Additional examples, video, and practice available at WileyPLUS
588
CHAPTE R 20 ENTROPY AND THE SECOND LAW OF THERMODYNAM ICS
rium.) Then slowly lower the temperature of the reservoir
and the block to 40ЊC. As the block’s temperature changes
by each increment dT during this process, energy dQ is
transferred as heat from the block to the reservoir. Using
Eq. 18-14, we can write this transferred energy as dQ ϭ
mc dT, where c is the specific heat of copper. According to
Eq. 20-1, the entropy change ⌬S L of block L during the full
temperature change from initial temperature T iL (ϭ 60ЊC ϭ
333 K) to final temperature T f (ϭ 40ЊC ϭ 313 K) is
Inserting the given data yields
Step 2: With the reservoir’s temperature now set at 20 C,
Њ
ϭ Ϫ35.86 J/K.
⌬S L ϭ (1.5 kg)(386 J/kgиK) ln
313 K
333 K
ϭ mc ln
T f
T iL
.
⌬S L ϭ ͵
f
i
dQ
T
ϭ ͵
T f
T iL
mc dT
T
ϭ mc ͵
T f
T iL
dT
T
put block R on the reservoir. Then slowly raise the temperature of the reservoir and the block to 40ЊC. With the
same reasoning used to find ⌬S L , you can show that the
entropy change ⌬S R of block R during this process is
38.23 J/K.
The net entropy change ⌬S rev of the two-block system undergoing this two-step reversible process is then
⌬S rev ϭ ⌬S L ϩ ⌬S R
ϭ Ϫ35.86 J/K ϩ 38.23 J/K ϭ 2.4 J/K.
Thus, the net entropy change ⌬S irrev for the two-block system undergoing the actual irreversible process is
⌬S irrev ϭ ⌬S rev ϭ 2.4 J/K.
(Answer)
This result is positive, in accordance with the entropy
postulate.
ϭ ϩ
⌬S R ϭ (1.5 kg)(386 J/kgиK) ln
313 K
293 K
in which n is the number of moles of gas present. From
Eq. 20-2 the entropy change for this reversible process in
which the temperature is held constant is
Substituting n ϭ 1.00 mol and V f /V i ϭ 2, we find
ϭ ϩ5.76 J/K.
Thus, the entropy change for the free expansion (and for all
other processes that connect the initial and final states
shown in Fig. 20-2) is
⌬S irrev ϭ ⌬S rev ϭ ϩ5.76 J/K.
(Answer)
Because ⌬S is positive, the entropy increases, in accordance
with the entropy postulate.
⌬S rev ϭ nR ln
V f
V i
ϭ (1.00 mol)(8.31 J/mol иK)(ln 2)
⌬S rev ϭ
Q
T
ϭ
nRT ln(V f /V i )
T
ϭ nR ln
V f
V i
.
Sample Problem 20.02 Entropy change of a free expansion of a gas
Suppose 1.0 mol of nitrogen gas is confined to the left side
of the container of Fig. 20-1a. You open the stopcock, and
the volume of the gas doubles. What is the entropy change
of the gas for this irreversible process? Treat the gas as ideal.
KEY IDEAS
(1) We can determine the entropy change for the irreversible process by calculating it for a reversible process
that provides the same change in volume. (2) The temperature of the gas does not change in the free expansion. Thus,
the reversible process should be an isothermal expansion —
namely, the one of Figs. 20-3 and 20-4.
Calculations: From Table 19-4, the energy Q added as heat
to the gas as it expands isothermally at temperature T from
an initial volume V i to a final volume V f is
,
Q ϭ nRT ln
V f
V i
The Second Law of Thermodynamics
Here is a puzzle. In the process of going from (a) to (b) in Fig. 20-3, the entropy
change of the gas (our system) is positive. However, because the process is reversible, we can also go from (b) to (a) by, say, gradually adding lead shot to the
piston, to restore the initial gas volume. To maintain a constant temperature, we
need to remove energy as heat, but that means Q is negative and thus the entropy
change is also. Doesn’t this entropy decrease violate the entropy postulate: en-
588
CHAPTE R 20 ENTROPY AND THE SECOND LAW OF THERMODYNAM ICS
rium.) Then slowly lower the temperature of the reservoir
and the block to 40ЊC. As the block’s temperature changes
by each increment dT during this process, energy dQ is
transferred as heat from the block to the reservoir. Using
Eq. 18-14, we can write this transferred energy as dQ ϭ
mc dT, where c is the specific heat of copper. According to
Eq. 20-1, the entropy change ⌬S L of block L during the full
temperature change from initial temperature T iL (ϭ 60ЊC ϭ
333 K) to final temperature T f (ϭ 40ЊC ϭ 313 K) is
Inserting the given data yields
Step 2: With the reservoir’s temperature now set at 20 C,
Њ
ϭ Ϫ35.86 J/K.
⌬S L ϭ (1.5 kg)(386 J/kgиK) ln
313 K
333 K
ϭ mc ln
T f
T iL
.
⌬S L ϭ ͵
f
i
dQ
T
ϭ ͵
T f
T iL
mc dT
T
ϭ mc ͵
T f
T iL
dT
T
put block R on the reservoir. Then slowly raise the temperature of the reservoir and the block to 40ЊC. With the
same reasoning used to find ⌬S L , you can show that the
entropy change ⌬S R of block R during this process is
38.23 J/K.
The net entropy change ⌬S rev of the two-block system undergoing this two-step reversible process is then
⌬S rev ϭ ⌬S L ϩ ⌬S R
ϭ Ϫ35.86 J/K ϩ 38.23 J/K ϭ 2.4 J/K.
Thus, the net entropy change ⌬S irrev for the two-block system undergoing the actual irreversible process is
⌬S irrev ϭ ⌬S rev ϭ 2.4 J/K.
(Answer)
This result is positive, in accordance with the entropy
postulate.
ϭ ϩ
⌬S R ϭ (1.5 kg)(386 J/kgиK) ln
313 K
293 K
in which n is the number of moles of gas present. From
Eq. 20-2 the entropy change for this reversible process in
which the temperature is held constant is
Substituting n ϭ 1.00 mol and V f /V i ϭ 2, we find
ϭ ϩ5.76 J/K.
Thus, the entropy change for the free expansion (and for all
other processes that connect the initial and final states
shown in Fig. 20-2) is
⌬S irrev ϭ ⌬S rev ϭ ϩ5.76 J/K.
(Answer)
Because ⌬S is positive, the entropy increases, in accordance
with the entropy postulate.
⌬S rev ϭ nR ln
V f
V i
ϭ (1.00 mol)(8.31 J/mol иK)(ln 2)
⌬S rev ϭ
Q
T
ϭ
nRT ln(V f /V i )
T
ϭ nR ln
V f
V i
.
Sample Problem 20.02 Entropy change of a free expansion of a gas
Suppose 1.0 mol of nitrogen gas is confined to the left side
of the container of Fig. 20-1a. You open the stopcock, and
the volume of the gas doubles. What is the entropy change
of the gas for this irreversible process? Treat the gas as ideal.
KEY IDEAS
(1) We can determine the entropy change for the irreversible process by calculating it for a reversible process
that provides the same change in volume. (2) The temperature of the gas does not change in the free expansion. Thus,
the reversible process should be an isothermal expansion —
namely, the one of Figs. 20-3 and 20-4.
Calculations: From Table 19-4, the energy Q added as heat
to the gas as it expands isothermally at temperature T from
an initial volume V i to a final volume V f is
,
Q ϭ nRT ln
V f
V i
The Second Law of Thermodynamics
Here is a puzzle. In the process of going from (a) to (b) in Fig. 20-3, the entropy
change of the gas (our system) is positive. However, because the process is reversible, we can also go from (b) to (a) by, say, gradually adding lead shot to the
piston, to restore the initial gas volume. To maintain a constant temperature, we
need to remove energy as heat, but that means Q is negative and thus the entropy
change is also. Doesn’t this entropy decrease violate the entropy postulate: en-