# 5.24: Untitled Page 121

- Page ID
- 18254

## Chapter 5

5‐31. In the previous problem we made use of a Taylor series expansion to obtain a simplified expression for the chemical potential as a function of the mole fraction. A Taylor series expansion is a powerful tool for predicting the value of a function at some position *x * *b * when information about the function is *only* available at *x * *a *. If we think about the definite integral given by *x * *b*

*f *( *b*)

*f *( *a*)

*df dx*

*dx *

(1)

*x * *a*

we see that we can determine *f *( *b*) , given *f *( *a*) , only if we know the derivative, *df / dx * *everywhere* between *a* and *b*. To see how we can use Eq. 1 to determine *f *( *b*) using *only* information at *x * *a *, we first make use of the change of variable, or *transformation*, defined by

*x * *b *

(2)

This leads to the relations

*dx * *d* *, *

*df dx * *df d*

(3)

which can be used to express Eq. 1 in the form

0

*f *( *b*)

*f *( *a*)

*df d *

*d*

(4)

*a * *b*

Remember that the rule for differentiating a product

*d*

*dV*

*dU*

*UV* *U*

*V*

*d*

*d*

*d*

(5)

is the basis for the technique known as *integration by parts* where it is employed in the form

*dV*

*d*

*dU*

*U*

*UV*

*d*

*d*

*V d*

(6)

If we make use of the two representations

*df*

*U *

*, *

*V*

*d*

(7)

Eq. 6 provides the following identity for the derivative of *f* with respect to :

*Two‐Phase Systems* & *Equilibrium Stages *

223

*df*

*d * *df *

*d * *df *

(8)

*d*

*d* *d*

*d*

*d*

Use this result in Eq. 4 to produce the second term in the Taylor series expansion for *f *( *b*) about *f *( *a*) . Repeat this entire procedure to extend the representation for *f *( *b*) to obtain the third term in the Taylor series expansion and thus illustrate how one obtains a representation for *f *( *b*) that involves only the function and its derivatives evaluated at *x * *a *. Keep in mind that the only mathematical tools used in this derivation are the definition of the definite integral and the rule for differentiating a product (Stein and Barcellos, 1992, pages 134 and 266).

5‐32. A gas mixture leaves a solvent recovery unit as illustrated in Figure 5.32.

*Figure 5.32*. Recovery‐condenser system

The partial pressure of benzene in this stream is 80 mm Hg and the total pressure is 750 mm Hg. The volumetric analysis of the gas, on a benzene‐free basis, is 15%

CO , 4% O and the remainder is nitrogen. This gas is compressed to 5 atm and 2

2

cooled to 100 F. Calculate the percentage of benzene condensed in the process.

Assume that CO , O and N are insoluble in benzene, thus the liquid phase is 2

2

2

pure benzene.