272 PROCEEDINGS OF THE AMERICAN ACADEMY. 



a binary mixture may be found by means of the apparatus shown in 

 Figure 1. A contains the mixture of Xi and Xj. D is a piston 

 which determines the pressure on A. E is a membrane permeable only 

 to Xi . B contains a solution of Xi in its ideal solvent. F is a piston 

 permeable only to the latter. Above F is the pure solvent. 



The pressure on the piston F is the osmotic pressure, IT, of the ideal 

 solution in B. In general if the pressure, P, on D is changed, the 

 equilibrium will be disturbed and the substance Xi will pass through 

 E, unless at the same' time the pressure on F is changed by a suitable 

 amount. Let us find the mathematical expression for the change in 11, 

 which just compensates a given change in P. 



Starting with the piston F at E and with a large {better, an infinite) 

 amount of the mixture in A, occupying the volume V, let us perform 

 isothermally the following cycle of reversible operations. 



(1) Keeping the pressure F constant on the piston D, and keeping 

 the pressure on F also constant and equal to the corresponding osmotic 

 pressure, n, raise F until one mol of Xi passes into B, where it occu- 

 pies the volume v' . The diminution in the volume of A we will denote 

 by the symbol v. The work done by the system by means of the pistons 

 F and D is, therefore, 



A^ = Uv' - Pv. . 



(2) Now increase the pressure on the piston J) to P + dP, and at 

 the same time increase the pressure on F to IT + dU, dll being the in- 

 crement in n which is necessary to prevent Xi from passing in either 

 direction through E. The volume of A will change from V~v to 

 (V — dV) — (v — dr), and the volume of the solution will change 

 fromw' to v' — dv'. The work done by the system by means of the 

 pistons F and D is, 



A. = -Udv' -P(dV-dv). 



(3) Keeping the pressures on the two pistons constant and equal to 

 U + dn and F -f dP respectively, lower F to E, forcing the mol of Xj 

 back into A. The work done by the two pistons is 



^3 = - (n -f dn)(:v' - dv') + (F + dP)(v - d7^). 



(4) Change the pressure in A back to P. The piston F is station- 

 ary, and the work done by the piston D is, 



A, = FdV. 



