PSE9e ISM chapter21 final tủ tài liệu training

21
The Kinetic Theory of Gases
CHAPTER OUTLINE
21.1

Molecular Model of an Ideal Gas

21.2

Molar Specific Heat of an Ideal Gas

21.3

The Equipartition of Energy

21.4

Adiabatic Processes for an Ideal Gas

21.5

Distribution of Molecular Speeds

* An asterisk indicates a question or problem new to this edition.

ANSWERS TO OBJECTIVE QUESTIONS
OQ21.1

Answer (c). The molecular mass of nitrogen (N2, 28 u) is smaller than
the molecular mass of oxygen (O2, 32 u), and the rms speed of a gas is
(3RT/M)1/2. Since the rms speeds are the same, the temperature of

nitrogen is smaller than the temperature of oxygen. The average
kinetic energy is proportional to the molecular mass and the square
1
2
of the rms speed ( K = mvrms
), so the average kinetic energy of
2
nitrogen is smaller.

OQ21.2

Answer (d). The rms speed of molecules in the gas is vrms = 3RT M.
Thus, the ratio of the final speed to the original speed would be

( vrms ) f
( vrms )0
OQ21.3

=

3RT f M
3RT0 M

=

Tf
T0

=

600 K
= 3
200 K

Answer (b). The gases are the same so they have the same molecular
mass, M. If the two samples have the same density, then their ratios
of number of moles to volume, n/V, are the same because their

1093
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1094

The Kinetic Theory of Gases
densities, (nM)/V, are the same. The pressures are the same; thus,
their temperatures are the same:
PV = nRT → p =

n
RT = constant → T = constant
V

Therefore the rms speed of their molecules, (3RT/M)1/2, is the same.
OQ21.4

OQ21.5

(i)

Answer (b). The volume of the balloon will decrease because
the gas cools.

(ii)

Answer (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 temperature. Call the process
isobaric contraction.

Answer (d). At 200 K,

1
3
2
m0 vrms0
= kBT0 . At the higher temperature,
2
2

1
3
2
m0 ( 2vrms0 ) = kBT
2
2

Then T = 4T0 = 4(200 K) = 800 K.

OQ21.6

Answer (c) > (a) > (b) > (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, the
average speed is (8RT/πM)1/2 ≅ (2.55RT/M)1/2, and the rms speed is
(3RT/M)1/2.

OQ21.7

(i)

Statements (a) and (e) are correct statements that describe the
temperature increase of a gas.

(ii)

Statement (f) is a correct statement but does not apply to the
situation. Statement (b) is true if the molecules have any size at
all, but molecular collisions with other molecules have nothing
to do with the temperature increase.

(iii) Statements (c) and (d) are 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.
OQ21.8

(i)

Answer (b). Average molecular kinetic energy, 3kT/2, increases
by a factor of 3.

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Chapter 21
(ii)

Answer (c). The rms speed, (3RT/M)1/2, increases by a factor of
3.

(iii) Answer (c). Average momentum change increases by
Δpavg = −2m0 vavg .
(iv) Answer (c). Rate of collisions increases by a factor of
Δtavg = 2d / vavg .
(v)

1095

3:
3:

Answer (b). Pressure increases by a factor of 3. See Equation
21.15:

2⎛ N ⎞⎛ 1
⎞ 2⎛ N ⎞

P = ⎜ i ⎟ ⎜ m0 v 2 ⎟ = ⎜ i ⎟ K
⎝
⎠
⎝
⎠ 3⎝ V ⎠
3 V
2

( )

OQ21.9

Answer (c). The kinetic theory of gases assumes that the molecules
do not interact with each other.

ANSWERS TO CONCEPTUAL QUESTIONS
CQ21.1

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.

CQ21.2

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.

CQ21.3

Alcohol evaporates rapidly, so that high-speed molecules leave the
liquid, reducing the average kinetic energy of the remaining
molecules of the liquid and therefore reducing the temperature of
the liquid. Then, because the alcohol is cool, energy transfers from
the skin, reducing its temperature.

CQ21.4

As the balloon rises into the air, 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 adiabatic expansion (see
Section 21.4), with P decreasing as V increases (PVγ = constant). If the
rubber wall is very strong it will eventually 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.

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1096

The Kinetic Theory of Gases

CQ21.5

The dry air is more dense. Since the air and the water vapor are at
the same temperature, the gases have the same average molecular
kinetic energy. Imagine a controlled experiment in which equalvolume containers, one with humid air and one with dry air, are at
the same pressure. The number of molecules must be the same for
both containers (PV = NkT). 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.

CQ21.6

The helium must have the higher rms speed. According to Equation
21.22 for the rms speed, (3RT/M)1/2, for the same temperature, the
gas with the smaller mass per atom must have the higher average
speed squared and thus the higher rms speed.

CQ21.7

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.
The partial pressure Pi of one of the gases can be expressed with
Equation 21.15:

2⎛ N ⎞⎛ 1

⎞ 2⎛ N ⎞
Pi = ⎜ i ⎟ ⎜ m0 v 2 ⎟ = ⎜ i ⎟ K
⎠ 3⎝ V ⎠
3⎝ V ⎠⎝ 2

( )

where Ni is the number of molecules of the ith gas and K is the
average kinetic energy of the molecules. Let us add up these
pressures for all the gases in the container:

2⎛ N ⎞
2K
2⎛ N⎞
P = ∑ Pi = ∑ ⎜ i ⎟ K  =
Ni = ⎜ ⎟ K
∑
3V i
3⎝ V ⎠
i
i 3⎝ V ⎠

( )

( )

where N is the total number of molecules of all types and we have
used the fact that the average kinetic energies of all types of
molecules are the same because all the gases have the same
temperature. The final expression for the pressure is the same as that

of a single gas with N molecules in the same volume V and at the
given temperature.

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Chapter 21

1097

SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Section 21.1
P21.1

(a)

Molecular Model of an Ideal Gas
 
4 3 4
3
π r = π ( 0.150 m ) = 1.41 × 10−2  m 3 . The
3
3
quantity of gas can be obtained from PV = nRT:

The volume is V =

5
2
−2

3
PV ( 1.013 × 10 N/m ) ( 1.41 × 10 m )
n=
=
= 0.588 mol
RT
( 8.314 N ⋅ m/mol ⋅ K )( 293 K )

The number of molecules is

N = nN A = ( 0.588 mol ) ( 6.02 × 1023  molecules/mol )
N = 3.54 × 1023 helium atoms

(b)

The kinetic energy is given by K =
K=

(c)

1
3
m0 v 2 = kBT:
2
2

3
1.38 × 10−23 J/K )( 293 K ) = 6.07 × 10−21 J
(
2

An atom of He has mass

m0 =

M
4.002 6 g/mol
=
N A 6.02 × 1023 molecules/mol

= 6.65 × 10−24 g = 6.65 × 10−27 kg
So the root-mean-square speed is given by

vrms = v 2 =
P21.2

(a)

Both kinds of molecules have the same average kinetic energy. It
is
K=

(b)

2K
2 × 6.07 × 10−27 J
=
= 1.35 km/s
m0
6.65 × 10−27 kg

3
3
kBT = ( 1.38 × 10−23 J K ) ( 423 K ) = 8.76 × 10−21 J
2
2

The root-mean square velocity can be calculated from the kinetic
energy:
vrms = v 2 =

2K
m0

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1098

The Kinetic Theory of Gases

so

vrms

1.75 × 10−20 J
=
m0

[1]

For helium,

m0 =

4.00 g mol
= 6.64 × 10−24 g molecule
6.02 × 1023 molecules mol

m0 = 6.64 × 10−27 kg molecule

Similarly for argon, m0 =

39.9 g mol
6.02 × 1023 molecules mol

= 6.63 × 10−23 g molecule
m0 = 6.63 × 10−26 kg molecule

Substituting into [1] above,
we find for helium,

vrms = 1.62 km s

and for argon, vrms = 514 m s
P21.3

(a)

From Newton’s second law, the average force is given by

F = Nm

Δv
= 500 ( 5.00 × 10−3 kg )
Δt
[ 8.00 sin 45.0° − ( −8.00 sin 45.0°)] m s
×
30.0 s

= 0.943 N
(b)

We find the pressure from

P=
P21.4

F
0.943 N
=
= 1.57 N m 2 = 1.57 Pa
2
A 0.600 m

The equation of state for an ideal gas can be used with the given
information to find the number of molecules in a specific volume.
⎛ N ⎞
PVN A
,

PV = ⎜
RT means N =
⎟
RT
⎝ NA ⎠

so that, suppressing units,

(1.00 × 10 )(133)(1.00)(6.02 × 10 )
−10

N=

23

(8.314)( 300 )

= 3.21 × 1012 molecules

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Chapter 21

P21.5

The gas temperature must be that implied by

1099

1
3
m0 v 2 = kBT for a
2
2

monatomic gas like helium.

⎛1
2⎞
2 ⎜ 2 m0 v ⎟ 2 ⎛ 3.60 × 10 –22 J ⎞
T= ⎜
=
= 17.4 K
kB ⎟ 3 ⎜⎝ 1.38 × 10 –23 J/K ⎟⎠
3
⎜⎝
⎟⎠
Now PV = nRT gives
5
2
–3
3
PV ( 1.20 × 10 N m ) ( 4.00 × 10 m )
n=
=
= 3.32 mol
RT
(8.314 J/mol ⋅ K)(17.4 K)

P21.6

P=

2N
K from the kinetic-theory account for pressure.
3V

3 PV
2 K
N
3 PV
n=
=
NA
2 KN A
N=

P21.7

Use the equation describing the kinetic-theory account for pressure:
2N ⎛ m0 v 2 ⎞
P=
⎜
⎟ . Then
3V ⎝ 2 ⎠
m0 v 2 3PV
=
, where N = nN A
2

2N
3 ( 8.00 atm ) ( 1.013 × 105 Pa atm ) ( 5.00 × 10−3 m 3 )
3PV
K=
=
2nN A
2 ( 2 mol ) ( 6.02 × 1023 molecules mol )
K=

K = 5.05 × 10−21 J

P21.8

The molar mass of diatomic oxygen is 32.0 g. The rms speed of oxygen
molecules is

vrms =

3RT
M

and
prms = mvrms =
=

M
NA

3RT
1

=
3RTM
M
NA

(

1
3 ( 8.314 J mol ⋅ K ) ( 350 K ) 32.0 × 10−3 kg
23
6.02 × 10

)

prms = 2.78 × 10−23 kg ⋅ m/s
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1100

The Kinetic Theory of Gases

P21.9

We use 1 u = 1.66 × 10−24 g.

P21.10

(a)

⎛ 1.66 × 10−24 g ⎞
=  6.64 × 10−27 kg
For He, m0 = 4.00 u ⎜
⎟
⎝
⎠
1u

(b)

⎛ 1.66 × 10−24 g ⎞
−26
For Fe, m0 = 55.9 u ⎜
⎟⎠ = 9.28 × 10 kg
⎝
1u

(c)

⎛ 1.66 × 10−24 g ⎞
= 3.44 × 10−25 kg
For Pb, m0 = 207 u ⎜
⎟
⎝
⎠
1u

The rms speed of molecules in a gas of molecular weight M and
absolute temperature T is vrms = 3RT M. Thus, if vrms = 625 m/s for
molecules in oxygen (O2), for which M = 32.0 g/mol =

32.0 × 10−3 kg/mol, the temperature of the gas is
2
32.0 × 10−3 kg mol ) ( 625 m s )
(
Mvrms
T=
=
= 501 K
3R
3 ( 8.31 J mol ⋅ K )
2

*P21.11

(a)

From the ideal gas law,

PV = nRT =

Nm0 v 2
3

The total translational kinetic energy is
Etrans =

Nm0 v 2
= Etrans :
2

3
3
PV = ( 3.00 × 1.013 × 105 Pa ) ( 5.00 × 10−3 m 3 )
2
2

= 2.28 kJ

P21.12

(b)

m0 v 2 3kBT 3RT 3 ( 8.314 J/mol ⋅ K ) ( 300 K )
=
=
=
= 6.21 × 10−21 J
2
2
2N A
2 ( 6.02 × 1023 )

(a)

The volume occupied by this gas is

V = 7.00 L ( 103 cm 3 1 L ) ( 1 m 3 106 cm 3 ) = 7.00 × 10−3 m 3
Then, the ideal gas law gives
6
−3

3
PV ( 1.60 × 10 Pa ) ( 7.00 × 10 m )
T=
=
= 385 K
nR
( 3.50 mol ) ( 8.31 J mol ⋅ K )

(b)

The average kinetic energy per molecule in this gas is
KE molecule =

3
3
kBT = ( 1.38 × 10−23 J K ) ( 385 K ) = 7.97 × 10−21 J
2
2

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Chapter 21
(c)

P21.13

1101

You would need to know the mass of the gas molecule to find its

average speed, which in turn requires knowledge of
the molecular mass of the gas .

To find the pressure exerted by the nitrogen molecules, we first
calculated the average force exerted by the molecules:
23
−26
Δv ( 5.00 × 10 ) ⎡⎣( 4.65 × 10 kg ) 2 ( 300 m s ) ⎤⎦
F = Nm0
=
= 14.0 N
Δt
1.00 s

the pressure is then

P=

F
14.0 N
=
= 17.4 kPa
A 8.00 × 10−4 m 2


 

Section 21.2
P21.14

Molar Specific Heat of an Ideal Gas

n = 1.00 mol, Ti = 300 K
(a)

Since V = constant, W = 0 .

(b)

ΔEint = Q + W = 209 J + 0 = 209 J

(c)

⎛3 ⎞
ΔEint = nCV ΔT = n ⎜ R ⎟ ΔT
⎝2 ⎠
so ΔT =

2 ( 209 J )
2ΔEint
=
= 16.8 K
3nR
3 ( 1.00 mol ) ( 8.314 J mol ⋅ K )

T = Ti + ΔT = 300 K + 16.8 K = 317 K
P21.15

Q = ( nCP ΔT )isobaric + ( nCV ΔT )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.
⎛7 ⎞
⎛5 ⎞
Q = n ⎜ R ⎟ ( 2Ti − Ti ) + n ⎜ R ⎟ ( 6Ti − 2Ti ) = 13.5nRTi
⎝2 ⎠
⎝2 ⎠
= 13.5PV

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1102
P21.16

The Kinetic Theory of Gases
(a)

Consider warming it at constant pressure. Oxygen and nitrogen
7R
are diatomic, so CP =
. Then,
2

Q = nCP ΔT =

7
7 ⎛ PV ⎞
nRΔT = ⎜
⎟ ΔT

2
2⎝ T ⎠

5
2
3
7 ( 1.013 × 10 N m ) ( 100 m )
Q=
(1.00 K ) = 118 kJ
2
300 K

(b)

We use the definition of gravitational potential energy,

U g = mgy
from which,
Ug

1.18 × 105 J
m=
=
= 6.03 × 103 kg
2
gy ( 9.80 m s )( 2.00 m )

P21.17

We use the tabulated values for CP and CV:

(a)

Since this is a constant-pressure process, Q = nCP ΔT.
The temperature rises by ΔT = 420 K – 300 K = 120 K:
Q = nCP ΔT = ( 1.00 mol ) ( 28.8 J mol ⋅ K )( 420 K − 300 K )
= 3.46 kJ

(b)

For any gas ΔEint = nCV ΔT, so

ΔEint = nCV ΔT = ( 1.00 mol ) ( 20.4 J mol ⋅ K ) ( 120 K ) = 2.45 kJ
(c)

The first law says ΔEint = Q + W, so

W = −Q + ΔEint = −3.46 kJ + 2.45 kJ = −1.01 kJ
P21.18

(a)

Molar specific heat is CV =

5
R.
2

Specific heat at constant volume per unit mass is given by
cV =
=

CV 5 ⎛ 1 ⎞
= R⎜ ⎟
M 2 ⎝ M⎠
⎛ 1.00 mol ⎞
5
8.314 J mol ⋅ K ) ⎜
(
2
⎝ 0.028 9 kg ⎟⎠

= 719 J kg ⋅ K = 0.719 kJ kg ⋅ K

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Chapter 21

(b)

1103

⎛ PV ⎞
m = Mn = M ⎜
⎝ RT ⎟⎠
⎡ ( 200 × 103 Pa ) ( 0.350 m 3 ) ⎤
m = ( 0.028 9 kg mol ) ⎢
⎥ = 0.811 kg
⎢⎣ ( 8.314 J mol ⋅ K ) ( 300 K ) ⎥⎦

(c)

We consider a constant-volume process where no work is done.
Q = mcV ΔT

= ( 0.811 kg ) ( 0.719 kJ kg ⋅ K ) ( 700 K − 300 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 = nCP ΔT =

m
m⎛5
m ⎛7 ⎞
⎞
CV + R ) ΔT =
(
⎜⎝ R + R ⎟⎠ ΔT =
⎜ R ⎟ ΔT
M
M 2
M⎝2 ⎠

⎛5 ⎞
R
⎡⎛ 7 ⎞ ⎛ C ⎞ ⎤
⎛ 7⎞ ⎜ 2 ⎟
= m⎜ ⎟ ⎜

ΔT = m ⎢⎜ ⎟ ⎜ V ⎟ ⎥ ΔT
⎝ 5⎠ M ⎟
⎣⎝ 5 ⎠ ⎝ M ⎠ ⎦
⎜⎝
⎟⎠

⎡7
⎤
→ Q = ( 0.811 kg ) ⎢ ( 0.719 kJ kg ⋅ K ) ⎥ ( 400 K ) = 327 kJ
⎣5
⎦
3
3
nRΔT = ( 3.00 mol ) ( 8.314 J mol ⋅ K ) ( 2.00 K ) = 74.8 J
2
2

*P21.19

ΔEint =

P21.20

Consider 800 cm3 of tea (flavored water) at 90.0°C mixing with 200 cm3
of diatomic ideal gas at 20.0°C:

Qcold = −Qhot
or

(

)
(T

mair c P, air T f − Ti, air = −mw cw ( ΔT )w

( ΔT )w =

−mair c P, air

f

mw cw

− Ti, air

) = − ( ρV )

air

c P, air ( 90.0°C − 20.0°C )

( ρwVw ) cw

where we have anticipated that the final temperature of the mixture
will be close to 90.0°C.
The molar specific heat of air is CP, air =

7
R.

2

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1104

The Kinetic Theory of Gases
So the specific heat per gram is
c P, air =

⎛ 1.00 mol ⎞
7⎛ R ⎞ 7
= 1.01 J g ⋅ °C
⎜⎝ ⎟⎠ = ( 8.314 J mol ⋅ K ) ⎜
2 M
2
⎝ 28.9 g ⎟⎠

and

( ΔT )w
or

⎡⎣( 1.20 × 10−3 g cm 3 ) ( 200 cm 3 ) ⎤⎦ ( 1.01 J g ⋅ °C ) ( 70.0°C )
=−
⎡⎣( 1.00 g cm 3 ) ( 800 cm 3 ) ⎤⎦ ( 4.186 J g ⋅ °C )

( ΔT )w ≈ −5.05 × 10−3°C

The change of temperature for the water is
between 10−3 °C and 10−2 °C .

*P21.21

(a)

The air is far from liquefaction so it behaves as an ideal gas. From
m
m
PV = nRT we have PV =
RT, or PM = RT = ρ RT. For the
M
V
samples of air in the balloon at 10.0°C (cold) and at the elevated
temperature (hot) we have PM = ρc RTc and PM = ρh RTh . Then
ρT
ρhTh = ρcTc and ρh = c c . For equilibrium of the balloon on the
Th
point of rising,

∑ Fy = may :

+ B − Fg hot air − Fg cargo = 0

+ ρcVg − ρhVg − mg = 0
ρT
+ ρcV − c c V − m = 0
Th

(1.25

⎛ 283 K ⎞
kg m 3 ) ( 400 m 3 ) − ( 1.25 kg m 3 ) ⎜
( 400 m3 )
⎝ Th ⎟⎠
− 200 kg = 0

⎛ 283 K ⎞
300 kg = ( 500 kg ) ⎜
⎝ Th ⎟⎠
Th =

( )(

500
283 K ) = 472 K
300

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Chapter 21

1105

The quantity of air that must be warmed is given by PV = nh RTh ,
PV
. The heat input required is
or nh =

RTh

Q = nCp ΔT
=−

PV 7
R (Th − Tc )
RTh 2

()

5
2
3
7 ( 1.013 × 10 N/m ) ( 400 m ) ( 472 K − 283 K )
=
2
472 K

= 5.66 × 107 J
(b)

Q = mH, so m =

Q
5.66 × 107 J
=
= 1.12 kg
H 5.03 × 107 J kg

 

Section 21.3
P21.22

P21.23

The Equipartition of Energy

(a)

⎛ k T⎞
⎛ nRT ⎞
Eint = Nf ⎜ B ⎟ = f ⎜
⎝ 2 ⎠
⎝ 2 ⎟⎠

(b)

CV =

(c)

CP = CV + R =

(d)

γ =

1 ⎛ dEint ⎞ 1
⎜
⎟ = fR
n ⎝ dT ⎠ 2
1
( f + 2) R
2

f +2
CP
=
CV
f

The rotational kinetic energy of the molecule
1
is given by K rot = Iω 2 . We determine the
2
moment of inertia from I = 2m0 r 2 , with
m0 = 35.0 × 1.67 × 10−27 kg and r = 10−10 m:

ANS. FIG. P21.23

I = 2m0 r 2 = 2 ( 35.0 × 1.67 × 10−27 kg ) ( 10−10 m )

2

= 1.17 × 10−45 kg ⋅ m 2

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1106

The Kinetic Theory of Gases
Then,
K rot =

2
1 2 1
Iω = ( 1.17 × 10−45 kg ⋅ m 2 ) ( 2.00 × 1012 s −1 )
2
2

= 2.34 × 10−21 J

P21.24

We must have the difference of molar specific heats given by Equation
21.31: CP − CV = R. The value of γ tells us that CP = 1.75CV, so
1.75CV  − CV  = R     →     CV  = 

R
4
 =  R
0.75 3

R
= 1.67 occurs
CV

3
for the lowest possible value for CV =  R. Therefore the
2
claim of γ = 1.75 for the newly discovered gas cannot be
The maximum possible value of γ =1+

true.
P21.25

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 five degrees of freedom.
1
Random collisions put equal amounts of energy kBT into all
2
five kinds of motion. The average energy of one molecule is
5
kBT. The internal energy of the two-mole sample is
2

⎛5
⎞
⎛5
⎞

⎛5 ⎞
N ⎜ kBT ⎟ = nN A ⎜ kBT ⎟ = n ⎜ R ⎟ T = nCV T
⎝2
⎠
⎝2
⎠
⎝2 ⎠
The molar heat capacity is CV =

5
R, and the sample’s heat
2

capacity is

⎛5 ⎞
⎡5
⎤
nCV = n ⎜ R ⎟ = ( 2.00 mol ) ⎢ ( 8.314 J mol ⋅ K ) ⎥
⎝2 ⎠
2
⎣
⎦
nCV = 41.6 J K

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 21
(b)

1107

For the heat capacity at constant pressure, we have

⎛5
⎞ 7
nCP = n ( CV + R ) = n ⎜ R + R ⎟ = nR
⎝2
⎠ 2
⎡7
⎤
= ( 2.00 mol ) ⎢ ( 8.314 J mol ⋅ K ) ⎥
⎣2
⎦
nCP = 58.2 J K
(c)

Vibration adds two more degrees of freedom for two more terms
in the molecular energy, for kinetic and for elastic potential
energy. We have

⎛7 ⎞
nCV = n ⎜ R ⎟ = 58.2 J K
⎝2 ⎠
⎛9 ⎞
and nCP = n ⎜ R ⎟ = 74.8 J K .
⎝2 ⎠

 

Section 21.4
P21.26

(a)

Adiabatic Processes for an Ideal Gas
In an adiabatic process PiViγ = Pf Vfγ :
γ

1.40
⎛V ⎞
⎛ 12.0 ⎞
Pf = Pi ⎜ i ⎟ = ( 5.00 atm ) ⎜
= 1.39 atm
⎝ 30.0 ⎟⎠
⎝ Vf ⎠

(b)

The initial temperature is
5
−3
3
PiVi 5.00 ( 1.013 × 10 Pa ) ( 12.0 × 10 m )
Ti =
=
= 366 K
nR
( 2.00 mol ) ( 8.314 J mol ⋅ K )

and similarly the final temperature is

Tf =
(c)

Pf Vf
nR

=

1.39 ( 1.013 × 105 Pa ) ( 30.0 × 10−3 m 3 )

( 2.00 mol ) ( 8.314

J mol ⋅ K )

= 253 K

The process is adiabatic, so by definition, Q = 0 .

(d) For any process,

ΔEint = nCV ΔT,

and for this diatomic ideal gas, CV =

R
5
= R

γ −1 2

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1108

The Kinetic Theory of Gases
Thus,

ΔEint = nCV ΔT
⎡5
⎤
= ( 2.00 mol ) ⎢ ( 8.314 J mol ⋅ K ) ⎥ ( 253 K − 366 K )
⎣2
⎦
= −4.66 kJ
(e)
P21.27

(a)

(b)
(c)

W = ΔEint − Q = −4.66 kJ − 0 = −4.66 kJ
γ
i

PiV = Pf V

Tf
Ti

=

γ
f

⎛P⎞
=⎜ i⎟
Vi ⎝ Pf ⎠

Vf

so

1γ

⎛ 1.00 ⎞
=⎜
⎝ 20.0 ⎟⎠

⎛ Pf ⎞ ⎛ Vf ⎞
= ⎜ ⎟ ⎜ ⎟ = ( 20.0 )( 0.118 )
PiVi ⎝ Pi ⎠ ⎝ Vi ⎠

Pf Vf

57

→

= 0.118 .
Tf
Ti

= 2.35

Since the process is adiabatic, Q = 0 .

(d) Since γ = 1.40 =

5
CP R + CV
=
, CV = R
2
CV
CV

and ΔT = 2.35Ti − Ti = 1.35Ti , then

ΔEint = nCV ΔT
⎛ 5⎞
= ( 0.016 0 mol ) ⎜ ⎟ ( 8.314 J mol ⋅ K )[ 1.35 ( 300 K )]
⎝ 2⎠
= 135 J

P21.28

(e)

W = −Q + ΔEint = 0 + 135 J = +135 J

(a)

The work done on the gas is
Vb

Wab = − ∫ PdV
Va

For the isothermal process,
V

b′
⎛ 1⎞
Wab ′ = −nRTa ∫ ⎜ ⎟ dV
⎝V⎠
V
a

⎛V ⎞
⎛V ⎞
Wab ′ = −nRTa ln ⎜ b ′ ⎟ = nRT ln ⎜ a ⎟
⎝ Va ⎠
⎝ Vb ′ ⎠

Thus,

Wab′ = ( 5.00 mol ) ( 8.314 J mol ⋅ K )( 293 K ) ln ( 10.0 ) = 28.0 kJ
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 21

1109

ANS. FIG. P21.28
(b)

For the adiabatic process, we must first find the final temperature,
Tb. Since air consists primarily of diatomic molecules, we shall use

γ air = 1.40

and

CV , air =

5R 5 ( 8.314 )
=
= 20.8 J mol ⋅ K
2
2

Then, for the adiabatic process,

⎛V ⎞

Tb = Ta ⎜ a ⎟
⎝V ⎠

γ −1

= ( 293 K ) ( 10.0 )

0.400

= 736 K

b

Thus, the work done on the gas during the adiabatic process is
Wab ( −Q + ΔEint )ab = ( −0 + nCV ΔT )ab = nCV (Tb − Ta )

or
(c)

Wab = ( 5.00 mol ) ( 20.8 J mol ⋅ K ) ( 736 K − 293 K ) = 46.0 kJ

For the isothermal process, we have Pb ′Vb ′ = PaVa .
⎛V ⎞
Thus, Pb′ = Pa ⎜ a ⎟ = ( 1.00 atm )( 10.0 ) = 10.0 atm .
⎝ Vb′ ⎠

(d) For the adiabatic process, we have PbVbγ = PaVaγ .
γ

⎛V ⎞

1.40
Thus, Pb = Pa ⎜ a ⎟ = ( 1.00 atm )( 10.0 ) = 25.1 atm .
⎝V ⎠
b

P21.29

Combining PVγ = constant with the ideal gas law gives one of the
textbook equations describing adiabatic processes, T1V1γ −1 = T2V2γ −1 .

⎛V ⎞
T2 = T1 ⎜ 1 ⎟
⎝V ⎠
2

γ −1

⎛ 1⎞
= ( 300 K ) ⎜ ⎟
⎝ 2⎠

(1.40−1)

= 227 K

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1110
P21.30

The Kinetic Theory of Gases
Use Equation 21.37 for an adiabatic process to find the temperature of
the compressed fuel-air mixture at the end of the compression stroke,
before ignition:
TiViγ −1  = T f Vfγ −1     

which gives
⎛V ⎞
T f  = Ti   ⎜ i ⎟
⎝ Vf ⎠

γ −1

 = ( 323 K )( 14.5 )

1.40−1

 = 941 K

This is equivalent to 668°C, which is higher than the melting
point of aluminum which is 660°C. Also, the temperature will
rise much more when ignition occurs. The engine will melt when
put into operation!Therefore, the claim of improved efficiency
using an engine fabricated out of aluminum cannot be true.
P21.31

We suppose the air plus burnt gasoline behaves like a diatomic ideal
gas. We find its final absolute pressure:

( 21.0 atm ) ( 50.0 cm 3 )

7 5

⎛ 1⎞
Pf = ( 21.0 atm ) ⎜ ⎟
⎝ 8⎠
Now Q = 0
so

= Pf ( 400 cm 3 )

7 5

7 5

= 1.14 atm

(

)

and W = ΔEint = nCV T f − Ti ,

W=

(

5
5

5
nRT f − nRTi = Pf Vf − PiVi
2
2
2

)

5
W = [( 1.14 atm ) (400 cm 3 ) − ( 21.0 atm ) (50.0 cm 3 )]
2
⎛ 1.013 × 105 N m 2 ⎞
= ( −1 485 atm ⋅ cm 3 ) ⎜
10−6 m 3 cm 3 ) 
(
⎟
1 atm
⎝
⎠
= −150 J

ANS. FIG. P21.31
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Chapter 21

1111

The output work is −W = +150 J

The time for this stroke is
P=

1 ⎛ 1 min ⎞ ⎛ 60 s ⎞
−3
⎜⎝
⎟⎠ ⎜⎝
⎟⎠ = 6.00 × 10 s
4 2 500 1 min

−W
150 J
=
=  25.0 kW
Δt
6.00 × 10−3 s
2

P21.32

(a)

⎛ 2.50 × 10−2 m ⎞
−4
3
Vi = π ⎜
⎟⎠ ( 0.500 m ) = 2.45 × 10 m
2
⎝

(b)

The quantity of air we find from PiVi = nRTi:
5
−4
3
PiVi ( 1.013 × 10 Pa ) ( 2.45 × 10 m )
n=
=
RTi
( 8.314 J mol ⋅ K )( 300 K )

n = 9.97 × 10−3 mol

(c)

Absolute pressure = gauge pressure + external pressure:

Pf = 101.3 kPa + 800 kPa = 901.3 kPa = 9.01 × 105 Pa
(d) Adiabatic compression: PiViγ = Pf Vfγ

⎛P⎞
Vf = Vi ⎜ i ⎟
⎝ Pf ⎠

1γ

= ( 2.45 × 10

−4

⎛ 101.3 ⎞
m )⎜
⎝ 901.3 ⎟⎠

57

3

Vf = 5.15 × 10−5 m 3
(e)

PfVf = nRTf

Pf ⎛ P ⎞
T f = Ti
= Ti ⎜ i ⎟
PiVi
Pi ⎝ Pf ⎠
Pf Vf

⎛ 101.3 ⎞
T f = 300 K ⎜
⎝ 901.3 ⎟⎠

(f)

( 5 7 −1)

1γ

⎛P⎞
= Ti ⎜ i ⎟
⎝ Pf ⎠

(1 γ −1)

= 560 K

The work done on the gas in compressing it is W = ΔEint = nCV ΔT:

ΔEint = W = nCV ΔT
= ( 9.97 × 10−3 mol )

5
( 8.314 J mol ⋅ K )( 560 K − 300 K )
2

ΔEint = 53.9 J

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1112

The Kinetic Theory of Gases
(g)

The pump wall has outer diameter 25.0 mm + 2.00 mm + 2.00 mm
= 29.0 mm, and volume

⎡π ( 14.5 × 10−3 m )2 − π ( 12.5 × 10−3 m )2 ⎤
⎣
⎦

× ( 4.00 × 10−2 m ) = 6.79 × 10−6 m 3

(h)

The mass of the pump is given by

ρV = ( 7.86 × 103 kg m 3 ) ( 6.79 × 10−6 m 3 ) = 53.3 g
(i)

Now imagine this energy being shared with the inner wall as the
gas is held at constant volume. The overall warming process is
described by

ΔEint  = W  = nCV ΔT  + mcΔT     →    ΔT  = 

W
nCV  + mc

Suppressing the units of R,

53.9 J
5
( 9.97 × 10−3  mol ) 2 ( 8.314) + ( 0.053 3 kg )( 448 J/kg ⋅°C)
         = 2.24°C =  2.24 K
  ΔT   = 

P21.33

(a)

See ANS. FIG. P21.33(a) on the
right.

(b)

PBVBγ = PCVCγ

3PiViγ = PiVCγ

( )

( )

VC = 31 γ Vi = 35 7 Vi = 2.19Vi
VC = 2.19 ( 4.00 L ) = 8.77 L
(c)

PBVB = nRTB = 3PiVi = 3nRTi
TB = 3Ti = 3 ( 300 K ) = 900 K

ANS. FIG. P21.33(a)

(d) After one whole cycle, TA = Ti = 300 K .
(e)

⎛5 ⎞
In AB, QAB = nCV ΔV = n ⎜ R ⎟ ( 3Ti − Ti ) = ( 5.00 ) nRTi
⎝2 ⎠
QBC = 0 as this process is adiabatic.
PCVC = nRTC = Pi ( 2.19Vi ) = ( 2.19 ) nRTi

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Chapter 21
so

1113

TC = 2.19Ti , and

⎛7 ⎞
QCA = nCP ΔT = n ⎜ R ⎟ (Ti − 2.19Ti ) = ( −4.17 ) nRTi
⎝2 ⎠
For the whole cycle,

QABCA = QAB + QBC + QCA = ( 5.00 − 4.17 ) nRTi = ( 0.829 ) nRTi

( ΔEint )ABCA = 0 = QABCA + WABCA

WABCA = −QABCA = − ( 0.829 ) nRTi = − ( 0.829 ) PiVi

WABCA = − ( 0.829 ) ( 1.013 × 105 Pa ) ( 4.00 × 10−3 m 3 ) = −336 J
P21.34

(a)

Refer to ANS. FIG. P21.34(a).

(b)

PBVBγ = PCVCγ

3PiViγ = PiVCγ

( )

VC = 31 γ Vi
(c)

PBVB = nRTB = 3PiVi = 3nRTi

TB = 3Ti
(d) After one whole cycle,
ANS. FIG. P21.34(a)

TA = Ti .
(e)

For AB,

QAB = nCv ΔT = n

R
R

2nRTi 2PiVi
ΔT = n
=
( 3Ti − Ti ) =
γ −1
γ −1
γ −1 γ −1

QBC = 0 as this process is abiabatic.

PCVC = nRTC = Pi ( 31 γ ) Vi = ( 31 γ ) nRTi so TC = ( 31 γ ) Ti
QCA = nCP ΔT = nγ CV ⎡⎣Ti − ( 31 γ ) Ti ⎤⎦ = γ

R
nT ⎡1 − ( 31 γ ) ⎤⎦
γ −1 i⎣

⎛ 1 ⎞
⎡⎣1 − ( 31 γ ) ⎤⎦
= PiViγ ⎜
⎟
⎝ γ − 1⎠
For the whole cycle,
QABCA = QAB + QBC + QCA =

⎛ 1 ⎞
2PiVi
⎡⎣1 − 31 γ ⎤⎦
+ 0 + PiViγ ⎜
⎟

γ −1
γ
−
1
⎝
⎠

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1114

The Kinetic Theory of Gases
⎡ 2
⎤
⎛ 1 ⎞
⎡⎣1 − 31 γ ⎤⎦ ⎥
QABCA = PiVi ⎢
+γ ⎜
⎟
⎝ γ − 1⎠
⎣γ − 1
⎦
⎡ 2
⎤
⎛ γ − 1 + 1⎞
1γ
⎡
⎤
= PiVi ⎢

+⎜
1
−
3
⎥
⎣
⎦
⎟
⎣γ − 1 ⎝ γ − 1 ⎠
⎦
QABCA

⎡ 2
⎛ 1 − 31 γ
1γ
= PiVi ⎢
+ 1− 3 + ⎜
⎝ γ −1
⎣γ − 1

(

)

⎡
⎛ 3 − 31 γ
1γ
= PiVi ⎢ 1 − 3 + ⎜
⎝ γ −1
⎣

(

)

⎞⎤
⎟⎠ ⎥
⎦

⎞⎤
⎟⎠ ⎥
⎦

( ΔEint )ABCA = 0 = QABCA + WABCA
⎡⎛ 1 ⎞
⎤
WABCA = −QABCA = −PiVi ⎢⎜
1 − 31 γ + 1 − 31 γ ⎥
⎟
⎣⎝ γ − 1 ⎠
⎦

(

) (

)


 

Section 21.5
*P21.35

Distribution of Molecular Speeds

The most probable speed is

vmp =
P21.36

(a)

2kBT
=
m0

2 ( 1.38 × 10−23 J K ) ( 4.20 K )
= 132 m s
6.64 × 10−27 kg

The average is

v=

∑n v
∑n

i i
i

=

1(2.00) + 2(3.00) + 3(5.00) + 4(7.00) + 3(9.00) + 2(12.0)
m/s
1+ 2+ 3+ 4+ 3+ 2

v = 6.80 m/s
(b)

To find the average squared speed we work out

v

2

∑n v
=
∑n

2
i i
i

⎛ 1⎞
v 2 = ⎜ ⎟ ⎡⎣1( 2.002 ) + 2 ( 3.002 ) + 3 ( 5.002 ) + 4 ( 7.002 )
⎝ 15 ⎠

+ 3 ( 9.002 ) + 2 ( 12.02 ) m 2 s 2 ⎤⎦

v 2 = 54.9 m 2 s 2
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Chapter 21

1115

Then the rms speed is

vrms = v 2 = 54.9 m 2 s 2 = 7.41 m/s

P21.37

(c)

More particles have vmp = 7.00 m/s than any other speed.

(a)

The ratio of the number at higher energy to the number at lower
energy is e − ΔE kBT , where ΔE is the energy difference. Here,

⎛ 1.60 × 10−19 J ⎞
−18
ΔE = ( 10.2 eV ) ⎜
⎟⎠ = 1.63 × 10 J
⎝
1 eV
and at 0°C,

kBT = ( 1.38 × 10−23 J K ) ( 273 K ) = 3.77 × 10−21 J
Since this is much less than the excitation energy, nearly all the
atoms will be in the ground state and the number excited is
⎛ −1.63 × 10−18 J ⎞
25
2.70
×
10
exp
(
) ⎜⎝ 3.77 × 10−21 J ⎟⎠ = ( 2.70 × 1025 ) e −433

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

At 10 000°C,

kBT = ( 1.38 × 10−23 J K )( 10 273 K ) = 1.42 × 10−19 J
The number excited is
× 10
( 2.70 × 10 ) exp ⎛⎜⎝ −1.63
1.42 × 10
25

−18

−19

J⎞
J ⎟⎠

= ( 2.70 × 1025 ) e −11.5 = 2.70 × 1020

P21.38

P21.39

(a)

Vrms, 35
Vrms, 37

=

3RT / M35
3RT / M37

(b)

The lighter atom,

(a)

From vavg =

T=

35

⎛ 37.0 g mol ⎞
=⎜
⎝ 35.0 g mol ⎟⎠

12

= 1.03

Cl , moves faster.

8kBT
we find the temperature as
π m0

π ( 6.64 × 10−27 kg ) ( 1.12 × 10 4 m s )
8 ( 1.38 × 10

−23

J mol ⋅ K )

2

= 2.37 × 10 4 K

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1116

The Kinetic Theory of Gases

(b)
P21.40

T=

π ( 6.64 × 10−27 kg ) ( 2.37 × 103 m s )
8 ( 1.38 × 10

−23

J mol ⋅ K )

2

= 1.06 × 103 K

For a molecule of diatomic nitrogen the mass is
m0 =

M
28.0 × 10−3 kg/mol
=
N A 6.02 × 1023 molecules/mol

= 4.65 × 10−26 kg/molecule

vmp =

2kBT
=
m0

2(1.38 × 10−23 J/molecule ⋅ K)(900 K)
= 731 m s
4.65 × 10−26 kg/molecule

vavg =

8kBT
=
π m0

8(1.38 × 10−23 J/molecule ⋅ K)(900 K)
= 825 m s
π ⋅ 4.65 × 10−26 kg/molecule

(c)

vrms =

3kBT
=
m0

3(1.38 × 10−23 J/molecule ⋅ K)(900 K)
= 895 m s
4.65 × 10−26 kg/molecule

(d)

The graph appears to be drawn correctly within about 10 m s.

(a)

(b)

P21.41

(a)

(b)

From the Boltzmann distribution law, the number density of
molecules with height y so that the gravitational potential energy
of the molecule-Earth system is m0gy is n0 e − m0 gy/kBT . These are the
molecules with height y, so this is the number per volume at
height y as a function of y.
n( y )
n0

= e −m0 gy

=e

kB T

= e − Mgy

(

)(

N A kB T

= e − Mgy RT

)(

− 28.9×10−3 kg mol 9.8 m s2 11×103 m

) (8.314 J mol⋅K )(293 K )

= e −1.279 = 0.278
P21.42

In the Maxwell-Boltzmann speed distribution function take

dN v
= 0 to
dv

find

⎛ m0 ⎞
4π N ⎜
⎝ 2π k T ⎟⎠
B

32

⎛ m0 v 2 ⎞ ⎛
2m0 v 3 ⎞
exp  ⎜ −
  2v −
=0
2k T ⎟⎠
⎝ 2k T ⎟⎠ ⎜⎝
B

B

and solve for v to find the most probable speed. Reject as solutions
v = 0 and v = ∞. They describe minimally probable speeds.

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Chapter 21

m0 v 2
= 0.
kBT

Retain only 2 −

Then,
P21.43

2kBT
.
m0

vmp =

It is convenient in the following to define a = 
(a)

1117

m0 g
.
k BT

We calculate
∞

∫e

−m0 gy kBT

0

∞

dy = ∫ e

− ay

∞

dy =

0

⎛ 1⎞
− ay
∫ e ( −a dy ) ⎜⎝ − a ⎟⎠

y=0

∞
1
⎛ 1⎞
⎛ 1⎞
= ⎜ − ⎟ e − ay 0 = ⎜ − ⎟ ( 0 − 1) =
⎝ a⎠
⎝ a⎠
a

Using Table B.6 in the appendix,
∞

∫0 ye

− ay

1!

⎛ 1⎞
dy = 2 = ⎜ ⎟
( a) ⎝ a ⎠

2

Then,
∞

yavg =

∫ ye

− ay

dy

0

∞

− ay
∫ e dy

2
1 a)
(
1
=
= =

1a

a

kBT
m0 g

0

(b)

yavg =

kBT
RT
( 8.314 J/mol ⋅ K ) ( 283 K ) = 8.31 km
=
=
( M N A ) g Mg ( 28.9 × 10−3 kg )( 9.8 m/s2 )


 

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