Physical Chemistry , 1st ed.

S 1 nRln 

p

p

f

i

(1.00 mol)8.314 

mo

J

lK

ln 

1

1

5

.5

.0

0

a

a

t

t

m

m



S 1 19.1 

K

J



For the second step, the isobaric step, we use equation 3.18:

S 2 Cln 

T

T

f

i

(1.00 mol)20.78 

mo

J

lK

ln 

1

2

0

9

0

8

.

.

0

0

K

K



S 2 22.7 

K

J



The overall change in entropy is the sum of the two, just as the overall process

is the combination of the two steps. We get S19.1 + (22.7) J/K 

41.8 J/K.

Consider the system illustrated in Figure 3.4a. A container is divided into

two systems having volumes V 1 and V 2 , but both systems have the same pres-

sure pand the same absolute temperature T. The number of moles of differ-

ent ideal gases in side 1 and side 2 are n 1 and n 2 , respectively. A barrier sepa-

rates the two sides. We will assume that the systems are isolated from the

surroundings so that qis zero for the following process (that is, it is adiabatic).

At some point, the barrier is removed while maintaining the overall pres-

sure and temperature. Since the process is adiabatic,q0. Since the tem-

perature is constant,U0 also. Therefore,w0. However, the two gases

mix so that our final system looks like Figure 3.4b: two mixed gases occupy-

ing the same volume. (This agrees with our conventional wisdom regarding

the behavior of gases: they expand to fill their container.) Since there is no

energy change to cause the mixing, then it must be entropy that is causing

the process.

Entropy is a state function, so the change in entropy is path-independent.

Consider that the mixing process can be broken down into two individual

steps, as illustrated in Figure 3.5. One process is the expansion of gas 1 from

V 1 to Vtot, and the other process is the expansion of gas 2 from V 2 to Vtot. Using

S 1 andS 2 to represent the changes in entropies for the steps, we have

S 1 n 1 Rln 

V

V

to

1

t

S 2 n 2 Rln 

V

V

to

2

t

Since Vtotis greater than V 1 or V 2 (because both gases are expanding), the log-

arithms of the volume fractions will always be positive. (Logarithms of num-

bers greater than 1 are positive.) The ideal gas law constant is always positive,

and the number of moles of each gas is also positive. Therefore, the individual

entropy changes will be positive overall, and the combination of the two com-

ponents to get Sfor the mixing process

SS 1 + S 2 (3.21)

will always be positive. Therefore, by the second law of thermodynamics, the

mixing of two (or more) gases is always a spontaneous process if it occurs in

an isolated system.

3.5 More on Entropy 77

Remove barrier:

gases mix

(a)

(b)

Vtot  V 1  V 2

ntot  n 1  n 2

V 1

n 1

V 2

n 2

T 1  T 2

p 1  p 2

p, T

Figure 3.4 The adiabatic mixing of two gases.
(a) On the left side is gas 1 with a certain volume
and amount, and on the right side is gas 2 with
its own volume and amount. (b) After mixing,
both gases occupy the complete volume. Since
there is no energy change to cause the gases to
mix, the mixing must have been caused by en-
tropy effects.