( 312 ) 



two unities side In side, we should have for the first ip t = — RTlogv, 

 for the second \]\ = - RT log v, so if, -J- *!\ = — 2RTlogv. For 

 the total quantity we should have i|> = — 2RT/og 2v -j- C; if now 

 we had again put C=0 (so that ip = if 2v = 1) then if' would 

 not have been = if?, -f- if>, . The C, however, has now been chosen 

 in such a way that tp = if 2v = 2 or C =2RT log 2 and hence 

 y = — 2RTlogv = ip i + «iv 



In connection with this we have the property that if two quan- 

 tities of different gases, being in equal volumes at the same T, are 

 mixed in the same volume at the same 7', the free energy remains 

 the same, whereas it decreases if this is done with two quantities 

 of the same gas. A similar property exists for the entropy (Gibus's 

 paradox). How is this now for the kineticallv defined entropy or 

 free energy? To answer this question we shall successively dis- 

 cuss: 1 is the entropy of an homogeneous gas mass in a volume 

 2v double that of half the quantity in a volume v ; 2 is the entropy 

 of an homogeneous gas mass in a volume v greater than the sum of the 

 entropies of two such masses forming together the first quantity 

 each in an equal volume v, 3 is the entropy of a mixture of 2 gases 

 equal to the sum of the entropies of the two gases separately? 

 In the entropy of BoLTZMANN the answer is ovvvy lime affirmative. 



If we consider H = Nllog 1- CL then in the 1 st case 



H, = iV x flog — + O), so 227, = 2A\ flog — + Ó) ; further for 

 the whole mass in the volume 2v. 



11= 2.V, ( log -^ + ( •} ■= 2.\\ ( loa — ? -f c\ = 2/7 \ . 

 In the second case : 



II x = TV, (log ^ + c\ II, = N a flog ^ + c\ 



so H x + E, = N, log JST, + AT log AT, — ( .V, + -V, ) ( log v + C) , 

 while ƒƒ= (A r , + N,) log W + A 7 ,) — (X, + A 7 ,) (logv+C), so that 

 //<//, + //, or the entropy of the whole is greater than the sum 

 of the entropies of the parts. 



In the 3 rd case the formula for H used here does not hold, but 

 now Boltzmann puts here H= H x -\- H. t by definition. For a mixture 

 of two gases Boltzmann puts viz. : 



// = J I f log f do dio -f I | ƒ, log/\ do dm. 



However, also the questions 1 and 2 might have been answered 



