight of the mercury filling the intervening 

 space, occupying at 19 Cels. 



13'55 



Consequently volume of the 60 grammes powder= 56-236 16-444 



V= 39-792 cuh. centhns. 

 p 



S = y = consequently specific weight of the 60 grammes gunpowder 



__ = 1-507 

 39-792 

 Example 2. 



If a powder prism weighs 42-0 grammes at 190 C. weight of reservoir, 

 powder prism and mercury =808-4 grammes. 



Reservoir empty 329-6 

 42 grammes powder 42-0 



371-6 

 consequently of 808-4 



371-6 deducted, 



= 436-8 grammes weight of mercury, 



436*8 



or - =32-236 cub. centim. occupied by these 436-8 grammes. 

 13*55 



Consequently volume of the 42 grammes weighing powder prism=56'236 

 32-236 = 24-0 cub. centim. 



Thus, specific weight of the powder prism= - = 1-75 



General Formula. 



g_P_' P P _ __ Q 



V I-I 1 T-R r "13-59 (1-*0-0018> 



13-59(1- t 0-0018) ~~ 



g _P [13-59 (1- t 0-0018)] Since Q 1 =T 1 -R-P 

 T-R-Qi 



S= P13'59(1 ! 0-0018) 

 T-T' + P 



