Tài liệu Physics exercises_solution: Chapter 20 pptx

b The magnitude of the entropy change is roughly five times the value found in Example 20.5.. The entropy per particle for any substance in a vapor state is expected to be roughly the sa

.

4

J 100

s) W)(1.00 10 180 (

b) 1(9.6) 0 40 60%, an increase of 2% If more figures are kept for the

efficiencies, the difference is 1.4%

20.9: a) 1.62 104J

10

210 J40

C C

Q

20.10: C 1 (Lf c T)

t

m K t K

Q t

kg0.88.2

J1080.9J1044

P H EER

W1633)

413.3(]s)(60J)108.9(s)(60J)1044

1

(

s)(60J)108.9(

4 5

K)K)(25.0kg

J4190(K)K)(5.0kg

J2100(kgJ10334kg)80.1(

210 J08 8

H C

|

T

T mL T

T Q

J,10088.3K)(273.15

K)(287.15kg)

J10kg)(3340

.85

20.16: a) From Eq (20.13), (320270KK)(415J)492J.b) The work per cycle is

J,77J

415

J

492   and P(2.75)1.0077 Js 212 W,keeping an extra figure

c) TC (TH  TC)(270K) (50K)5.4

20.17: For all cases, |W ||QH ||QC | a) The heat is discarded at a higher

temperature, and a refrigerator is required; | || |(( )1)(5.00103J)

C H

Q W

J

665)1)15

K)J10kg)(334(5.00

K)20(4190kg)(5.00

6

3 K

kgJ

F water

freeze C

0 water to Cool in

Carnot cycle:

K293K

268

J1009

hot

out cold

T

Q T

J1009

W W

20.20: For a heat engine, QH QC/1e(3000J) 10.6007500J,

J

4500)

J7500)(

600.0( then

and W  eQH   This does not make use of the given value

of TH IfTHisused, then for aCarnot engine,TC TH1e  800K10.600320K

,/andQH QCTH TC which gives the same result

J10825.1K15.273K373.15J

10336.1

3 H

C

4 4

C C H H

H

C H

W

Q T T Q

T

T Q

Q

20.22: The claimed efficiency of the engine is 58%

J 10 60 2 J 10 51 1 8 8

1

/

H C

H C

e

e e

e T

T

T T

C

f C

Q

K298.15

J1017

used,

20.25: a) Heat flows out of the 80.0 water into the ocean water and the C 80.0Cwater cools to 20.0 (the ocean warms, very, very slightly) Heat flow for an isolated Csystem is always in this direction, from warmer objects into cooler objects, so this

process is irreversible

b) 0.100kg of water goes form 80.0C to20.0Cand theheat flowisQmcT 

J102.154)

60.0CK)(

kgJkg)(4190

This Q comes out of the 0.100 kg of water and goes into the ocean.

For the 0.100 kg of water,

KJ78.02353.15)

293.15ln(

K)kgJkg)(4190100

.0()

293.15

J10154

J02.78

ocean water

20.26: (a) Irreversible because heat will not spontaneously flow out of 15 kg of water into

a warm room to freeze the water

(b) S Sice Sroom

room

F ice

F

T

mL T

mL 



K293

kg)J10kg)(3340

.15(K

273

)kgJ10kg)(3340

.15



1,250J K

This result is consistent with the answer in (a) because S 0 for irreversible processes

20.27: The final temperature will be

C, 60 kg)

(3.00

C) kg)(80.0 (2.00

C) kg)(20.0 00

1

and so the entropy change is

K

J4.47K

353.15

K333.15

ln kg)00.2(K293.15

K333.15

ln kg)(1.00K)

20.28: For an isothermal expansion,

K

J31.6is

entropy of

changeThe

.and

0,

Q U

216.4(

)kgJ10kg)(2.0913

.0

) kg J 10 kg)(2256 00 1

of entropy of the water as it changes to steam b) The magnitude of the entropy change

is roughly five times the value found in Example 20.5 Water is less ordered (more random) than ice, but water is far less random than steam; a consideration of the density changes indicates why this should be so

K)15.373(

)kgJ10kg)(225610

0.18

Q S

K)34.77(

kg)J10kg)(20110

0.28(:N

3 3

K)2466(

kg)J10kg)(233610

(107.9:

K)630(

kg)J10kg)(27210

(200.6:

c) The results are the same order or magnitude, all around 100J K The entropy change is a measure of the increase in randomness when a certain number (one mole) goes from the liquid to the vapor state The entropy per particle for any substance in a vapor state is expected to be roughly the same, and since the randomness is much higher

in the vapor state (see Exercise 20.30), the entropy change per molecule is roughly the same for these substances

20.32: a) The final temperature, found using the methods of Chapter 17, is

C,94.28K)kgJkg)(4190(0.800

K)kgJkg)(39050

.3(

)CK)(100kg

Jkg)(39050

.3

K273.15

K302.09

ln K)kgJkg)(4190800

.0(

K373.15

K302.09

ln K)kgJkg)(39050

.3(

m0.0420

ln K)molJ5mol)(8.31400

.2(

3 1

S

20.34: a) On the average, each half of the box will contain half of each type of

molecule, 250 of nitrogen and 50 of oxygen b ) See Example 20.11 The total change in entropy is

ln(2))(

c) See also Exercise 20.36 The probability is  1 2 500 12100  1 2 600 2.410 181,and is not likely to happen

The numerical result for part (c) above may not be obtained directly on some

standard calculators For such calculators, the result may be found by taking the log base ten of 0.5 and multiplying by 600, then adding 181 and then finding 10 to the power of the sum The result is then 10 181100 87 2.410 181

20.35: a) No; the velocity distribution is a function of the mass of the particles, the

number of particles and the temperature, none of which change during the isothermal expansion b) As in Example 20.11, 31 2

w  N (the volume has increased, and w2 w1); and

ln(3),)

(3ln

)

ln(w2 w1  N N S kNln(3)knN Aln(3)nRln(3)18.3J K.c) As in Example 20.8, SnRlnV2 V1nRln(3),the same as the expression used in part (b), and S 18.3J K

20.36: For those with a knowledge of elementary probability, all of the results for this

exercise are obtained from

2

1)!

4(

!4)

1()

p k

!

14!

4

! 1

 The number of heads must be one of 0, 1, 2, 3 or

4, and there must be unit probability of one and only one of these possibilities

C

T

T Q

34.3cycle

J

100

J10

3.34

isrequiredcycles

ofnumber the

so J,100iscycleone

for

J1034.3kgJ10334kg0.10is

requiredTotal

b)

4 6

6 3

f C

20.38: a) Solving Eq (20.14) for

,1

155.0

1K15.1831

11

1

C H

T T

T

b) Similarly, TC TH1e,andifTH TH,

600.0

050.0K15.1831

C C

20.39: The initial volume is 8.62 10 3m3

is given as atmospheric, and 1.01 105Pa,

p with the volume found above, V1 

Pa1003.22and

,m1062.8

3 3 1

2 3

isefficiency thermal

enginecycle

Carnota

-For

%

4.10104.0

e)

J.227J1956J

2183iscycleonefor engine theintoflowheat the2,-

1

process

the

for nscalculatio

in thefiguresextra

Keepingd)

J

227J559J786isdone

net work

Thec) J

1956)

K192)(

KmolJ3145.8)(

27)(

mol350

0

(

andJ1397)

K192)(

KmolJ3145.8)(

25(

J,

559

)K192)(

KmolJ3145.8)(

mol350.0(isobaric;

(J786)

K108)(

KmolJ3145.8)(

25)(

mol350.0(

and ,0adiabatic,is

3-2processThe

J

1018.2K300KJ/mol

3145

8

25mol350.0

0so0isochoric,

is2

Pa1001.1Pa)

10013

1

using

(

600 300 J

2183227J

3

3 2 1

3

5 1

3 5

1 3

nC

Q n

T nC U

T nR V p W

W T

nC

W

U Q

T nC Q U W

V

V V p

p p

p

V V

V T T a

20.40: (a) The temperature at point c is T c 1000K sincefrompV nRT, the maximum temperature occurs when the pressure and volume are both maximum So

8.3145J mol K1000K 2.16mol.

m0300.0Pa1000

V p n (b) Heat enters the gas along paths ab and bc, so the heat input per cycle is QH 

ac ac

Q   Path ab has constant volume and path bc has constant pressure, so

J.1020.1)m0100.0m0300.0)(

Pa1000.6()(

46.28,

COFor using

,)(

)T

00.2()mPa)(0.030010

00.6((

KmolJ3145

8

KmolJ46

00.2()(

105.48J100.400so

J,1048

105.86J1068.6

or ),)(

20.41: a) W 1.00J, TC 268.15K,TH290.15K

For the heat pump QC 0andQH 0

H

C H

C H

C ;combining this with

T

T Q

Q Q

Q

J2.13290.15)268.15

(1

J00.1

max T b T c   

T

K200K)600(3

K

Pa103.0

K)600)(

1moles)(8.32

nRT

V

P

%6767.01

1

)

(

%2121.0J102.10

J104.4

J104.4J101.66J1010.20

(b)

J101.66K)400(Kmole

J31.82

5moles)(2.00

J1010.2

J101.103

ln K)600(Kmole

J31.8moles)(2.00

J109.97K)400(Kmole

J31.82

3moles)2

(

and:

gas

Monatomic

m0997.01

3)m0332.0(

max

4

3 in

3 4

4 out

in

4 out

4 in

4

3 2

5 2

3

3 3

ca p ca

bc ab

c

b

c b b

c b bc

bc

ab V

ab

P V

a c

b b c c

c c b

b b

U

Q

T nC Q

Q

Q Q

Q

nRT V

V nRT dV

V

nRT PdV

W

Q

T nC

Q

R C R C

V P

P V V T

V P T V P

20.43: a)

Pa1057.2m

1000.5

K)K)(773mol

J5mol)(8.31400

.2(,

volume

minimumand

pressuremaximum

thehasstate(a),part

From

e)

cycleeach heat ofJ206 wastesJ;

C45K318J)]

500(J)206(K)[

773()(

J206J

500J294

,

J294m)00.2)(

smkg)(9.800

.15(

J500

b)

6 3

3

C

H

H C H

H C

2 H

pV

a Q

Q

W

e

Q Q

e p

e

MW

2.8kW)

hr)s3600( W)108.2

dt Q d dt

20.45: There are many equivalent ways of finding the efficiency; the method presented

here saves some steps The temperature at point 3 is T3 4T0, and so

,2

192

)3(2

5)2

)(

2()(

QH U13W13 nC V T3T0  p0 V0V0  nRT0  p0V0  p0V0

where nRT0  p0V0has been used for an ideal gas The work done by the gas during one cycle is the area enclosed by the blue square in Fig (20.22), W  p0V0,and so the efficiency is 192 10.5%

 Q W

e

20.46: a) p2  p1 2.00 atm,

L,00.6

L

6.00L)(3/2)00

.4

3 3

),(p1 V1 (2.00 atm, 4.00 L) (p2,V2)(2.00 atm, 6.00 L)

),(p3 V3 (1.111 atm, 6.00 L) (p4,V4)(1.67 atm, 4.00 L)

b) The number of moles of oxygen is ,

1

1

RT V p

n and the heat capacities are those in Table (19.1) The product p1V1has the valuex810.4J; using this and the ideal gas law,

J,14222)

1J)(

4.810)(

508.3(1

C T nC

0

WJ,13553)

2J)(

4.810)(

508.2(

C T nC

V

W Q V

V T

T x V

V nRT

3 3

4 3

0

WJ,3396)1J)(

4.810)(

508.2(1

C T nC

V

In the above, the terms are given to nearest integer number of joules to reduce roundoff error

c) The net work done in the cycle is 405J-274J131J

d) Heat is added in steps i and iv, and the added heat is 1422J339J1761J and the efficiency is 1761131 JJ 0.075,or 7.5% The efficiency of a Carnot-cycle engine operating between 250K and450K is1450250 0.4444%

20.47: a) U 1657kJ1005kJ6.52105J,W  pV (363103Pa)

J,1039.8)m2202.0

J 10 1.97 J 10 36 7 5

5 5

00.3)(

5.2()

00.2)(

5.1()

0

(V  U QW   5

ca: The easiest way to do this is to find the work done first; W will be the negative of area

in the p-V plane bounded by the line representing the process ca and the verticals from points a and c The area of this trapezoid is (3.00105 Pa1.00105 Pa)

2 1

J,1000.6)m500

0

m

(0.800 3 3   4 and so the work is 0.60105J.U must be

0(since

 

Q W

20.49: a) ab: For the isothermal process, T 0andU 0.W nRT1ln(V b V a)

Q

U   V 

 b) The values of Q for the processes are the negatives of each

other c) The net work for one cycle is Wnet nR(T2 T1)ln(r),and the heat added (neglecting the heat exchanged during the isochoric expansion and compression, as mentioned in part (b)) is Qcd nRT2ln(r), and the efficiency is 1 (T1 T2)

e   and that of the second is C,

T T

e  and the overall efficiency is

C H

H 2

T T e e e

The first term in the product is necessarily less than the original efficiency since TTC,and the second term is less than 1, and so the overall efficiency has been reduced

20.51: a) The cylinder described contains a mass of air m ρ(πd2 4)L, and so the total kinetic energy is K ρ(π 8)d2Lv2. This mass of air will pass by the turbine in a time

ρ t

ms W5.0(

)25.0( W)102.3

2 5

4

6 3

/ 1

v

c) Wind speeds tend to be higher in mountain passes

20.52: a)   9.89L h.

gal1

L788.3km1.609

mi1mi25

gal1hkm

3600

hL

20.53: (Extra figures are given in the numerical answers for clarity.) a) The efficiency

is e1r 0 400.611, so the work done is Q H e122Jand|QC|78J b) Denote the

length of the cylinder when the piston is at point a by L0and the stroke as s Then,

is volumeand

6.101

3 4 2

3 3

L

c) The calculations are presented symbolically, with numerical values substituted at the

end At point a, the pressure is p a 8.50104 Pa, the volumeisV a 5.1010 4 m3as found in part (b) and the temperature is T a 300K.At point b, the volumeisV b V a r,the pressure after the adiabatic compression is γ

H H

H

a a

a V a a V a a a V

T f T

T C V p

RQ C

RT V p

Q nC

Q T

c    1 The pressure is found from the volume and

,)1()

nC

Q

a d a V

isochoric, so ,and (1 (1 ) ).Asacheck,note that γ

c d a

d a

)(

1 1

1

f e T

V f

e p

d

f r T r V f

p

b

T V

p

a

T V

p

a a

a

γ a a

γ

a

γ a a

γ

a

a a

20.54: (a) for furnaceand water

L

T A k t

0

K313

1K523

1K

210cm

100

m1cm15m

65

0

)Km W

2 2

w f

w f

water furnace

T

kA

T

L T kA T

L T

kA

t

S t

20.55: a) Consider an infinitesimal heat flow dQ that occurs when the temperature of H

the hot reservoir is T :

(

H

H C

C

H C

C

H C

C

S T T

dQ

dQ T

J1057.3

so engine, the

ofout

comes

J1057.3)KJ1308)(

K273(

|

|gives(a)part

of

result

The

J/K1308)

373273ln(

)KkgJ4190)(

kg00.1()(

ln

S

J1019.4)K100)(

KkgJ4190)(

kg00.1(

4 5

5 H

C

5 C

C

5 C

1 2

5 H

W

Q Q

Q

T T

J1085.3

|

|

KJ1041.1)323/273(ln)KkgJ4190)(

kg00.2()(

ln

J1019.4)K50)(

KkgJ4190)(

kg00.2(

4 H

C

5 h

C

C

3 1

2

5 H

20.56: See Figure (20.15(c)), and Example 20.8.

a) For the isobaric expansion followed by the isochoric process, follow a path fromT to2T toT UsedQnC VdT or dQnC pdT toget S nC pln2nC V ln 21 

)(

2

ln ln

Then

20.57: The much larger mass of water suggests that the final state of the system will be

water at a temperature between 0Cand60.0C.This temperature would be

C,83.34)

KkgJ4190)(

kg650.0(

)kgJ10334

C0.15KkgJ2100kg

0500.0

C0.45KkgJ4190kg0.600

kgJkg)(4190(0.600

S

K

J5.10273.15

307.98

ln K)kgJ4190(

K273.15

kgJ10334

258.15

273.15

ln K)kgJ(2100kg)0500.0(

(Some precision is lost in taking the logarithms of numbers close to unity.)

20.58: a) For constant-volume processes for an ideal gas, the result of Example 20.10

may be used; the entropy changes are nC V ln(T c T b)and nC V ln(T a T d) b) The total entropy change for one cycle is the sum of the entropy changes found in part (a); the

other processes in the cycle are adiabatic, with Q = 0 and S 0.The total is then

.ln

a c V d

a V b

c V

T T

T T nC T

T nC T

T nC S

From the derivation of Eq (20.6), 1 and 1 ,

d

γ c a

γ

T     and so the argument of the logarithm in the expression for the net entropy change is 1 identically, and the net entropy change is zero c) The system is not isolated, and a zero change of entropy for an

irreversible system is certainly possible

20.59: a)

b) From Eq (20.17), dS  dQ T ,andsodQT dS,and

QdQT dS

which is the area under the curve in the TS plane c) Q is the area under the rectangle H

bounded by the horizontal part of the rectangle at T and the verticals H |QC | is the area bounded by the horizontal part of the rectangle at TCand the verticals The net work is then QH|QC |, the area bounded by the rectangle that represents the process The ratio

of the areas is the ratio of the lengths of the vertical sides of the respective rectangles, and

H C H

T T

Q W

e   d) As explained in problem 20.49, the substance that mediates the heat exchange during the isochoric expansion and compression does not leave the system, and the diagram is the same as in part (a) As found in that problem, the ideal efficiency is the same as for a Carnot-cycle engine

Q S

K

J196Δ

b)   f  (0.160kg)(334(273.15K)103Jkg) 

T

mL T Q

S

c) From the time equilbrium has been reached, there is no heat exchange between the rod and its surroundings (as much heat leaves the end of the rod in the ice as enters at the end of the rod in the boiling water), so the entropy change of the copper rod is zero d) 196J K143J K53 J K

20.61: a) S mcln(T2 T1)

K

J150

K)293.15K

5K)ln(338.1kg

Jkg)(419010

250(

parts (a) and (b) is Ssystem 30J K d) Heating a liquid is not reversible Whatever the energy source for the heating element, heat is being delivered at a higher temperature than that of the water, and the entropy loss of the source will be less in magnitude than the entropy gain of the water The net entropy change is positive