Homework Solutions - Physics 8C
chpt. 18 – 4,9,11,25,30,36,41,46,49,60
18.4. Solve: (a) Air is primarily comprised of diatomic molecules, so r ≈ 1.0 × 10−10 m. Using the ideal-gas
law in the form pV = NkBT , we get
1.013 × 105 Pa
1.0 × 10−10 mm of Hg ×
N
p
760 mm of Hg
=
=
=3.30 × 1012 m −3
V kBT
(1.38 × 10−23 J/K)(293 K)
(b) The mean free path is
1
1
=
=1.71 × 106 m
12 −3
4 2π ( N/V )(r ) 4 2π (3.30 × 10 m )(1.0 × 102 10 m) 2
λ=
2
Assess: The pressure p in the vacuum chamber is 1.33 × 10−8 Pa =
1.32 × 10−13 atm. A mean free path of
1.71 × 106 m is large but not unreasonable.
18.9. Solve: (a) In tabular form we have
Particle
v x (m/s)
v y (m/s)
1
20
30
2
70
−40
3
−80
−10
4
60
−20
5
0
−50
6
40
−20
Average
0
0



ˆ
The average velocity is vavg
= 0 i + 0 ˆj .
vx2 (m/s)2
v y2 (m/s)2
v2 (m/s)2
v (m/s)
400
1600
6400
3600
0
1600
900
4900
100
400
2500
400
1300
6500
6500
4000
2500
2000
3800
36.06
80.62
80.62
63.25
50.00
44.72
59.20
(b) The average speed is vavg = 59 m/s .
(c) The root-mean-square speed=
is vrms
=
( v2 )
avg
3800=
m 2 / s 2 62 m/s .
18.11. Solve: (a) The atomic mass number of neon is 20. This means the mass of a neon atom is
m=
20 u =
20(1.661 × 10−27 kg) =.
3 322 × 10−26 kg
The pressure of the gas is
p=
1 N
3 V

 2
 mvrms =

1 (5.00 × 1025
3
m −3 )(3.322 × 10−26 kg)(600 m/s) 2 = 19.93 × 104 Pa
The temperature of the gas in the container can be obtained from the ideal-gas equation in the form pV = NkBT :
=
T
pV
19.93 × 104 Pa
=
= 289 K
NkB (5.00 × 1025 m −3 )(1.38 × 10−23 J/K)
(b) We already computed the pressure to be 200 kPa.
18.25. Solve: (a) The average kinetic energy of a proton at the center of the sun is
ε avg = kBT ≈ 32 (1.38 × 10−23 J/K )( 2.0 × 107 K ) =
4.1 × 10−16 J
3
2
(b) The root-mean-square speed of the proton is
1
3 (1.38 × 10−23 J/K )( 2.0 × 107 K )
3kBT
vrms =
7.0 × 105 m/s
≈
=
m
1.67 × 10−27 kg
18.30. Solve: (a) For a monatomic gas,
∆Eth= nCV ∆T= 1.0 J=
(1.0 mol )(12.5 J/mol K ) ∆T
⇒=
∆T 0.080°C or 0.080 K
(b) For a diatomic gas,
1.0 J = (1.0 mol)(20.8 J/mol K)∆T ⇒ ∆T = 0.048°C or K
(c) For a solid,
1.0 J = (1.0 mol)(25.0 J/mol K) ∆T ⇒ ∆T = 0.040°C or K
18.36. Solve: (a) The thermal energy of a monatomic gas is
E=
th
3
2
NkB=
T
3
nRT
2
⇒=
T
2 Eth 1
3 n R
1
 2   5000 J 
=
⇒ TA   =
201 K

 3   2.0 mol  (8.31 J/mol K)
1
 2   8000 J 
=
TB =
214 K


 3   3.0 mol  (8.31 J/mol K)
Thus, gas B has the higher initial temperature.
(b) The equilibrium condition is (e=
=
A )avg (e
B )avg (etot )avg . This means
EAf EBf
Etot
= =
nA
nB nA + nB
⇒
=
EAf


nA
2 mols
=
Etot 
=
J) 5200 J
 (5000 J + 8000
nA + nB
2
mols
3
mols
+


 3 mols 
nB
=
Etot 
=
 (13,000 J) 7800 J
nA + nB
 5 mols 
=
EBf
18.41. Solve: (a) The number density is =
N / V 1=
cm −3 106 m −3 . Using the ideal-gas equation,
N
kBT ≈ (1 × 106 m −3 )(1.38 × 10−23 J/K ) ( 3 K )
V
1 atm
=
4 × 10−17 Pa ×
=
4 × 10−22 atm
1.013 × 105 Pa
p
=
(b) For a monatomic gas,
vrms
=
3kBT
=
m
3 (1.38 × 10−23 J/K ) ( 3 K )
1.67 × 10−27 kg
= 270 m/s
(c) The thermal energy is Eth = 32 NkBT , where N = (106 m −3 )V . Thus
Eth 1.0
J
=
=
3
2
(10
6
m −3 )V (1.38 × 10−23 J/K ) ( 3 K ) ⇒ V = 1.6 × 1016 m3 = L3
⇒ L = 2.5 × 105 m
18.46. Model: Pressure is due to random collisions of gas molecules with the walls.
Solve: According to Equation 18.9, the collision rate with one wall is
rate of collisions
=
2
N coll 1 N
=
Avx
∆tcoll 2 V
However, this equation assumed that all molecules are moving in the x-direction with constant speed. The rms
speed vrms is for motion in three dimensions at varying speeds. Consequently, we need to replace vx not with
(vx )avg , which is zero, but with
v2
v
vx → (vx2 )avg = rms =rms
3
3
With this change,
1 N
Avrms
2 3V
The molecular mass of nitrogen is A = 28 u, thus the rms speed of the molecules at 20°C is
rate of collisions =
3(1.38 × 10−23 J/K)(293 K)
= 510 m/s
28(1.661 × 10−27 kg)
3kBT
=
m
=
vrms
22
molecules, the number density is
With=
N 0.10N
=
A 6.02 × 10
N
6.02 × 1022
=
=6.02 × 1025 m −3
V 0.10 m × 0.10 m × 0.10 m
Thus
rate of collisions =
1
(6.02 × 1025 m −3 )(0.10 m × 0.10 m)(510 m/s) = 8.9 × 1025 s −1
2 3
18.49. Solve: (a) The number of moles of helium and oxygen are
=
nhelium
2.0 g
8.0 g
= 0.50 mol =
noxygen = 0.25 mol
4.0 g/mol
32.0 g/mol
Since helium is a monoatomic gas, the initial thermal energy is
=
K)
( 0.50 mol ) ( 32 ) (8.31 J/mol K )( 300
=
Ehelium i = nhelium ( 32 RThelium
)
1870 J ≈ 1900 J
Since oxygen is a diatomic gas, the initial thermal energy is
Eoxygen i = noxygen ( 52 RToxygen
=
)
=
K)
( 0.25 mol ) ( 52 ) (8.31 J/mol K )( 600
3116 J ≈ 3100 J
(b) The total initial thermal energy is
E tot = E helium i + E oxygen i = 4986 J
As the gases interact, they come to equilibrium at a common temperature T f . This means
=
4986 J
=
⇒ Tf
(
1
2
4986 J
=
R ) ( 3nhelium + 5noxygen )
1
2
nhelium ( 32 RTf ) + noxygen ( 52 RTf )
4986 J
=
(8.31 J/mol K )( 3 × 0.50 mol + 5 × 0.25 mol )
436.4 K = 436 K
The thermal energies at the final temperature T f are
Ehelium f = nhelium ( 32 RTf ) = ( 32 ) ( 0.50 mol )( 8.31 J/mol K )( 436.4 K ) = 2700 J
Eoxygen f = noxygen ( 52 RTf ) = ( 52 ) ( 0.25 mol )( 8.31 J/mol K )( 436.4 K ) = 2300 J
(c) The change in the thermal energies are
Eoxygen f − Eoxygen i =
2266 J − 3116 J =
−850 J
Ehelium f − Ehelium i = 2720 J − 1870 J = 850 J
The helium gains energy and the oxygen loses energy.
(d) The final temperature can also be calculated as follows:
=
Ehelium f
( nhelium ) 32 RTf ⇒
2720 J = (0.50 mol)(1.5)(8.31 J/mol K)T f ⇒
=
Tf 436.4 K ≈ 436 K
18.60. Solve: (a) The thermal energy of a monatomic gas of n 1 moles is E1 = 32 n1RT . The thermal energy of a
diatomic gas of n 2 moles is E2 = 52 n2 RT . The total thermal energy of the mixture is
3
=
Eth
1
2
( 3n1 + 5n2 ) RT
⇒ ∆=
Eth
1
2
( 3n1 + 5n2 ) R∆T
Comparing this expression with
∆Eth = ntotal CV ∆T =
we get
CV =
( n1 + n2 ) CV ∆T
( 3n1 + 5n2 ) R
2 ( n1 + n2 )
(b) For a diatomic gas, n 1 → 0, and CV = 52 R . For a monotomic gas, n 2 → 0, and CV = 32 R .
4