Page 63 - Bruno Linder Elementary Physical Chemistry

August 18, 2010 11:36 9in x 6in b985-ch06 Elementary Physical Chemistry

48 Elementary Physical Chemistry

is called an ideal dilute solution. If Henry’s Law constant is equal to P ,
∗
B
then the ideal dilute solution becomes ideal.
Example 6.1. Methane (CH 4) and benzene (C 6 H 6 ) form an ideal dilute
solution. The pressure of pure benzene is P benz = 300 Torr at 298 K. Henry’s
4
Law constant of the solute (methane) is K meth =4.27 × 10 Torr at 298 K.
If the mole fraction of CH 4 in the liquid phase is x meth =1.01 × 10 −2 , find

(a) the partial pressure of CH 4 at 298 K;
(b) the partial pressure of C 6H 6;
(c) the mole fraction of CH 4 in the gaseous phase.

Solution

4
(a) By Henry’s Law, P meth = x methK meth =1.01×10 −2 ×4.27×10 Torr =
431.27Torr
(b) The solvent obeys Raoult’s Law

P benz = x benzP o =(1 − 1.01 × 10 −2 ) × 300 Torr = 296.97 Torr
benz

(c) Assuming ideal behavior in the gaseous phase,

x meth = P meth/(P met + P benz) = 431.27/(431.27 + 296.97) = 0.529

Example 6.2. Predict whether natural water can support life. It is known
that to support aquatic life, the concentration of O 2 must be 4 mg/L. What
must the partial pressure of O 2 in air be to achieve that concentration?

In a liter of water, there are 55.5 moles of H 2 O and a negligible amount
of O 2 .The O 2 molar fraction is

x O2 =(4 × 10 −3 gL −1 /32.00 gmol −1 )/55.5molL −1
=2 × 10 −6 (6.13)

7
Henry’s Law constant for O 2 is 3.79 × 10 Torr in water at 25 C. Thus,
◦
7
P O2 = x O2 K O2 =2×10 −6 ×3.60×10 Torr = 70 Torr. This is the minimum
pressure O 2 must have in the air. The actual pressure of O 2 in the air is