398 



Mr. R. A. Lehfeldt on the 



rise we may evaporate a little of A without altering the 

 quantity of vapour of B present, nor the total pressure on the 

 liquid : similarly, by raising the second piston we may eva- 

 porate B alone. 



Let <J> be the thermodynamic potential of the liquid, 



<3X „ „ „ of the vapour ; 



m and n the masses of A and B present as liquid, 



m 



and 



n 



jj jj j) 



Then the conditions of equilibrium are that 



vapour. 



and 



0771 Qm 



> 



(1) 



(2) 



=■— bn+ ^r- 7 bn / = {); 

 but as Sm= — 8m' and Bn= — Sn f , we have 



~dn 



We shall endeavour to evaluate the thermodynamic poten- 

 tial of the vapour ; to do this we assume that each of the two 

 constituents of the vapour behaves as a perfect gas, and that 

 the two constituents have no mutual action. We may then 

 set the potential of the vapour equal to the sura of the poten- 

 tials that each constituent would possess if it existed alone 

 under the same temperature, and a pressure equal to its 

 partial pressure in the mixture. We will express the specific 

 potential of a simple liquid by <£(A, p, T), where A denotes 

 the substance, p the pressure, and T the temperature ; and 

 by <p' '(A, p,T) that of the vapour. The second of the above 

 assumptions may then be expressed algebraically by saying 



& = m'4>'{A,p A ,T)+n'<l>'{B,p B ,T), ... (3) 



where p A , ps are the partial pressures. 



For the liquid we cannot make a similar assumption, we 

 therefore put 



* = m 4>(A,p,T) + fup{B,p, T) + (m + »)/(A, B,™,p, t), (4) 



where p is the total pressure (p=Pa+Pb) 9 an d / is a 

 quantity expressing the mutual action of the two liquids ; it 

 is a function of m/n, which we will write q, and its value is 

 the subject of the investigation. 



