Solucionario serway cap 21

21The Kinetic Theory of Gases

CHAPTER OUTLINE

21.1 Molecular Model of an Ideal Gas21.2 Molar Specifi c Heat of an Ideal Gas21.3 Adiabatic Processes for an Ideal Gas21.4 The Equipartition of Energy21.5 Distribution of Molecular Speeds

ANSWERS TO QUESTIONS

Q21.1 The molecules of all different kinds collide with the walls of the container, so molecules of all different kinds exert partial pressures that contribute to the total pressure. The molecules can be so small that they collide with one another relatively rarely and each kind exerts partial pressure as if the other kinds of molecules were absent. If the molecules collide with one another often, the collisions exactly conserve momentum and so do not affect the net force on the walls.

Q21.2 The helium must have the higher rms speed. According to Equation (21.4), the gas with the smaller mass per atom must have the higher average speed-squared and thus the higher rms speed.

Q21.3 The alcohol evaporates, absorbing energy from the skin to lower the skin temperature.

*Q21.4 (i) Statements a, d, and e are correct statements that describe the temperature increase of a gas.

(ii) Statement b is true if the molecules have any size at all, but molecular collisions with other molecules have nothing to do with temperature.

(iii) Statement c is incorrect. The molecular collisions are perfectly elastic. Temperature is determined by how fast molecules are moving through space, not by anything going on inside a molecule.

*Q21.5 (i) b. The volume of the balloon will decrease.

(ii) c. The pressure inside the balloon is nearly equal to the constant exterior atmospheric pressure. Snap the mouth of the balloon over an absolute pressure gauge to demonstrate this fact. Then from PV nRT= , volume must decrease in proportion to the absolute tempera-ture. Call the process isobaric contraction.

*Q21.6 At 200 K, 1

2

3

20 0m k TBvrms02 = . At the higher temperature,

1

22

3

20

2m k TBvrms0( ) =

Then T T= = ( ) =4 4 200 8000 K K. Answer (d).

*Q21.7 Answer c > a > b > e > d. The average vector velocity is zero in a sample macroscopically at rest. As adjacent equations in the text note, the asymmetric distribution of molecular speeds makes the average speed greater than the most probable speed, and the rms speed greater still. The most probable speed is (2RT�M)1�2 and the speed of sound is (γRT�M)1�2, necessarily smaller. Sound represents an organized disturbance superposed on the disorganized thermal motion of molecules, and moving at a lower speed.

543

13794_21_ch21_p543-570.indd 54313794_21_ch21_p543-570.indd 543 12/27/06 12:04:44 PM12/27/06 12:04:44 PM

544 Chapter 21

*Q21.8 Answer (b). The two samples have the same temperature and molecular mass, and so the same rms molecular speed. These are all intrinsic quantities. The volume, number of moles, and sample mass are extrinsic quantities that vary independently, depending on the sample size.

Q21.9 The dry air is more dense. Since the air and the water vapor are at the same temperature, they have the same kinetic energy per molecule. For a controlled experiment, the humid and dry air are at the same pressure, so the number of molecules per unit volume must be the same for both. The water molecule has a smaller molecular mass (18.0 u) than any of the gases that make up the air, so the humid air must have the smaller mass per unit volume.

Q21.10 Suppose the balloon rises into air uniform in temperature. The air cannot be uniform in pressure because the lower layers support the weight of all the air above them. The rubber in a typical balloon is easy to stretch and stretches or contracts until interior and exterior pressures are nearly equal. So as the balloon rises it expands. This is an isothermal expansion, with P decreasing as V increases by the same factor in PV nRT= . If the rubber wall is very strong it will eventu-ally contain the helium at higher pressure than the air outside but at the same density, so that the balloon will stop rising. More likely, the rubber will stretch and break, releasing the helium to keep rising and “boil out” of the Earth’s atmosphere.

Q21.11 A diatomic gas has more degrees of freedom—those of molecular vibration and rotation—than a monatomic gas. The energy content per mole is proportional to the number of degrees of freedom.

*Q21.12 (i) Answer (b). Average molecular kinetic energy increases by a factor of 3.

(ii) Answer (c). The rms speed increases by a factor of 3.

(iii) Answer (c). Average momentum change increases by 3.

(iv) Answer (c). Rate of collisions increases by a factor of 3 since the mean free path remains unchanged.

(v) Answer (b). Pressure increases by a factor of 3. This is the product of the answers to iii and iv.

Q21.13 As a parcel of air is pushed upward, it moves into a region of lower pressure, so it expands and does work on its surroundings. Its fund of internal energy drops, and so does its temperature. As mentioned in the question, the low thermal conductivity of air means that very little energy will be conducted by heat into the now-cool parcel from the denser but warmer air below it.

*Q21.14 Answer (a), temperature 900 K. The area under the curve represents the number of molecules in the sample, which must be 100 000 as labeled. With a molecular mass larger than that of nitrogen by a factor of 3, and the same speed distribution, krypton will have (1�2)mov2 = (3�2)k

BT average

molecular kinetic energy larger by a factor of 3. Then its temperature must be higher by a factor of 3 than that of the sample of nitrogen at 300 K.


SOLUTIONS TO PROBLEMS

Section 21.1 Molecular Model of an Ideal Gas

P21.1 Use 1 u . g= × −1 66 10 24

(a) For He, m0244 00 6 64 10=

×⎛⎝⎜

⎞⎠⎟

= ×. .u1.66 10 g

1 ug

24−−

(b) For Fe, m02355 9 9 29 10=

×⎛⎝⎜

⎞⎠⎟

= × −. .u1.66 10 g

1 ug

24−

(c) For Pb, m0

2422207

1 66 103 44 10= ×⎛

⎝⎜⎞⎠⎟

= ×−

−ug

1 ug

..

*P21.2 Because each mole of a chemical compound contains Avogadro’s number of molecules, the number of molecules in a sample is N

A times the number of moles, as described by N = nN

A,

and the molar mass is NA times the molecular mass, as described by M = m

0N

A. The defi nition

of the molar mass implies that the sample mass is the number of moles times the molar mass, as described by m = nM. Then the sample mass must also be the number of molecules times the molecular mass, according to m = nM = nN

Am

0 = Nm

0. The equations are true for chemical

compounds in solid, liquid, and gaseous phases—this includes elements. We apply the equa-tions also to air by interpreting M as the mass of Avogadro’s number of the various molecules in the mixture.

P21.3 F Nmt

= = ×( ) − −−∆∆

v500 5 00 10

8 00 45 0 8 03.. sin . .

kg° 00 45 0

30 00 943

sin .

..

°( )[ ] =m s

sN

PF

A= = =1 57 1 57. .N m Pa2

P21.4 F Nmt

= =×( ) ×( )−

0

23 265 00 10 4 68 10 2 300∆∆

v . . kg m s(( )⎡⎣ ⎤⎦ =1 00

14 0.

.s

N and

PF

A= =

×=−

14 0

1017 64

..

N

8.00 mkPa2

*P21.5 PVN

NRT

A

=⎛⎝⎜

⎞⎠⎟

and NPVN

RTA= so that

N =×( )( )( ) ×( )

( )

−1 00 10 133 1 00 6 02 10

8 314

10 23. . .

. 33003 21

( ) = ×. 10 molecules12

P21.6 Use the equation describing the kinetic-theory account for pressure: PN

V

m=

⎛⎝⎜

⎞⎠⎟

2

3 20

2v. Then

Km PV

NN nN N

KPV

N

av A A

avA

where= = = =

= (

02

2

3

22

3

2 2

v

)) =( ) ×( ) × −3 8 00 1 013 10 5 00 105 3. . .atm Pa atm m3(( )

( ) ×( )=

2 2 6 02 10

5 05

23mol molecules mol

av

.

.K ×× −10 21 J molecule

The Kinetic Theory of Gases 545


P21.7 PN

VKE= ( )2

3 from the kinetic-theory account for pressure.

NPV

KE= ( ) =

×( ) ×( )×

−

−

3

2

3

2

1 20 10 4 00 10

3 60 10

5 3

2

. .

. 2224

24

2 00 10

2 00 10

( ) = ×

= = ×

.

.

molecules

mol

A

nN

N

eecules

molecules molmol

6 02 103 3223.

.×

=

P21.8 (a) PV nRTNm

= = 02

3

v

The total translational kinetic energy is Nm

E02

2

v= trans:

E PVtrans = = × ×( ) ×( ) =−3

2

3

23 00 1 013 10 5 00 10 25 3. . . ..28 kJ

(b) m k T RT

NB0

2

232

3

2

3

2

3 8 314 300

2 6 02 10

v= = = ( )( )

×(A

.

. )) = × −6 21 10 21. J

P21.9 (a) PV Nk TB= : NPV

k TB

= =× ( )⎡⎣ ⎤⎦

×1 013 10 0 150

1 38 1

5 43

3. .

.

Pa mπ00 293

3 54 102323

−( )( )= ×

J K Katoms.

(b) K k TB= = ×( )( ) = ×− −3

2

3

21 38 10 293 6 07 1023 21. .J J

(c) For helium, the atomic mass is m0 23

4 00

6 02 106 64 10=

×= × −.

..

g mol

molecules mol224 g molecule

m0276 64 10= × −. kg molecule

1

2

3

202m k TBv = : ∴ = =vrms km s

31 35

0

k T

mB .

P21.10 (a) 1 11

1

11Pa Pa

N m

Pa

J

1 N mJ m

2

= ( )⎛⎝⎜

⎞⎠⎟ ⋅

⎛⎝

⎞⎠ = 33

(b) For a monatomic ideal gas, E nRTint = 3

2 For any ideal gas, the energy of molecular translation is the same, E nRT PVtrans = =3

2

3

2

Thus, the energy per volume is E

VPtrans = 3

2

546 Chapter 21


P21.11 (a) K k TB= = ×( )( ) = ×− −3

2

3

21 38 10 423 8 76 1023 21. .J K K J

(b) K m rms= = × −1

28 76 100

2 21v . J

so vrms m= × −1 75 10 20. J

(1)

For helium, m0 23

4 00

6 02 106 64 10=

×= × −.

..

g mol


m0276 64 10= × −. kg molecule

Similarly for argon, m0 23

39 9

6 02 106 63 10=

×= × −.

..

g mol


m0266 63 10= × −. kg molecule

Substituting in (1) above, we fi nd for helium, vrms = 1 62. km s

and for argon, vrms = 514 m s

Section 21.2 Molar Specifi c Heat of an Ideal Gas

P21.12 n = 1 00. mol, Ti = 300 K

(b) Since V = constant, W = 0

(a) ∆E Q Wint J J= + = + =209 0 209

(c) ∆ ∆ ∆E nC T n R TVint = = ⎛⎝

⎞⎠

3

2

so ∆∆

TE

nR= = ( )

( ) ⋅2

3

2 209

3 1 00 8 314int J

mol J mol. . KKK( ) = 16 8.

T T Ti= + = + =∆ 300 16 8 317K K K.

P21.13 We use the tabulated values for CP and CV

(a) Q nC TP= = ⋅( ) −( ) =∆ 1 00 420 300 3. .mol 28.8 J mol K K 446 kJ

(b) ∆ ∆E nC TVint mol 20.4 J mol K K= = ⋅( )( ) =1 00 120 2. .445 kJ

(c) W Q E= − + = − + = −∆ int kJ kJ kJ3 46 2 45 1 01. . .



P21.14 (a) Consider warming it at constant pressure. Oxygen and nitrogen are diatomic, so CR

P = 7

2

Q nC T nR TPV

TT

Q

P= = = ⎛⎝

⎞⎠

=×

∆ ∆ ∆7

2

7

2

7

2

1 013 105. N m2(( )( ) ( ) =100

3001 00 118

m

KK kJ

3

.

(b) U mgyg =

mU

gyg= = ×

( ) = ×1 18 10

2 006 03 10

53.

..

J

9.80 m s m2 kkg

P21.15 Consider 800 cm3 of (fl avored) water at 90.0°C mixing with 200 cm3 of diatomic ideal gas at 20.0°C:

Q Qcold hot= −

or m c T T m c TP f iair , air air−( ) = − ( ), w w w∆

∆Tm c T T

m c

VP f i( ) =− −( )

=−( )

ww w

air , air air air, ρ cc

V cP , air C C90 0 20 0. .° °−( )

( )ρw w w

where we have anticipated that the fi nal temperature of the mixture will be close to 90.0°C.

The molar specifi c heat of air is C RP , air = 7

2

So the specifi c heat per gram is cR

MP , air J mol Kmol

28= ⎛

⎝⎞⎠ = ⋅( )7

2

7

28 314

1 00.

.

..9 gJ g C

⎛⎝⎜

⎞⎠⎟

= ⋅1 01. °

∆T( ) = −×( )( )⎡⎣ ⎤⎦

−

w

1 20 10 200 1 013. .g cm cm J g3 3 ⋅⋅( )( )( )( )⎡⎣ ⎤⎦

° °C C

g cm cm3 3

70 0

1 00 800 4 186

.

. . J kg C⋅( )°

or ∆T( ) ≈ − × −w 5 05 10 3. °C

The change of temperature for the water is between C and C10 103 2− −° ° .

P21.16 (a) C RV = = ⋅( )⎛⎝⎜

5

2

5

28 314

1 00.

.J mol K

mol

0.028 9 kg

⎞⎞⎠⎟

= ⋅ = ⋅719 0 719J kg K kJ kg K.

(b) m Mn MPV

RT= = ⎛

⎝⎞⎠

m = ( ) × ( )0 028 9

200 10

8 314

3

..

kg molPa 0.350 m

J

3

mmol K Kkg

⋅( )( )⎛

⎝⎜

⎞

⎠⎟ =

3000 811.

(c) We consider a constant volume process where no work is done.

Q mC TV= = ⋅( ) −( )∆ 0 811 700 300. kg 0.719 kJ kg K K K == 233 kJ

(d) We now consider a constant pressure process where the internal energy of the gas is increased and work is done.

Q mC T m C R T mR

T mC

T

Q

P VV= = +( ) = ⎛

⎝⎞⎠ = ⎛

⎝⎞⎠

=

∆ ∆ ∆ ∆7

2

7

5

0.. .811 0 719 400 327kg7

5kJ kg K K k⋅( )⎡

⎣⎢⎤⎦⎥( ) = JJ

548 Chapter 21


P21.17 Q nC T nC TP V= ( ) + ( )∆ ∆isobaric isovolumetric

In the isobaric process, V doubles so T must double, to 2Ti.

In the isovolumetric process, P triples so T changes from 2Ti to 6Ti.

Q n R T T n R T T nRi i i i= ⎛⎝

⎞⎠ −( ) + ⎛

⎝⎞⎠ −( ) =7

22

5

26 2 13 5. TT PVi = 13 5.

Section 21.3 Adiabatic Processes for an Ideal Gas

P21.18 (a) PV P Vi i f fγ γ= so

V

V

P

Pf

i

i

f

=⎛

⎝⎜⎞

⎠⎟= ⎛

⎝⎞⎠ =

15 71 00

20 00 118

γ.

..

(b) T

T

P V

PV

P

P

V

Vf

i

f f

i i

f

i

f

i

= =⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= ( )20 0 0. .1118( ) T

Tf

i

= 2 35.

(c) Since the process is adiabatic, Q = 0

Since γ = = = +1 40.

C

C

R C

CP

V

V

V

, C RV = 5

2 and ∆T T T Ti i i= − =2 35 1 35. .

∆ ∆E nC TVint mol J mol K= = ( )⎛⎝

⎞⎠ ⋅(0 016 0

5

28 314. . )) ( )[ ] =1 35 300 135. K J

and W Q E= − + = + = +∆ int J J0 135 135 .

P21.19 (a) PV P Vi i f fγ γ=

P PV

Vf ii

f

=⎛

⎝⎜⎞

⎠⎟= ⎛

⎝⎜⎞⎠⎟

γ

5 0012 0

30 0

1 40

..

.

.

atm == 1 39. atm

(b) TPV

nRii i= =

×( ) ×( )−5 00 1 013 10 12 0 10

2

5 3. . .Pa m3

..00366

mol 8.314 J mol KK

⋅( ) =

TP V

nRff f= =

×( ) ×( )−1 39 1 013 10 30 0 10

2

5 3. . .Pa m3

..00253

mol 8.314 J mol KK

⋅( ) =

(c) The process is adiabatic: Q = 0

γ = = = +1 40.

C

C

R C

CP

V

V

V

, C RV = 5

2

∆ ∆E nC TVint mol

5

2J mol K= = ⋅( )⎛

⎝⎞⎠2 00 8 314 253. . K K kJ

kJint

−( ) = −

= − = − − = −

366 4 66

4 66 0 4

.

.W E Q∆ ..66 kJ



P21.20 Vi = ×⎛⎝⎜

⎞⎠⎟

= ×−

−π 2 50 100 500 2 45 10

2 2

4.. .

m

2m m3

The quantity of air we fi nd from PV nRTi i i=

n

PV

RTi i

i

= =×( ) ×( )−1 013 10 2 45 10

8 314

5 4. .

.

Pa m3

J mol K K

mol

⋅( )( )= × −

300

9 97 10 3n .

Adiabatic compression: Pf = + =101 3 800 901 3. .kPa kPa kPa

(a) PV P Vi i f fγ γ=

V V

P

Pf ii

f

=⎛

⎝⎜⎞

⎠⎟= × ⎛

⎝⎞−

1

42 45 10101 3

901 3

γ

..

.m3

⎠⎠

= × −

5 7

55 15 10Vf . m3

(b) P V nRTf f f=

T T

P V

PVT

P

P

P

PT

P

Pf if f

i ii

f

i

i

fi

i

f

= =⎛

⎝⎜⎞

⎠⎟=

⎛

⎝⎜⎞

1 γ

⎠⎠⎟

= ⎛⎝⎜

⎞⎠⎟ =

−( )

−( )

1 1

5 7 1

300101 3

901 3560

γ

Tf K.

.K

(c) The work put into the gas in compressing it is ∆ ∆E nC TVint =

W = ×( ) ⋅( ) −( )−9 97 105

28 314 560 3003. .mol J mol K KK

JW = 53 9.

Now imagine this energy being shared with the inner wall as the gas is held at constant volume. The pump wall has outer diameter 25 0 2 00 2 00 29 0. . . .mm mm mm mm+ + = , and volume

π π14 5 10 12 5 10 4 00 103 2 3 2. . .×( ) − ×( )⎡

⎣⎤⎦ ×− − −m m 22 66 79 10m m3= × −. and mass

ρV = ×( ) ×( ) =−7 86 10 6 79 10 53 33 6. . .kg m m g3 3

The overall warming process is described by

53 9

53 9 9 97 105

283

.

. . .

J

J mol

= +

= ×( )−

nC T mc TV ∆ ∆

3314 300

53 3 10 443

J mol K K

kg

⋅( ) −( )+ ×( )−

Tff

. 88 300

53 9 0 207 23 9

J kg K K

J J K J

⋅( ) −( )= +

Tff

. . . KK K

K K

( ) −( )− =

T

T

ff

ff

300

300 2 24.

P21.21 T

T

V

Vf

i

i

f

=⎛

⎝⎜⎞

⎠⎟= ⎛

⎝⎞⎠

−γ 10 4001

2

.

If Ti = 300 K, then Tf = 227 K .

550 Chapter 21


P21.22 We suppose the air plus burnt gasoline behaves like adiatomic ideal gas. We fi nd its fi nal absolute pressure:

21 0 400

21 0

7 5 7 5.

.

atm 50.0 cm cm

a

3 3( ) = ( )=

P

P

f

f ttm1

8atm⎛

⎝⎞⎠ =

7 5

1 14.

Now Q = 0

and W E nC T TV f i= = −( )∆ int

∴ = − = −( )W nRT nRT P V PVf i f f i i

5

2

5

2

5

2

W = ( ) − ( )⎡⎣ ⎤5

21 14 21 0. .atm 400 cm atm 50.0 cm3 3

⎦⎦×⎛

⎝⎜⎞⎠⎟

( )= −

−1.013 10 N m

atmm cm

5 23 3

110

15

6

W 00 J

The output work is − = +W 150 J

The time for this stroke is 1

4

1 606 00 10 3min

2 500

s

1 mins

⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠ = × −.

P =−

=×

=−

W

t∆150

25 03

J

6.00 10 skW.

P21.23 (a) See the diagram at the right.

(b) P V P VB B C Cγ γ=

3

3 3 2 19

2 1

1 5 7

PV PV

V V V V

V

i i i C

C i i i

C

γ γ

γ

=

= ( ) = ( ) =

=

.

. 99 4 00 8 77. .L L( ) =

(c) P V nRT PV nRTB B B i i i= = =3 3

T TB = = ( ) =3 3 300 900i K K

(d) After one whole cycle, T TA i= = 300 K .

(e) In AB, Q nC V n R T T nRTAB V i i i= = ⎛⎝⎜

⎞⎠⎟ −( ) = ( )∆

5

23 5 00.

QBC = 0 as this process is adiabatic

P V nRT P V nRTC C C i i i= = ( ) = ( )2 19 2 19. . so T TC i= 2 19.

Q nC T n R T T nRTCA P i i i= = ⎛⎝

⎞⎠ −( ) = −( )∆ 7

22 19 4 17. .

For the whole cycle,

Q Q Q Q nRT nRABCA AB BC CA i= + + = −( ) = ( )5 00 4 17 0 829. . . TT

E Q W

W Q

i

ABCA ABCA ABCA

ABCA ABCA

∆ int( ) = = +

= − = −

0

0.. .

. .

829 0 829

0 829 1 013

( ) = −( )= −( )

nRT PV

W

i i i

ABCA ××( ) ×( ) = −−10 4 00 10 3365 3Pa m J3.


FIG. P21.22

B

AC

VCVi = 4L

Pi

Pi3

P

V(L)

Adiabatic

FIG. P21.23


P21.24 (a) See the diagram at the right.

(b) P V P VB B C Cγ γ=

3

3 3 2 191 5 7

PV PV

V V V V

i i i C

C i i i

γ γ

γ

=

= = = .

(c) P V nRT PV nRTB B B i i i= = =3 3

T TB i= 3

(d) After one whole cycle, T TA i=

(e) In AB, Q nC T n R T T nRTAB V i i i= = ⎛⎝

⎞⎠ −( ) = ( )∆ 5

23 5 00.

QBC

= 0 as this process is adiabatic

P V nRT P V nRT T T

Q

C C C i i i C i

CA

= = ( )= =2 19 2 19 2 19. . .so

== =⎛⎝⎜

⎞⎠⎟

−( )= −nC T n R T T nRTP i i i∆ 7

22 19 4 17. .

For the whole cycle,

Q Q Q Q nRT nRTABCA AB BC CA i i= + + = −( ) =5 00 4 17 0 830. . .

∆∆E Q W

W Q

ABCA ABCA ABCA

ABCA ABCA

int( ) = = +

= − = −

0

0 8. 330 0 830nRT PVi i i= − .

P21.25 (a) The work done on the gas is

W PdVab

V

V

a

b

= −∫ For the isothermal process,

W nRTV

dV

W nRTV

V

ab aV

V

ab ab

a

a

b

′

′′

= −⎛⎝⎜

⎞⎠⎟

= −

′

∫1

ln⎛⎛

⎝⎜⎞

⎠⎟=

⎛

⎝⎜⎞

⎠⎟′

nRTV

Va

b

ln

Thus,

Wab′ = ⋅( )( ) ( )5 00 293 10 0. ln .mol 8.314 J mol K K

Wab′ = 28 0. kJ

552 Chapter 21

P

B

Pi

Adiabatic

A C

Vi VC

3Pi

V (L)

FIG. P21.24

FIG. P21.25

continued on next page


(b) For the adiabatic process, we must fi rst fi nd the fi nal temperature, Tb. Since air consists primarily of diatomic molecules, we shall use

γ air = 1 40. and CR

V , air J mol K= = ( )= ⋅5

2

5 8 314

220 8

..

Then, for the adiabatic process

T TV

Vb aa

b

=⎛⎝⎜

⎞⎠⎟

= ( ) =γ −1

0 400293 10 0 736K K. .

Thus, the work done on the gas during the adiabatic process is

W Q E nC T nC T Tab ab V ab V b a− +( ) = − +( ) = −( )∆ ∆int 0

or Wab = ⋅( ) −( ) =5 00 736 293 46 0. .mol 20.8 J mol K K kJJ

(c) For the isothermal process, we have

P V P Vb b a a′ ′ =

Thus

P PV

Vb aa

b′

′

=⎛⎝⎜

⎞⎠⎟

= ( ) =1 00 10 0. .atm 10.0 atm

For the adiabatic process, we have P V P Vb b a aγ γ=

Thus

P PV

Vb aa

b

=⎛⎝⎜

⎞⎠⎟

= ( ) =γ

1 00 25 11 40. ..atm 10.0 atmm

Section 21.4 The Equipartition of Energy

P21.26 (1) E Nfk T

fnRTB

int = ⎛⎝

⎞⎠ = ⎛

⎝⎞⎠2 2

(2) Cn

dE

dTfRV = ⎛

⎝⎞⎠ =1 1

2int

(3) C C R f RP V= + = +( )1

22

(4) γ = = +C

C

f

fP

V

2




P21.27 The sample’s total heat capacity at constant volume is nCV. An ideal gas of diatomic molecules has three degrees of freedom for translation in the x, y, and z directions. If we take the y axis along the axis of a molecule, then outside forces cannot excite rotation about this axis, since they have no lever arms. Collisions will set the molecule spinning only about the x and z axes.

(a) If the molecules do not vibrate, they have fi ve degrees of freedom. Random collisions put equal

amounts of energy 1

2k TB into all fi ve kinds of motion. The average energy of one molecule is

5

2k TB . The internal energy of the two-mole sample is

N k T nN k T n R T nC TB A B V

5

2

5

2

5

2⎛⎝

⎞⎠ = ⎛

⎝⎞⎠ = ⎛

⎝⎞⎠ =

The molar heat capacity is C RV = 5

2 and the sample’s heat capacity is

nC n R

nC

V

V

= ⎛⎝

⎞⎠ = ⋅( )⎛

⎝⎞⎠

=

5

22 8 314

4

mol5

2J mol K.

11 6. J K

For the heat capacity at constant pressure we have

nC n C R n R R nRP V= +( ) = +⎛

⎝⎞⎠ = =5

2

7

22 8 314mol

7

2J m. ool K

J K

⋅( )⎛⎝

⎞⎠

=nCP 58 2.

(b) In vibration with the center of mass fi xed, both atoms are always moving in opposite direc-tions with equal speeds. Vibration adds two more degrees of freedom for two more terms in the molecular energy, for kinetic and for elastic potential energy. We have

nC n RV = ⎛⎝

⎞⎠ =7

258 2. J K and nC n RP = ⎛

⎝⎞⎠ =9

274 8. J K

P21.28 Rotational Kinetic Energy = 1

22Iω

I m r= 2 02, m0

2735 0 1 67 10= × × −. . kg, r = −10 10 m

I = × ⋅−1 17 10 45. kg m2 ω = × −2 00 1012 1. s

∴ = = × −K Irot J1

22 33 102 21ω .

*P21.29 Sulfur dioxide is the gas with the greatest molecular mass of those listed. If the effective spring constants for various chemical bonds are comparable, SO

2 can then be expected to have low

frequencies of atomic vibration. Vibration can be excited at lower temperature than for the other gases. Some vibration may be going on at 300 K. With more degrees of freedom for molecular motion, the material has higher specifi c heat.

554 Chapter 21

Cl

Cl

FIG. P21.28


*P21.30 (a) The energy of one molecule can be represented as

(1�2)m0vx

2 + (1�2)m0vy

2 + (1�2)m0vz

2 + (1�2)Iωx2 + (1�2)Iωz

2

Its average value is (1�2)kBT + (1�2)k

BT + (1�2)k

BT + (1�2)k

BT + (1�2)k

BT = (5�2)k

BT

The energy of one mole is obtained by multiplying by Avogadro’s number, Eint

�n = (5� 2)RT

And the molar heat capacity at constant volume is Eint

�nT = (5/2)R

(b) The energy of one molecule can be represented as

(1�2)m0vx

2 + (1�2)m0vy

2 + (1�2)m0vz

2 + (1�2)Iωx2 + (1�2)Iωz

2 + (1�2)Iωy2


BT + (1�2)k

BT + (1�2)k

BT + (1�2)k

BT + (1�2)k

BT = 3k

BT


�n = 3RT


�nT = 3R

(c) Let the modes of vibration be denoted by 1 and 2. The energy of one molecule can be represented as

0.5m0[vx

2 + vy2 + vz

2] + 0.5Iωx2 + 0.5Iωz

2 + [0.5µ vrel2 + 0.5kx2]

1 + [0.5µv

rel2 + 0.5kx2]

2

Its average value is

(3� 2)kBT + (1�2)k

BT + (1�2)k

BT + (1�2)k

BT + (1�2)k

BT + (1�2)k

BT + (1�2)k

BT = (9�2)k

BT


�n = (9�2)RT


�nT = (9/2)R

(d) The energy of one molecule can be represented as

0.5m0

[vx2 + vy

2 + vz2] + 0.5Iωx

2 + 0.5Iωz2 + 0.5Iωy

2 + [0.5µvrel2 + 0.5kx2]

1 + [0.5µv

rel2 + 0.5kx2]

2


BT + (1�2)k

BT + (1�2)k

BT + (1�2)k

BT + (1�2)k

BT = (5)k

BT


�n = 5RT


�nT = 5R

(e) Measure the constant-volume specifi c heat of the gas as a function of temperature and look for plateaus on the graph, as shown in Figure 21.7. If the fi rst jump goes from 3

2 R to 5

2 R, the molecules can be diagnosed as linear. If the fi rst jump goes from 32 R to 3R, the

molecules must be nonlinear. The tabulated data at one temperature are insuffi cient for the determination. At room temperature some of the heavier molecules appear to be vibrating.

Section 21.5 Distribution of Molecular Speeds

P21.31 (a) vv

av = = ( ) + ( ) + ( ) + ( ) + ( ) + (∑ n

Ni i 1

151 2 2 3 3 5 4 7 3 9 2 12))[ ] = 6 80. m s

(b) vv2

2

54 9( ) = =∑av

2 2m sn

Ni i .

so v vrms avm s= ( ) = =2 54 9 7 41. .

(c) vmp m s= 7 00.



P21.32 (a) The ratio of the number at higher energy to the number at lower energy is e E k TB− ∆ where ∆E is the energy difference. Here,

∆E = ( ) ×⎛⎝⎜

⎞⎠⎟

= ×−

−10 21 60 10

1 63 1019

..

.eV1 eV

J 118 J

and at 0°C,

k TB = ×( )( ) = ×− −1 38 10 273 3 77 1023 21. .K JJ K

Since this is much less than the excitation energy, nearly all the atoms will be in the ground state and the number excited is

2 70 101 63 1025

18

21. exp.

×( ) − ××

⎛⎝⎜

⎞−

−3.77 10 J

J

⎠⎠⎟= ×( ) −2 70 1025 433. e

This number is much less than one, so almost all of the time no atom is excited .

(b) At 10 000°C,

k TB = ×( ) = ×− −1 38 10 10 273 1 42 1023 19. .J K K J

The number excited is

2 70 101 63 1025

18

19. exp.×( ) − ×

×⎛⎝⎜

⎞−

−

J

1.42 10 J ⎠⎠⎟= ×( ) = ×−2 70 10 2 70 1025 11 5 20. ..e

P21.33 In the Maxwell Boltzmann speed distribution function take dN

dv

v= 0 to fi nd

42 2

220

3 2

02

0ππ

Nm

k T

m

k T

m

B B

⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

−expv

vv33

20

k TB

⎛⎝⎜

⎞⎠⎟

=

and solve for v to fi nd the most probable speed. Reject as solutions v = 0 and v = � They describe minimally probable speeds.

Retain only 2 002

− =m

k TB

v

Then vmp = 2

0

k T

mB

P21.34 (a) V

V

RT M

RT Mrms, 35

rms, 37

g mol= =3

3

37 0

3535

37

/

/

.

...

01 03

1 2

g mol

⎛⎝⎜

⎞⎠⎟

=

(b) The lighter atom, 35Cl , moves faster.

P21.35 (a) From vav = 8

0

k T

mB

π

we fi nd the temperature as T =×( ) ×( )

×

−

−

π 6 64 10 1 12 10

8 1 38 10

27 4 2

23

. .

.

kg m s

J mol KK

⋅( ) = ×2 37 104.

(b) T =×( ) ×( )

×

−

−

π 6 64 10 2 37 10

8 1 38 10

27 3 2

23

. .

.

kg m s

J mol KK

⋅( ) = ×1 06 103.

556 Chapter 21


*P21.36 For a molecule of diatomic nitrogen the mass is

m0 = M �N

A = (28.0 × 10–3 kg�mol)�(6.02 × 1023 molecules�mol) = 4.65 × 10–26 kg�molecule

(a) vmpBk T

m= = × ⋅−2 2 1 38 10 900

0

23( . (J/molecule K) K))

4.65 10 kg/moleculem/s

×=−26 731

(b) vavgBk T

m= =

× ⋅−8 8 1 38 10 900

0

23

π( . (J/molecule K) K)


π ⋅ ×=−26 825

(c) vrmsBk T

m= = × ⋅−3 3 1 38 10 900

0

23( . (J/molecule K) KK)


×=−26 895

(d) The graph appears to be drawn correctly withhin about 10 m/s.

P21.37 (a) From the Boltzmann distribution law, the number density of molecules with gravitational

energy m0gy is n e m gy k TB

00−

. These are the molecules with height y, so this is the number

per volume at height y as a function of y.

(b) n y

ne e em gy k T Mgy N k T Mgy RTB A B

( ) = − − −= =0

0

=

− × ×−( )( )( )e28 9 10 9 8 11 10 8 3143 3. . .kg mol m s m2 J mol K K⋅( )( )

−= =

293

1 279 0 278e . .

P21.38 (a) We calculate

e em gy k T m gy k TB Bdym gdy

k T

k T

B

B− −∫ = −⎛⎝⎜

⎞⎠⎟

−0 0

0

0�

mm g

k T

m g

k T

m g

y

B Be m gy k TB

00

0 0 0

0 0

⎛⎝⎜

⎞⎠⎟

= − = −

=∫

−

�

�

−−( ) =10

k T

m gB

Using Table B.6 in the appendix,

ye dym gy k TB

m g k T

k T

m gB

B−∫ =( )

=⎛⎝⎜

⎞⎠⎟

0

0

1

0

20

2�!

Then

y

y dyk T m g

k

e

e dy

m gy k T

m gy k T

B

B

B

B

= =( )

−

−

∫

∫

0

0

0 0

2

0

�

� TT m g

k T

m gB

0 0

=

(b) yk T

M N g

RT

MgB

A

= ( ) = =⋅

8 314. J 283 K s

mol K 28.9

2

××= ×−10

8 31 1033

kg 9.8 mm.



Additional Problems

P21.39 (a) Pf = 100 kPa Tf = 400 K

VnRT

Pff

f

= =⋅( )( )

×2 00 400

100

. mol 8.314 J mol K K

1100 066 5 66 53 Pa

m L3= =. .

∆ ∆E nR Tint mol J mol K= ( ) = ( ) ⋅3 50 3 50 2 00 8 314. . . .(( )( ) =100 5 82K kJ.

W P V nR T= − = − = −( ) ⋅( )∆ ∆ 2 00 8 314 100. .mol J mol K K(( ) = −1 66. kJ

Q E W= − = + =∆ int kJ kJ kJ5 82 1 66 7 48. . .

(b) Tf = 400 K

V VnRT

Pf ii

i

= = =⋅( )( )2 00 300

1

. mol 8.314 J mol K K

000 100 049 9 49 93×

= =Pa

m L3. .

P PT

Tf if

i

=⎛⎝⎜

⎞⎠⎟

= ⎛⎝

⎞⎠ =100

400133kPa

K

300 KkPPa W PdV= − =∫ 0

since V = constant

∆Eint kJ= 5 82. as in part (a) Q E W= − = − =∆ int kJ kJ5 82 0 5 82. .

(c) Pf = 120 kPa Tf = 300 K

V VP

Pf ii

f

=⎛

⎝⎜⎞

⎠⎟= ⎛

⎝⎜⎞⎠⎟ =49 9. L

100 kPa

120 kPa441 6. L ∆ ∆E nR Tint = ( ) =3 50 0. since T = constant

W PdV nRTdV

VnRT

V

VnRi

V

V

if

i

f

= − = − = −⎛⎝⎜

⎞⎠⎟

= −∫ ∫i

ln TTP

P

W

ii

f

ln

. .

⎛

⎝⎜⎞

⎠⎟

= −( ) ⋅( )2 00 8 314 3mol J mol K 000100

909KkPa

120 kPaJ

int

( ) ⎛⎝⎜

⎞⎠⎟ = +

=

ln

Q E∆ −− = − = −W 0 909 909J J

(d) Pf = 120 kPa γ = = + = + = =C

C

C R

C

R R

RP

V

V

V

3 50

3 50

4 50

3 50

9

7

.

.

.

.

P V PVf f i iγ γ= : so V V

P

Pf ii

f

=⎛

⎝⎜⎞

⎠⎟= ⎛

⎝⎞⎠

17

49 9

γ

. L100 kPa

120 kPa

99

43 3= . L

T TP V

PVf if f

i i

=⎛⎝⎜

⎞⎠⎟

= ⎛⎝

⎞⎠300

120K

kPa

100 kPa

443 3312

3 50 3 5

.

. .

L

49.9 LK

int

⎛⎝

⎞⎠ =

= ( ) =∆ ∆E nR T 00 2 00 8 314 12 4 722

0

. . .mol J mol K K J( ) ⋅( )( ) =

=Q adiabatic process

Jint

( )= − + = + = +W Q E∆ 0 722 7222 J

558 Chapter 21


*P21.40 (a) nPV

RT= = × × ×( . )( . . . )1 013 10 4 20 3 00 2 505 Pa m m m

(( . )( )8 314 2931 31 103

J mol K K. mol

⋅= ×

N nNA= = ×( ) ×(1 31 10 6 02 103 23. mol . molecules mol))

= ×N 7 89 1026. molecules

(b) m nM= = ×( )( ) =1 31 10 0 028 9 37 93. mol . kg mol kg.

(c) 1

2

3

2

3

21 38 10 293 6 070

2 23m k TBv = = ×( )( ) = ×−. .J k K 110 21− J molecule

(d) For one molecule,

mM

NA0 23

0 028 9

6 02 104= =

×=.

.

kg mol

molecules mol.. kg molecule

J molerms

80 10

2 6 07 10

26

21

×

=×

−

−

v. ccule

kg moleculem s

( )×

=−4 80 1050326.

(e), (f) E nC T n R T PVVint = = ⎛⎝

⎞⎠ =5

2

5

2

Eint . . .= ×( )( ) =5

21 013 10 31 5 7 985 3Pa m MJ

The smaller mass of warmer air at 25°C contains the same internal energy as the cooler air. When the furnace operates, air expands and leaves the room.

*P21.41 For a pure metallic element, one atom is one molecule. Its energy can be represented as

(1� 2)m0vx

2 + (1� 2)m0vy

2 + (1� 2)m0vz

2 + (1� 2)kxx2 + (1� 2)k

yy2 + (1�2)k

zz2

Its average value is (1� 2)kBT + (1� 2)k

BT + (1� 2)k

BT + (1� 2)k

BT + (1� 2)k

BT + (1� 2)k

BT = 3k

BT


�n = 3RT


�nT = 3R

(b) 3(8.314 J�mole ⋅ K) = 3 × 8.314 J�[55.845 × 10–3 kg] ⋅ K = 447 J�kg ⋅ K = 447 J/kg °C. This agrees with the tabulated⋅ value of 448 J/kg °C within 0.3%.⋅

(c) 3(8.314 J�mole ⋅ K) = 3 × 8.314 J�[197 × 10–3 kg] ⋅ K = 127 J�kg ⋅ K = 127 J/kg °C. This agrees with the tabulated⋅ value of 129 J/kg °C within 2%.⋅

P21.42 (a) The average speed vavg is just the weighted average of all the speeds.

vv v v v v v v

avg =( ) + ( ) + ( ) + ( ) + ( ) + ( ) +2 3 2 5 3 4 4 3 5 2 6 1 7(( )[ ]

+ + + + + +( ) =2 3 5 4 3 2 1

3 65. v

(b) First fi nd the average of the square of the speeds,

( )v

v v v v v2

2 2 2 2 22 3 2 5 3 4 4 3 5 2avg =

( ) + ( ) + ( ) + ( ) + ( ) + 66 1 7

2 3 5 4 3 2 115 95

2 2

2v v

v( ) + ( )⎡⎣ ⎤⎦

+ + + + + += .

The root-mean square speed is then v v vrms = =( ) .avg2 3 99

(c) The most probable speed is the one that most of the particles have;

i.e., fi ve particles have speed 3 00. v .




(d) PV Nm= 1

3 0vav2

Therefore,

Pm

V

m

V=

( )⎡⎣ ⎤⎦ =⎛⎝⎜

⎞⎠⎟

20

3

15 951060

2

02. v v

(e) The average kinetic energy for each particle is

K m m m= = ( ) =1

2

1

215 95 7 980 0

20

2v v vav2 . .

*P21.43 (a) PV kγ = . So W PdV k dVV

kV

i

f

V

V

V

V

i

f

i

f

= = =∫ ∫– – ––

–

γ

γ

γ1

1

For k we can substitute PiV

iγ and also P

fV

fγ to have

WP V V PV V P V PV

f f f i i i f f i i= =––

–

–

–

– –γ γ γ γ

γ γ

1 1

1 1

(b) dE dQ dWint = + and dQ = 0 for an adiabatic process.

Therefore,

W E nC T TV f i= + = −( )∆ int

To show consistency between these two equations, consider that γ = C

CP

V

and C C RP V− = .

Therefore, 1

1γ −= C

RV

Using this, the result found in part (a) becomes W P V PVC

Rf f i iV= −( )

Also, for an ideal gas PV

RnT= so that W nC T TV f i= −( ), as found in part (b).

P21.44 (a) W nC T TV f i= −( ) − = ⋅ −( )2 500 1 8 314 500J mol

3

2J mol K K. Tf

Tf = 300 K

(b) PV P Vi i f fγ γ=

PnRT

PP

nRT

Pii

if

f

f

⎛⎝⎜

⎞⎠⎟

=⎛

⎝⎜⎞

⎠⎟

γ γ

T P T Pi i f fγ γ γ γ1 1− −=

T

P

T

Pi

i

f

f

γ γ γ γ−( ) −( )=

1 1

P PT

Tf if

i

=⎛⎝⎜

⎞⎠⎟

−( )γ γ 1

P PT

Tf if

i

=⎛⎝⎜

⎞⎠⎟

= ⎛⎝

⎞⎠

( )( )5 3 3 2 5

3 60300

500. atm

22

1 00= . atm

560 Chapter 21


P21.45 Let the subscripts ‘1’ and ‘2’ refer to the hot and cold compartments, respectively. The pressure is higher in the hot compartment, therefore the hot compartment expands and the cold compartment contracts. The work done by the adiabatically expanding gas is equal and opposite to the work done by the adiabatically compressed gas.

nR

T TnR

T Ti f i fγ γ−−( ) = −

−−( )

1 11 1 2 2

Therefore

T T T Tf f i i1 2 1 2 800+ = + = K (1)

Consider the adiabatic changes of the gases.

P V P Vi i f f1 1 1 1γ γ= and P V P Vi i f f2 2 2 2

γ γ=

∴ =P V

P V

P V

P Vi i

i i

f f

f f

1 1

2 2

1 1

2 2

γ

γ

γ

γ

∴ =⎛

⎝⎜⎞

⎠⎟P

P

V

Vi

i

f

f

1

2

1

2

γ

, since V Vi i1 2= and P Pf f1 2=

∴ =⎛

⎝⎜⎞

⎠⎟nRT V

nRT V

nRT P

nRT Pi i

i i

f f

f f

1 1

2 2

1 1

2 2

γ

, using the ideal gas law

∴ =⎛

⎝⎜⎞

⎠⎟T

T

T

Ti

i

f

f

1

2

1

2

γ

, since V Vi i1 2= and P Pf f1 2=

∴ =⎛⎝⎜

⎞⎠⎟

= ⎛⎝

⎞⎠ =

T

T

T

Tf

f

i

i

1

2

1

2

1 1 1 45501

γK

250 K

.

..756 (2)

Solving equations (1) and (2) simultaneously gives T Tf f1 2510 290= =K, K

P21.46 The net work done by the gas on the bullet becomes the bullet’s kinetic energy:

1

2

1

21 1 10 7 922 3 2

mv = × ( ) =−. .kg 120 m s J

The air in front of the bullet does work

P∆V = (1.013 × 105 N�m2)(−0.5 m)(0.03 × 10−4 m2) = −0.152 J

The hot gas behind the bullet then must do output work +W in +W − 0.152 J = 7.92 JW = 8.07 J. The input work on the hot gas is −8.07 J

1

18 07

γ −−( ) = −P V PVf f i i . J

Also P V PVf f i iγ γ= P P

V

Vf ii

f

=⎛

⎝⎜⎞

⎠⎟

γ

So − =⎛

⎝⎜⎞

⎠⎟−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

8 071

0 40.

.J P V

V

VVi f

i

fi

γ

And Vf = + =12 50 13 5cm cm 0.03 cm cm3 2 3.

Then Pi = − ( )( )

8 07 10

13 5 12 13 5

6

1

.

. / .

J 0.40 cm m

cm

3 3

3 ... .

406

125 85 10 57 7

−⎡⎣ ⎤⎦= × =

cmPa atm

3



P21.47 The pressure of the gas in the lungs of the diver must be the same as the absolute pressure of the water at this depth of 50.0 meters. This is:

P P gh= + = + ×( )( )031 00 1 03 10 9 80ρ . . .atm kg m m s3 2 550 0. m( )

or P = + ××

⎛1 00 5 05 105. .atm Pa1.00 atm

1.013 10 Pa5⎝⎝⎞⎠ = 5 98. atm

If the partial pressure due to the oxygen in the gas mixture is to be 1.00 atmosphere (or the fraction

1

5 98. of the total pressure) oxygen molecules should make up only

1

5 98. of the total number of

molecules. This will be true if 1.00 mole of oxygen is used for every 4.98 mole of helium. The ratio by weight is then

4 98 4 003

1 00 2

. .

.

mol He g mol He

mol O2

( )( )( )

g

××( ) =15 999

0 623.

.g mol O2 g

P21.48 (a) Maxwell’s speed distribution function is

N Nm

k TB

e m k TBv v v=

⎛⎝⎜

⎞⎠⎟

−42

0

3 2

2 02 2π

π

With N = ×1 00 104. ,

mM

NA0 23

26

0 032

6 02 10

5 32 10

= =×

= × −

.

.

.

kg

kg

T = 500 K

and kB = × ⋅−1 38 10 23. J molecule K

this becomes N evvv= ×( )− − ×( )−

1 71 10 4 2 3 85 10 6 2

..

To the right is a plot of this function for the range 0 1 500≤ ≤v m s.

(b) The most probable speed occurs where Nv is a maximum.

From the graph, vmp m s≈ 510

(c) vav = =×( )( )

×( )−

−

8 8 1 38 10 500

5 32 100

23

26

k T

mB

π π.

.== 575 m s

Also,

vrms = =×( )( )

×=

−

−

3 3 1 38 10 500

5 32 1062

0

23

26

k T

mB

.

.44 m s

(d) The fraction of particles in the range 300 600m s m s≤ ≤v

is N d

N

v v300

600

∫

where N = 104

and the integral of N v is read from the graph as the area under the curve. This is approximately

(11 + 16.5 + 16.5 + 15)(1�4)(300) = 4 400 and the fraction is 0.44 or 44% .

562 Chapter 21

FIG. P21.48(a)


P21.49 (a) Since pressure increases as volume decreases (and vice versa),

dV

dP< 0 and − ⎡

⎣⎢⎤⎦⎥

>10

V

dV

dP

(b) For an ideal gas, VnRT

P= and κ1

1= − ⎛⎝

⎞⎠V

d

dP

nRT

P If the compression is isothermal, T is constant and

κ1 2

1 1= − −⎛⎝

⎞⎠ =nRT

V P P

(c) For an adiabatic compression, PV Cγ = (where C is a constant) and

κγ γ

γ γ

γ

γ

2

1 1

1 1

11 1 1= − ⎛⎝

⎞⎠ =

⎛⎝⎜

⎞⎠⎟

=( )+V

d

dP

C

P V

C

P

P

PP P1 1

1γ γ+ =

(d) κ111 1

2 000 500= =

( ) = −

P ..

atmatm

γ = C

CP

V

and for a monatomic ideal gas, γ = 5

3, so that

κγ2 5

3

11 1

2 000 300= =

( ) = −

P ..

atmatm

*P21.50 (a) The speed of sound is v = B

ρ where B V

dP

dV= −

According to Problem 49, in an adiabatic process, this is B P= =1

2κγ

Also,

ρ = = = ( )( ) =m

V

nM

V

nRT M

V RT

PM

RTs

where ms is the sample mass. Then, the speed of sound in the ideal gas is

v = = ⎛⎝

⎞⎠ =B

PRT

PM

RT

Mργ γ

(b) v =⋅( )( )

=1 40 8 314 293

0 028 9344

. .

.

J mol K K

kg molm s

This agrees within 0.2% with the 343 m/s lissted in Table 17.1.

(c) We use kR

NBA

= and M m NA= 0 : v = = =γ γ γRT

M

k N T

m N

k T

mB A

A

B

0 0

The most probable molecular speed is 2

0

k T

mB , the average speed is 8

0

k T

mB

π,

and the rms speed is 3

0

k T

mB .

The speed of sound is somewhat less than each measure of molecular speed. Sound propagation is orderly motion overlaid on the disorder of molecular motion.



*P21.51 (a) The latent heat of evaporation per molecule is

2430 243018 0 1J

g

J

g

g

1 mol

mol

6.02 1023= ⎛⎝

⎞⎠ ×

.

mmoleculeJ/molecule⎛

⎝⎞⎠ = × −7 27 10 20.

If the molecule has just broken free, we assume that it possesses the energy as translational kinetic energy.

(b) Consider one gram of these molecules: K = (1� 2)mv2

2430 J = (1�2)(10–3 kg) v2 v = (4 860 000 m2�s2)1�2 = 2.20 10 m s3× �

(c) The total translational kinetic energy of an ideal gas is (3�2)nRT, so we have

(2430 J�g)(18.0 g�mol) = (3�2)(1 mol)(8.314 J�mol ⋅ K)T T = 3.51 10 K3×

The evaporating molecules are exceptional, at the high-speed tail of the distribution of molecular speeds. The average speed of molecules in the liquid and in the vapor is appropriate just to room temperature.

*P21.52 (a) Let d = 2r represent the diameter of the particle. Its mass is

m V rd d= = = ⎛

⎝⎞⎠ =ρ ρ π ρ π ρπ4

3

4

3 2 63

3 3

. Then 1

2

3

2m kTvrms

2 = gives ρπ d

kT3

63vrms

2 = so

vrms

J/K 29=

⎛⎝⎜

⎞⎠⎟

=× × ×−18 18 1 38 10

3

1 2 23kT

dρπ. 33 K

1000 kg/mm3 π

⎛⎝⎜

⎞⎠⎟

= ×− −1 2

3 2 124 81 10/

/ .d 55/2 1s− −d 3 2/

(b) v = d�t [4.81 × 10−12 m5�2�s]d−3�2 = d�t

td

d=

[4.81 10 m /s]s m12 5/2 3/2

5

×= × ⋅− −

−2 08 1011. //2d5 2/

(c) vrms

J K=

⎛⎝⎜

⎞⎠⎟

=×( )−

18 18 1 38 10 2933

1 2 23kT

dρπ. KK

kg m m3

( )( ) ×( )

⎛

⎝⎜

⎞

⎠⎟ = ×

−1 000 3 109 26 10

6 3

1 2

π. −−4 m s

v = x

t t

x= = ××

=−

−v3 10

3 246

4

m

9.26 10 m sms.

(d) 70 1 0006

3

kg kg m3= π d d = 0 511. m

vrms

J K=

⎛⎝⎜

⎞⎠⎟

=×( )−

18 18 1 38 10 2933

1 2 23kT

dρπ. KK

kg m .511 m3

( )( ) ( )

⎛

⎝⎜

⎞

⎠⎟ = × −

1 000 01 32 103

1 2

π. 111 m s

t =×

= × =−

0 5113 88 10 1 23011

10..

m

1.32 10 m ss yr This motion is too slow to observe.

(e) 18

13

1 2kT

d

d

ρπ⎛⎝⎜

⎞⎠⎟

=s

18

1

5kT d

ρπ=

s2

d =×( )( )( )( )

−18 1 38 10 293 1

1 000

23. J K K s

kg m

2

3 ππ⎛

⎝⎜

⎞

⎠⎟ = × −

1 5

52 97 10. m

(f ) Brownian motion is best observed with pollen grains, smoke particles, or latex spheres

smaller than this 29.7-µm size. Then they can jitter about convincingly, showing rela-tively large accelerations several times per second. A simple rule is to use the smallest particles that you can clearly see with some particular microscopic technique.

564 Chapter 21


P21.53 nm

M= = =1 20

41 5.

.kg

0.028 9 kg molmol

(a) VnRT

Pii

i

= =( ) ⋅( )( )41 5 8 314 298

20

. .mol J mol K K

00 100 5143×

=Pa

m3.

(b) P

P

V

Vf

i

f

i

= so V VP

Pf if

i

=⎛⎝⎜

⎞⎠⎟

= ( )⎛⎝

⎞⎠ =

2 2

0 514400

2002. m .063 m3

(c) TP V

nRff f= =

×( )( )( )

400 10 2 06

41 5 8

3 Pa m

mol

3.

. .33142 38 103

J mol KK

⋅( ) = ×.

(d) W PdV C V dVP

V

V

V

V

V

V

i

ii

f

i

f

= − = − = −⎛⎝⎜

⎞⎠⎟∫ ∫ 1 2

1 2

3 22

3VV

V

i

if i

i

f P

VV V= −

⎛⎝⎜

⎞⎠⎟

−( )2

3 1 23 2 3 2

W = − ×⎛⎝⎜

⎞⎠⎟

( ) −2

3

200 102 06 0 5

33 2Pa

0.514 mm3. . 114 4 80 103 2 5m J( )⎡

⎣⎤⎦ = − ×.

(e) ∆ ∆E nC TVint mol J mol K= = ( ) ⋅( )⎡⎣⎢

⎤⎦

41 55

28 314. . ⎥⎥ × −( )2 38 10 2983. K

∆

∆

E

Q E W

int

int

J

J

= ×

= − = × + ×

1 80 10

1 80 10 4 80 1

6

6

.

. . 00 2 28 10 2 285 6J J MJ= × =. .

P21.54 The ball loses energy 1

2

1

2

1

20 142 47 2 42 52 2 2 2m mi fv v− = ( ) ( ) − ( )⎡⎣. . .kg ⎤⎤⎦ =m s J2 2 29 9.

The air volume is V = ( ) ( ) =π 0 037 0 19 4 0 0832

. . .m m 4 m3

and its quantity is nPV

RT= =

× ( )⋅(

1 013 10

8 314

5.

.

Pa 0.083 4 m

J mol K

3

))( ) =293

3 47K

mol.

The air absorbs energy as if it were warmed over a stove according to Q nC TP= ∆

So

∆TQ

nCP

= = ( ) ⋅( ) =29 9

3 47 8 3140

72

.

. ..

J

mol J mol K2296°C



P21.55 N Nm

k T

m

k TB Bv v v

v( ) =⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠

42 2

0

3 2

2 02

ππ

exp ⎟⎟ where exp(x) represents e x

Note that vmp =⎛⎝⎜

⎞⎠⎟

2

0

1 2k T

mB

Thus, N Nm

k TBv

v vv v( ) =⎛⎝⎜

⎞⎠⎟

−( )42

0

3 2

22 2

ππ

e mp

And N

Nev

v

v vvv

vv

( )( ) =

⎛

⎝⎜⎞

⎠⎟−( )

mp mp

mp

2

1 2 2

For vv

= mp

50

N

Nev

v

vv( )

( ) = ⎛⎝

⎞⎠ = ×−( )⎡⎣ ⎤⎦ −

mp

1

501 09 10

21 1 50 2

. 33

The other values are computed similarly, with the vv

vv

v

vmp mp

N

N

( )( )×

×

−

−

1

501 09 10

1

102 69 10

1

20

3

2

.

.

..

.

.

.

.

529

1 1 00

2 0 199

10 1 01 10

50 1 25 10

41

1 082

×

×

−

−

following results:

To fi nd the last value, we note:

50 2 500

10

2 1 2 500 2 499

2 500 10 2 49

( ) =− −

( ) −

e e

elog ln 99 10 2 500 2 499 10 2 500 2 4910 10 10ln log ln log( ) − −= = 99 10 1 081 904 0 096 108210 10 10ln . .= = ×− −

P21.56 (a) The effect of high angular speed is like the effect of a very high gravitational fi eld on an atmo-

sphere. The result is: The larger-mass molecules settle to the outsside while the region at

smaller r has a higher concentration of low-mass molecules.

(b) Consider a single kind of molecules, all of mass m0. To cause the centripetal acceleration

of the molecules between r and r + dr, the pressure must increase outward according to F m ar r∑ = 0 . Thus,

PA P dP A nm A dr r− +( ) = −( )( )02ω

where n is the number of molecules per unit volume and A is the area of any cylindrical surface. This reduces to dP nm rdr= 0

2ω .

But also P nk TB= , so dP k TdnB= . Therefore, the equation becomes

dn

n

m

k Trdr

B

= 02ω

giving dn

n

m

k Trdr

n

n

B

r

0

02

0∫ ∫=

ω or ln n

m

k T

rn

n

B

r

( ) =⎛⎝⎜

⎞⎠⎟0

02 2

02

ω

lnn

n

m

k Tr

B0

02

2

2

⎛⎝⎜

⎞⎠⎟

=ω

and solving for n gives n n em r k TB= 020

2 2ω

566 Chapter 21


P21.57 First fi nd vav2 as v v vvav

2 = ∫1 2

0NN d

�

. Let am

k TB

= 0

2

Then,

v v v vav2 =

⎡⎣ ⎤⎦ = ⎡⎣ ⎤−

− −∫4

41 2 3 2

4

0

3 2 1 22N a

Ne aa d

ππ

�

⎦⎦ =3

8

32a a

k T

mBπ

The root-mean square speed is then v vrms av2= = 3

0

k T

mB

To fi nd the average speed, we have

v v v v vvv

av = =( )

=∫ ∫−

−1 4 4

0

3 2 1 23

0

32

NN d

Na

Ne d

aa� �π 22 1 2

202

8ππ

−

=a

k T

mB

P21.58 We want to evaluate dP

dV for the function implied by PV nRT= = constant, and also for the different

function implied by PV γ = constant. We can use implicit differentiation:

From PV = constant PdV

dVV

dP

dV+ = 0

dP

dV

P

V⎛⎝

⎞⎠ = −

isotherm

From PV γ = constant P V VdP

dVγ γ γ− + =1 0

dP

dV

P

V⎛⎝

⎞⎠ = −

adiabat

γ

Therefore, dP

dV

dP

dV⎛⎝

⎞⎠ = ⎛

⎝⎞⎠

adiabat isotherm

γ

The theorem is proved.

*P21.59 (a) nPV

RT= =

×( ) ×( )−1 013 10 5 00 10

8 314

5 3. .

.

Pa m

J m

3

ool K K

mol

⋅( )( )=

300

0 203.

(b) T TP

PB AB

A

=⎛⎝⎜

⎞⎠⎟

= ⎛⎝

⎞⎠ =300

3 00

1 00900K K

.

.

T T

V VT

T

C B

C AC

A

= =

=⎛⎝⎜

⎞⎠⎟

= ⎛⎝

⎞⎠

900

5 00

K

L900

300. == 15 0. L

(c) E nRTA Aint, mol J mol K= = ( ) ⋅( )3

2

3

20 203 8 314 3. . 000 760K J( ) =

E E nRTB C Bint, int, mol J= = = ( )3

2

3

20 203 8 314. . mmol K K kJ⋅( )( ) =900 2 28.

(d) P (atm) V(L) T(K) Eint

(kJ)

A 1.00 5.00 300 0.760

B 3.00 5.00 900 2.28

C 1.00 15.00 900 2.28


FIG. P21.59



(e) For the process AB, lock the piston in place and put the cylinder into an oven at 900 K. For BC, keep the sample in the oven while gradually letting the gas expand to lift a load on the piston as far as it can. For CA, carry the cylinder back into the room at 300 K and let the gas cool without touching the piston.

(f ) For AB: W = 0 ∆E E EB Aint int, int, kJ= − = −( ) =2 28 0 760 1 52. . . kkJ

Q E W= − =∆ int kJ1 52.

For BC: ∆Eint = 0 , W nRTV

VBC

B

= −⎛⎝⎜

⎞⎠⎟

ln

W = −( ) ⋅( )( ) ( )0 203 8 314 900 3 00. . ln .mol J mol K K == −

= − =

1 67

1 67

.

.

kJ

kJintQ E W∆

For CA: ∆E E EA Cint int, int, kJ= − = −( ) = −0 760 2 28 1 52. . . kJ

W P V nR T= − = − = −( ) ⋅( ) −∆ ∆ 0 203 8 314 600. .mol J mol K K kJ

kJ kJint

( ) =

= − = − − = −

1 01

1 52 1 01 2

.

. . .Q E W∆ 553 kJ

(g) We add the amounts of energy for each process to fi nd them for the whole cycle.

Q

W

ABCA

ABC

= + + − =1 52 1 67 2 53 0 656. . . .kJ kJ kJ kJ

AA

ABCAE

= − + = −

( ) = +

0 1 67 1 01 0 656. . .kJ kJ kJ

int∆ 11 52 0 1 52 0. .kJ kJ+ − =

P21.60 (a) 10 0001 00 6 02 1023

gmol

18.0 g

mole( )⎛⎝⎜

⎞⎠⎟

×. . ccules

1.00 molmolecules

⎛⎝⎜

⎞⎠⎟

= ×3 34 1026.

(b) After one day, 10 1− of the original molecules would remain. After two days, the fraction

would be 10 2− , and so on. After 26 days, only 3 of the original molecules would likely

remain, and after 27 days , likely none.

(c) The soup is this fraction of the hydrosphere: 10 0. kg

1.32 10 kg21×⎛⎝⎜

⎞⎠⎟

Therefore, today’s soup likely contains this fraction of the original molecules. The number of original molecules likely in the pot again today is:

10 0

3 34 1026..

kg

1.32 10 kgmolecule21×

⎛⎝⎜

⎞⎠⎟

× ss molecules( ) = ×2 53 106.

P21.61 (a) For escape, 1

2 02 0m

Gm M

Rv =

E

. Since the free-fall acceleration at the surface is gGM

R=

E2 , this

can also be written as:

1

2 02 0

0mGm M

Rm gRv = =

EE

568 Chapter 21



(b) For O2, the mass of one molecule is

m0 23

0 032 0

6 02 105 32=

×= ×.

..

kg mol

molecules mol110 26− kg molecule

Then, if m gRk TB

0 103

2E = ⎛⎝

⎞⎠ , the temperature is

Tm gR

kB

= =×( )( ) ×−

0

26

15

5 32 10 9 80 6 37 1E

2kg m s. . . 00

15 1 38 101 60 10

6

234m

J mol KK

( )× ⋅( ) = ×−.

.

P21.62 (a) For sodium atoms (with a molar mass M = 32 0. g mol)

1

2

3

2

1

2

3

2

3 3 8

02

2

m k T

M

Nk T

RT

M

B

AB

v

v

v

=

⎛⎝⎜

⎞⎠⎟

=

= =rms

.. .

..

314 2 40 10

23 0 100 5

4

3

J mol K K

kg

⋅( ) ×( )×

=−

− 110 m s

(b) td= = =

vrms

m

0.510 m sms

0 01020

.

ANSWERS TO EVEN PROBLEMS

P21.2 Because each mole of a chemical compound contains Avogadro’s number of molecules, the number of molecules in a sample is N

A times the number of moles, as described by N = nN

A, and the molar

mass is NA times the molecular mass, as described by M = m

0N

A. The defi nition of the molar mass

implies that the sample mass is the number of moles times the molar mass, as described by m = nM. Then the sample mass must also be the number of molecules times the molecular mass, according to m = nM = nN

Am

0 = Nm

0. The equations are true for chemical compounds in solid, liquid, and

gaseous phases—this includes elements. We apply the equations also to air by interpreting M as the mass of Avogadro’s number of the various molecules in the mixture.

P21.4 17.6 kPa

P21.6 5 05 10 21. × − J molecule

P21.8 (a) 2.28 kJ (b) 6 21 10 21. × − J

P21.10 see the solution

P21.12 (a) 209 J (b) 0 (c) 317 K

P21.14 (a) 118 kJ (b) 6 03 103. × kg

P21.16 (a) 719 J kg K⋅ (b) 0.811 kg (c) 233 kJ (d) 327 kJ

P21.18 (a) 0.118 (b) 2.35 (c) 0; +135 J; +135 J

P21.20 (a) 5 15 10 5. × − m3 (b) 560 K (c) 2.24 K

P21.22 25.0 kW

P21.24 (a) see the solution (b) 2.19 Vi (c) 3T

i (d) T

i (e) − 0.830P

i V

i



570 Chapter 21


P21.28 2.33 × 10–21 J

P21.30 (a) 5R�2 (b) 3R (c) 9R � 2 (d) 5R (e) Measure the constant-volume specifi c heat of the gas as a function of temperature and look for plateaus on the graph, as shown in Figure 21.7. If the fi rst jump goes from 3

2 R to 52 R, the molecules can be diagnosed as linear. If the fi rst jump goes

from 32 R to 3R, the molecules must be nonlinear. The tabulated data at one temperature are insuf-

fi cient for the determination. At room temperature some of the heavier molecules appear to be vibrating.

P21.32 (a) No atom, almost all the time (b) 2 70 1020. ×

P21.34 (a) 1.03 (b) 35Cl

P21.36 (a) 731 m �s (b) 825 m �s (c) 895 m �s (d) The graph appears to be drawn correctly within about 10 m �s.

P21.38 (a) see the solution (b) 8.31 km

P21.40 (a) 7 89 1026. × molecules (b) 37.9 kg (c) 6 07 10 21. × − J molecule (d) 503 m s (e) 7.98 MJ (f ) 7.98 MJ The smaller mass of warmer air contains the same internal energy as the cooler air.

When the furnace operates, air expands and leaves the room.

P21.42 (a) 3.65v (b) 3.99v (c) 3.00v (d) 106 02m

V

v⎛⎝⎜

⎞⎠⎟

(e) 7 98 02. m v

P21.44 (a) 300 K (b) 1.00 atm

P21.46 5.85 MPa

P21.48 (a) see the solution (b) 5 1 102. × m s (c) vav m s= 575 ; vrms m s= 624 (d) 44%

P21.50 (a) see the solution (b) 344 m �s, in good agreement with Table 17.1 (c) The speed of sound is somewhat less than each measure of molecular speed. Sound propagation is orderly motion overlaid on the disorder of molecular motion.

P21.52 (a) [18 kBT�πρ d 3]1� 2 = [4.81 × 10−12 m5�2�s]d −3�2 (b) [2.08 × 1011 s ⋅ m−5�2]d5�2 (c) 0.926 mm�s and

3.24 ms (d) 1.32 × 10−11 m �s and 3.88 × 1010 s (e) 29.7µm (f ) It is good to use the smallest particles that you can clearly see with some particular microscopic technique.

P21.54 0.296°C

P21.56 (a) The effect of high angular speed is like the effect of a very high gravitational fi eld on an atmo-sphere. The result is that the larger-mass molecules settle to the outside while the region at smaller r has a higher concentration of low-mass molecules. (b) see the solution


P21.60 (a) 3 34 1026. × molecules (b) during the 27th day (c) 2 53 106. × molecules

P21.62 (a) 0 510. m s (b) 20 ms


Engineering

Solucionario serway cap 21