LEWIS. — A NEW SYSTEM OF THERMODYNAMIC CHEMISTRY. 



277 



Finally we must determine how the activities of the components of a 

 mixture vary when the composition is changed at constant temperature 

 and pressure. In order to solve this problem we may employ the ap- 

 paratus shown in Figure 2. A contains a mixture of Xi and X 2 . E x is 

 a membrane permeable only to X 1} E 2 one permeable only to X 2 . In Bi 

 and B 2 are ideal solutions of Xi and X 2 . The two pure ideal solvents 

 lie above the pistons, Fi and F 2 , which are permeable only to these 

 solvents. D is a piston which exerts a constant pressure on A. The 

 pressure at Ci and C 2 will also be held constant. We may perform the 

 following isothermal cycle of reversible operations, starting with N~i 

 mols of Xi and X 2 mols of X 2 in A, and none of these substances in B t 

 and B 2) the pistons F x and F 2 being at E x and E 2 . 



(1) Keeping the pressures on F x and F 2 constant and at such values, 

 ITj and fl 2 , as to maintain equilibrium with the mixture in A, raise 

 these two pistons at such 

 rates that as X : and X 2 

 pass into Bi and B 2 the 

 remaining mixture in A 

 still keeps its original com- 

 position. Finally, when all 

 the mixture has disap- 

 peared from A there will 

 be A 7 ! mols of X x in Bi 

 where it exerts the os- 

 motic pressure U u and oc- 

 cupies a volume which we 

 will call }\, and there will 

 be X 2 mols of X, in B 2 

 volume l'o. 



the osmotic pressure being I7 2 , and the 



(2) By simultaneous movements of the pistons Fi and F 2 change the 

 volumes in B x and B 2 to T, — d V x and V 2 — dV 2 . The osmotic pres- 

 sures will change to Uj + dn l anil n 2 + d\l 2 . The solutions in B t and 

 B 2 are now able to exist in equilibrium, not with the original mixture, 

 but with a mixture containing Xi and X 2 in another proportion, say X 1 

 mols to X 2 — dX 2 mols. 



(3) Form a mixture of this composition in A by lowering the pistons 

 Fi and F 2 . This operation will be just the reverse of (1), except that 

 Xi and X 2 enter the mixture in the constant proportion, not of A\ to 

 A" 2 but of X t to X 2 — dX 2 At the end of this process all of Xi and 

 all but dX 2 of X 2 will have passed into A. 



(4) Finally force into A the remaining. dN 2 mols of X 2 , whereby the 

 whole system returns to its original condition. 



