CHAPTER 14 THE IDEAL GAS LAW
AND KINETIC THEORY
CONCEPTUAL QUESTIONS
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1.
REASONING AND SOLUTION
a. Avogadro's number NA is the number of particles per mole of substance. Therefore, one
mole of hydrogen gas (H2) and one mole of oxygen gas (O2) contain the same number
(Avogadro's number) of molecules.
b. One mole of a substance has a mass in grams that is equal to the atomic or molecular
mass of the substance. The molecular mass of oxygen is greater than the molecular mass of
hydrogen. Therefore, one mole of oxygen has more mass than one mole of hydrogen.
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3.
REASONING AND SOLUTION A tightly sealed house has a large ceiling fan that blows
air out of the house and into the attic. The fan is turned on, and the owners forget to open
any windows or doors. As the fan transports air molecules from the house into the attic, the
number of air molecules in the house decreases. Since the house is tightly sealed, the
volume of the house remains constant. If the temperature of the air inside the house remains
constant, then from the ideal gas law, PV  nRT , the pressure in the house must decrease.
The air pressure in the attic, however, increases. The fan must now blow air from a lower
pressure region to a higher pressure region. Thus, it becomes harder for the fan to do its job.
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7.
REASONING AND SOLUTION Atmospheric pressure decreases with increasing altitude.
Helium filled weather balloons are underinflated when they are launched. As the balloon
rises, the pressure exerted on the outside of it decreases. The number of helium molecules in
the balloon is fixed. If we assume that the temperature of the atmosphere remains constant,
then from the ideal gas law, PV  nRT , we see that the volume of the helium in the balloon
will increase, thereby further inflating the balloon. If the balloon is fully inflated when it is
launched from earth, it would burst when it reaches an altitude where the expanded volume
of the helium is greater than the maximum volume of the balloon.
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10. REASONING AND SOLUTION The kinetic theory of gases assumes that a gas molecule
rebounds with the same speed after colliding with the wall of a container. From the impulsemomentum theorem, Equation 7.4, we know that Ft  mv f  v 0 , where the magnitude
of F is the magnitude of the force on the wall. This implies that in a time interval t, a gas
molecule of mass m, moving with velocity v before the collision and –v after the collision,
exerts a force of magnitude F  2mv / t on the walls of the container, since the magnitude of
v f  v 0 is 2v. If the speed of the gas molecule after the collision is less than that before the
collision, the force exerted on the wall of the container will be less than F  2mv / t , since
the magnitude of v f  v 0 is then less than 2v. Therefore, the pressure of the gas will be less
than that predicted by kinetic theory.
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11. REASONING AND SOLUTION The relationship between the average kinetic energy per
particle in an ideal gas and the Kelvin temperature T of the gas is given by Equation 14.6:
1 mv 2  3 kT
.
rms
2
2
If the translational speed of each molecule in an ideal gas is tripled, then the root-meansquare speed for the gas is also tripled. From Equation 14.6, the Kelvin temperature is
proportional to the square of the root-mean-square speed. Therefore, the Kelvin temperature
will increase by a factor of 9.
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13. REASONING AND SOLUTION According to Equation 14.7, the internal energy of a
sample consisting of n moles of a monatomic ideal gas at Kelvin temperature T is
U  32 nRT ; therefore, the internal energy of such an ideal gas depends only on the Kelvin
temperature. If the pressure and volume of this sample is changed isothermally, the internal
energy of the ideal gas will remain the same. Physically, this means that the experimenter
would have to change the pressure and volume in such a way, that the product PV remains
the same. This can be verified from the ideal gas law (Equation 14.1), PV  nRT ; if the
values of P and V are varied so that the product PV remains constant, then T will remain
constant and, from Equation 14.7, the internal energy of the gas remains the same.
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PROBLEMS
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2.
REASONING AND SOLUTION
a.
The molecular mass of C55H72MgN4O5 is
55(12.011 u) + 72(1.00794 u) + 24.305 u + 4(14.0067 u) + 5(15.9994 u) = 893.51 u
b.
The mass per mol is 893.51 g/mol. The mass of 3.00 mol is
(893.51 g/mol) (3.00 mol) = 2680 g
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4.
REASONING The number of molecules in a known mass of material is the number n
of moles of the material times the number NA of molecules per mole (Avogadro's number). We
can find the number of moles by dividing the known mass m by the mass per mole.
SOLUTION Using the periodic table on the inside of the text’s back cover, we find that
the molecular mass of Tylenol ( C8H 9NO2 ) is
M olecular mass  8 (12.011 u)  9 (1.00794 u)
of Tylenol
Mass of 8
carbon atoms
Mass of 9
hydrogen atoms
+ 14.0067 u + 2 (15.9994) =151.165 u
Mass of
nitrogen atom
Mass of 2
oxygen atoms
The molecular mass of Advil ( C13H18O2 ) is
Molecular mass
 13 (12.011 u)  18 (1.00794 u) + 2 (15.9994) = 206.285 u
of Advil
Mass of 13
carbon atoms
a.
Tylenol is
Mass of 18
hydrogen atoms
Mass of 2
oxygen atoms
Therefore, the number of molecules of pain reliever in the standard dose of
m


Number of molecules = n N A  
 NA
 Mass per mole 
 325  103 g 
23
1
21

  6.022 10 mol   1.29 10
 151.165 g/mol 
b.
Similarly, the number of molecules of pain reliever in the standard dose of Advil
is
m


Number of molecules = n N A  
 NA
 Mass per mole 
 2.00  10 g 
23
1
20

  6.022  10 mol   5.84  10
 206.285 g/mol 
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9.
REASONING We can use the ideal gas law, Equation 14.1 (PV = nRT) to find the
number of moles of helium in the Goodyear blimp, since the pressure, volume, and temperature
are known. Once the number of moles are known, we can find the mass of helium in the blimp.
SOLUTION The number n of moles of helium in the blimp is, according to Equation
14.1,
n
PV (1.1  10 5 Pa)(5400 m 3 )

 2.55  10 5 mol
RT [8.31 J/(mol  K)](280 K)
According to the periodic table on the inside of the text’s back cover, the atomic mass of
helium is 4.002 60 u. Therefore, the mass per mole is 4.002 60 g/mol. The mass m of helium in
the blimp is, then,
 1 kg 
3
m  2.55 105 mol  4.002 60 g/mol  
  1.0 10 kg
 1000 g 
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

12.
REASONING The number N of molecules in an ideal gas is given by Equation 14.2 as
N  PV / kT , where P, V, and T are the absolute pressure, volume, and Kelvin temperature of
the gas, and k is Boltzmann’s constant. The ratio of the number of air molecules in the two rooms
is equal to the ratio of the ideal gas laws for the air in each room.
SOLUTION The number of air molecules in rooms A and B is
NA 
PAVA
PV
and N B  B B
kTA
kTB
(14.2)
Dividing NA by NB, and noting that PA = PB and VA = VB, we see that
PAVA
NA
kTA
T
19 + 273 K  1.10

 B 
PBVB
NB
TA
 8.0  273 K
kTB
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20.
REASONING
a.
The volume V of gas in the tank is related to the Kelvin temperature T and
absolute pressure P of the gas by Equation 14.1, V  nRT / P , where n is the number of moles of
gas and R is the universal gas constant.
b.
The mass of chlorine gas that has leaked out of the tank is equal to the initial mass
(11.0 g) minus the mass remaining in the tank at the later time. The mass at the later time is equal
to the mass per mole (in g/mol) of chlorine times the number n of moles remaining in the tank.
The number of moles of gas remaining can be found from the ideal gas law, n  PV / RT
(Equation 14.1)
SOLUTION
a.
The number n of moles of chlorine originally in the tank is equal to the mass of
chlorine (11.0 g) divided by the mass per mole of chlorine (70.9 g/mol), so n = (11.0 g)/(70.9
g/mol). The temperature of the gas is 82 C, which, when converted to the Kelvin temperature
scale, is (82 + 273 )K. The volume of the tank is
 11.0 g 

 8.31 J/  mol  K   273  82  K 
nRT  70.9 g /mol  
4
3
V

 8.17  10 m
5
P
5.60  10 Pa
(14.1)
b.
The mass mCl of chlorine gas that has leaked out of the tank is mCl = 11.0 g – mR, where
mR is the mass of the gas remaining in the tank. The remaining mass, in turn, is equal to the
number nR of moles of gas times the mass per mole (70.9 g/mol) of chlorine. Thus,
mCl  11.0 g  mR = 11.0 g  nR  70.9 g/mol 
The number of remaining moles of gas can be found from the ideal gas law,
nR  PV / RT , so that
 PV 
mCl  11.0 g  
  70.9 g/mol 
 RT 
 11.0 g 
3.80  10
5

Pa 8.17  10
4
m
3
  70.9 g/mol   2.3 g
8.31 J/  mol  K    273  31 K 
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26.
REASONING AND SOLUTION The volume of the cylinder is V = AL where A is the
cross- sectional area of the piston and L is the length. We know P1V1 = P2V2 so that the new
pressure, P2 can be found. We have
P2 = P1(V1/V2) = P1A1L1/A2L2 = P1L1/L2
(since A1 = A2)
P2 = (1.01  105 Pa)(L/2L) = 5.05  104 Pa
The force on the piston and spring is, therefore,
F = P2A = (5.05  104 Pa) (0.0500 m)2 = 397 N
The spring constant is k = F/x so
3
k = F/x = (397 N)/(0.200 m) = 1.98 10 N/m
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28.
REASONING AND SOLUTION From Equation 14.5  PV  23 N ( KE )  , we obtain
KE 
3PV
3(4.5 105 Pa)(8.5 103 m3 )
–21

 4.8 10 J
23
1
2N
2 (2.0 mol)(6.022  10 mol ) 
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35.
REASONING AND SOLUTION Since the argon atom has a kinetic energy equal to its
average translational kinetic energy,
1
mv 2rms
2
3
 2 kT
(1)
Using conservation of mechanical energy, we have Eearth's surface = Ehighest altitude, or
1
mv 2rms
2
 mgh , where the gravitational potential energy near the earth's surface has been
chosen to be zero. Using Equation (1), this statement of energy conservation becomes
3
kT  mgh . Solving for h gives
2
h
3kT
2mg
(2)
In order to use Equation (2), we must find the mass of the argon atom:
m
39.948 g/mol
 6.634 10 –23 g = 6.634  10 –26 kg
6.022  10 23 mol –1
Substituting into Equation (2) gives
3(1.38  10 –23 J/K)(295 K)
 9400 m
2(6.634  10 –26 kg)(9.80 m/s 2 )
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h
37.
SSM WWW REASONING AND SOLUTION The average force exerted by one
electron on the screen is, from the impulse-momentum theorem (Equation 7.4),
F  p/ t  mv / t . Therefore, in a time t , N electrons exert an average force
F  Nmv / t  (N / t)mv . Since the pressure on the screen is the average force per unit area
(Equation 10.19), we have
P
F (N / t)mv

A
A
(6.2  1016 electrons/s)(9.11  10–31 kg)(8.4  107 m/s – 0 m/s)
 4.0  101 Pa
–7
2
1.2  10 m
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