Download University Physics AI No. 12 The Second Law of Thermodynamics

University Physics AI
No. 12 The Second Law of Thermodynamics
Class
Number
Name
I．Choose the Correct Answer
1. A real engine has an efficiency of 33%. The engine has a work output of 24J per cycle. How
much heat energy is extracted from the high- temperature reservoir per cycle?
( D )
(A) 8J
(B) 16J
(C) 48J
(D) 72J
(E) The question can be answered only of the engine is a Carnot engine.
Solution:
Using the definition of the efficiency
ε=
W
W
24
, we have QH =
=
= 72.7(J ) .
QH
ε 0.33
2. A real engine has an efficiency of 33%. The engine has a work output of 24J per cycle. For this
（ C ）
engine TL=27°C. What can be concluded about TH?
(A) TH =450°C. (B) TH =177°C (C) TH >177°C (D) TH <177°C. (E) 177°C <TH <450°C.
Solution:
Since real heat engines have efficiencies less than a Carnot heat engine operating between the same
two temperatures.
The efficiency of the Carnot engine is
ε =1−
ε =1−
TC
, so we have
TH
27 + 273
TC
TC
> 33% ⇒ TH >
=
= 175o
1 − 33%
67%
TH
3. Ten identical particles are to be divided up into two different containers. How many different
configurations are possible?
（ B ）
(A) 1
(B) 11
(C) 120
(D) 1024
(E) 3628800
Solution:
The possible configurations are
configuration I
Ⅱ
Ⅲ
Ⅳ
Ⅴ
Ⅵ
Ⅶ
Ⅷ
Ⅸ
Ⅹ
ⅩI
N1
0
1
2
3
4
5
6
7
8
9
10
N2
10
9
8
7
6
5
4
3
2
1
0
4. Ten identical particles are to be divided up into two containers. Which configuration has the
largest number of microstates?
（ D ）
(A) 0, 10
(B) 3, 7
(C) 4, 6
(D) 5, 5
Solution:
The most probable state is the ten particles divided into two equal parts, in this situation, the number
⎛10 ⎞ 10!
⎟⎟ =
= 252 ，therefore the answer is (D).
5
⎝ ⎠ 5!5!
of microstates is largest, ⎜⎜
5. For which of the following processes is the entropy change zero?
（ C ）
(A) Isobaric
(B) Isothermal
(C) Adiabatic
(D) Constant volume
(E) None of these, since ∆S>0 for all processes.
Solution:
If these processes are quasi-static reversible, according to the definition of entropy ∆S =
∫
dQ
,
T
the answer is (C).
6. One mole of an ideal gas is originally at P0, V0, and T0. The gas is heated at constant volume to
2T0, then allowed to expand at constant temperature to 2V0, and finally it is allowed to cool at
（ E ）
constant pressure to T0. The net entropy change for this ideal gas is
(A) ∆S=(5R/2)ln2
(B) ∆S=5R/2
(C) ∆S=Rln2
(D) ∆S=3R/2
(E) ∆S=0
Solution:
For the entropy is a state variable of a thermodynamic system, the net entropy change of the ideal
gas is (E).
7. A block of aluminum originally at 80°C is places into an insulated container of water originally at
25°C. After a while the system reaches an equilibrium temperature of 31°C. During this process
（ C ）
(B) ∆Saluminum=0.
(C) ∆Saluminum<0.
(A) ∆Saluminum>0.
Solution:
The total entropy change of the block of aluminum is ∆S =
∫
304 K
353 K
mcalu min um dT
<0 J , the answer
T
should be (C).
8. Which of the following is a consequence of the second law of thermodynamics?
( C )
(A) Heat can flow only from high temperature to low temperature.
(B) Objects in contact will tend toward having the same temperature.
(C) Any system that produces order from disorder must have an external influence.
Solution:
The second law of thermodynamics tells that if there is no external effect, an isolated system will
always change from order to disorder. The answer is (C)
II. Filling the Blanks
1. An ideal gas undergoes a reversible isothermal expansion at 132°C. The entropy of the gas
increases by 46.2 J/K. The heat absorbed is 18711 J .
Solution
∆S =
Q
T
⇒ Q = ∆ST = 46.2 × (273 + 132) = 18711 ( J )
2. In Fig. 1, suppose that the change in entropy of the system in
P
passing from state a to state b along path 1 is +0.60 J/K. The
entropy change in passing from state a to b along path 2 is
0.6 J/K , and the entropy change in passing from state b to a
along path 2 is
-0.6 J/K .
Solution:
T(K)
The entropy is a state variable of a thermodynamic system.
It is settled by the initial and final states.
400
3. For the Carnot cycle shown in Fig.2, the heat that enters
22.5 J .
is
60 J , the work done on the system is
Solution:
According to the entropy change ∆S =
1
a
b
2
Fig.1
300
200
100
0
S(J/K)
0.2
Q
,
T
0.4
Fig.2
0.6
We have Q = ∆ST . So the heat enters is
Q = QH = (0.6 − 0.2) × 400 = 160(J )
According to the first law of thermodynamics, the work done on the system is
W = εQ = (1 −
250
) × 160 = 60J
400
4. A Carnot heat engine has an efficiency of 0.300 operating between a high-temperature reservoir at
temperature TH and a low-temperature reservoir at 20 °C. To increase the efficiency of the Carnot
heat engine to 0.400, the temperature of the hot reservoir be increased is 69.76 K (in Kelvin).
Solution:
The efficiency of the Carnot heat engine is
ε =1−
TC
TH
TC
273 + 20
=
= 418.57(K )
1 − ε1
1 − 0.3
ε1 = 1 −
TC
TH 1
⇒ TH 1 =
ε2 = 1−
TC
TH 2
⇒ TH 2 =
TC
273 + 20
=
= 488.33(K )
1− ε2
1 − 0.4
Then we have
TH 2 − TH 1 = 488.33 − 418.57 = 69.76 (K )
5. One (1.00) kilogram of steam at 100 °C condenses into water at 100 °C. The entropy change of
the mass is -0.61×104 J/K . Does your result violate the second law of thermodynamics? Explain
your answer. No. For a system which is not isolated, the entropy change of the system could be
less than zero. According to the entropy change ∆S=Q/T , since the heat is negative, the entropy is
negative too.
Solution:
The entropy change of the mass is ∆S =
mL
1 × 22.57 × 105
=−
= −0.61 × 104 (J/K )
273 + 100
T
6. Two (2.00) moles of a monatomic ideal gas is warmed slowly from 300 K to 400 K at constant
volume. The entropy change of the gas is 7.17 J/K .
Solution:
The
entropy
change
of
the
gas
is
∆S = ∫
T2 nC dT
T
dQ
3R 400
v
=∫
=nCv ln 2 = 2 ×
ln
= 7.17(J/K )
T
1
T
T
T1
2 300
7. A heat engine with an efficiency of 0.25 operates between two reservoirs at 1000K and 400K.
The power output of the heat engine is 1.00kW. The heat flow to the heat engine from the
high-temperature reservoir is 4000 J .
Solution:
The efficiency of the heat engine is
high-temperature reservoir is QH =
W
ε
ε=
=
W
, so the heat flow to the heat engine from the
QH
1000
= 4000 (J)
0.25
8. An ideal gas undergoes an isothermal expansion at 77°C increasing its volume from 1.3 to 3.4 L.
The entropy change of the gas is 24 J/K. The quantity of gas present is 3mole (in mole).
Solution
For isothermal process ∆S =
∫
V
V
dQ 1
1
= ∫ dQ = nRT ln 2 = nR ln 2
T
T
T
V1
V1
So the quantity of gas present is n =
24
∆S
=
= 3(mole)
V2
3.4
R ln
8.315 × ln
V1
1.3
III. Give the Solutions of the Following Problems
1. In Carnot cycle, the isothermal expansion of an ideal gas takes place at 412K and the isothermal
compression at 297K. During the expansion, 2090 J of heat energy are transferred to the gas.
Determine (a) the work performed by the gas during the isothermal expansion, (b) the heat rejected
from the gas during the isothermal compression, and (c) the work done on the gas during the
isothermal compression.
Solution:
(a) According to the first law of thermodynamics, the work performed by the gas during the
isothermal expansion is W = Q = 2090 J .
(b) The efficiency of the Carnot heat engine is
ε=
Q
W
T
=1− C =1− C
QH
QH
TH
so the heat rejected from the gas during the isothermal compression is
QC = QH ⋅
297
TC
= 2090 ×
= 1506.63 (J )
412
TH
According to the first law of thermodynamics, the work done on the gas during the isothermal
compression is W = QC = 1506.63 J
2. One (1.00) mole of an ideal diatomic gas (with γ =
Pressure (atm)
1.40) initially at 20.0 °C and 1.00 atm pressure is taken
B Isothermal process
P
around the following cycle (see Figure 3).
Path (1): an isochoric increase in pressure until the
1
2
temperature of the gas is 150 °C and the pressure is
1.00
C
P;
3
A
Path (2): an isothermal expansion until the pressure
Volume
returns to 1.00 atm.
V
Path (3): an isobaric compression until the gas
Fig.3
reaches its original volume.
(a) What is the original volume of the gas at the beginning and end of the cycle?
(b) What is the pressure of the gas at the completion of path (1)? the volume of the gas at the
completion of path (2)? And the temperature of the gas at the completion of path (2)?
(c)Calculate the work done by the gas during each path of the cycle and the total work done by the
gas. And the heat transfer to the gas during each path of the cycle and the total heat transfer to the
gas over the cycle.
(d) Find the efficiency of the cycle.
(e) Calculate the maximum efficiency that a heat engine could have if it operated between the
hottest and coldest temperature encountered by this gas in this cycle.
Solution:
i +1
R
CP
i +1
2
Since γ =
=
=
= 1.4 , so we get i=2.5
i
CV
i
R
2
(a) Using the equation of state for an ideal gas: PV = nRT
We have the original volume of the gas at the beginning and end of the cycle
VA =
nRTA 1 × 8.315 × (273.15 + 20)
=
= 2.41 × 10− 2 (m3 )
1 × 105 × 1.013
PA
(b) For isochoric process:
PA PB
=
TA TB
The pressure of the gas at the completion of path (1) is
PB =
PATB 1× 1.013 × 105 × (273.15 + 150)
=
= 1.46 × 105 (Pa )
273.15 + 20
TA
For isothermal process: PBVB = PCVC
The volume of the gas at the completion of path (2) is
VC =
PBVB 1.46 × 105 × 2.41 × 10−2
=
= 3.47 × 10− 2 (m3 )
1 × 105 × 1.013
PC
The temperature of the gas at the completion of path (2) is TB = 273 + 150 = 423K
(c) (1) The work
Path 1: W1 = 0J
Path 2: W2 = RT ln
3.47 × 10−2
VC
= 8.315 × (273 + 150) × ln
= 1282.13 (J )
2.41 × 10− 2
VB
Path 3:
W3 = − PA (VC − VA ) = −1 × 1.013 × 105 × (3.47 × 10−2 − 2.41 × 10−2 ) = −1073.78(J)
The total work done by the gas is
Wtotal = W1 + W2 + W3 = 0 + 1282.13 − 1073.78 = 208.35 (J)
(2) The heat
Path 1: Q1 =
M
µ
CV (TB − TA ) = 1 × 2.5 × 8.315 × (423 − 293) = 2702.38(J )
Path 2: Q2 = nRTB ln
VC
3.47 × 10−2
= 1 × 8.315 × 423 × ln
= 1282.13(J)
VB
2.41 × 10− 2
Path 3: Q3 = nCP (TA − TC ) = 1 × 3.5 × 8.315 × (293 − 423) = −3783.33(J )
The total heat transfer to the gas over the cycle is
Qtotal = Q1 + Q2 + Q3 = 2702.38 + 1282.13 − 3783.33 = 201.18(J)
(d) The efficiency of the cycle is
ε=
Wtotal
Wtotal
208.35
=
=
= 0.05
QH
Q1 + Q2 2702.38 + 1282.59
(e) TA = 293K(coldest); TB = 423(hottest); TC = 423K
The maximum efficiency that a heat engine is the carnort engine’s efficiency, that is
ε =1−
TC
293
= 1−
= 0.31
TH
423
3. A gasoline internal combustion engine can be approximated
by the cycle shown in Fig.4. Assume an ideal diatomic gas and
use a compression ratio of 4:1 (Vd=4Va). Assume that Pb=3Pa.
(a) Determine the pressure and temperature of each of the
vertex points of the PV diagram in terms of Pa and Ta. (b)
Calculate the efficiency of the cycle.
Solution:
P
b
Pb
Adiabatic
c
Spark
i +1
= 1.2
For the ideal diatomic gas: i=5, so γ =
i
a
Intake
Adiabatic
Vd
Va
Fig.4
(a) For the isochoric process a→b : Pb=3Pa
Tb Pb
P
3P
= ⇒ Tb = b ⋅ Ta = a ⋅ Ta = 3Ta
Ta Pa
Pa
Pa
For the adiabatic process b→c : Vc = Vd = 4Va
PbVbγ = PcVcγ ⇒ Pc = Pb (
Vb γ
V
1
) = 3Pa ( a )γ = 3Pa ( )1.2 = 0.57 Pa
Vc
Vd
4
Vbγ −1Tb = Vcγ −1Tc ⇒ Tc = (
For the isochoric process c→d : Td = Ta
d
Vb γ −1
1
) Tb = ( )1.2 −1 3Ta = 2.27Ta
Vc
4
V
Td Pd
T
Ta
=
⇒ Pd = d ⋅ Pc =
⋅ 0.57 Pa = 0.25Ta
Tc Pc
Tc
2.27Ta
(b) For the isochoric process a→b :
Q1 = nCV (Tb − Ta ) = nCV (3Ta − Ta ) = 2nCV Ta
For the isochoric process c→d :
Q2 = nCV (Td − Tc ) = nCV (Ta − 2.27Ta ) = −1.27 nCV Ta
The efficiency of the cycle is
η = 1−
− 1.27 nCvTa
Q2
=1−
= 0.365
Q1
2nCvTa
4. One (1.00) mole of an ideal monatomic gas is taken around the
P
cycle shown in Figure 5.
3
(a) What is the work done by the gas in going from point (1) to
2P0
point (2)? and the work done by the gas in going from point (2) to
point (3)?
P0
2
(b)What is the change in the internal energy of the gas in going from
1
point (1) to point (3) along the path (1) → (2) → (3)?
V
(c) What is the entropy change of the gas in going form point (1) to
2V0
V0
point (3)?
Fig.5
(d) What is the change in the internal energy and entropy of the gas
in one complete cycle?
Solution:
(a) The work done by the gas in going from point (1) to point (2) is the area under the curve (1) →
(2). That is W1 = P0 (2V0 − V0 ) = P0V0
The work done by the gas in going from point (2) to point (3) is zero.
(b) Using the equation of state for an ideal gas: PV = nRT
The temperature at point (1) is T1 =
P1V1 P0V0
=
R
R
The temperature at point (3) is T3 =
P3V3 2 P0 ⋅ 2V0 4 P0V0
=
=
R
R
R
The change in the internal energy of the gas in going from point (1) to point (3) along the path (1)
→ (2) → (3) is
∆E = nCV (T3 − T1 ) =
iR 4 P0V0 P0V0
i
3
(
) = P0V0 = P0V0
−
2
R
R
2
2
(c) Using the equation: dS =
dQ CV dT + PdV
=
T
T
The entropy change of the gas in going form point (1) to point (3) is
3
∆S = ∫ dS = ∫
1
T3
T1
V3 dV
CV dT
T
V
+ R∫
= CV ln 3 + R ln 3
V
1
T
V
T1
V1
4 P0V0
3R
2V
3R
=
ln 4 + R ln 2
ln R + R ln 0 =
P
V
V0
2
2
0 0
R
= 8.315 × (1.5 × 1.37 + 0.69) = 22.82(J/K )
(d) The entropy change and the internal energy change of the gas in going form point (1) to point (3)
are zero, since they are state variables.