Chapter 19
The Second Law of Thermodynamics
31 ••
[SSM] The working substance of an engine is 1.00 mol of a
monatomic ideal gas. The cycle begins at P1 = 1.00 atm and V1 = 24.6 L. The gas
is heated at constant volume to P2 = 2.00 atm. It then expands at constant pressure
until its volume is 49.2 L. The gas is then cooled at constant volume until its
pressure is again 1.00 atm. It is then compressed at constant pressure to its
original state. All the steps are quasi-static and reversible. (a) Show this cycle on
a PV diagram. For each step of the cycle, find the work done by the gas, the heat
absorbed by the gas, and the change in the internal energy of the gas. (b) Find the
efficiency of the cycle.
Picture the Problem To find the heat added during each step we need to find the
temperatures in states 1, 2, 3, and 4. We can then find the work done on the gas
during each process from the area under each straight-line segment and the heat
that enters the system from Q = CV ΔT and Q = CP ΔT . We can use the 1st law of
thermodynamics to find the change in internal energy for each step of the cycle.
Finally, we can find the efficiency of the cycle from the work done each cycle
and the heat that enters the system each cycle.
(a) The cycle is shown to the right:
Apply the ideal-gas law to state 1 to find T1:
T1 =
P1V1
=
nR
(1.00 atm )(24.6 L )
(1.00 mol)⎛⎜ 8.206 ×10 − 2 L ⋅ atm ⎞⎟
mol ⋅ K ⎠
⎝
The pressure doubles while the
volume remains constant between
states 1 and 2. Hence:
T2 = 2T1 = 600 K
The volume doubles while the
pressure remains constant between
states 2 and 3. Hence:
T3 = 2T2 = 1200 K
1867
= 300 K
1868 Chapter 19
T4 = 12 T3 = 600 K
The pressure is halved while the
volume remains constant
between states 3 and 4. Hence:
For path 1→2:
W12 = PΔV12 = 0
and
J ⎞
⎛
Q12 = C V ΔT12 = 32 RΔT12 = 32 ⎜ 8.314
⎟ (600 K − 300 K ) = 3.74 kJ
mol ⋅ K ⎠
⎝
The change in the internal energy of
the system as it goes from state 1 to
state 2 is given by the 1st law of
thermodynamics:
ΔEint = Qin + Won
Because W12 = 0 :
ΔEint,12 = Q12 = 3.74 kJ
For path 2→3:
⎛ 101.325 J ⎞
Won = −W23 = − PΔV23 = −(2.00 atm )(49.2 L − 24.6 L )⎜
⎟ = − 4.99 kJ
⎝ L ⋅ atm ⎠
J ⎞
⎛
Q23 = C P ΔT23 = 52 RΔT23 = 52 ⎜ 8.314
⎟ (1200 K − 600 K ) = 12.5 kJ
mol ⋅ K ⎠
⎝
Apply ΔEint = Qin + Won to obtain:
ΔEint, 23 = 12.5 kJ − 4.99 kJ = 7.5 kJ
For path 3→4:
W34 = PΔV34 = 0
and
J ⎞
⎛
Q34 = ΔEint,34 = C V ΔT34 = 32 RΔT34 = 32 ⎜ 8.314
⎟ (600 K − 1200 K ) = − 7.48 kJ
mol ⋅ K ⎠
⎝
Apply ΔEint = Qin + Won to obtain:
For path 4→1:
ΔEint, 34 = −7.48 kJ + 0 = − 7.48 kJ
The Second Law of Thermodynamics 1869
⎛ 101.325 J ⎞
Won = −W41 = − PΔV41 = −(1.00 atm )(24.6 L − 49.2 L )⎜
⎟ = 2.49 kJ
⎝ L ⋅ atm ⎠
and
J ⎞
⎛
Q41 = C P ΔT41 = 52 RΔT41 = 52 ⎜ 8.314
⎟ (300 K − 600 K ) = − 6.24 kJ
mol ⋅ K ⎠
⎝
Apply ΔEint = Qin + Won to obtain:
ΔEint, 41 = −6.24 kJ + 2.49 kJ = − 3.75 kJ
For easy reference, the results of the preceding calculations are summarized in the
following table:
Process Won , kJ Qin , kJ ΔEint (= Qin + Won ) , kJ
1→2
2→3
3→4
4→1
0
−4.99
0
2.49
3.74
12.5
−7.48
−6.24
(b) The efficiency of the cycle is
given by:
ε=
Substitute numerical values and
evaluate ε:
ε=
3.74
7.5
−7.48
−3.75
Wby
Qin
=
− W23 + (− W41 )
Q12 + Q23
4.99 kJ − 2.49 kJ
≈ 15%
3.74 kJ + 12.5 kJ
Remarks: Note that the work done per cycle is the area bounded by the
rectangular path. Note also that, as expected because the system returns to its
initial state, the sum of the changes in the internal energy for the cycle is zero.
33 ••
An engine using 1.00 mol of an ideal gas initially at a volume of
24.6 L and a temperature of 400 K performs a cycle consisting of four steps: (1)
an isothermal expansion at 400 K to twice its initial volume, (2) cooling at
constant volume to a temperature of 300 K (3) an isothermal compression to its
original volume, and (4) heating at constant volume to its original temperature of
400 K. Assume that Cv = 21.0 J/K. Sketch the cycle on a PV diagram and find its
efficiency.
Picture the Problem We can find the efficiency of the cycle by finding the work
done by the gas and the heat that enters the system per cycle.
1870 Chapter 19
2
P (atm)
The PV diagram of the cycle is
shown to the right. A, B, C, and D
identify the four states of the gas and
the numerals 1, 2, 3, and 4 represent
the four steps through which the gas
is taken.
A
1.5
4
1
1
D
3
0.5
0
C
0
10
20
30 40
V (L)
B
2
50
Express the efficiency of the cycle:
ε=
W1 + W2 + W3 + W4
W
=
Qh Qh,1 + Qh, 2 + Qh,3 + Qh, 4
Because steps 2 and 4 are constantvolume processes, W2 = W4 = 0:
ε=
W1 + 0 + W3 + 0
W
=
Qh Qh,1 + Qh, 2 + Qh,3 + Qh, 4
Because the internal energy of the
gas increases in step 4 while no work
is done, and because the internal
energy does not change during step 1
while work is done by the gas, heat
enters the system only during these
processes:
ε=
W + W3
W
= 1
Qh Qh,1 + Qh, 4
The work done during the isothermal
expansion (1) is given by:
⎛V
W1 = nRT ln⎜⎜ B
⎝ VA
The work done during the isothermal
compression (3) is given by:
⎛V ⎞
W3 = nRTc ln⎜⎜ D ⎟⎟
⎝ VC ⎠
Because there is no change in the
internal energy of the system during
step 1, the heat that enters the system
during this isothermal expansion is
given by:
⎛V
Q1 = W1 = nRTh ln⎜⎜ B
⎝ VA
The heat that enters the system
during the constant-volume step 4 is
given by:
(1)
⎞
⎟⎟
⎠
⎞
⎟⎟
⎠
Q4 = C V ΔT = C V (Th − Tc )
400 K
300 K
60
The Second Law of Thermodynamics 1871
Substituting in equation (1) yields:
Noting the
⎛V ⎞
⎛V ⎞
nRTh ln⎜⎜ B ⎟⎟ + nRTc ln⎜⎜ D ⎟⎟
⎝ VA ⎠
⎝ VC ⎠
ε=
⎛V ⎞
nRTh ln⎜⎜ B ⎟⎟ + C V (Th − Tc )
⎝ VA ⎠
VB
V
1
= 2 and D = , substitute and simplify to obtain:
VA
VC 2
⎛1⎞
Th ln (2) + Tc ln⎜ ⎟
Th − Tc
⎝ 2 ⎠ = Th ln (2) − Tc ln (2) =
ε=
C
C
CV
(Th − Tc )
Th ln (2) + V (Th − Tc ) Th ln (2) + V (Th − Tc ) Th +
nR
nR
nR ln (2)
Substitute numerical values and evaluate ε:
ε=
400 K − 300 K
= 13.1%
J
21.0
K
(400 K − 300 K )
400 K +
J ⎞
⎛
(1.00 mol)⎜ 8.314
⎟ ln (2)
mol ⋅ K ⎠
⎝
37 ••
″As far as we know, Nature has never evolved a heat engine″—Steven
Vogel, Life’s Devices, Princeton University Press (1988). (a) Calculate the
efficiency of a heat engine operating between body temperature (98.6ºF) and a
typical outdoor temperature (70ºF), and compare this to the human body’s
efficiency for converting chemical energy into work (approximately 20%). Does
this efficiency comparison contradict the second law of thermodynamics?
(b) From the result of Part (a), and a general knowledge of the conditions under
which most warm-blooded organisms exist, give a reason why no warm-blooded
organisms have evolved heat engines to increase their internal energies.
Picture the Problem We can use the efficiency of a Carnot engine operating
between reservoirs at body temperature and typical outdoor temperatures to find
an upper limit on the efficiency of an engine operating between these
temperatures.
(a) Express the maximum efficiency
of an engine operating between body
temperature and 70°F:
Use T = 59 (tF − 32) + 273 to obtain:
εC = 1 −
Tc
Th
Tbody = 310 K and Troom = 294 K
1872 Chapter 19
Substitute numerical values and
evaluate ε C :
ε C = 1−
294 K
= 5.16%
310 K
The fact that this efficiency is considerably less than the actual efficiency of a
human body does not contradict the second law of thermodynamics. The
application of the second law to chemical reactions such as the ones that supply
the body with energy have not been discussed in the text but we can note that we
don’t get our energy from heat swapping between our body and the environment.
Rather, we eat food to get the energy that we need.
(b) Most warm-blooded animals survive under roughly the same conditions as
humans. To make a heat engine work with appreciable efficiency, internal body
temperatures would have to be maintained at an unreasonably high level.
51 •
A refrigerator is rated at 370 W. (a) What is the maximum amount of
heat it can absorb for the food compartment in 1.00 min if the temperature in the
compartment is 0.0ºC and it releases heat into a room at 35ºC? (b) If the COP of
the refrigerator is 70% of that of a reversible pump, how much heat can it absorb
from the food compartment in 1.00 min? Is the COP for the refrigerator greater
when the temperature of the room is 35ºC or 20ºC? Explain.
Picture the Problem We can use the definition of the COP to relate the heat
removed from the refrigerator to its power rating and operating time. By
expressing the COP in terms of Tc and Th we can write the amount of heat
removed from the refrigerator as a function of Tc, Th, P, and Δt.
(a) Express the amount of heat the
refrigerator can remove in a given
period of time as a function of its
COP:
Express the COP in terms of Th
and Tc and simplify to obtain:
Qc = (COP )W
= (COP )PΔt
COP =
=
=
Substituting for COP yields:
Qc
Q
Q −W
= c = h
εQh
W εQh
1− ε
ε
=
1
ε
−1
Tc
1
−1 =
T
Th − Tc
1− c
Th
⎛ Tc ⎞
⎟⎟ PΔt
Qc = ⎜⎜
T
T
−
⎝ h c⎠
The Second Law of Thermodynamics 1873
Substitute numerical values and evaluate Qc:
273 K
60 s ⎞
⎞
⎛
⎛
Qc = ⎜
⎟ = 173 kJ = 0.17 MJ
⎟ (370 W )⎜1.00 min ×
min ⎠
⎝
⎝ 308 K − 273 K ⎠
(b) If the COP is 70% of the
efficiency of an ideal pump:
Qc' = (0.70 )(173 kJ ) = 0.12 MJ
Because the temperature difference increases when the room is warmer, the COP
decreases.
Consider the freezing of 50.0 g of water once it is placed in the freezer
55 •
compartment of a refrigerator. Assume the walls of the freezer are maintained at
–10ºC. The water, initially liquid at 0.0ºC, is frozen into ice and cooled to
–10ºC. Show that even though the entropy of the water decreases, the net entropy
of the universe increases.
Picture the Problem The change in the entropy of the universe resulting from the
freezing of this water and the cooling of the ice formed is the sum of the entropy
changes of the water-ice and the freezer. Note that, while the entropy of the water
decreases, the entropy of the freezer increases.
The change in entropy of the
universe resulting from this freezing
and cooling process is given by:
ΔS u = ΔS water + ΔS freezer
(1)
Express ΔS water :
ΔS water = ΔS freezing + ΔS cooling
(2)
Express ΔS freezing :
ΔS freezing =
− Qfreezing
Tfreezing
(3)
where the minus sign is a consequence
of the fact that heat is leaving the water
as it freezes.
Relate Qfreezing to the latent heat of
Qfreezing = mLf
fusion and the mass of the water:
Substitute in equation (3) to obtain:
ΔS freezing =
− mLf
Tfreezing
1874 Chapter 19
Express ΔS cooling :
⎛T
ΔS cooling = mC p ln⎜⎜ f
⎝ Ti
Substitute in equation (2) to obtain:
Noting that the freezer gains heat
(at 263 K) from the freezing
water and cooling ice, express
ΔS freezer :
ΔS water =
⎞
⎟⎟
⎠
⎛T
− mLf
+ mCp ln⎜⎜ f
Tfreezing
⎝ Ti
ΔS freezer =
=
ΔQice ΔQcooling ice
+
Tfreezer
Tfreezer
mC p ΔT
mLf
+
Tfreezer
Tfreezer
Substitute for ΔS water and ΔS freezer in equation (1):
ΔS u =
⎛T
− mLf
+ mCp ln⎜⎜ f
Tfreezing
⎝ Ti
mCp ΔT
⎞ mLf
⎟⎟ +
+
Tfreezer
⎠ Tfreezer
⎡ − Lf
⎛T
= m⎢
+ C p ln⎜⎜ f
⎝ Ti
⎣⎢ Tfreezing
⎞ Lf + C p ΔT ⎤
⎟⎟ +
⎥
Tfreezer ⎦⎥
⎠
⎞
⎟⎟
⎠
The Second Law of Thermodynamics 1875
Substitute numerical values and evaluate ΔSu:
⎡
3 J
⎢ 333.5 ×10 kg ⎛
J ⎞ ⎛ 263 K ⎞
⎟⎟
⎟⎟ ln⎜⎜
+ ⎜⎜ 2100
ΔS u = (0.0500 kg ) ⎢−
⋅
273
K
kg
K
273 K
⎢
⎝
⎠ ⎝
⎠
⎢⎣
⎤
J ⎞
J ⎛
⎟⎟ (273 K − 263 K ) ⎥
+ ⎜⎜ 2100
333.5 ×10 3
kg ⋅ K ⎠
kg ⎝
⎥ = 2.40 J/K
+
⎥
263 K
⎥
⎦
and, because ΔSu > 0, the entropy of the universe increases.
57 ••
[SSM] A system completes a cycle consisting of six quasi-static
steps, during which the total work done by the system is 100 J. During step 1 the
system absorbs 300 J of heat from a reservoir at 300 K, during step 3 the system
absorbs 200 J of heat from a reservoir at 400 K, and during step 5 it absorbs heat
from a reservoir at temperature T3. (During steps 2, 4 and 6 the system undergoes
adiabatic processes in which the temperature of the system changes from one
reservoir’s temperature to that of the next.) (a) What is the entropy change of the
system for the complete cycle? (b) If the cycle is reversible, what is the
temperature T3?
Picture the Problem We can use the fact that the system returns to its original
state to find the entropy change for the complete cycle. Because the entropy
change for the complete cycle is the sum of the entropy changes for each process,
we can find the temperature T3 from the entropy changes during the 1st two
processes and the heat released during the third.
(a) Because S is a state function of
the system, and because the system’s
final state is identical to its initial
state:
ΔSsystem
(b) Relate the entropy changes for
each of the three heat reservoirs and
the system for one complete cycle of
the system:
ΔS1 + ΔS 2 + ΔS 3 + ΔS system = 0
Substitute numerical values. Heat is
rejected by the two high-temperature
= 0
1 complete cycle
or
Q1 Q2 Q3
+
+
+0 = 0
T1 T2 T3
− 300 J − 200 J 400 J
+
+
=0
300 K 400 K
T3
1876 Chapter 19
reservoirs and absorbed by the cold
reservoir:
Solving for T3 yields:
T3 = 267 K
58 ••
In this problem, 2.00 mol of an ideal gas initially has a temperature of
400 K and a volume of 40.0 L. The gas undergoes a free adiabatic expansion to
twice its initial volume. What is (a) the entropy change of the gas and (b) the
entropy change of the universe?
Picture the Problem The initial and final temperatures are the same for a free
expansion of an ideal gas. Thus, the entropy change ΔS for a free expansion from
Vi to Vf is the same as ΔS for an isothermal process from Vi to Vf. We can use the
definition of entropy change and the 1st law of thermodynamics to express ΔS for
the ideal gas as a function of its initial and final volumes.
(a) The entropy change of the gas is
given by:
ΔS gas =
Apply the first law of thermodynamics
to the isothermal process to express Q:
Q = ΔEint − Won
Q
T
(1)
or, because ΔEint = 0 for a free
expansion of a gas,
Q = −Won
The work done on the gas is given
by:
⎛V
Won = nRT ln⎜⎜ i
⎝ Vf
⎛V
⎞
⎟⎟ ⇒ Q = −nRT ln⎜⎜ i
⎝ Vf
⎠
Substitute for Q in equation (1) to
obtain:
⎛V
ΔS gas = −nR ln⎜⎜ i
⎝ Vf
⎞
⎟⎟
⎠
Substitute numerical values and evaluate ΔS:
J ⎞ ⎛ 40.0 L ⎞
J
⎛
⎟⎟ = 11.5
ΔS gas = −(2.00 mol)⎜ 8.314
⎟ ln⎜⎜
mol ⋅ K ⎠ ⎝ 80.0 L ⎠
K
⎝
(b) The change in entropy of the
universe is the sum of the entropy
changes of the gas and the
surroundings:
ΔS u = ΔS gas + ΔS surroundings
⎞
⎟⎟
⎠
The Second Law of Thermodynamics 1877
For the change in entropy of the
surroundings we use the fact that,
during the free expansion, the
surroundings are unaffected:
The change in entropy of the universe
is the change in entropy of the gas:
ΔS surroundings =
ΔS u = 11.5
Qrev 0
= =0
T
T
J
K
63 ••
[SSM] A a reservoir at 300 K absorbs 500 J of heat from a second
reservoir at 400 K. (a) What is the change in entropy of the universe, and (b) how
much work is lost during the process?
Picture the Problem We can find the entropy change of the universe from the
entropy changes of the high- and low-temperature reservoirs. The maximum
amount of the 500 J of heat that could be converted into work can be found from
the maximum efficiency of an engine operating between the two reservoirs.
(a) The entropy change of the
universe is the sum of the
entropy changes of the two
reservoirs:
ΔS u = ΔS h + ΔSc = −
Substitute numerical values and
evaluate ΔSu:
⎛ 1
1 ⎞
⎟⎟
ΔS u = (− 500 J ) ⎜⎜
−
400
K
300
K
⎠
⎝
Q Q
+
Th Tc
⎛1 1⎞
= −Q⎜⎜ − ⎟⎟
⎝ Th Tc ⎠
= 0.42 J/K
(b) Relate the heat that could have
been converted into work to the
maximum efficiency of an engine
operating between the two
reservoirs:
W = ε max Qh
The maximum efficiency of an
engine operating between the two
reservoir temperatures is the
efficiency of a Carnot device
operating between the reservoir
temperatures:
ε max = ε C = 1 −
Tc
Th
1878 Chapter 19
Substitute for εmax to obtain:
⎛ T ⎞
W = ⎜⎜1 − c ⎟⎟Qh
⎝ Th ⎠
Substitute numerical values and
evaluate W:
⎛ 300 K ⎞
⎟⎟ (500 J ) = 125 J
W = ⎜⎜1 −
⎝ 400 K ⎠