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Catégorie :Category: mViewer GX Creator Lua TI-Nspire
Auteur Author: paolagonzalez
Type : Classeur 3.6
Page(s) : 33
Taille Size: 1.92 Mo MB
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28 Chapter 20 Solutions

Chapter 21 Solutions

*21.1 One mole of helium contains Avogadro's number of molecules and has a mass of 4.00 g. Let us
call m the mass of one atom, and we have

NAm = 4.00 g/mol

4.00 g/mol
or m= = 6.64 × 10–24 g/molecule
6.02 × 1023 molecules/mol

m = 6.64 × 10–27 kg

*21.2 We first find the pressure exerted by the gas on the wall of the container.

NkT 3NA kB T 3RT 3(8.315 N · m/mol · K)(293 K)
P= = = = = 9.13 × 105 Pa
V V V 8.00 × 10–3 m 3

Thus, the force on one of the walls of the cubical container is

F = PA = (9.13 × 105 Pa)(4.00 × 10–2 m2) = 3.65 × 104 N

– ∆v [8.00 sin 45.0° – (–8.00 sin 45.0°)]m/s
21.3 F = Nm = 500(5.00 × 10–3 kg) = 0.943 N
∆t 30.0 s

F
P = = 1.57 N/m2 = 1.57 Pa
A

21.4 Consider the x axis to be perpendicular to the plane of the window. Then, the average force
exerted on the window by the hail stones is

– ∆v [vxf – vxi] [v sin θ – (–v sin θ)] 2v sin θ
F = Nm = Nm = Nm = Nm
∆t t t t

Thus, the pressure on the window pane is

F 2v sin θ
P = = Nm
A At

– (5.00 × 1023)(2 × 4.68 × 10–26 kg × 300 m/s)
*21.5 F = = 14.0 N
1.00 s

F 14.0 N
and P= = = 17.6 kPa
A 8.00 × 10–4 m 2

2 Chapter 21 Solutions

2N  mv 2 
21.6 Use Equation 21.2, P = , so that
3V  2 

mv2 3PV
K av = = where N = nNA = 2NA
2 2N

3PV 3(8.00 atm)(1.013 × 105 Pa/atm)(5.00 × 10–3 m3)
Kav = =
2(2N A ) 2(2 mol)(6.02 × 1023 molecules/mol)

Kav = 5.05 × 10–21 J/molecule

2N —
21.7 P= (KE) Equation 21.2
3V

3 PV 3 (1.20 × 10 5)(4.00 × 10 –3)
N= = = 2.00 × 1024 molecules
2 — 2 (3.60 × 10 –22 )
(K E)

N 2.00 × 1024 molecules
n= = = 3.32 mol
N A 6.02 × 1023 molecules/mol

Goal Solution
G: The balloon has a volume of 4.00 L and a diameter of about 20 cm, which seems like a
reasonable size for a typical helium balloon. The pressure of the balloon is only slightly more
than 1 atm, and if the temperature is anywhere close to room temperature, we can use the
estimate of 22 L/mol for an ideal gas at STP conditions. If this is valid, the balloon should
contain about 0.2 moles of helium.

O: The average kinetic energy can be used to find the temperature of the gas, which can be used
with PV=nRT to find the number of moles.

1 3
A: The gas temperature must be that implied by mv2 = kBT for a monatomic gas like He.
2 2
1
2  2 mv  2  3.6 × 10–22 J 
2

T=   = = 17.4 K
3  ka  3 1.38 × 10–23 J/K
Now PV = nRT gives

PV (1.20 × 105 N/m2)(4.00 × 10–3 m3)
n= =
RT (8.315 J/mol ⋅ K)(17.4 K)

n = 3.32 mol

L: This result is more than ten times the number of moles we predicted, primarily because the
temperature of the helium is much colder than room temperature! In fact, T is only slightly
above the temperature at which the helium would liquify (4.2 K at 1 atm). We should hope
this balloon is not being held by a child; not only would the balloon sink in the air, it is cold
enough to cause frostbite!

Chapter 21 Solutions 3

3kBT
21.8 v=
m

vO MHe 4.00 1
= = =
vHe MO 32.0 8.00

1350 m/s
vO = = 477 m/s
8.00

21.9 (a) PV = NkBT

PV (1.013 × 105 Pa) 43 π (0.150 m)3
N= = = 3.53 × 1023 atoms
kBT (1.38 × 10–23 J/K)(293 K)

— 3 3
(b) K = kBT = (1.38 × 10–23)(293) J = 6.07 × 10–21 J
2 2

1 –2 3 3kBT
(c) mv = kBT ∴ vrms = = 1.35 km/s
2 2 m

Nmv2 Nmv 2
21.10 (a) PV = nRT = K= = Etrans
3 2

3PV 3
Etrans = = (3.00 × 1.013 × 105)(5.00 × 10–3) = 2.28 kJ
2 2

mv 2 3kBT 3RT 3 (8.315)(300)
(b) = = = = 6.22 × 10–21 J
2 2 2N A 2 (6.02 × 10 23)

– 3 3
21.11 (a) K = kBT = (1.38 × 10–23 J/K)(423 K) = 8.76 × 10–21 J
2 2

– 1 2
(b) K = mv rms = 8.76 × 10–21 J
2

1.75 × 10–20 J
so vrms = (1)
m

For helium,

4.00 g/mol
m= = 6.64 × 10–24 g/molecule
6.02 × 1023 molecules/mol

m = 6.64 × 10–27 kg/molecule

4 Chapter 21 Solutions

Similarly for argon,

39.9 g/mol
m= = 6.63 × 10–23 g/molecule
6.02 × 1023 molecules/mol

m = 6.63 × 10–26 kg/molecule

Substituting in (1) above, we find

for helium, vrms = 1.62 km/s ; and for argon, vrms = 514 m/s

1 Pa = (1 Pa) 
1 N/m2  1 J  J
*21.12 (a) = 1 3
 1 Pa  1 N ⋅ m m

3
(b) For a monatomic ideal gas, Eint = nRT
2

For any ideal gas, the energy of molecular translation is the same,

3 3
Etrans = nRT = PV
2 2

Etrans 3
Thus, the energy per volume is = P
V 2

3
21.13 Eint = nRT
2

3 3
∆Eint = nR ∆T = (3.00 mol)(8.315 J/mol ⋅ K)(2.00 K) = 75.0 J
2 2

nRT
21.14 The piston moves to keep pressure constant. Since V = , then
P

nR ∆T
∆V = for a constant pressure process.
P

Q Q 2Q
Q = nCP ∆T = n(CV + R)∆T so ∆T = = =
n(C V + R) n(5R/2 + R) 7nR

nR  2Q  2Q 2 Q V
and ∆V = = =
P 7nR 7P 7 nRT

2 (4.40 × 10 3 J)(5.00 L)
∆V = = 2.52 L
7 (1.00 mol)(8.315 J/mol ⋅ K)(300 K)

Thus, Vf = Vi + ∆V = 5.00 L + 2.52 L = 7.52 L

...

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