Physical Chemistry Third Edition

256 6 The Thermodynamics of Solutions

EXAMPLE 6.9

At 25◦C, water at equilibrium with air at 1.000 atm contains about 8.3 ppm (parts per million)

of dissolved oxygen by mass. Compute the Henry’s law constant.

Solution

The mole fraction of oxygen in air saturated with water vapor at 25◦C is 0.203. The mole

fraction of oxygen in the water phase is

x 2 

(

8 .3g

32 .0 g mol−^1

)(

18 .0 g mol−^1

1 × 106 g

)

 4. 7 × 10 −^6

The Henry’s law constant is

k 2 

0 .203 atm

4. 7 × 10 −^6

 4. 3 × 104 atm

 4. 3 × 104 atm

(

101325 Pa

1 atm

)

 4. 4 × 109 Pa

The Solvent in a Dilute Solution Obeys Raoult’s Law

We now show from thermodynamics that if the solute in a two-component solution

obeys Henry’s law for some dilute range of composition, the solvent obeys Raoult’s

law over this range of composition. We begin with the Gibbs–Duhem relation for

constant pressure and temperature:

x 1

(

∂μ 1

∂x 2

)

T,P

+x 2

(

∂μ 2

∂x 2

)

T,P

 0 (6.2-20)

The vapor phase must have an additional gas or gases (such as air) present to keep

the pressure constant if the mole fractions are changed. The small amount of air that

dissolves in the solution can be ignored, and the small effect of the change in pressure

on the vapor pressures is negligible (see Example 5.7).

Assume that component 2 obeys Henry’s law over the range of composition from

x 2 0tox 2 x′ 2. In this range

(

∂μ 2

∂x 2

)

T,P

RT

(

dln(x 2 )

dx 2

)



RT

x 2

(6.2-21)

Since x 1 +x 2 1,dx 2 −dx 1 , and∂μ 1 /∂x 1 −∂μ 1 /∂x 2. Using this fact and

Eq. (6.2-20), we obtain

−x 1

(

∂μ 1

∂x 2

)

T,P

x 1

(

∂μ 1

∂x 1

)

T,P

x 2

(

∂μ 2

∂x 2

)

T,P



x 2 RT

x 2

RT (6.2-22)

Dividing this equation byx 1 and multiplying bydx 1 , we obtain

(

∂μ 1

∂x 1

)

T,P

dx 1 

(

RT

x 1

)

dx 1 (6.2-23)

Integrating this equation fromx 1 1tox 1 x′ 1  1 −x′ 2 , we obtain

μ 1 (x′ 1 )−μ 1 (1)RTln(x′ 1 )−ln(1) (6.2-24a)