278 PROCEEDINGS OF THE AMERICAN ACADEMY, 



The work done during this cycle at D, Ci, and C2, is zero, since in 

 each case the final position is the same as the initial, and the pressure 

 is constant throughout the cycle. Therefore the total work done by 

 the system during the cycle is that done by the pistons Fi and F2, 

 which is as follows : 



In operation (1), 



^i = niFi + n2F2. 



In operation (2), 



In operations (3) and (4), except for a differential of the second 

 order, 



As + A, = -(n,+ cUhXV, - dW) - (Ho + dn.^(V, - dV,). 



By the second law of thermodynamics the sum of these terms, the total 

 work of a reversible isothermal process, r ust be zero. Hence, 

 neglecting differentials of the second order, 



VidUi + Vodlh = 0. 

 Since we are dealing with ideal solutions, 



^. N,BT , .. N,RT 



hence N^d In Hi + N.d In Ho = 0. 



Now the activity ^1 of Xi in A is always the same as in Bi, and fa in A 

 is the same as in B2 ; hence, applying equation III (p and T being 

 constants) we have, 



N^d In ^1 + Nd In ^^ = 0, XV 



which may also be written 



( 



dNi ) p,T 



OjSz JP,T 



