Physical Chemistry Third Edition

262 6 The Thermodynamics of Solutions

Solution

a( 2 I)

48 .31 torr

59 .03 torr

 0. 8184

γ( 2 I)

0. 8184

0. 2748

 2. 978

Exercise 6.14

Find the value of the activity and the activity coefficient of ethanol in the solution of the previous

example according to convention I. The partial vapor pressure is 46.91 torr and the vapor pressure

of pure ethanol at 25◦C is equal to 49.31 torr.

Convention II

The mole fractions of all components are used as composition variables in convention II,

as in convention I. One substance, component number 1, is designated as the solvent

and is treated just as in convention I:

a(II) 1 a(I) 1 

P 1

P 1 ∗

(component 1solvent) (6.3-26)

γ 1 (II)γ 1 (I)

a( 1 I)

x 1



P 1 /P◦

P 1 (ideal)/P◦



P 1

P 1 ∗x 1

(component 1solvent) (6.3-27)

Solutes are treated differently from convention I. The standard state for a solute in

convention II is the same as that for a dilute solute: the hypothetical pure substance

that obeys Henry’s law. The chemical potential of the solute in its standard state is equal

to the chemical potential of an ideal gas at pressure equal toki, which is the same as

the Henry’s law standard state:

μ◦i(II)μ◦i(H)μ◦i(g)+RTln

(

ki

P◦

)

(componentisolute) (6.3-28)

The activity takes on the value that is needed to produce a version of Eq. (6.3-6):

μiμ◦i(II)+RTln(a(II)i ) (6.3-29)

Equation (6.3-28) is substituted into Eq. (6.3-29) and the chemical potential of sub-

stanceiin the solution is equated to its chemical potential in the gas phase, which is

assumed to be an ideal gas mixture. This gives

μ◦i(g)+RTln

(

ki

P◦

)

+RTln

(

a(II)i

)

μ◦i(g)+RTln

(

Pi

P◦

)

(6.3-30)

Cancellation and taking antilogarithms lead to

Pikia(II)i (componentisolute) (6.3-31)