Physical Chemistry Third Edition

1264 D Some Derivations of Formulas and Methods

The subscriptn′means that all of then′’s are held fixed in the differentiation except

forni. We use the fact that

(

∂(λni)

∂(λ)

)

n′

ni (D-20)

and setλequal to unity in Eq. (D-19) to obtain Eq. (D-18), and the theorem is proved.

D.4 The Method of Intercepts

The value of the derivative (∂Ym/∂x 1 )T,Pgives the slope of the desired tangent line

when evaluated atx 1 x′ 1. We write the analogue of Eq. (4.6-15) for a general extensive

quantityYand differentiate with respect tox 1 :

(

∂Ym

∂x 1

)

T,P

Y ̄ 1 +x 1

(

∂Y ̄ 1

∂x 1

)

T,P

−Y ̄ 2 +x 2

(

∂Y ̄ 2

∂x 1

)

T,P

(D-21)

where we have used the fact that (∂x 2 /∂x 1 )−1, which follows from the fact that

x 1 +x 2 1. The second and fourth terms on the right-hand side of Eq. (D-21) sum to

zero by the analogue of Eq. (4.6-13), giving

(

∂Ym

∂x 1

)

T,P

Y ̄ 1 −Y ̄ 2 (D-22)

If this derivative is evaluated atx 1 x 1 ′, it gives the slope of the tangent line at that

point. If we letystand for the ordinate of a point on the line, then

y[Y ̄ 1 (x 1 ′)−Y ̄ 2 (x′ 1 )]x 1 +b (D-23)

wherebis the intercept of the tangent line atx 1 0, and where we consider both of

the partial molar quantities to be functions ofx 1 , and omit mention of the dependence

onPandT.

The line and the curve must coincide atx 1 x′ 1 , so that from Eqs. (D-21) and (D-23)

Y ̄ 2 (x 1 ′)+x 1 ′[Y ̄ 1 (x 1 ′)−Y ̄ 2 (x′ 1 )][Y ̄ 1 (x 1 ′)−Y ̄ 2 (x′ 1 )]x′ 1 +b (D-24)

Canceling equal terms on both sides of the equation, we get

Y ̄ 2 (x′ 1 )b (D-25)

One can repeat the entire argument with the roles of components 1 and 2 reversed

to show that the intercept at the right side of the figure is equal to the value ofY ̄ 1 at

x 1 x′ 1. However, it can more easily be shown by evaluating the function represented

by the line atx 1 1.

y(1)[Y ̄ 1 (x′ 1 )−Y ̄ 2 (x′ 1 )]+Y ̄ 2 (x′ 1 )Y ̄ 1 (x′ 1 ) (D-26)

Thus, the intercept atx 1 1 is equal toY ̄ 1 (x′ 1 ).