The problem of physical chemistry for chemical engineers

BSP (= BPC), and next PD'Q to the first and to the second PCD' (= PD'Q). We then obtain the two triangles BDD' and BB'D' which are obviously equal. He...
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The Geometric Interpretation of the ~ e t h o dof Intercepts H. S. VAN KLOOSTER Remselaer Polytechnic Institute, Troy, New York

I .

T HAS long been known that the thermodynamically nnportant extensive properties of a system, as energy (E), entropy (8,total heat content (H),free energy (F), work content ( A ) , and volume ( V ) are homogeneous functions of the first degree with respect to the rnol numbers of the components of the system-that is to say, if all rnol numbers are made n times as great, the value of the function under discussion will be n.' times as great as before. The French scientist Duhem, after whom the familiar Duhem eauation has been named. was one of the f i s t t o noint out the significance of homogeneous functions in the field of physical chemistry. Let us consider, for instance, the free energy function F. At constant temperature and pressure F is a function of the rnol numbers XI,m,na . . . of the various components A, B, C . . . of the system, viz., F = fh,n. m . . .) and according t o the theorem of homogeneous functions:

and we can also write for equation (4) : dF JN,

-F,

+z

(6)

bF -

= 31 represents the partial mold free an, energy of t h e first component A , viz., the change in F caused by the addition of one rnol of component A to a large amount of the solution, such that the amounts of all other components remain constant. Similarly we write

where

dF JN,

bF = F*, = %, etc., &k bnr

c

and equation (1) becomes: d ~ = F ; d ~ , + F ; d ~ + F ; d... ~+

on integration,

for constantrelative

have : F

=

En,+ Z m + Ens + ...

(2)

we (3)

=

nr n' and Nn = nl + m n ~+ nr

we obtain the expressions: dF = Z d N ,

and F = 7,N,

+ Fm,

(4)

+7 * N 2

(5)

where Nl f Na = 1. Since N,

= 1

- Nz, dN, =

-dN,

.

.

I f we represent, for a binary system of consolute liquids, the variation in the (mean) molar free energy as a function of the rnol fraction, we obtain a smooth curve which is always convex toward the concentration axis. Frequently (his curve exhibits a minimum. For the sake of simplicity the eurve in Figure 1 is drawn without a minimum.' If we.draw the tangent BPD' to the c w e KPL in P.then according to gakhuis Roozeboom, the intercept AB equals 31and A'D' equals F2. The proof for this so-called method of intercepts is frequently omitted in textbooks and the reader is referred to Lewis and Randall2 for the solution. I t is the purpose of this note to visualize with the aid of a few additional lines e r r e c t n e s s of this procedure. ~~

Applying these equations t o the particular case of a binary system and introducing mol fractions: N,

FIGURE 1

~

d~ 'Since- for z = 0 equals - m and for z = 1 equals + m , dX the shape of the curve is not quite correct. LEWIS AND RANDALL, "Thermodynamics and the free energy of chemical substances." McGraw-Hill Book Co., Inc.. New York City, 1923, p. 38.

283

In the first place, by drawing BB' parallel to AA' we note that

AA'C'C = RA'B'S

d- F = - D'B' dN, BB'

or (since BB' = AA' = unity) =

D'B' = A'D' - A'B' = - A B

+ A'D'

which fits equation (G), assuming A B ( = A'B') =

Fl and A'D'

PR X AA' or the area AA'C'C, N Z equals RA'B'S and N& equals ARQD or:

=

-

F

That equation ( 5 ) is also satisfied becomes evident when we draw DD' parallel to AA' and the vertical QPSR through P . The value of F is represented by

+ ARQD

This last equation results directly from the co-linearity of B, P , and D' and can be readily verified, if we state that the rectangle CPQD which lies outside AA'C'C must be equal to the rectangle SB'C'P which is inside AA'C'C. That this is the case becomes clear if we add to the first rectangle BPC and to the second ESP (= BPC), and next PD'Q to the 6rst and to the second PC'D' (= PD'Q). We then obtain the two triangles BDD' and BB'D' which are obviously equal. Hence: KN,

+ EN* = RA'B'S + ARQD

=

AA'C'C = F