Download POP4e: Ch. 1 Problems

PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-12
HOMEWORK PROBLEMS Chapter 12: THE LAWS OF THERMODYNAMICS
PART-A: Hand in your answers in class on scantron on Monday -2010. Write your name, class
(1401) and HW # 12 on the scantron.
1.
An ideal gas is maintained at a constant pressure of 70.0 kPa during an isobaric process
while its volume decreases by 0.20 m3. What is the work done by the system on its environment?
(a) 14 kJ
(b) 35 kJ
(c) −14 kJ
(d) −35 kJ
(e) −72 kJ
The work done by the system on the environment is
Wenv   P  V   70.0  103 Pa 0.20 m 3  14  10 3 J  14 kJ



and (c) is the correct choice.
2.
A 2.0-mole ideal gas system is maintained at a constant volume of 4.0 liters. If 100 J of
thermal energy is transferred to the system, what is the change in the internal energy of the
system?
(a) 0 (b) 400 J
(c) 70 J
(d) 100 J
(e) −0.100 J
When volume is constant, the work done on the gas is zero so the first law of thermodynamics gives the
change in internal energy as U  Q  W  100 J  0  100 J , and (d) is the correct answer for this
question.
3.
A monatomic ideal gas expands from 1.00 m3 to 2.50 m3 at a constant pressure of 2.00 X
105 Pa. Find the change in the internal energy of the gas.
(a) 7.50 X 105 J
(b) 1.05 X 106 J
(c) 4.50 X 105 J
(d) 3.00 X 105 J
For a monatomic ideal gas, the internal energy is U 
3
2
energy is U 
3
2
nRT and the change in internal
nR(T ) . From the ideal gas law, PV  nRT , observe that
nR (T )  nRT f  nRTi  Pf V f  PV
i i . When the pressure is constant, Pf  Pi  P , this
reduces to U 
U 
3
2
3
2
[ P (V f  Vi )] 
3
2
P ( V ) , so the change in internal energy for this gas is
(2.00  105 Pa)(1.50 m3 )  4.50  105 J making (c) the correct answer.
4.
An ideal gas drives a piston as it expands from 1.00 m3 to 2.00 m3 at a constant
temperature of 850 K. If there are 390 moles of gas in the piston, how much work does the gas
do in displacing the piston?
(a) 1.9 X 106 J
(b) 2.5 X 106 J
(c) 4.7 X 106 J
(d) 2.1 X 105 J
The work an ideal gas does on the environment during an isothermal expansion is


Wenv  nRT ln V f Vi , so for the given process,


Wenv  3.9  102 mol  8.31 J mol  K   850 K  ln  2   1.9  10 6 J
and (a) is the correct choice.
5.
A diatomic ideal gas expands adiabatically from a volume of 1.00 m3 to a final volume of
3.50 m3. If the initial pressure is 1.00 X 105 Pa, what is the final pressure?
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PHYS-1401: College Physics-I
CRN 55178
(a) 6.62 X 104 Pa
(c) 3.54 X 104 Pa
(b) 1.24 X 105 Pa
Khalid Bukhari
HW-12
(d) 1.73 X 104 Pa
For an adiabatic process, PV   constant , where   c p cv  1.4 for diatomic gases (See Table 12.1
1.4
in the textbook.) Thus, Pf V f1.4  PV
i i , or the final pressure will be
V 
Pf  Pi  i 
 Vf 
1.4
 1.00 m3 
 1.00  105 Pa 
 3.50 m3 


1.4
 1.73  10 4 Pa
6.
How much net work is done by the gas undergoing the cyclic process illustrated in Figure
MCQ12.6? Choose the best estimate.
(a) 1 X 105 J (b) 2 X 105 J (c) 3 X 105 J (d) 4 X 105 J (e) 5 X 105 Pa
In a cyclic process, the net work done equals the area enclosed by the process curve in a PV diagram.
Thus,
Wnet    4.00  1.00   105 Pa    2.00  1.00  m 3 
   2.00  1.00   105 Pa    3.00  2.00  m 3   4.00  10 5 J
and (d) is the correct answer.
7.
An engine does 15 kJ of work while rejecting 37 kJ to the cold reservoir. What is the
efficiency of the engine?
(a) 0.15
(b) 0.29
(c) 0.33
(d) 0.45
(e) 1.2
From conservation of energy, the energy input to the engine must be
Qh  Weng  Qc  15 kJ  37 kJ  52 kJ
so the efficiency is
Weng
15 kJ
e 

 0.29
Qc
52 kJ
and the correct choice is (b).
or
29%
8.
A refrigerator does 18 kJ of work while moving 115 kJ of thermal energy from inside the
refrigerator. What is its coefficient of performance?
(a) 3.4
(b) 2.8
(c) 8.9
(d) 6.4
(e) 5.2
The coefficient of performance of this refrigerator is
Q
115 kJ
which is choice (d).
COP  c 
 6.4
W
18 kJ
2
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-12
9.
A steam turbine operates at a boiler temperature of 450 K and an exhaust temperature of
3.0 X 102 K. What is the maximum theoretical efficiency of this system?
(a) 0.24
(b) 0.50
(c) 0.33
(d) 0.67
(e) 0.15
The maximum theoretical efficiency (the Carnot efficiency) of a device operating between absolute
temperatures Tc  Th is eC  1  Tc Th . For the given steam turbine, this is
3.0  102 K
 0.33
450 K
and (c) is the correct answer.
eC  1 
or
33%
10.
A 1.00-kg block of ice at 0° C and 1.0 atm melts completely to water at 0°C. Calculate
the change of the entropy of the ice during the melting process. (For ice, Lf = 3.33 X 105 J/kg.)
(a) 3 340 J/K (b) 2 170 J/K
(c) −3 340 J/K
(d) 1 220 J/K
(e) −1 220 J/K
At a pressure of 1.0 atm, ice melts at absolute temperature TK  0°  273.15  273.15 K . The thermal
energy this block of ice must absorb to fully melt is
Qr  mL f   1.00 kg  3.33  10 5 J kg  3.33  10 5 J


so the change in entropy of the ice is
Q
3.33  105 J
S  r 
 1.22  103 J K  1.22 kJ K
TK
273.15 K
and (d) is the correct choice.
11.
If an ideal gas is compressed isothermally, which of the following statements is true? (a)
Energy is transferred to the gas by heat. (b) No work is done on the gas. (c) The temperature of
the gas increases. (d) The internal energy of the gas remains constant. (e) The pressure remains
constant.
In an ideal gas, the internal energy is directly proportional to the absolute temperature. Thus, in an
isothermal process (constant temperature process) the internal energy of the ideal gas is constant. Choice
(d) is the only true statement among the listed choices. In the compression, work is done on the gas while
the internal energy is constant. The first law of thermodynamics then says that energy must be transferred
from the gas by heat. Also, by the ideal gas law, when the volume decreases while temperature is constant,
the pressure must increase.
12.
When an ideal gas undergoes an adiabatic expansion, which of the following statements
is true? (a) The temperature of the gas doesn’t change. (b) No work is done by the gas. (c) No
energy is delivered to the gas by heat. (d) The internal energy of the gas doesn’t change. (e) The
pressure increases.
By definition, in an adiabatic process, no energy is transferred to or from the gas by heat. Thus, (c) is a
true statement. All other choices are false. In an expansion process, the gas does work on the environment.
Since, there is no energy input by heat, the first law of thermodynamics says that the internal energy of the
ideal gas must decrease, meaning the temperature will decrease. Also, in an adiabatic process,
PV   constant meaning that the pressure must decrease as the volume increases in this adiabatic
expansion.
3
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-12
13.
If an ideal gas undergoes an isobaric process, which of the following statements is true?
(a) The temperature of the gas doesn’t change. (b) Work is done on or by the gas. (c) No energy
is transferred by heat to or from the gas. (d) The volume of the gas remains the same. (e) The
pressure of the gas decreases uniformly.
In an isobaric process on an ideal gas, pressure is constant while the gas either expands or is compressed.
Since the volume of the gas is changing, work is done either on or by the gas, so choice (b) is a true
statement. Also, from the ideal gas law with pressure constant, P  V   nR  T  . Thus, the gas must
undergo a change in temperature having the same sign as the change in volume. If V  0 , then both T
and the change in the internal energy of the gas are positive  U  0  . However, when V  0 , the
work done on the gas is negative  W  0  , and the first law of thermodynamics says that there must be a
positive transfer of energy by heat to the gas  Q  U  W  0  . When V  0 , a similar argument
shows that  U  0  ,  W  0  , and Q  U  W  0 . Thus, all of the other listed choices are false
statements.
14.
Of the following, which is not a statement of the second law of thermodynamics? (a) No
heat engine operating in a cycle can absorb energy from a reservoir and use it entirely to do
work. (b) No real engine operating between two energy reservoirs can be more efficient than a
Carnot engine operating between the same two reservoirs. (c) When a system undergoes a
change in state, the change in the internal energy of the system is the sum of the energy
transferred to the system by heat and the work done on the system. (d) The entropy of the
Universe increases in all natural processes. (e) In all real processes, the resulting energy available
for doing work decreases.
Choice (c) is a statement of the first law of thermodynamics, not the second law, and hence is the correct
answer to this question. Choices (a), (b), (d) and (e) are alternative statements of the second law, (a) being
the Kelvin-Planck formulation, (b) the Carnot statement, (d) the Clausius statement, and (e) summarizes
the primary consequence of all these various statements.
15.
If an ideal gas is compressed to half its initial volume, which of the following statements
is true regarding the work done on the gas? (a) The isothermal process involves the most work.
(b) The adiabatic process involves the most work. (c) The isobaric process involves the most
work. (d) The isovolumetric process involves the most work. (e) The work done is independent
of the process.
First, eliminate choice (d) since the work involved
in an isovolumetric process is zero, and by
definition, one cannot have an isovolumetric
process when the volume is changing. In the other
processes, the work done on the gas equals the
area under the process curve in a PV diagram. In
an isobaric process, the pressure is constant, so
Pf = Pi and the work done is the area under
curve 1 in the sketch at the right. For an isothermal
4
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
process, the ideal gas law gives Pf Vf  PV
i i , so Pf  (Vi Vf ) Pi  2Pi and the work
HW-12
done is the area

under curve 2 in the sketch. Finally, for an adiabatic process, Pf Vf  PV
i i  constant , so
Pf  (Vi V f ) Pi and Pf  2 Pi  2 Pi since   1 for all ideal gases (See Table 12.1 in the textbook).
The work done in an adiabatic process is the area under curve 3, which exceeds that done in either of the
other processes. Thus, the correct choice is (b), the adiabatic process involves the most work.
16.
A window air conditioner is placed on a table inside a well-insulated apartment, plugged
in and turned on. What happens to the average temperature of the apartment? (a) It increases. (b)
It decreases. (c) It remains constant. (d) It increases until the unit warms up and then decreases.
(e) The answer depends on the initial temperature of the apartment.
With this method of using an air conditioner, the average temperature in the room will increase, and
choice (a) is the correct answer. The air conditioner operates on a cyclic process so the change in the
internal energy of the refrigerant is zero. Then, the conservation of energy gives the thermal energy
exhausted to the room as Qh  Qc  Weng where Qc is the thermal energy the air conditioner removes
from the room and Weng is the work done to operate the device. Since Weng  0 , the air conditioner is
returning more thermal energy to the room than it is removing, so the average temperature in the room will
increase.
17.
The second law of thermodynamics implies that the coefficient of performance of a
refrigerator must be what? (a) less than 1 (b) less than or equal to 1 (c) greater than or equal to 1
(d) finite (e) greater than 0
The correct choice is (d). The second law basically says that you must put in some work to transfer energy
by heat from a lower-temperature to a higher-temperature location. But it can be very little work if the two
temperatures are very nearly equal. Still, in the coefficient of performance for a refrigerator,
COP  Qc W , W can never be zero and the result must be finite in all cases.
18.
A thermodynamic process occurs in which the entropy of a system changes by −6 J/K.
According to the second law of thermodynamics, what can you conclude about the entropy
change of the environment? (a) It must be +6 J/K or less. (b) It must be equal to 6 J/K. (c) It must
be between 6 J/K and 0. (d) It must be 0. (e) It must be +6 J/K or more.
The Clausius statement of the second law of thermodynamics says that in any thermodynamics process,
reversible or irreversible, the total entropy of the universe must either remain constant (reversible process)
or increase (irreversible process). Thus, if in a thermodynamics process, the entropy of a system changes
by –6 J/K, the entropy of the environment (i.e., the rest of the universe) must increase by +6 J/K or more.
The correct choice here is (e).
19
A gas changes in volume from 0.750 m3 to 0.250 m3 at a constant pressure of 1.50 X 105
Pa. (a) How much work is done on the gas?
(a) 7.50 x104 J
(b) 5.50 x104 J
(c) 3.50 x104 J
(d) 1.50 x104 J
The work done on the gas is



W  P  V    1.50  105 Pa 0.250 m2  0.750 m2  7.50  10 4 J
5
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-12
20.
A thermodynamic system undergoes a process in which its internal energy decreases by
500 J. If at the same time 220 J of work is done on the system, find the energy transferred to or
from it by heat.
(a) -750 J
(b) -550 J
(c) -720 J
(d) -580 J
From the first law,
Q  U  W  500 J  220 J  720 J
The negative sign in the result means that energy is transferred from the system by heat.
21.
A heat engine operates between a reservoir at 25° C and one at 375° C. What is the
maximum efficiency possible for this engine?
(a) 32 %
(b) 38 %
(c) 54 %
(d) 65 %
The maximum possible efficiency for a heat engine operating between reservoirs with absolute
temperatures of Tc  25  273  298 K and Th  375  273  648 K is the Carnot efficiency
T
298 K
eC  1  c  1 
 0.540 or 54.0%
Th
648 K
22.
A heat engine is being designed to have a Carnot efficiency of 65% when operating
between two heat reservoirs. (a) If the temperature of the cold reservoir is 20° C, what must be
the temperature of the hot reservoir?
(a) 564 C
(b) 438 C
(c) 862 C
(d) 485 C
The absolute temperature of the cold reservoir is Tc  20  273  293 K . If the Carnot efficiency is to
be eC  0.65 , it is necessary that
Tc
 0.65 or
Th
Thus,
293 K
Th 
 837 K
0.35
1
Tc
 0.35
Th
or
Th 
and
Tc
0.35
Th  837  273  564C
6