Tài liệu Physics exercises_solution: Chapter 19 pptx

b For the process in reverse, the pressures are the same, but the volume changes are all the negatives of those found in part a, so the total work is negative of the work found in part a

ln )K)(400.15K

molJ3145.8)(

3(

19.4: At constant pressure, W  pV nRT,so

C.62.1C

35.1C0.27so

,Tand

K1.35K)molJ(8.3145mol)

(6

J1075.1

2 C K

nR

W T

19.5: a)

b) At constant volume, dV 0 andsoW 0

19.6:

b) p V (1.50105 Pa)(0.0600 m3 0.0900 m3)4.50103 J

19.7: a)

b) In the first process, W1  pV 0 In the second process,

J

1000.4)m0.080(

Pa)10005

W

19.8: a) W13  p1(V2 V1),W32 0,W24  p2(V1V2) and W41 0 The total work done

by the system is W13 W32 W24 W41 (p1 p2)(V2 V1), which is the area in the p- V

plane enclosed by the loop b) For the process in reverse, the pressures are the same, but the volume changes are all the negatives of those found in part (a), so the total work is negative of the work found in part (a)

19.9: Q254 J,W 73J (work is done on the system), and so U QW 327 J

19.10: a) p V (1.80105 Pa)(0.210m2)3.78104 J

b) U QW 1.15105J3.78104 J7.72104 J

c) The relations W  pV andU QW hold for any system

19.11: The type of process is not specified We can use U QW because this applies

to all processes

Q is positive since heat goes into the gas; Q1200J

Wpositive since gas expands; W 2100J

J900J

2100-J

J)900(23

2

nR

U T

C113C

14.4C1271

c) No; the first law of thermodynamics is valid for any system

19.14: a) The greatest work is done along the path that bounds the largest area above the

V-axis in the p- V plane (see Fig (19.8)), which is path 1 The least work is done along

path 3 b) W in all three cases; 0 QU W,soQ0 for all three, with the greatest

Q for the greatest work, that along path 1 When Q0,heat is absorbed

19.15: a) The energy is

kcal,139g)kcalg)(9.0(7.0g)kcalg)(4.00.17()gkcal0.4)(

10

139

(

19.16: a) The container is said to be well-insulated, so there is no heat transfer b) Stirring

requires work The stirring needs to be irregular so that the stirring mechanism moves

against the water, not with the water c) The work mentioned in part (b) is work done on

the system, so W 0, and since no heat has been transferred, U W 0

19.17: The work done is positive from a to b and negative from b to a; the net work is the

area enclosed and is positive around the clockwise path For the closed path U 0,so

Q positive value for Q means heat is absorbed

b) Q 7200J,andfrompart (a),Q0andsoQW 7200J

c) For the counterclockwise path, Q = W < 0 W=7200J, soQ7200Jand heat

is liberated, with |Q|=7200 J.

19.18: a), b) The clockwise loop (I) encloses a larger area in the p-V plane than the

counterclockwise loop (II) Clockwise loops represent positive work and

counterclockwise loops negative work, so WI 0andWII 0 Over one complete cycle, the net work WI WII 0, and the net work done by the system is positive c) For the complete cycle, U 0and soW Q. From part (a), W > 0 so Q > 0, and heat flows

into the system d) Consider each loop as beginning and ending at the intersection point

of the loops Around each loop, U 0,soQW; then, QI WI 0andQII WII 0.Heat flows into the system for loop I and out of the system for loop II

19.19: a) Yes; heat has been transferred form the gasses to the water (and very likely the

can), as indicated by the temperature rise of the water For the system of the gasses, 0

.1(

b)

6 5

19.21: a) Using Equation (19.12), d d (0.185mol)(20.76645 J JmolK) 167.9K,or

b) Using Equation (19.14), dT  d  (0.185mol)(29.07645 J Jmol.K) 119.9K,or T 900K

p

nC Q

19.22: a)nC V T (0.0100mol)(12.47J molK)(40.0C)4.99J

19.23: n5.00mol.T 30.0C

a) For constant p,

J3120)

CK)(30.0mol

Jmol)(20.7800

.5

CK)(30.0mol

Jmol)(12.4700

.5

CK)(30.0mol

Jmol)(36.9400

.5

Q > 0 so heat goes into gas.

19.24: For an ideal gas, U C VT, and at constant pressure,

R C

00.4(2

32

32

V p nC

19.27: a) For an isothermal process,

W nRTln (V2 V1)(0.150mol)(8.3145J molK)(350.15K)ln(14)

605J

b) For an isothermal process for an ideal gas, T 0andU 0 c) For a process with U 0,Q  W 605J; 605 J are liberated

19.28: For an isothermal process, U 0,soW Q335 J

19.29: For an ideal gas γCp CV 1R CV,and so CV R ( 1)

KmolJ5.65)127.0(K)mol

c) The work is done on the piston

d) Since Eq (19.13) holds for any process,

19.31: a) C p R 1 1γ ,and so

J

553220

.111

C0.5KmolJ3145.8mol40.2

m0800.0Pa1050

3

3 5

2

1 1 2

0800.01m0800.0Pa1050.123

111

4 3

5

1 2

1 1

p γ W

c) From Eq (19.22),  2 1  2 1 1 0.0800 0.0400 3 1.59,

V V T

the final temperature is higher than the initial temperature, the gas is heated (see the note

in Section 19.8 regarding “heating” and “cooling.”)

19.33: a)

)19.6Example

in as,400

and from Eq (19.24), 2  1 1 2   1.00atm 11.11.400 29.1atm

V V p p

19.34: γ1.4for idealdiatomicgasQU W 0for adiabatic process

γ i i

PV

PdV W

L)(30atm)2.1(

VPdVU

3

4 1 1 1.4 - 1 L) 30 ( 1.4 - 1 L) 10 ( 1.4

10L1 L 30 1

1 V i i V

V P 10L L

The internal energy increases because work is done on the gas (U 0)

The temperature increases because the internal energy has increased

19.35: For an ideal gas U nC VT Thesign of Uis thesameas thesign of T

negative;

issonegativeis

.so

positiveis

1and

)(

and

so,and

1 2 1

2

1 1 2 1 2

T T p

p

p p T T p

T

p

T

p nRT V

19.36: Equations (19.22) and (19.24) may be re-expressed as

.,

2

1 1 2 1 2

1 1 2

γ γ

V

V p

p V

V T

32(K)350(atm,27.2)32(atm)00.4(,b)

K.267)

32K)(

350(atm,04.2)32atm)(

00.4(,)

5 5

3 3

2 2

5 7

2 2

3 5

γ

T p

an For

J

19.38: a)   ( 1(.000.1mol)105Pa)(8.3145(2.50 J10mol3Km)3) 301K.

nR pV

T

b) i) Isothermal: If the expansion is isothermal, the process occurs at constant

temperature and the final temperature is the same as the initial temperature, namely

100.0(

)m10(5.00Pa)1000.1

pV T

iii) Adiabatic: Using Equation (19.22), (301K)( ) 189

)2(

))(

301

2

1 67

1

67 1

2  p V V   

p

b) Using Equation (19.26),

,)

129.1(

)]

m100.1)(

mN105.4()m100.5)(

mN101.1[(

1

)(

3 2 3

4 3

3 3

5

2 2 1 1

.818.0))m1000.1()m1000.5((

)()

(

1 2 1

19.40: a) The product pV increases, and even for a non-ideal gas, this indicates a

temperature increase b) The work is the area in the p plane bounded by the blue line V

representing the process and the vericals at V a and Vb. The area of this trapeziod is

J

4800)

m(0.0400Pa)

1040.2(2

1))(

(2

19.41: W is the area under the path from A to B in the pV -graph The volume

U  V 



nR

V p V p T T T nR

V P T nR

V

p

1 2 2

2 2 1

1

))(

(C R p2V2 p1V1



10015.5)]

mPa)(0.8010

150()mPa)(0.2010

500)[(

315.885.20

101.95J10015

19.42: (a) Q abc U acW abc nC vT acW abc

get T ac:PV nRT T PV nR

nR

V P V P nR

V P nR

V P T T

a c ac

moleJ8.31mole)(

(

)mPa)(0.002010

0.1()mPa)(0.01010

0.1(

3 1

3 5

3 5

Pa)105.2(m)002.0010.0(2

1graph under

Area

5 3

5 3

108.1J1020.1

J1020.1K289Kmole

J31.82

3mole3

1

23

J1080.1

3 3

3 ac

v abc

Q

T R n T nC U

800J1200

J800Pa100.1m002.0010.0

ac

W U Q

W

(c) More heat is transfered in abc than in ac because more work is done in abc.

19.43: a)U QW 90.0J  60.0J30.0J for any path between a and b If W=15.0 J along path abd, then QU W 30.0J15.0J45.0J b) Along the return path, U 30.0J,andQU W 30.0J  35.0J65.0J;the negative sign indicates that the system liberates heat

c) In the process db,dV 0and so the work done in the process ad is

  8.00J 15.0J 23.0J.In theprocess ,J;

0.8J0.30so

and

W

19.44: For each process, QU W No work is done in the processes ab and dc, and

so W bc W abc andW ad W adc, and the heat flow for each process is: for ab,Q90J:for

J

350,

for :J300J120J180,

for :J890J450J

440

= 350 each process, heat is absorbed in each process Note that the arrows representing

the processes all point the direction of increasing temperature (increasing U).

19.45: We will need to use Equations (19.3), W  pV2 V1and(17-4),U QW

a) The work done by the system during the process: Along ab or cd, W=0 Along bc,

c c b c a b abc a c c

abc

d c dc d

c dc

a c a a d ad a

d ad

a c c b c bc b

c bc

a b ab a

b ab

V V p U U Q

V V p

W

adc c

a

V V p U U V V p U U U U Q V V p

W

abc c

a

U U Q U U

U

V V p U U Q U U

U

V V p U U Q U U

U

U U Q U U

tostate

From

:path alongstate

tostate

From

c)

.0so

,

.so

,

.so

,

.0so

,

Assuming p c  p a,Q abc Q adc,andW abc W adc

d) To understand this difference, start from the relationship QW U The internal energy change U is path independent and so it is the same for path abc and path adc The work done by the system is the area under the path in the pV-plane and is not the same for the two paths Indeed, it is larger for path abc Since U is the same and W is different, Q must be different for the two paths The heat flow Q is path dependent.

19.46: a)

0 since gas the J; 28,000

Pa) 6000 )( m 7.00 ( Pa) 2000 )( m (7.00

) area( 0 ) area( 0

b)

3 3





























W on

da bc

W W W W

W ab bc cd da

c) QU W 0(28,000 J)28,000 J

Heat comes out of the gas since Q < 0.

19.47: a) We aren’t told whether the pressure increases or decreases in process bc The

cycle could be

In cycle I, the total work is negative and in cycle II the total work is positive For a cycle,

tot tot so

,



The net heat flow for the cycle is out of the gas, so heat

0 and

b) Wtot  Qtot 800J

ca bc

W

Wtot   

0 since

W bc

p V

p

W ab   since is constant But since it is an ideal gas, pV nRT

J 1660 )

W

J 2460 J

1660 J

800

W

19.48: Path achas constant pressure, so W ac  pV nRT, and

(3mol)(29.1J molK8.3145J molK)(600K492K)6.735103 J

Path ba has constant volume, so W ba 0 So the total work done is

p

n

29.07 20.76

,

dV

19.51: U 0,andsoQW  pV and

Pa)10(9.50

J)1015.2

with the negative sign indicating a decrease in volume

19.52: a)

b) At constant temperature, the product pV is constant, so

L)5.1()( 2.501.001010 PaPa

2 1

3 5

208J,

keeping an extra figure For the second process,

))(

1()

()

Pa1000.11)m10Pa)(1.510

50.2

5 3

19.53: a) The fractional change in volume is

.m104.32K)0.30)(

K1020.1)(

m1020.1

J1051.2)(

mkg791)(

m1020.1(

J390)(

m1000.8)(

mkg109

19.55: For a mass m of ejected spray, the heat of reaction L is related to the temperature

rise and the kinetic energy of the spray by mLmCT (1 2)mv2,or

kg

J104.3)sm19(2

1)C(80)KkgJ4190(2

1

1 2

1 2 1 2

1 2 1

1 1

γ

p

p T T T p T

10 80 2 10 1.60 1

5

γ

19.57: a) As the air moves to lower altitude its density increases; under an adiabatic

compression, the temperature rises If the wind is fast-moving, Q is not as likely to be

signigicant, and modeling the process as adiabatic (no heat loss to the surroundings) is more accurate b) See Problems 19.59 and 19.56: The temperature at the higher

pressure is

andC13.9is which K,

1.287Pa))

10Pa)/(5.6010

R

C V V p

soand)4

Note that p is the absolute pressure c) The most direct way to find the temperature is to 0

find the ratio of the final pressure and volume to the original and treat the air as an ideal gas;

  γ

γ γ

T T

V

V V

V T V p

V p T

1

3 3

2 0 1 1

3 3 0

21

0

0 0 0

V p Q RT

V p

heat flows into the gas

19.59: a) From constant cross-section area, the volume is proportional to the length, and

Eq (19.24) becomes L L p p 1 /γ

2 1 1

2  and the distance the piston has moved is

Pa105.21

Pa1001.11m250.01

400 1 / 1 5 5 /

1 2

1 1

p L

L L

0.173m

b) Raising both sides of Eq (19.22) to the power γand both sides of Eq (19.24) to the power γ1, dividing to eliminate the terms ( 1 )

2 )

1 (

Pa1021.5K15.300

400 1 1 5

5 /

1 1 1

2 1

Using the result of part (a) to find L and then using Eq (19.22) gives the same result.2

c) Of the many possible ways to find the work done, the most straightforward is to use the result of part (b) in Eq (19.25),

5

5 3

6 3

a 0

5 3

6 3

1 a 0

Pa101.01

Pa1045.1)m10575(mkg23.1

0,and

U W c) For isobaric, W  pdV nRdT,or dT  nR W Then, QnCpdT

and substituting for dT gives

duse

,findToJ

750Thus,

J)

300(or

2

Q C

nC

R

W p

19.62: a)

b) The isobaric process doubles the temperature to 710 K, and this must be the temperature of the isothermal process c) After the isothermal process, the oxygen is at its original volume but twice the original temperature, so the pressure is twice the original pressure, 4.80105 Pa d) Break the process into three steps

J;

738K)

K)(335mol

J5mol)(8.31425

.0(o

W

)693)(.K710)(

KmolJ5mol)(8.314250

.0()21ln(

)2()(p

K)(355mol

Jmol)(29.17

and

T

19.64: a)

b) At constant pressure, halving the volume halves the Kelvin temperature, and the temperature at the beginning of the adiabatic expansion is 150K.The volume doubles during the adiabatic expansion, and from Eq (19.22), the temperature at the end of the expansion is (150K)(1 2)0.40 114K c) The minimum pressure occurs at the end of the adiabatic expansion During the heating the volume is held constant, so the minimum

pressure is proportional to the Kelvin temperature,

Pa

106.82K)

300KPa)(113.710

654K)

150K)(

molJmol)(29.07150

.0

J5mol)(8.314150

.0(40.0

in Problem 19.64 and part (b),

andJ,580K)113.7-KK)(300mol

Jmol)(20.76150

.0





21 (γ 1 )  0 33 and the final volume is (1410 3 m3)2 0 33 0.114m3

19.68: a) The difference between the pressure, multiplied by the area of the piston,

must be the weight of the piston The pressure in the trapped gas is 0 0 πr2

mg A

p   

b) When the piston is a distance h above the cylinder, the pressure in the trapped y

2 0 2

0

mg πr p h y

mg πr

p h

y πr

mg p

This form shows that for positive h, the net force is down; the trapped gas is at a lower

pressure than the equilibrium pressure, and so the net force tends to restore the piston to equilibrium c) The angular frequency of small oscillations would be

0

2 0 2

g m

h mg πr p ω

If the displacements are not small, the motion is not simple harmonic This can be seen be

considering what happens if y ~ h; the gas is compressed to a very small volume, and

the force due to the pressure of the gas would become unboundedly large for a finite

displacement, which is not characteristic of simple harmonic motion If y >> h (but not so

large that the piston leaves the cylinder), the force due to the pressure of the gas becomes small, and the restoring force due to the atmosphere and the weight would tend toward a constant, and this is not characteristic of simple harmonic motion

19.69: a) Solving for p as a function of V and T and integrating with respect to V,

.11ln

1 2

2 1

2 2 2

nb V nRT pdV

W

V

an nb V

nRT p

V V

When ab0,W nRT lnV2 V1,as expected b) Using the expression found in part

ii)

J

1080

2

m1000.2

1m

1000.4

1mol

80.1molmJ554.0

mol/m1038.6mol80.1m1000.2

mol/m1038.6mol80.1m1000.4

ln

K300KmolJ3145.8mol80.1

i)

3 3

3 3 3

3 2

2 3

2 5 3

3

2 5 3

c) 300 J to two figures, larger for the ideal gas For this case, the difference due to

nonzero a is more than that due to nonzero b The presence of a nonzero a indicates that

the molecules are attracted to each other and so do not do as much work in the expansion