Chapter 20
Questions: 2, 5, 7, 9, 12, 16, 17, 19, 23, 28, 31
Problems: 5, 10, 11, 15, 17, 21, 30, 47, 49, 57, 59
Questions
2. According to the ideal gas law when T = 0, the P V product vanishes.
If the pressure vanishes then V can be …nite, however if P is nonzero then V
must vanish for an ideal gas. This is clearly absurd for a real gas.
5. Relative to the bottle nothing has changed. The gas molecules are
simply in a slightly di¤erent frame relative to the ground. So the pressure
and temperature of the gas is unchanged. That being said the frame of the
bottle is now moving relative to the ground with some velocity vx ; hence
the average velocity of the gas molecules relative to the ground satis…es,
hvx i = vx :
7. A sound wave transports energy via a pressure wave. It is the collisions
between gas molecules that transmit this energy. Since the rate of these
collisions depends on the speeds of the molecules the pressure wave can only
move through the gas approximately as fast as the molecules are traveling.
9. The escape velocity for Jupiter is much larger than that for the smaller
planets. Additionally Jupiter is much colder than the Earth which reduces
the thermal speed. Thus from the Maxwell distribution, the relative number
of molecules with a speed higher than the escape velocity is too small for the
planet to lose a signi…cant amount of even the light elements.
12. Just under the surface of the ice, the temperature is just below or at
freezing. At the bottom of the lake is where the water that is the most dense
resides. This occurs at 4 C.
16. The ice and water have come to equilibrium and are the same temperature, T = 0 C:
17. For water the amount of heat to melt a gram of water is 334J. The
amount of heat to heat water from 0 C to 100 C is 100 4:18J = 418J.
Thus it takes more heat to raise the temperature from freezing to boiling
that it does to melt the ice.
19. The melting and boiling point are both pressure dependent, while
the triple point only occurs at a unique value of pressure and temperature.
23. You lower the pressure when you remove the cap, which lowers the
boiling point.
28. The larger pressure at the core of the Earth raises the melting point
of the core material enough so that the core is solid.
1
31. The brass expands faster as the bimetallic strip is heated causing the
strip to bend or arch.
Problems
5. (a) From the ideal gas law, P V = nRT; the volume is
V =
nRT
2 8:314 250
=
= :0273m3 = 27:3L
5
P
1:5 1:013 10
(b) Since the number of moles remains the same the new temperature is
Tnew =
Pnew Vnew
4
Pnew Vnew
=
Told =
nR
Pold Vold
1:5
1
2
250 = 333K
10. (a) Using the ideal gas law, P V = nRT , so that n = P V =RT: For
a volume V = (:1)2 1 = 0:0314; a pressure P = 1:01 105 180 =
1: 818 107 ; the number of moles is (Troom = 300K)
n=
1: 818 107 0:0314
PV
'
' 229moles.
RT
8:314 300
(b)
T
= 229
273
11. From the ideal gas law
V = nVmole
Pcrit = 4:40
22:4
300
' 5637L.
273
323
' 515kP a:
276
105
This is a little more than 5atms.
15. From kinetic theory we know that hv 2 i = 3kT =m: So the ratio of the
velocity squared for hydrogen at 75K and SO2 at 350K is
2
vH
75 64
2
=
' 7:
2
350 2
vSO2
So hydrogen molecules are moving faster (on average).
17. The VdW constants for He are a = :0341L2 -atm=mol2 and b =
:0237L=mol: From the Van der Waals equation
(P + n2 a=V 2 ) (V
nR
= 268K:
TV dW =
TV dW
nb)
=
(90 + 9
2
:0341=:64) 1:013 105 (:8
3 8:314
3
:0237)
10
3
For an ideal gas the temperature is
T =
1:013 105 :8
3 8:314
PV
90
=
nR
10
3
= 292K
Clearly at this temperature and pressure the ideal gas law is not a good
model.
21. The Maxwell-Boltzmann distribution for an ideal gas is
N (v) v = 4 N
m
2 kT
3=2
v2e
mv 2 =2kT
v=4 N
mmol
2 RT
3=2
v2e
mmol v 2 =2RT
For the conditions given in this problem with T = 100K we …nd
mmol
=
2RT
2
2 10 3
= 1:203
8:314 100
10
6
Substituting this result into the Maxwell-Boltzmann distribution yields
1:203
24
N (v) v = 4
10
N (v) v = 9:1
1020
10
6
3=2
(900:5)2 e (1:203
10
6
(900:5)2 )
For the conditions given in this problem with T = 450K we …nd
mmol
=
2RT
2
2 10 3
= 2:673
8:314 450
10
7
Substituting this result into the Maxwell-Boltzmann distribution yields
2:673
24
N (v) v = 4
10
N (v) v = 2:0
1020
10
7
3=2
(900:5)2 e (2:673
10
7
(900:5)2 )
30. The total energy emitted by a 625W microwave oven in 1 minute
is Q = 625 60 = 37500J. The heat required to melt 1g of ice at 0 C is
Lf = 334J=g. Thus the total amount of ice that can be melted is
M = 37500:=334 ' 112g
47. Water requires 4:18J to change its temperature 1 degree K( C). This
means that for 16kg of water to cool down 25K requires
E = 16
103
25
3
4:18 = 1672kJ:
v:
of heat to be removed. Since the heat required to melt 1g of ice at 0 C is
Lf = 334J=g, the total amount of ice is
M=
1672
334
103 ' 5kg
of ice.
49. The amount of heat required to heat ice from 10 C to 0 C is
20:5J=g while the amount of heat (removed) required to cool water from
T C to 0 C is 4:18T =1gm:. To maintain the same mass ratio, no ice melts
or water freezes, hence
4:18T = 20:5 ! T = 4:9 C.
57. The coe¢ cients of linear expansion for Pyrex and steel are pyr =
3:2 10 6 and st = 12 10 6 : Hence as the Pyrex and steel are cooled, the
radius of the steel ball bearing will contract faster than the Pyrex. From the
conditions of the problem (1 = 10 4 cm) we have
d
=
d
d =
st
(T
st
(T
x
= pyr (T 330)
x
330) d + 2 10 4 = x = pyr (T
330) and
330) x:
Initially both the diameter of the steel ball, d, and the side of the Pyrex cube,
x, are both 1cm: This allows us to divide by d and/or x and solve for T .
12
10
6
(T 330) + 2 10 4
12 10 2 T 39:6 + 2
8:8 10 2 T
T
=
=
=
=
3:2 10 6 (T 330)
3:2 10 2 T 10:56
37:6 10:56 = 27:04
307K
59. The new length of the rod from …gure 20-12 is given by
p
L = 2 L2o =4 + d2 :
Assuming a linear expansion coe¢ cient of
L=L
Lo =
then
T Lo :
Solving for d we …nd
L2o + 4d2 = L2o (1 +
T )2 = L2o 1 + 2
Lo p
d =
2 T + 2 T 2:
2
4
T+
2
T2