Physical Chemistry Third Edition

3.3 The Calculation of Entropy Changes 131

wherenis the total amount of all gases,

n

∑s

j 1

nj (3.3-18)

From Eq. (3.3-13) we have

xi

ni

n



PVi/RT

PV /RT



V

Vi

(3.3-19)

so that the entropy change on mixing the ideal gases is

∆SmixR

∑s

i 1

niln(1/xi)−R

∑s

i 1

niln(xi) (3.3-20)

This equation applies to the mixing of substances in other kinds of systems besides

ideal gases, if the mixing is random. It can be used to calculate the entropy change of

mixing of isotopes of a single element. In a later chapter we will show that it can also

be used for the entropy of change of mixing of a class of liquid or solid solutions called

ideal solutions.

EXAMPLE3.12

Assume that 1.000 mol of dry air consists of 0.780 mol of nitrogen, 0.210 mol of oxygen, and

0.010 mol of argon. Find the entropy change of mixing of 1.000 mol of dry air. Disregard the

fact that each substance has more than one isotope.

Solution

∆S−(8.3145 J K−^1 mol−^1 )

[

(0.780 mol) ln(0.780)

+(0.21 mol) ln(0.210)+(0.01 mol) ln(0.01)

]

 4 .72JK−^1

Exercise 3.13

a.Find the entropy change of mixing for 1.000 mol of the normal mixture of bromine atoms,

with 50.69%^79 Br and 49.31%^81 Br.

b.Find the entropy change of mixing in 0.500 mol of naturally occurring Br 2. Note that there

are three kinds of Br 2 molecules if there are two isotopes.

PROBLEMS

Section 3.3: The Calculation of Entropy Changes

3.13 a.Calculate∆Sfor each step of the following cycle and
sum the results to obtain∆Sfor the cycle: Step 1:
1.000 mol of helium is expanded reversibly and

isothermally at 298.15 K from 10.00 L to 20.00 L.

Step 2: The gas is heated reversibly at a constant

volume from 298.15 K and 20.00 L to a temperature of

473.15 K. Step 3: The gas is compressed reversibly and

isothermally at 473.15 K from 20.00 L to 15.00 L.