PHYSICAL CHEMISTRY IN BRIEF

CHAP. 6: THERMODYNAMICS OF HOMOGENEOUS MIXTURES [CONTENTS] 168

Example

The activity coefficient of ethanol in an ethanol (1)—water (2) mixture containing 17 mole

percent ethanol at temperature 298.15 K isγ 1 = 2.485. The differential heat of solution of

ethanol in this mixture is−783 J mol/^1. Estimate the activity coefficient of ethanol at the given

composition and temperature 303.15 K. Assume independence ofH

E

i on temperature.

Solution

We integrate (6.106) to obtain

ln

γ 1 (T 2 )

γ 1 (T 1 )

= −

H

E

1

R

(

1

T 1

−

1

T 2

)

=

= −

− 783

8. 314

(

1

298. 15

−

1

303. 15

)

= 0. 00521.

Hencelnγ 1 = ln 2.485 + 0.00521 = 0. 9155 andγ 1 = 2. 498.

Thedependence of the activity coefficient on pressureis expressed by the relation

lnγi(p 2 ) = lnγi(p 1 ) +

∫p 2

p 1

(

∂lnγi

∂p

)

T,n

dp , (6.107)

where (
∂lnγi
∂p

)

T,n

=

Vi−Vist

RT

. (6.108)

Note: The dependence of the activity coefficient on composition is dealt with in section

6.5.6.

6.5.5.1 Relation betweenγ[ix]andγi

The relation between the activity coefficientsγi[x]andγiis

lnγi[x]= lnγi−lnγi∞, (6.109)