Homework Chapter 19 Solutions!
19.15
19.26
19.28
19.33
19.47
19.52
19.57
19.60
page 1
Problem 9.15
A hollow aluminum cylinder 20 cm deep as an internal capacity of 2 L at 20 °C. It is completely
filled with turpentine and 20 °C. The turpentine and the aluminum cylinder are then slowly warm
together to 80 °C.
(a)!How much turpentine overflows?
(b)!What is the volume of the turpentine remain in the cylinder at 80°C?
(c)!If the combination with this amount of turpentine is then cooled back to 20 °C, how far below
the cylinders rim does the turpentine’s surface recede?
Solution
The volume of the aluminum container is 2.000 L at 20 °C. The volume of turpentine is 2.000 L at
20 °C.
(a)!When the entire system is heated to 80 °C. The volume of the aluminum container is
(
)
Vf = Vi (1 + 3αΔT ) = 2.000 1 + (3)(24 ×10−6 °C −1 )(60 °C ) = 2.00864 L
The volume of the turpentine is
(
)
Vf = Vi (1 + βΔT ) = 2.000 L 1 + (9.0 ×10−4 °C −1 )(60 °C ) = 2.108 L
The overflow is
Vturpentine −Valuminum = 2.108 L − 2.00864 L = 0.09936 L = 0.099 L
(b)!The turpentine volume remaining is the same as the aluminum container volume which is
2.009 L.
(c)!Cooling the system back to 20 °C means the aluminum container is back to 2.000 L. The
volume of turpentine is
(
)
Vf = Vi (1 + βΔT ) = 2.00864 L 1 + (9.0 ×10−4 °C −1 )(−60 °C ) = 1.9002 L
This is the following fraction of the volume of the container.
Vturpentine
Vcontainer
=
1.9002 L
= 0.95009
2.000 L
It is also the following fraction of the 20 cm height of the container.
hturpentine
hcontainer
=
hturpentine
20.0 cm
= 0.95009
⇒ hturpentine = 19.9002 cm
This is 0.0998 cm from the top.
page 2
Problem 19.26
The pressure gauge on a tank registers the gauge pressure which is the difference between the
interior pressure and exterior pressure. When the tank is full of oxygen it contains 12 kg of the
gas at a gauge pressure of 40 atm. Determine the mass of oxygen that has been withdrawn from
the tank when the pressure reading is 25 atm. Assume the temperature of the tank remained
constant.
Solution
The number of moles of oxygen in the initial state is
⎛ 1000 g ⎞⎟⎛⎜ 1 mol O2 ⎞⎟
⎟ = 375 mol O2
⎟
12 kg O2 ⎜⎜
⎜⎝ 1 kg ⎟⎟⎠⎜⎜⎝ 32 g O ⎟⎟⎠
2
The absolute pressure is 41 atm inside the tank. The temperature is
PV = nRT
⇒ (41 atm)V = (375 mol)RT
In the final state, the ideal gas law says this.
PV = nRT
⇒ (26 atm)V = (375 mol − x mol)RT
Dividing the second equation by the first,
(26 atm)V
(375 mol − x mol)RT
=
(41 atm)V
(375 mol)RT
⇒
26
x
= 1−
41
375
⇒ x = 137.2 mol
The mass of the extracted oxygen is
⎛ 32 g O ⎞⎟⎛ 1000 g ⎞⎟
2 ⎟⎜
⎟ = 4.39 kg O2
137.2 mol O2 ⎜⎜
⎜
⎜⎝ 1 mol O2 ⎟⎟⎠⎜⎝ 1 kg ⎟⎟⎠
page 3
Problem 9.28
To measure how far below the ocean surface a bird dives to catch a fish scientists uses a
message originated by Lord Kelvin. He dust the interiors of plastic tubes with powdered sugar
and then seals one end of each tube. He captures the bird at nighttime in its nest and attaches a
tube to it’s back. He then catches the same bird the next night and removed the tube. In one trial,
using a tube 6.5 cm long, water washes away the sugar over a distance of 2.7 cm from the open
end of the tube. Find the greatest depth to which the bird dove, assuming the air in the tube stays
at constant temperature.
Solution
When the cup comes in contact with the water surface, the air inside is described by the ideal gas
model.
PiVi = nRT
⇒ (1 atm)(A ⋅ 6.50 cm) = nRT
At the maximum depth, the air is describe by
PfVf = nRT
⇒ Pf (A ⋅ (6.50 cm − 2.70 cm)) = nRT
Both “nRT”s are the same, so
(1 atm)(A ⋅ 6.50 cm) = Pf (A ⋅ (6.50 cm − 2.70 cm))
(1 atm)(6.50 cm) = Pf (3.80 cm)
Pf = (1 atm)
(6.50 cm)
= 1.7105 atm
(3.80 cm)
This pressure of the air inside the tube equals the pressure of the top of the water inside the tube.
pressure = Pf
pressure = Patm + Pwater
This pressure comes from the atmosphere and water.
Pf = 1 atm + ρwater gh = 1.7105 atm
ρwater gh = 0.7105 atm
101, 325 Pa
= 71991.4 Pa = (1000
1 atm
kg
)(9.8 m2 )h
m3
s
h = 7.3461 m
The final depth is 7.35 m.
page 4
Problem 19.33
At 25 m below the surface of the sea where the temperature is 5 °C, a diver exhales an air bubble
having a volume of 1 cm³. If the surface temperature of the sea is 20 °C, what is the volume of
the bubble just before it breaks the surface?
Solution
When the bubble was created, the number of molecules is fixed. The volume is 1 x 10-6 m3. The
temperature is 278 K. The pressure of the bubble depends on its depth.
Pi = 1 atm + ρwater gh = 101325 Pa + (1000
kg
)(9.8 m2 )(25
m3
s
m) = 346, 330 Pa
Treat the air in the bubble as an ideal gas. The ideal gas model states that
(346, 330 Pa)(1×10−6 m 3 ) = nR(278 K)
nR = 1.2458 ×10−3
J
K
At the surface of the water, the number of molecules is still the same. The temperature is now
293 K. The pressure reduces to 1 atm. The ideal gas model states that
(101, 325 Pa)Vf = (1.2458 ×10−3
J )(293
K
K)
Vf = 3.60 ×10−6 m 3 = 3.60 cm 3
page 5
Problem 9.47
A clock with a brass pendulum has a period of one second at 20 °C. If the temperature increases
to 30 °C, how much does the period change and how much time does the clock gain or lose in
one week?
Solution
The period of a pendulum with the small angle approximation is
T = 2π
l
g
At 20 °C, it is
1 s = 2π
l
9.8
m
s2
⇒ l = 0.24824 m
At 30 °C, the length of the pendulum becomes
l f = l(1 + αΔT ) = 0.24824 m(1 + (19 ×10−6 °C −1 )(+10 °C )) = 0.24828 m
The new period is
T = 2π
0.24828 m
= 1.0001 s
9.8 m2
s
The actual time of one week is 6.0480 x 105 s. The time for the warmer pendulum to register one
week is 6.0486 x 105 s. It takes an extra 57.5 seconds to complete a week. Therefore, at 30 °C,
the clock is 57.5 seconds slower.
You can also look at the frequency of the warmer pendulum as being lower so that it takes more
oscillations to complete the number of cycles in a week.
page 6
Problem 9.52
A vertical cylinder of cross-sectional area A is fitted with a tight-fitting frictionless piston of mass m.
The piston is not restricted in its motion in any way and is supported by the gas at pressure P
below it. Atmospheric pressure is Po. Suppose n moles of an ideal gas are in the cylinder at a
temperature of T. Find the height h of the piston above the bottom of the cylinder.
Solution
The net force on the piston must be zero.
mg + P0A = PA
Assuming we know n and T, we can describe the gas using the ideal gas model.
P=
nRT
V
The force balance is now
mg + P0A =
nRT
nRT
A=
V
h
The height is
h=
nRT
mg + P0A
page 7
Problem 19.57
Write an expression for the buoyant force on a spherical balloon submerged in water as a function
of the depth h below the surface, the volume Vi of the balloon at the surface, the pressure Po at
the surface, and the density of the water ρo. Assume the water temperature does not change with
depth.
Does the buoyant force increase or decrease out of the balloon is submerged?
At what depth is the buoyant force one-half its surface value?
Solution
The buoyant force is, from Archimedes principle, the weight of the fluid displaced. The weight of
the fluid displaced is the mass (the density times the volume of the fluid) times the acceleration
due to gravity.
Fbuoyant = ρw gV(h)
We know that the volume of the balloon at the surface (uncompressed) is based on the ideal gas
equation.
PiVi = nRT
At depth, the volume decreases as the pressure increases.
temperature stay the same. The ideal gas model says
The number of moles and the
PfVf = nRT
The final pressure is function of the depth ,h.
Pf = Pi + ρw gh
Therefore, the volume as a function of depth is
Vf =
nRT
nRT
=
Pf
Po + ρw gh
And the buoyant force is
F =
ρw gPoVi
Po + ρw gh
As the depth is increased, the force decreases since the volume of displaced fluid decreases.
The depth at which the force is halved its value at the surface is
ρw gPoVi
1
= ρw gVi
Po + ρw gh
2
ρw gh = Po
⇒ h=
⇒
Po
1
=
Po + ρw gh
2
⇒
2Po = Po + ρw gh
Po
101, 325 Pa
=
= 10.3 m
ρw g
(1000 kg3 )(9.8 m2 )
m
s
page 8
Problem 19.60
A cylinder is closed by a piston connected to a spring of constant 2x103 N/m. With the spring
relaxed, the cylinder is filled with 5 L of gas at a pressure of 1 atm and temperature of 20 °C.
If the piston has a cross-sectional area of 0.01 m² and negligible mass, how high will the piston
rise when the temperature is raised to 250°C?
What is the pressure of the gas at 250°C?
Solution
The spring starts out relaxed so it applies no force on the piston. The only pressure is from the
atmosphere. The initial and final states of the gas are
state
P
V
n
T
initial
101,325 Pa
0.005 m3
n
293 K
final
Pf
Vf
n
523 K
We can find the number of moles from the initial values.
PiVi = nRTi
state
⇒ n=
PiVi
(101, 325 Pa)(0.005 m 3 )
=
= 0.20797 mol
J )(293 K)
RTi
(8.314 mol⋅K
P
V
initial
101,325 Pa
final
Pf
0.005
m3
Vf
n
T
0.20797 mol
293 K
0.20797 mol
523 K
The final pressure is going to be the atmosphere plus the force the spring applies to the gas. Say
the piston is displaced upward by a distance h.
Pf = Pi +
F
kh
= Pi +
A
A
The final volume is going to be this.
Vf = Vi + Ah
The ideal gas equation for the final state is
PfVf = nRTf
⇒ (Pi +
kh
)(V + Ah) = nRTf
A i
⇒ (k)h 2 + (Pi A +
Vik
)h + (PiVi − nRTf ) = 0
A
This is quadratic in h. Not going to do this analytically.
Pi A +
(0.005 m 3 )(2000
Vik
= (101, 325 Pa)(0.01 m 2 ) +
A
(0.01 m 2 )
nR(Ti −Tf ) = (0.20797 mol)(8.314
J )(293 − 523
mol⋅K
N)
m
= 2, 013.3 N
K) = −397.68 J
The quadratic equation is this.
2000h 2 + 2013.3h − 397.68 = 0
The solutions are 0.16912 m, and –1.1758 m. The correct answer is 0.169 m.
page 9
The pressure is
Pf = Pi +
kh
(2, 000 N/m)(0.16912 m)
= 101, 325 Pa +
= 1.3515 ×105 Pa
A
(0.01 m 3 )
page 10