Ideal Solution

7. Ideal Solution

a. Volume

A liquid mixture in which all of the components are miscible at the molecular level is called an ideal solution, provided the following condition is satisfied, i.e.,

(59) In an ideal solution, the forces between the unlike molecules are assumed to be the same as

ˆv id

id

k = v k , and v = Σ k X k v k = Σ k X k v k .

those between like molecules.

b. Internal Energy and Enthalpy In ideal mixture of liquids or real gases,

h id = Σ

k X k h k , and

id id

u id = h – P v = Σ

k X k ( h k –P v k ).

c. Gibbs Function At a specified temperature, the change in the Gibbs free energy of a pure component (i.e. when it is alone at T and P) is given as (Chapter 7)

d g k = v k (T, P) dP, i.e., (a) If the composition and temperature in a mixture are held fixed and the pres-

sure is altered then applying Eq.(31) for k in a mixture

d ˆg k = ˆv k (T, P, X k ) dP (62a) Then from Eqs. (a) and (62a) d( ˆg k – g k )=( ˆv k (T, P, X k )– v k (T, P))dP.

(62b) If a mixture of fixed composition is subjected to an incremental pressure dP, and the pure

component is also subjected to the same pressure increment, this expression provides the dif- ference between two Gibbs function ˆg k and g k due to difference in ˆv k and v k . For example,

H 2 O in the solution is compressed from 200 to 250 kPa, with ˆv HO 2 = 0.015 m 3 kmole –1 , dg ˆ HO

and d g H2O = 0.9 kJkmole . Integrating

2 = 0.75 kJ kmole -1 . On the other hand,

v H2O = 0.018 m kmole

3 -1

ˆg k – g k = ∫( ˆv k – v k )dP + f(T, X k ). In an ideal solution at any temperature and pressure, ˆv k = v k and hence

– g k = f(T, X k ). As P →0 at a specified temperature and composition the same relation should hold good.

ˆg id

Hence id ˆg

and g k approach their corresponding values in an ideal gas. Therefore, f(T, X k )= R T ln X k . consequently, in an ideal solution ˆg id

= g k (T, P) + R T ln X k or ˆg k id - g k (T, P) = R T ln X k . (62c)

d. Entropy Since,

= ˆh k –T ˆs k (T, P, X k )= ˆh k –T s k (T, P) + R T ln X k i.e., then ˆs k (T, P, X k )= s k (T, P)– R ln X k .

ˆg id

(63) The entropy of the ideal mixture is S id (T, P, N) =

(T, P, X )N , and s Σ id k s k k k =S id /N = Σ k s k (T, P, X k )X k . (64)

g. Example 8 Lake water at 25ºC and 1 bar absorbs air from the atmosphere. If the air mole fraction in the liquid is 0.001, what is the entropy of H 2 O in the lake water?

Solution Air consists of O 2 and N 2 molecules that have weak attractive forces between them- selves in the atmosphere. However once these molecules enter liquid water, they are surrounded by H 2 O molecules which exert strong attractive forces and hold the gas molecules in the liquid phase. Thereby, small amounts of air become dissolved in liq- uid water.

ˆs HO 2 (25ºC, 1 bar, 0.999) = s HO 2 (25ºC, 1 bar) – R ln X HO 2 .

Pure water exists at 25ºC and 1 bar as a compressed liquid. We will assume that the liquid is incompressible and that s HO 2 (25ºC, P sat = 0.032 bar) = s HO 2 (25ºC, 1 bar) =

6.621 kJ kmole –1 K . Therefore, ˆs –1

HO 2 (25ºC, 1 bar, 0.999) = 6.621 – 8.314 × ln 0.999 = 6.629 kJ kmole .