Physical Chemistry , 1st ed.

particular phase at a specified temperature and standard pressure is repre-

sented by °i. In the last chapter, we found that









p

i

TVi

where Viis the molar volume of the ith material. We rearrange this into

diVidp

The differential of equation 5.11 at constant temperature is

diRT (dln ai)

Combining the last two equations and solving for dln ai:

dln ai

V

R

i

T

dp



Integrating both sides from the standard state ofai1 and p1:



1

a

dln ai

1

p



V

R

i

T

dp



ln ai

R

1

T



1

p

Vidp

If the molar volume Viis constant over the pressure interval (and it usually

is to a good approximation unless the pressure changes are severe), this inte-

grates to

ln ai

R

V

T

i(p1) (5.14)

Example 5.7

Determine the activity of liquid water at 25.0°C and 100 bar pressure. The

molar volume of H 2 O at this temperature is 18.07 cm^3.

Solution

Using equation 5.14, we set up the following:

ln ai (100 bar 1 bar)

A conversion factor between liters and cubic centimeters is included in the

numerator. Solving:

ln ai0.0722

ai1.07

Notice that the activity of the liquid is close to 1, even at a pressure that is

100 times that of standard pressure. This is generally true for condensed phases

at pressures that are typically found in chemical environments. Therefore, in

most cases the activities of condensed phases can be approximated as 1and

they make no numerical contribution to the value of the reaction quotient or

18.07 

c

m

m

o

3

l



100

1

0

L

cm^3





0.08314 m

L

o

b

l

a

K

r

(298 K)

130 CHAPTER 5 Introduction to Chemical Equilibrium