Concise Physical Chemistry

c12 JWBS043-Rogers September 13, 2010 11:27 Printer Name: Yet to Come

VAPOR PRESSURE 187

found for component B dissolved in solvent A is true if we change the AB ratio and

start calling A the solute and B the solvent (on the right-hand side of Fig. 12.3).

Therefore we can obtain Henry’s law activities of both components, one from the

partial pressure of very dilute solutions of A in B and the other from very dilute

solutions of B in A.

12.6 VAPOR PRESSURE

Henry’s original (1803) paper dealt with the solubility of CO 2 in water, for which he

found that the partial pressure ofpCO 2 confined in a closed container over water is

directly proportional to the amount of CO 2 dissolved in the water. It was many years

later that his lawpCO 2 =kXCO 2 whereXCO 2 is the mole fraction of CO 2 dissolved

in water, was extended to the binary solutions of mutually miscible components just

discussed.

Let us go back now to Henry’s point of view and look atpCO 2 =kXCO 2 as a rela-

tionship telling us what the partial pressure of CO 2 , which we shall callcomponent 2,

will be over a solution that we have made up to have a mole fraction ofXCO 2. Notice

that we are looking at the same problem as Henry but from the opposite side. He was

thinking about the amount of gas dissolved in a solvent (water), and we are looking

at the pressure of solute leaving a solution that we have made up to some value ofX 2

and going into the gas phase above the solution. At equilibrium, it’s the same thing:

Component 2(solution)→←Component 2(g)

The critical condition for all the relatedcolligative properties(Section 12.8) is

a balance between the Gibbs free energy functions of two systems—for example,

gas-phase CO 2 and dissolved CO 2. The Gibbs free energy relationship at equilibrium

between dissolved component 2 and gas phase 2 is

μ 2 (sol)=μ 2 (g)

whereμ 2 (sol) designates the Gibbs free energy function of component 2 in solution.

We know from Chapters 6 and 7 that, for an ideal gas at constant temperature, we

have

dG=Vdp=

RT

p

dp

Integrating over the difference between the Gibbs function in an arbitrary state 2 and

a standard state pressure arbitrarily chosen as 1 bar, we get

G 2 =

∫Gp

2

G1bar

dG=RT

∫Gp

2

G1bar

dp

p

=RTln

p 2

1