528 Prof. Thomson on the Dynamical Theory of Heat. 



91. Let lis now suppose the mechanical energy of a particular 

 fluid mass in various states to have been determined in any way, 

 and let us find what results regarding its pressure, and thermal 

 capacities may be deduced. In the first place, by integrating 

 equation (8), considered as a differential equation with reference 

 to t for p, we find 



p = e }J » \ fi'ge jJo dt + y}r(v)e 3 ' /o * . . (10), 



where yfr(v) denotes a constant with reference to t, which may 

 vary with v, and cannot be determined without experiment. 

 Again, we have from (5), (4), and (1), 



N-I- 

 J dt 



dp 



„ \de 1 (de \ dt \ 



K =Jdt + j\dv + P)Z±\ 



do J 



(11). 



From the first of these equations we infer, that, with a complete 



knowledge of the mechanical energy of a particular fluid, we 



have enough of data for determining for every state its thermal 



capacity in constant volume. From equation (9) we infer, that 



with, besides, a knowledge of the pressure for all volumes and a 



particular temperature, or for all volumes and a particular series 



of temperatures, we have enough to determine completely the 



pressure, and consequently also, according to equation (11), to 



determine the two thermal capacities, for all states of the fluid. 



92. For example, let these equations be applied to the case of 



de 

 a fluid subject to the gaseous laws. If we use for y its value 



derived from (9), in equation (10), we find 



p = P -^{\+^t)+ X {v)e^ dt . . . . (12), 



where %(t>), denoting an arbitrary function of v, is used instead 



of -^(v) — -^-5. We conclude that the same expression for the 



mechanical energy holds for any fluid whose pressure is expressed 



by this equation, as for one subject to the gaseous laws. Again, 



dc dc 



by using for -=- and -p, their values derived from (9), in equa- 

 dt dv 



tion (11), we have 



