Tài liệu Physics exercises_solution: Chapter 18 doc

b The net effect of the two changes is to keep the pressure the same while decreasing the Kelvin temperature by a factor of third, resulting in a decrease in volume by a factor of one-th

In doing the numerical calculations for the exercises and problems for this chapter, the values of the ideal-gas constant have been used with the precision given on page 501 of the text,

K

molatmL0.08206K

molJ3145

to find the pressure, Eq (18.3) gives

Pa

106.81atm67.2

L)(20.0

K)K)(291.15mol

atmL06mol)(0.0823

.56(

18.2: a) The final temperature is four times the initial Kelvin temperature, or 4(314.15

K) –273.15=983 to the nearest degree.C

)K15.314)(

Katm/molL

08206.0(

)L60.2)(

atm30.1)(

kg/mol10

00.4(

m

18.3: For constant temperature, Eq (18.6) becomes

atm

96.0)390.0110.0)(

atm40.3()( 1 21

p

18.4: a) Decreasing the pressure by a factor of one-third decreases the Kelvin

temperature by a factor of one-third, so the new Celsius temperatures is 1/3(293.15 K) –273.15=175Crounded to the nearest degree b) The net effect of the two changes is

to keep the pressure the same while decreasing the Kelvin temperature by a factor of third, resulting in a decrease in volume by a factor of one-third, to 1.00 L

one-18.5: Assume a room size of 20 ft X 20 ft X 20 ft

V 4000ft3 113m3.Assumea temperatureof 20C

molecules10

8.2

mol4685)

K293)(

KJ/mol315.8(

)m113)(

Pa1001.1(so

27

3 5

RT

pV n nRT pV

3 6

27

cmmolecules/

105.2cm

10113

molecules10

8.2

18.6: The temperature is T 22.0C295.15K (a) The average molar mass of air is

somol,kg10

atmL08206.0(

mol)kg10L)(28.8atm)(0.900

00.1

m

(b) For Helium M 4.0010 3kg mol,so

kg.1049.1K)K)(295.15mol

atmL08206.0(

mol)kg10L)(4.00atm)(0.900

00.1

m

18.7: From Eq (18.6),

)cmPa)(49910

01.1(

)cmPa)(46.210

(2.821K)

15.300

3 6

1 1

2 2 1

V p

b) Using the final pressure of 2.813105 Paand temperature of

kg,0.275K,

m750.0K300.15

K430.15Pa)

1050.1

3

3 5

2

1 1

2 1

T

p

18.10: a) 5.78 10 mol.

K)(295.15K)

molatmL(0.08206

L)10atm)(14000

.1

b) (32.010 3kg mol)(5.78103mol)185kg

18.11: V2 V1(T2 T1)(0.600L)(77.3 292.15)0.159L

18.12: a) nRT V 7.28106Pa whileEq.(18.7)gives5.87106 Pa b) The van der Waals equation, which accounts for the attraction between molecules, gives a pressure that is 20% lower

c) 7.28105 Pa,7.13105Pa, 2.1% d) As n V decreases, the formulas and the numerical values are the same

18.13: At constant temperature, p2  p1(V1 V2)(1.0atm)(6.05.7)1.1atm

18.14: a) (3.50)(296277KK) 3.74

1 2 2 1 1

p

p V

change; breathing is a good idea

K)molatmL0.08206mol)(

0.11(

L)atm)(3.10100

(2

b) This is a very small temperature increase and the thermal expansion of the tank may be neglected; in this case, neglecting the expansion means not including expansion

in finding the highest safe temperature, and including the expansion would tend to relax safe standards

18.16: (a) The force of any side of the cube is F  pA(nRT V)A(nRT) L, since the ratio of area to volume is A V 1 L.For T 20.0C293.15K.

m200.0

K)(293.15K)

molJ(8.3145mol)

b) For T 100.00C373.15K,

m200.0

)KK)(373.15mol

J5mol)(8.3143

18.17: Example 18.4 assumes a temperature of 0 at all altitudes and neglects the C

variation of g with elevation.

With these approximations, / RT

0

Mgy

e p

m850)90.0

(We have used M 28.810 3 kg/molfor air.)

18.18: From example 18.4, the pressure at elevation y above sea level is / RT

0e Mgy p

The average molar mass of air is M 28.810 3 kg/mol,so at an altitude of 100 m,

,01243.0K)

K)(273.15mol

J3145.8(

m)100)(

sm80.9)(

molkg108.28

and the percent decrease in pressure is 1 1 0 01243 0.0124 1.24%

0

1e 0 1243   These answers differ by a factor of

,44.91.24%

11.7%  which is less than 10 because the variation of pressure with altitude

is exponential rather than linear

18.19: p p0eMyg RT from Example 18.4.

Eq (18.5) says p(ρ M)RT. Example 18.4 assumes a constant

ρ p

T 273K,so and are directly proportional and we can write

RT Mgy

e ρ

ρ 0 

0 0.988so

,0124.0m,

RT

Mgy

The density at sea level is 1.2% larger than the density at 100m

18.20: Repeating the calculation of Example 18.4 (and using the same numerical values

for R and the temperature gives) (0.537) 5.44 104Pa

10mol)(1.41molecules

1002.6

kg108.9

kg1906

mol)molecules10

023.6(K)K)(300mol

J3145

8

(

)m10Pa)(1.0010

18.25: a)

mol)molecules10

(6.023

K)K)(7500mol

atmL08206.0(L

molecules10

N V

18.26: Since this gas is at standard conditions, the volume will be

3 16 3

3 m ) 2.23 10 m10

g1000

 which is (55.6mol)(6.0231023molecules mol)=molecules

10

18.28: a) The volume per molecule is

.m10091.4

Pa)10mol)(1.013molecules

10(6.023

K)K)(300.15mol

J3145.8(

3 26

5 23

p nRT N

V

If this volume were a cube of side L,

1 3

3 5 3

/ 1

A

3 / 1

molmolecules/

10023.6mol00.5

m1000.9

V

3.1010 10m

c) This is comparable to the size of a water molecule

18.30: a) From Eq (18.16), the average kinetic energy depends only on the

temperature, not on the mass of individual molecules, so the average kinetic energy is the same for the molecules of each element b) Equation (18.19) also shows that the rms speed is proportional to the inverse square root of the mass, and so

and301.000.222

18.20,

491.080.83

18.20

Ne rms

Rn rms Ne

v

.614.000.222

80.83

Kr rms

Rn

v v

18.31: a) At the same temperature, the average speeds will be different for the different

isotopes; a stream of such isotopes would tend to separate into two groups

.004

1

b) 00..349352 

18.32: (Many calculators have statistics functions that are preprogrammed for such

calculations as part of a statistics application The results presented here were done on such a calculator.) a) With the multiplicity of each score denoted by

.1.61150

1b)6.54150

1isaverage

the

,

2 / 1 2

18.33: We known that V A V B and that T A T B.

a)pnRT/V; we don’t know n for each box, so either pressure could be higher.

A

N RT

pVN N

RT N

how the pressures compare, so either N could be larger.

c)pV m MRT We don’t know the mass of the gas in each box, so they could contain the same gas or different gases

18.34: Box A has higher pressure than B This could be due to higher temperature

and/or higher particle density in A Since we know nothing more about these gases, none

of the choices is necessarily true, although each of them could be true.

3

b)

10 7

mms

2 rms

k mv T

18.36: From pV nRT, the temperature increases by a factor of 4 if the pressure and volume are each doubled Then the rms speed vrms  3RT M increases by a factor of

,

2

4  so the final rms speed is 2(250 m s)500m s

J)1021.6(22

23 3

K)K)(300mol

J3145.8(33

which is of course the square root of the result of part (b)

mol)molecules10

(6.023

mol)kg100.32(s

A

2.5710 23kgm

This may also be obtained from

mol)molecules10

(6.023

mol)kg10J)(32.010

21.6(2

3 21

N

1024.1m)(0.100

J)1021.6(2s2ss

mv v

L

mv F

f) 2 1.24 10 17Pa

ave ave

h) A NA

RT

pV N

molmolecules23

10023.6K300K/molatmL08206.0

L00.1atm00.1

i) The result of part (g) was obtained by assuming that all of the molecules move

in the same direction, and that there was a force on only two of the sides of the cube

18.38: This is the same calculation done in Example 16-9, but with

.m106.1givingatm,10

)K300)(

KJ10381.1(33

in thermal equilibrium with its surroundings, its motion will depend only on the

surrounding temperature, not the mass of the individual particles

18.41: a) The six degress of freedom would mean a heat capacity at constant volume of

K

J/mol9.243

mol kg 10 0 18

K mol J 3145 8 3 2

contribute to the heat capacity

18.42: a) C v  C   molar mass ,so 833J/kgC0.018kg/mol15.0J molC

18.43: a) UsingEq.(18.26),Q2.50mol20.79J molK30.0K1.56kJ.

b) From Eq (18.25), 53of the result of part (a), 936 J

18.44: a) 741 J/kg K,

kg/mol10

28.0

KJ/mol 76.20

w w N N N

C

C m m T

C m T C

nRT pV

atm1

K293Katm/molL

08206.0kg/mol028

.0/

) 60 1 (

3 3

M

RT m

kT

v   

(see Exercise 18.48), and so the temperature is

.)msK10385.4(K)molJ3145.8(3(1.60)

mol)kg100.28()

2

3 2

2

v v

m1500)(

msK10

m1000)(

msK10

m500)(

msK10

82

24)(

2 2

kT ε kT

πkT

m m

π e

m

ε πkT

m π v

18.47: Express Eq (18.33) as f  Aε eε kT, with A a constant Then,

e kT

ε e

A de

N R m

J3145

J3145

18.49: Ice crystals will form if T 0.0C; using this in the given relation for

temperature as a function of altitude gives y 2.5103 m2.5km

18.50: a) The pressure must be above the triple point, p1 610Pa.If p p1, the water cannot exist in the liquid phase, and the phase transition is from solid to vapor

18.52: The atmospheric pressure is below the triple point pressure of water, and there can

be no liquid water on Mars The same holds true for CO2

K)K)(295.15mol

J3145.8(

)m103000(Pa)10mol)(2.026kg

100.28



kg

213.0K)

15.295(K)molJ3145.8(

mol)kg1010.44)(

)m(0.060m)

00.1)((

p h T

T p

18.57: The change in the height of the column of mercury is due to the pressure of the

air The mass of the air is

RT

hV g ρ M RT

PV nM

)m10620.0))(

m690.0m900.0((

m)060.0)(

sm80.9)(

mkg106

13

(

2 4

2 3

ρ  where ρ is the density of the

ambient air and m is the load The density is inversely proportional to the temperature, so

11

)(

V m ρ

ρ ρ

ρ T T

)m500)(

mkg23.1(

)kg290(1

K)15.288(

1 3

m0159.0(

)K15.318)(

m0150.0()atm72.2

3

1 2

2 1 1

T V p p

so the gauge pressure is 1.92atm

18.60: (Neglect the thermal expansion of the flask.) a) p2  p1(T2 T1)

Pa

108.00380)300Pa)(

(30.1g mol) 1.45g

)KK)(300mol

J(8.3145

L)Pa)(1.5010

00.8

1

1030

m90.1(

m750

5 6

Vg ρ

ρ    2 

2

H air

H air )(

mol)kg10Pa)(2.0210

01.1(mkg23.1

3 5

18.62: If the original height is h and the piston descends a distance y, the final pressure

of the air will be atm 

mkg106.13(

Pa)10013.1(m)

900.0(

y

18.63: a) The tank is given as being “large,” so the speed of the water at the top of the

surface in the tank may be neglected The efflux speed is then obtained from

,2

Pa)1020.3()m50.2(sm80.9(2

ρ

p h g v

m00.4

h gives v16.1m s and with h2.00m, v5.44m s c) Setting v2 0 in

the above expression gives a quadratic equation in h which may be re-expressed as

.m4.00

m50.0g

m)00.1

h ρg

p ρ

p h

0

ρg

p y

ρ

p

this quadratic becomes

,0))m00.4()m00.4((

)m00.5

18.64:

s) 3600 ( K) 15 293 ( K) mol atm L 08206 0 (

) mol molecules 10

023 6 )(

L 5 14 )(

atm 00 1 (

a) 23          RT t pVN t nN t N A A 1.011020molecule36pt 10 min ) 163 0 210 0 )( L (0.5 min) 60 ) L 5 14 (



c) The density of the air has decreased by a factor of (0.72atm 1.00atm)

0.773, K)

273

K

293

(  and so the respiration rate must increase by a factor of 0.7331 ,

min breaths

13

to If the breathing rate is not increased, one would experience “shortness

of breath.”

18.65:

mol) kg 10 0 18 (

) kg mol)(50 molecules

10 023 6 ( 3 )

( 3 3

23 A







N

5.01027 atoms

18.66: The volume of gas per molecule (see Problem 18.28) is

Ap

N RT , and the volume of a molecule is about (2.0 10 m) 3.4 10 m

3

0





 π

volumes as f,

) m 10 mol)(3.4 molecules

10 023 6 (

K) K)(300 mol

J 3145 8

3 29 23

0 A

f f

V N

RT

f









“Noticeable deviations” is a subjective term, but f on the order of unity gives a pressure

of 108 Pa.Deviations from ideality are likely to be seen at values of f substantially lower

than this

18.67: a) Dividing both sides of Eq (18.7) by the product RTV gives the result b) The

algorithm described is best implemented on a programmable calculator or computer; for a calculator, the numerical procedure is an interation of

1 (4.29 10 ) .)

15.400)(

3145.8(

)448.0()

15.400)(

3145.8(

)108.9

represents the attraction of the molecules, and hence more molecules will be in a given volume for a given pressure

18.68:

J.1082.1)m400)(

sm80.9(molmolecules10

023.6

molkg100.28

U

A

KJ1038.1

J1082.13

2 ,2

all likely for a molecule to rise to that altitude This altitude is much larger than the mean free path

18.69: a), b) (See figure.) The solid curve is U (r), in units of U and with 0, xr R0.The dashed curve is F (r) in units of U0 R0 Note that r1 r2.

0 1 6

1 0 12

1

r

R r

3 2

gas is MpV RT , and so the ratio of the energies is

8.3145J mol K300K 2.42 10 0.0242%.

sm0.30molkg10016.23

13

12

18.72: a) 1.20 10 m s.

kg)1067.1(

K)5800()KJ1038.1(3

m)1096.6(

kg)1099.1(kgmN10673.6(2

8

30 2

2 11

c) The escape speed is about 50 times the rms speed, and any of Fig 18.20, Eq

18.32 or Table 18.2 will indicate that there is a negligibly small fraction of molecules with the escape speed

18.73: a) To escape, the total energy must be positive, K  U 0 At the surface of the earth, U GmM RmgR, so to escape K mgR b) Setting the average kinetic energy equal to the expression found in part (a),  3 2 kT mgR,or T  2 3 mgR k For nitrogen, this is

K)J10mol)(1.381molecules

10023.6(

m)1038.6(smmol)(9.80kg

100.28(3

2

23 - 23

6 2

18.74: (See Example 12.5 for calculation of the escape speeds)

a) Jupiter:

.(0.0221)s

m1031.1mol)kg1002.2(K)K)(140mol

J3(8.3145

m1065.1mol)kg1002.2((

K)K)(220mol

J3(8.3145

2GM R  so there can be no such atmosphere

18.75: a) From Eq (18.19),

s)m001.0(

K)K)(300J

10381.1(3s

2

23 2

b) mN A M (1.2410 14 kg)(6.0231023molecules mol) (18.010 3 kg mol) 4.161011 molecules

c)

3 3

4

324

32

V r

D

)mkg920(4

kg)1024.1(3

which is too small to see

18.76: From x Acosωt,vωAsinωt,

2

1

,)(cos2

ave ave

2 2

2 ave

18.77: a) In the same manner that Eq (18.27) was obtained, the heat capacity of the

two-dimensional solid would be 2R = 16.6 J molK b) The heat capcity would behave qualitatively like those in Fig (18.18), and heat capacity would decrease with decreasing temperature

18.78: a) The two degrees of freedom associated with the rotation for a diatomic

molecule account for two-fifths of the total kinetic energy, so Krot  nRT (1.00)

J

1049.2)KK)(300mol

023.6

molkg100.162

2)L(

023 6 )(

m kg 10 (1.94

J) 10 49 2 ( 2 2

3 rot

 6.52 1012 rad s,

much larger than that of machinery

18.79: For CO2,the contribution to CVother than vibrationis

and the respective fractions of C are 0.25 and 0.039.V

18.80: a)

12

2(4

12

4

24)(

2 0

2 / 2 2

π KT

m πkT

m π

dv e

v πkT

m π dv v

where the tabulated integral (given in Problem18.81) has been used b) f )(v dv is the probability that a particle has speed between vandvdv;the probability that the particle has some speed is unity, so the sum (integral) of f )(v dvmust be 1

18.81: With n2andam/2kT, theintegralis

,

3)2()2(2

32

2

m

kT kT

m

π kT

m πkT

2

.2

4)

πkT

m π dv v

2)(

2

dx xe

πkT

m π dv v

m π

2 2 8 ,

m π

KT m

KT



which is Eq (18.35)

18.83: a) See Problem 18.80 Because f )(v dv is the probability that a particle has a speed between vandvdv, f(v)dv is the fraction of the particles that have speed in that range The number of particles with speeds between v andvdv is therefore

mp

1 mp

2

v π e

e m

kT πkT

m π v

J3145.8(

)mPa)(1.0010

mol)(1.40kg

100.18

g RT

18.85: The partial pressure of water in the room is the vapor pressure at which

condensation occurs The relative humidity is 41..2581 42.6%