Solution to HW#13
CJ5 14.CQ.009.
REASONING AND SOLUTION Packing material consists of
"bubbles" of air trapped between bonded layers of plastic. The packing material
exerts normal forces on the packed object at each place where the bubbles make
contact with the object. The motion of the packed object is thereby restricted. The
magnitude of the normal force that any given bubble can exert depends on the
pressure of the air trapped inside the bubble. The magnitude of the normal force is
equal to the pressure of the air times the area of contact between the bubble and the
object. Since the air is trapped in the bubbles, the number of air molecules and the
volume of the gas are fixed. The pressure exerted by the bubbles, then, depends on
the temperature according to the ideal gas law: PV = nRT . On colder days, T is
smaller than on warmer days; therefore, on colder days, the pressure P of the air in
the bubbles is less than on warmer days. When the pressure is smaller, the magnitude
of the normal force that each bubble can exert on a packed object is smaller.
Therefore, the packing material offers less protection on cold days.
CJ5 14.CQ.013.
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 nR T ; 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.
CJ5 14.P.004. 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 9NO 2 ) is
Molecular 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
The molecular mass of Advil (C13H18O2 ) is
Mass of 2
oxygen atoms
Molecular mass
= 13 (12.011 u) + 18 (1.00794 u) + 2 (15.9994) = 206.285 u
of Advil
Mass of 13
carbon atoms
Mass of 18
hydrogen atoms
Mass of 2
oxygen atoms
a.
Therefore, the number of molecules of pain reliever in the standard dose
of Tylenol is
m


Number of molecules = n N A = 
 NA
 Mass per mole 
 325 × 10−3 g 
23
21
−1
=
 ( 6.022 × 10 mol ) = 1.29 ×10
151.165
g/mol


b.
Advil is
Similarly, the number of molecules of pain reliever in the standard dose of
m


Number of molecules = n N A = 
 NA
 Mass per mole 
 2.00 × 10−1 g 
−1
23
20
=
 ( 6.022 × 10 mol ) = 5.84 × 10
206.285
g/mol


CJ5 14.P.010. REASONING AND SOLUTION To find the temperature T2, use the
ideal gas law with n and V constant. Thus, P1/T1 = P2/T2. Then,
T2 = T1(P2/P1) = (284 K)[(3.01 × 105 Pa)/(2.81 × 105 Pa)] = 304 K
CJ5 14.P.021. REASONING AND SOLUTION
present for each case.
Find the number of moles of gas
5
(1.01 X 10 Pa)(60.0 m3)
PV
n =
=
1 RT
R(289 K)
1
5
(1.01 X 10 Pa)(60.0 m3)
PV
=
n =
2 RT
R(302 K)
2
The number of moles lost is n1 – n2. The mass that is lost is, therefore,
mlost = (n1 – n2 )(28.01 g/mol) = 3.0 × 103 g = 3.0 kg
CJ5 14.P.023. SSM WWW REASONING Since the temperature of the confined air
is constant, Boyle's law applies, and PsurfaceVsurface = PhVh , where Psurface and Vsurface
are the pressure and volume of the air in the tank when the tank is at the surface of the
water, and Ph and Vh are the pressure and volume of the trapped air after the tank has
been lowered a distance h below the surface of the water. Since the tank is completely
filled with air at the surface, Vsurface is equal to the volume Vtank of the tank. Therefore,
the fraction of the tank's volume that is filled with air when the tank is a distance h below
the water's surface is
Vh
Vtank
=
Vh
Vsurface
=
Psurface
Ph
We can find the absolute pressure at a depth h using Equation 11.4. Once the
absolute pressure is known at a depth h, we can determine the ratio of the pressure at the
surface to the pressure at the depth h.
SOLUTION According to Equation 11.4, the trapped air pressure at a depth
h = 40.0 m is
Ph = Psurface + ρ gh = (1.01× 105 Pa) +  (1.00 ×103 kg/m3 )(9.80 m/s 2 )(40.0 m) 
= 4.93 ×105 Pa
where we have used a value of ρ = 1.00 × 103 kg/m3 for the density of water.
The desired volume fraction is
Psurface 1.01× 10 5 Pa
Vh
=
=
= 0.205
Ph
Vtank
4.93 × 10 5 Pa
CJ5 14.P.029. REASONING AND SOLUTION Using KE = (1/2) mvrms2 = (3/2) kT,
we can solve for vrms.
3kT =
m
vrms =
-23
3(1.38 x 10
3
J/K)(6.0 x 10 K)
-27
1.67 x 10
kg
= 1.2 × 104 m/s
(14.6)
CJ5 14.P.035. 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
1
2
altitude, or 2 mv rms
= mgh , where the gravitational potential energy near the earth's
surface has been chosen to be zero. Using Equation (1), this statement of energy
3
conservation becomes kT = mgh . Solving for h gives
2
3kT
h=
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
23
–1
6.022 × 10 mol
Substituting into Equation (2) gives
3(1.38 × 10 –23 J/K)(295 K)
h=
=
2(6.634 × 10 –26 kg)(9.80 m/s 2 )
9400 m