Ch19_SSM

Chapter 19 The Second Law of Thermodynamics Conceptual Problems 5 An air conditioners COP is mathematically identical to that of a refrigerator, that is, WQcrefAC COPCOP == . However a heat pumps COP is defined differently, as WQhhpCOP = . Explain clearly why the two COPs are defined differently. Hint: Think of the end use of the three different devices. Determine the Concept The COP is defined so as to be a measure of the effectiveness of the device. For a refrigerator or air conditioner, the important quantity is the heat drawn from the already colder interior, Qc. For a heat pump, the ideas is to focus on the heat drawn into the warm interior of the house, Qh. 9 Explain why the following statement is true: To increase the efficiency of a Carnot engine, you should make the difference between the two operating temperatures as large as possible; but to increase the efficiency of a Carnot cycle refrigerator, you should make the difference between the two operating temperatures as small as possible. Determine the Concept A Carnot-cycle refrigerator is more efficient when the temperatures are close together because it requires less work to extract heat from an already cold interior if the temperature of the exterior is close to the temperature of the interior of the refrigerator. A Carnot-cycle heat engine is more efficient when the temperature difference is large because then more work is done by the engine for each unit of heat absorbed from the hot reservoir. 17 Sketch an SV diagram of the Carnot cycle for an ideal gas. Determine the Concept Referring to Figure 19-8, process 12 is an isothermal expansion. In this process heat is added to the system and the entropy and volume increase. Process 23 is adiabatic, so S is constant as V increases. Process 34 is an isothermal compression in which S decreases and V also decreases. Finally, process 41 is adiabatic, that is, isentropic, and S is constant while V decreases. During the isothermal expansion (from point 1 to point 2) the work done by the gas equals the heat added to the gas. The change in entropy of the gas from point 1 (where the temperature is T1) to an arbitrary point on the curve is given by:

1TQS =

385

Chapter 19

386

For an isothermal expansion, the work done by the gas, and thus the heat added to the gas, are given by:

==

11 ln V

VnRTWQ

Substituting for Q yields:

=

1

lnVVnRS

Since , we have: SSS += 1

+=

11 ln V

VnRSS

The graph of S as a function of V for an isothermal expansion shown to the right was plotted using a spreadsheet program. This graph establishes the curvature of the 12 and 34 paths for the SV graph.

V

S

An SV graph for the Carnot cycle (see Figure 19-8) is shown to the right.

V

S

1

2 3

4

Estimation and Approximation 23 Estimate the maximum efficiency of an automobile engine that has a compression ratio of 8.0:1.0. Assume the engine operates according to the Otto cycle and assume = 1.4. (The Otto cycle is discussed in Section 19-1.) Picture the Problem The maximum efficiency of an automobile engine is given by the efficiency of a Carnot engine operating between the same two temperatures. We can use the expression for the Carnot efficiency and the equation relating V and T for a quasi-static adiabatic expansion to express the Carnot efficiency of the engine in terms of its compression ratio.

The Second Law of Thermodynamics

387

Express the Carnot efficiency of an engine operating between the temperatures Tc and Th:

h

cC 1 T

T=

Relate the temperatures Tc and Th to the volumes Vc and Vh for a quasi-static adiabatic compression from Vc to Vh:

1hh

1cc

= VTVT 1

c

h1

c

1h

h

c

==

VV

VV

TT

Substitute for h

c

TT to obtain:

1

c

hC 1

=

VV

Express the compression ratio r:

h

c

VVr =

Substituting for r yields:

1C11 = r

Substitute numerical values for r and (1.4 for diatomic gases) and evaluate C:

( ) %560.811 14.1C =

25 The average temperature of the surface of the Sun is about 5400 K, the average temperature of the surface of Earth is about 290 K. The solar constant (the intensity of sunlight reaching Earths atmosphere) is about 1.37 kW/m2. (a) Estimate the total power of the sunlight hitting Earth. (b) Estimate the net rate at which Earths entropy is increasing due to this solar radiation. Picture the Problem We can use the definition of intensity to find the total power of sunlight hitting Earth and the definition of the change in entropy to find the changes in the entropy of Earth and the Sun resulting from the radiation from the Sun. (a) Using its definition, express the intensity of the Suns radiation on Earth in terms of the power P delivered to Earth and Earths cross sectional area A:

API =

Solve for P and substitute for A to obtain:

2RIIAP == where R is the radius of Earth.

Substitute numerical values and evaluate P:

( )( )W1075.1W10746.1

m1037.6kW/m37.11717

262

=== P

Chapter 19

388

(b) Express the rate at which Earths entropy SEarth changes due to the flow of solar radiation:

Earth

Earth

TP

dtdS =

Substitute numerical values and

evaluate dt

dSEarth : sJ/K1002.6

K290W10746.1

14

17Earth

=

=dt

dS

Heat Engines and Refrigerators 27 A heat engine with 20.0% efficiency does 0.100 kJ of work during each cycle. (a) How much heat is absorbed from the hot reservoir during each cycle? (b) How much heat is released to the cold reservoir during each cycle? Picture the Problem (a) The efficiency of the engine is defined to be hQW= where W is the work done per cycle and Qh is the heat absorbed from the hot reservoir during each cycle. (b) Because, from conservation of energy,

, we can express the efficiency of the engine in terms of the heat Qch QWQ += c released to the cold reservoir during each cycle.

(a) Qh absorbed from the hot reservoir during each cycle is given by:

J5000.200

J100h ===

WQ

(b) Use to obtain: ch QWQ += J 400J 100J 500hc === WQQ

31 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 TCQ = V and .P TCQ = 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.


389

(a) The cycle is shown to the right:

Apply the ideal-gas law to state 1 to find T1: ( )( )

( )K300

KmolatmL108.206mol1.00

L24.6atm1.002

111 =

==

nRVPT

The pressure doubles while the volume remains constant between states 1 and 2. Hence:

KTT 6002 12 ==

The volume doubles while the pressure remains constant between states 2 and 3. Hence:

KTT 12002 23 ==

The pressure is halved while the volume remains constant between states 3 and 4. Hence:

KTT 6003214 ==

For path 12:

0 1212 == VPW and

( ) kJ74.3K300K600Kmol

J8.314 23

1223

12V12 =

=== TRTCQ

Chapter 19

390

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:

oninint WQE +=

Because : 012 =W kJ 74.3 1212int, == QE

For path 23:

( )( ) kJ99.4atmL

J101.325L24.6L49.2atm2.00 2323on =

=== VPWW

( ) kJ5.12K600K1200Kmol

J8.314 25

2325

23P23 =

=== TRTCQ

Apply to obtain: oninint WQE +=

kJ 5.7kJ 99.4kJ 5.12 23 int, ==E For path 34:

03434 == VPW and

( ) kJ48.7K0021K600Kmol

J8.314 23

3423

34V34int,34 =

==== TRTCEQ

Apply to obtain: oninint WQE +=

kJ 48.70kJ 48.7 34 int, =+=E For path 41:

( )( ) kJ49.2atmL

J101.325L2.94L24.6atm1.00 4141on =

=== VPWW

and

( ) kJ24.6K600K003Kmol

J8.314 25

4125

41P41 =

=== TRTCQ

Apply to obtain: oninint WQE += kJ 75.3

kJ 49.2kJ 24.6 41 int,=

+=E


391

onW inQ

For easy reference, the results of the preceding calculations are summarized in the following table:

Process , kJ , kJ ( )oninint WQE += , kJ12 0 3.74 3.74 23 4.99 12.5 7.5 34 0 7.48 7.48 41 2.49 6.24 3.75

(b) The efficiency of the cycle is given by:

( )2312

4123

in

by

QQWW

QW

++==

Substitute numerical values and evaluate : %15kJ5.12kJ3.74

kJ2.49kJ4.99 +=

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. Second Law of Thermodynamics 39 A refrigerator absorbs 500 J of heat from a cold reservoir and releases 800 J to a hot reservoir. Assume that the heat-engine statement of the second law of thermodynamics is false, and show how a perfect engine working with this refrigerator can violate the refrigerator statement of the second law of thermodynamics. Determine the Concept The following diagram shows an ordinary refrigerator that uses 300 J of work to remove 500 J of heat from a cold reservoir and releases 800 J of heat to a hot reservoir (see (a) in the diagram). Suppose the heat-engine statement of the second law is false. Then a perfect heat engine could remove energy from the hot reservoir and convert it completely into work with 100 percent efficiency. We could use this perfect heat engine to remove 300 J of energy from the hot reservoir and do 300 J of work on the ordinary refrigerator (see (b) in the diagram). Then, the combination of the perfect heat engine and the ordinary refrigerator would be a perfect refrigerator; transferring 500 J of heat from the cold reservoir to the hot reservoir without requiring any work (see (c) in the diagram).This violates the refrigerator statement of the second law.

Chapter 19

392

500 JCold reservoir at temperature Tc

Hot reservoir at temperature Th800 J

300 J 300 J

300 J

Ordinaryrefrigerator

Perfectrefrigerator

500 J

500 J

a b c( ) ( ) ( )

Perfectheatengine

Carnot Cycles 41 A Carnot engine works between two heat reservoirs at temperatures Th = 300 K and Tc = 200 K. (a) What is its efficiency? (b) If it absorbs 100 J of heat from the hot reservoir during each cycle, how much work does it do each cycle? (c) How much heat does it release during each cycle? (d) What is the COP of this engine when it works as a refrigerator between the same two reservoirs? Picture the Problem We can find the efficiency of the Carnot engine using

hc /1 TT= and the work done per cycle from ./ hQW= We can apply conservation of energy to find the heat rejected each cycle from the heat absorbed and the work done each cycle. We can find the COP of the engine working as a refrigerator from its definition.

(a) The efficiency of the Carnot engine depends on the temperatures of the hot and cold reservoirs:

%3.33K300K20011

h

cC === T

T

(b) Using the definition of efficiency, relate the work done each cycle to the heat absorbed from the hot reservoir:

( )( ) J33.3J1000.333hC === QW

(c) Apply conservation of energy to relate the heat given off each cycle to the heat absorbed and the work done:

J67

J 66.7J33.3J100hc=

=== WQQ


393

(d) Using its definition, express and evaluate the refrigerators coefficient of performance:

0.2J33.3J66.7COP c ===

WQ

47 In the cycle shown in Figure 19-19, 1.00 mol of an ideal diatomic gas is initially at a pressure of 1.00 atm and a temperature of 0.0C. The gas is heated at constant volume to T2 = 150C and is then expanded adiabatically until its pressure is again 1.00 atm. It is then compressed at constant pressure back to its original state. Find (a) the temperature after the adiabatic expansion, (b) the heat absorbed or released by the system during each step, (c) the efficiency of this cycle, and (d) the efficiency of a Carnot cycle operating between the temperature extremes of this cycle. Picture the Problem We can use the ideal-gas law for a fixed amount of gas and the equations of state for an adiabatic process to find the temperatures, volumes, and pressures at the end points of each process in the given cycle. We can use

and to find the heat entering and leaving during the constant-volume and isobaric processes and the first law of thermodynamics to find the work done each cycle. Once weve calculated these quantities, we can use its definition to find the efficiency of the cycle and the definition of the Carnot efficiency to find the efficiency of a Carnot engine operating between the extreme temperatures.

TQ = VC TQ = PC

(a) Apply the ideal-gas law for a fixed amount of gas to relate the temperature at point 3 to the temperature at point 1:

3

33

1

11

TVP

TVP =

or, because P1 = P3,

1

313 VVTT = (1)

Apply the ideal-gas law for a fixed amount of gas to relate the pressure at point 2 to the temperatures at points 1 and 2 and the pressure at 1:

2

22

1

11

TVP

TVP =

12

2112 TV

TVPP =

Because V1 = V2:

( ) atm1.55K273K423atm1.00

1

212 === T

TPP

Apply an equation for an adiabatic process to relate the pressures and volumes at points 2 and 3:

3311 VPVP =

1

3

113

=

PPVV

Chapter 19

394

Noting that V1 = 22.4 L, evaluate V3: ( ) L30.6atm1

atm1.55L22.41.41

3 =

=V

Substitute numerical values in equation (1) and evaluate T3 and t3:

( ) K373L22.4L30.6K2733 ==T

and C10027333 == Tt

(b) Process 12 takes place at constant volume (note that = 1.4 corresponds to a diatomic gas and that CP CV = R):

( )kJ3.12

K273K423Kmol

J8.314

C

25

1225

12V12

=

=== TRTQ

Process 23 takes place adiabatically:

023 =Q

Process 31 is isobaric (note that CP = CV + R): ( )

kJ2.91

K373K732Kmol

J8.314

C

27

1227

31P31

=

=== TRTQ

(c) The efficiency of the cycle is given by:

inQW= (2)

Apply the first law of thermodynamics to the cycle:

oninint WQE += or, because (the system

begins and ends in the same state) and

0cycle int, =E

ingas by theon QWW == .

Evaluating W yields: kJ0.21kJ2.910kJ3.12

312312

=+=++== QQQQW

Substitute numerical values in equation (2) and evaluate : %7.6kJ3.12

kJ0.21 ==


395

(d) Express and evaluate the efficiency of a Carnot cycle operating between 423 K and 273 K:

35.5%K234K73211

h

cC === T

T

*Heat Pumps 49 As an engineer, you are designing a heat pump that is capable of delivering heat at the rate of 20 kW to a house. The house is located where, in January, the average outside temperature is 10C. The temperature of the air in the air handler inside the house is to be 40C. (a) What is maximum possible COP for a heat pump operating between these temperatures? (b) What must the minimum power of the electric motor driving the heat pump be? (c) In reality, the COP of the heat pump will be only 60 percent of the ideal value. What is the minimum power of the electric motor when the COP is 60 percent of the ideal value? Picture the Problem We can use the definition of the COPHP and the Carnot efficiency of an engine to express the maximum efficiency of the refrigerator in terms of the reservoir temperatures. We can apply the definition of power to find the minimum power needed to run the heat pump.

(a) Express the COPHP in terms of Th and Tc:

ch

h

h

c

h

c

h

hhHP

1

1

1

1

COP

TTT

TT

QQ

QQQ

WQ

c

==

=

==

Substitute numerical values and evaluate COPHP:

3.6

26.6K263K313

K133COPHP

=

==

(b) The COPHP is also given by: motor

outHPCOP P

P= HP

outmotor COP

PP =

Substitute numerical values and evaluate Pmotor:

kW2.36.26

kW20motor ==P

Chapter 19

396

(c) The minimum power of the electric motor is given by: ( )maxHP,

c

HP

c

min COPdt

dQdt

dQ

P == where HP is the efficiency of the heat pump.

Substitute numerical values and evaluate Pmin: ( )( ) kW3.56.2660.0

kW20min ==P

Entropy Changes 53 You inadvertently leave a pan of water boiling away on the hot stove. You return just in time to see the last drop converted into steam. The pan originally held 1.00 L of boiling water. What is the change in entropy of the water associated with its change of state from liquid to gas? Picture the Problem Because the water absorbed heat in the vaporization process

its change in entropy is positive and given byT

QS OHby

absorbed

OH2

2 = . See Table 18-2

for the latent heat of vaporization of water. The change in entropy of the water is given by:

T

QS OHby

absorbed

OH2

2 =

The heat absorbed by the water as it vaporizes is the product of its mass and latent heat of vaporization:

vvOHby

absorbed2

VLmLQ ==

Substituting for yields: OHby

absorbed2

QTVLS vOH2

=

Substitute numerical values and evaluate : OH2S

( )

KkJ05.6

K 373kgkJ2257L 00.1

Lkg 00.1

OH 2

=

=S

57 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


397

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 reservoirs 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 systems final state is identical to its initial state:

0cyclecomplete 1

system =S

(b) Relate the entropy changes for each of the three heat reservoirs and the system for one complete cycle of the system:

0 system321 =+++ SSSS or

003

3

2

2

1

1 =+++TQ

TQ

TQ

Substitute numerical values. Heat is rejected by the two high-temperature reservoirs and absorbed by the cold reservoir:

0J400K400

J200K300

J300

3

=++T

Solving for T3 yields: K2673 =T

61 A 1.00-kg block of copper at 100C is placed in an insulated calorimeter of negligible heat capacity containing 4.00 L of liquid water at 0.0C. Find the entropy change of (a) the copper block, (b) the water, and (c) the universe. Picture the Problem We can use conservation of energy to find the equilibrium temperature of the water and apply the equations for the entropy change during a constant pressure process to find the entropy changes of the copper block, the water, and the universe.

Chapter 19

398

(a) Use the equation for the entropy change during a constant-pressure process to express the entropy change of the copper block:

=

i

fCuCuCu ln T

TcmS (1)

Apply conservation of energy to obtain:

0i

i =Q or

0waterwarmingblockcopper =+QQ

Substitute to relate the masses of the block and water to their temperatures, specific heats, and the final temperature Tf of the water:

( ) ( )( ) ( ) 0K273

KkgkJ 4.18

Lkg1.00L4.00

K373Kkg

kJ0.386kg1.00

f

f

=

+

T

T

Solve for Tf to obtain: K275.26 f =T

Substitute numerical values in equation (1) and evaluate : CuS

( )KJ117

K373K275.26ln

KkgkJ0.386kg1.00 Cu =

=S

(b) The entropy change of the water is given by:

=

i

fwaterwaterwater ln T

TcmS

Substitute numerical values and evaluate : waterS

( )KJ138

K273K26.275ln

KkgkJ18.4kg00.4 water =

=S

(c) Substitute for and and evaluate the entropy change of the universe:

CuS waterS

KJ12

KJ138

KJ117 waterCuu

=

+=+= SSS

Remarks: The result that Su > 0 tells us that this process is irreversible.


399

Entropy and Lost Work 63 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:

=

+=+=

ch

chchu

11TT

Q

TQ

TQSSS

Substitute numerical values and evaluate Su: ( )

J/K0.42

K3001

K4001J500 u

=

=S

(b) Relate the heat that could have been converted into work to the maximum efficiency of an engine operating between the two reservoirs:

hmaxQW =

The maximum efficiency of an engine operating between the two reservoir temperatures is the efficiency of a Carnot device operating between the reservoir temperatures:

h

cCmax 1 T

T==

Substitute for max to obtain: h

h

c1 QTTW

=

Substitute numerical values and evaluate W: ( ) J125J500K400

K3001 =

=W

Chapter 19

400

General Problems 67 An engine absorbs 200 kJ of heat per cycle from a reservoir at 500 K and releases heat to a reservoir at 200 K. Its efficiency is 85 percent of that of a Carnot engine working between the same reservoirs. (a) What is the efficiency of this engine? (b) How much work is done in each cycle? (c) How much heat is released to the low-temperature reservoir during each cycle? Picture the Problem We can use the definition of efficiency to find the work done by the engine during each cycle and the first law of thermodynamics to find the heat released to the low-temperature reservoir during each cycle.

(a) Express the efficiency of the engine in terms of the efficiency of a Carnot engine working between the same reservoirs:

==

h

cC 185.085.0 T

T

Substitute numerical values and evaluate : %51510.0K500

K200185.0 ==

=

(b) Use the definition of efficiency to find the work done in each cycle:

( )( )MJ0.10

kJ 102kJ200.5100h=

=== QW

(c) Apply the first law of thermodynamics to the cycle to obtain: kJ89

kJ021kJ002cycleh,cyclec,

=== WQQ

73 (a) Which of these two processes is more wasteful? (1) A block moving with 500 J of kinetic energy being slowed to rest by sliding (kinetic) friction when the temperature of the environment is 300 K, or (2) A reservoir at 400 K releasing 1.00 kJ of heat to a reservoir at 300 K? Explain your choice. Hint: How much of the 1.00 kJ of heat could be converted into work by an ideal cyclic process? (b) What is the change in entropy of the universe for each process? Picture the Problem All 500 J of mechanical energy are lost, i.e., transformed into heat in process (1). For process (2), we can find the heat that would be converted to work by a Carnot engine operating between the given temperatures and subtract that amount of work from 1.00 kJ to find the energy that is lost. In Part (b) we can use its definition to find the change in entropy for each process. (a) For process (2): inCrecoveredmax,2 QWW ==


401

The efficiency of a Carnot engine operating between temperatures Th and Tc is given by:

h

cC 1 T

T= and hence

inh

crecovered 1 QT

TW

=

Substitute for C to obtain:

( ) J250kJ1.00K 400K 3001recovered =

=W or 750 J are lost.

Process (1) produces more waste heat. Process (2) is more wasteful of available work.

(b) Find the change in entropy of the universe for process (1):

J/K1.67K300J500

1 === TQS

Express the change in entropy of the universe for process (2):

=

+=+=

hc

chch2

11TT

Q

TQ

TQSSS

Substitute numerical values and evaluate S2: ( )

J/K833.0

K4001

K3001kJ1.00 2

=

=S

75 A heat engine that does the work of blowing up a balloon at a pressure of 1.00 atm absorbs 4.00 kJ from a reservoir at 120C. The volume of the balloon increases by 4.00 L, and heat is released to a reservoir at a temperature Tc, where Tc < 120C. If the efficiency of the heat engine is 50% of the efficiency of a Carnot engine working between the same two reservoirs, find the temperature Tc. Picture the Problem We can express the temperature of the cold reservoir as a function of the Carnot efficiency of an ideal engine and, given that the efficiency of the heat engine is half that of a Carnot engine, relate Tc to the work done by and the heat input to the real heat engine. Using its definition, relate the efficiency of a Carnot engine working between the same reservoirs to the temperature of the cold reservoir:

h

cC 1 T

T= ( )Chc 1 = TT

Chapter 19

402

Relate the efficiency of the heat engine to that of a Carnot engine working between the same temperatures:

C21

in

==QW

inC

2QW=

Substitute for C to obtain:

=

inhc

21QWTT

The work done by the gas in expanding the balloon is:

( )( )Latm4.00

L4.00atm1.00=

== VPW

Substitute numerical values and evaluate Tc:

( ) K313kJ4.00

LatmJ101.325Latm4.002

1K393c =

=T

79 In a heat engine, 2.00 mol of a diatomic gas are carried through the cycle ABCDA shown in Figure 19-21. (The PV diagram is not drawn to scale.) The segment AB represents an isothermal expansion, the segment BC an adiabatic expansion. The pressure and temperature at A are 5.00 atm and 600 K. The volume at B is twice the volume at A. The pressure at D is 1.00 atm. (a) What is the pressure at B? (b) What is the temperature at C? (c) Find the total work done by the gas in one cycle. Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points B, C, and D. We can then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the various thermodynamic processes of the cycle. (a) Apply the ideal-gas law for a fixed amount of gas to the isothermal process AB to find the pressure at B:

( )

kPa253

kPa253.3atm1

kPa101.325atm2.50

2atm00.5

A

A

B

AAB

=

==

==V

VVVPP


403

(b) Apply the ideal-gas law for a fixed amount of gas to the adiabatic process BC to express the temperature at C:

BB

CCBC VP

VPTT = (1)

Use the ideal-gas law to find the volume of the gas at B:

( ) ( )

L39.39kPa253.3

K600Kmol

J8.314mol2.00

B

BB

=

=

=P

nRTV

Use the equation of state for an adiabatic process and = 1.4 to find the volume occupied by the gas at C:

( )L75.78

atm1.00atm2.50L39.39

1.411

C

BBC

=

=

=

PPVV

Substitute numerical values in equation (1) and evaluate TC:

( ) ( )( )( )( )K462

L39.39atm2.50L75.78atm1.00K600C

=

=T

(c) The work done by the gas in one cycle is given by:

DACDBCAB WWWWW +++=

The work done during the isothermal expansion AB is:

( ) ( ) kJ6.915V2VlnK600

KmolJ8.314mol2.00ln

A

A

A

BAAB =

=

=VVnRTW

The work done during the adiabatic expansion BC is:

( ) ( )kJ5.737

K006K624Kmol

J8.314mol2.00 25

BC25

BCVBC

=

=== TnRTCW

Chapter 19

404

The work done during the isobaric compression CD is:

( ) ( )( )kJ5.680

LatmJ101.325Latm56.09L75.78L19.7atm1.00CDCCD

==== VVPW

Express and evaluate the work done during the constant-volume process DA:

0DA =W

Substitute numerical values and evaluate W: kJ97.6kJ972.6

0kJ5.680kJ5.737kJ915.6

==++=W

83 A common practical cycle, often used in refrigeration, is the Brayton cycle, which involves (1) an adiabatic compression, (2) an isobaric (constant pressure) expansion,(3) an adiabatic expansion, and (4) an isobaric compression back to the original state. Assume the system begins the adiabatic compression at temperature T1, and transitions to temperatures T2, T3 and T4 after each leg of the cycle. (a) Sketch this cycle on a PV diagram. (b) Show that the efficiency of the

overall cycle is given by = 1 T4 T1( )

T3 T2( ) . (c) Show that this efficiency, can be written as = 1 r 1( ) , where r is the pressure ratio Phigh/Plow of the maximum and minimum pressures in the cycle. Picture the Problem The efficiency of the cycle is the ratio of the work done to the heat that flows into the engine. Because the adiabatic transitions in the cycle do not have heat flow associated with them, all we must do is consider the heat flow in and out of the engine during the isobaric transitions. (a) The Brayton heat engine cycle is shown to the right. The paths 12 and 34 are adiabatic. Heat Qh enters the gas during the isobaric transition from state 2 to state 3 and heat Qc leaves the gas during the isobaric transition from state 4 to state 1. 1

2 3

4

P

V

hQ

cQ


405

(b) The efficiency of a heat engine is given by: in

ch

in QQQ

QW == (1)

During the constant-pressure expansion from state 1 to state 2 heat enters the system:

( )23PPh23 TTnCTnCQQ ===

During the constant-pressure compression from state 3 to state 4 heat enters the system:

( )41PPc41 TTnCTnCQQ ===

Substituting in equation (1) and simplifying yields:

( ) ( )( )( )

( ) ( )( )( )( )23

14

23

4123

23P

41P23P

1TTTT

TTTTTT

TTnCTTnCTTnC

=

+=

=

(c) Given that, for an adiabatic transition, , use the ideal-gas law to eliminate V and obtain:

constant 1 =TV

constant 1 =

PT

Let the pressure for the transition from state 1 to state 2 be Plow and the pressure for the transition from state 3 to state 4 be Phigh. Then for the adiabatic transition from state 1 to state 2:

1high

21

low

1 =

PT

PT 2

1

high

low1 TP

PT

=

Similarly, for the adiabatic transition from state 3 to state 4:

3

1

high

low4 TP

PT

=

Subtract T1 from T4 and simplify to obtain:

( )231

high

low

2

1

high

low3

1

high

low14

TTPP

TPPT

PPTT

=

=

Chapter 19

406

Dividing both sides of the equation by T3 T2 yields:

1

high

low

23

14

=

PP

TTTT

Substitute in the result of Part (b) and simplify to obtain:

( )

=

=

=

1

1

low

high

1

high

low

1

11

r

PP

PP

where low

high

PP

r = 89 The English mathematician and philosopher Bertrand Russell (1872-1970) once said that if a million monkeys were given a million typewriters and typed away at random for a million years, they would produce all of Shakespeares works. Let us limit ourselves to the following fragment of Shakespeare (Julius Caesar III:ii): Friends, Romans, countrymen! Lend me your ears. I come to bury Caesar, not to praise him. The evil that men do lives on after them, The good is oft interred with the bones. So let it be with Caesar. The noble Brutus hath told you that Caesar was ambitious, And, if so, it were a grievous fault, And grievously hath Caesar answered it . . . Even with this small fragment, it will take a lot longer than a million years! By what factor (roughly speaking) was Russell in error? Make any reasonable assumptions you want. (You can even assume that the monkeys are immortal.) Picture the Problem There are 26 letters and four punctuation marks (space, comma, period, and exclamation point) used in the English language, disregarding capitalization, so we have a grand total of 30 characters to choose from. This fragment is 330 characters (including spaces) long; there are then 30330 different possible arrangements of the character set to form a fragment this long. We can use this number of possible arrangements to express the probability that one monkey will write out this passage and then an estimate of a monkeys typing speed to approximate the time required for one million monkeys to type the passage from Shakespeare. Assuming the monkeys type at random, express the probability P that one monkey will write out this passage:

330301=P


407

Use the approximation 5.110100030 = to obtain: ( )( )

4954953305.1 10101

101 ===P

Assuming the monkeys can type at a rate of 1 character per second, it would take about 330 s to write a passage of length equal to the quotation from Shakespeare. Find the time T required for a million monkeys to type this particular passage by accident:

( )( )( )

y10

s103.16y1s1030.3

1010s330

484

7491

6

495

=

=T

Express the ratio of T to Russells estimate:

4786

484

Russell

10y10y10 ==

TT

or Russell

47810 TT

Chapter 19

408

Ch19_SSM

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