96 Prof. Clausius on the Internal Work of a Mass of Matter. 

 two quantities, we may write 



dl = -in dT + -j-dv. 

 fll dv 



The external work, assuming it to consist merely in overcoming 

 an external pressure, is represented by pdv. Hence, if we further 

 decompose the differential dZ into its two parts, we may write the 

 above equation thus : — 



T dZ 

 AdT 

 whence we have 



jrr T dZ , dl ,_ (dl \ , 

 dT + Ach dv== dT dT+ \ch + ?) dv > 



TdZ dl 

 AdT~ dT 

 TdZ_dl 

 A dv dv 



r" 



(13) 



Differentiating the first of these equations according to v } and 

 the second according to T, we get 



T d*Z _ d*l 



A dTdv ~ dTdv' 



IdZ T d 2 Z _ d 9 l ^dp 



A dv + A dTdv ~ dTdv ~~ dT 



The first of these equations subtracted from the second, gives 



1 dZ dp 



A~dv~a^' 



The differential coefficient of Z according to v consequently fulfils 

 the condition given in (12) ; the second of the equations (13) 

 gives at the same time the differential coefficient accordiug to T; 

 and puttiug these two together, we obtain the complete differen- 

 tial equation 



To obtain the quantity -r-Z, we must integrate this equation. 



It is easy to see that this integral will in general differ by a 

 function of T from that which would be obtained by integrating 

 only the last term. The two integrals can only be regarded as 



directly equal if -^ =0, whence also, in order that the foregoing 



dl d z 



equation maybe integrable, it follows that -™- 2 == 0, a case which 

 occurs in perfect gases. 



