# THE IDEAL GAS LAW AND KINETIC THEORY

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```					                                                                             Chapter 14 Problems       1

CHAPTER 14 THE IDEAL GAS LAW
AND KINETIC THEORY
PROBLEMS
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19. REASONING AND SOLUTION
a. Since the heat gained by the gas in one tank is equal to the heat lost by the gas in the
other tank, Q1 = Q2, or (letting the subscript 1 correspond to the neon in the left tank, and
letting 2 correspond to the neon in the right tank) cm1∆T1 = cm2 ∆T2 ,

cm1 (T − T1 ) = cm2 (T2 − T )

m1 (T − T1 ) = m2 (T2 − T )
Solving for T gives

m2T2 + m1T1
T=                                                       (1)
m2 + m1

The masses m1 and m2 can be found by first finding the number of moles n1 and n2. From
the ideal gas law, PV = nRT, so

PV1  ( 5.0 ×105 Pa )( 2.0 m3 )
n1 =    1 =                             = 5.5 ×102 mol
RT1 [8.31 J/(mol ⋅ K)] ( 220 K )

This corresponds to a mass m1 = (5.5×102 mol) ( 20.179 g/mol ) = 1.1×104 g = 1.1×101 kg .
Similarly, n2 = 2.4 × 102 mol and m2 = 4.9 × 103 g = 4.9 kg. Substituting these mass values
into Equation (1) yields

( 4.9 kg ) ( 580 K ) + (1.1×101 kg ) ( 220 K )
T=                                                  =   3.3 ×102 K
( 4.9 kg ) + (1.1×101 kg )
b. From the ideal gas law,

nRT ⎡( 5.5 ×102 mol ) + ( 2.4 × 102 mol ) ⎤ [8.31 J/ ( mol ⋅ K )] ( 3.3 ×102 K )
⎣                                   ⎦
P=     =                                                                            = 2.8 × 105 Pa
V                               2.0 m + 5.8 m
3          3

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2   THE IDEAL GAS LAW AND KINETIC THEORY

35. REASONING AND SOLUTION Since we are treating the air as an ideal gas, PV = nRT,
so that U = 5 nRT = 5 PV = 5 ( 7.7×106 Pa ) (5.6 ×105 m3 ) = 1.1×1013 J . The number of
2       2      2
joules of energy consumed per day by one house is

⎛           J ⋅ h ⎞ ⎛ 3600 s ⎞
30.0 kW ⋅ h = ⎜ 30.0 ×103       ⎟⎜         ⎟ = 1.08 × 10 J
8
⎝             s ⎠⎝ 1 h ⎠

The number of homes that could be served for one day by 1.1 × 1013 J of energy is

⎛               ⎞
(1.1×1013 J ) ⎜ 1.08home8 J ⎟ = 1.0 ×105 homes
1
⎝          ⎠      ×10
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37. REASONING AND SOLUTION The average force exerted by one electron on the screen
is, from the impulse-momentum theorem (Equation 7.4), F = ∆p/ ∆t = m∆v / ∆t . Therefore,
in a time ∆ t , N electrons exert an average force F = Nm∆v / ∆t = (N / ∆t)m∆v . Since the
pressure on the screen is the average force per unit area (Equation 10.19), we have

F (N / ∆t)m∆v
P=     =
A       A
16                        –31              7
electrons/s)(9.11 × 10
(6.2 × 10                        kg)(8.4 × 10 m/s – 0 m/s)
=                                                              = 4.0 × 101 Pa
1.2 × 10 –7 m 2
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43. REASONING Since mass is conserved, the mass flow rate is the same at all points, as
described by the equation of continuity (Equation 11.8). Therefore, the mass flow rate at
which CCl4 enters the tube is the same as that at point A. The concentration difference of
CCl4 between point A and the left end of the tube, ∆C , can be calculated by using Fick's
law of diffusion (Equation 14.8). The concentration of CCl4 at point A can be found from
CA = Cleft end – ∆C.

SOLUTION
a. As discussed above in the reasoning, the mass flow rate of CCl4 as it passes point A is
the same as the mass flow rate at which CCl4 enters the left end of the tube; therefore, the
mass flow rate of CCl4 at point A is 5.00 × 10–13 kg/s .

b. Solving Fick's law for ∆C , we obtain
Chapter 14 Problems   3

mL (m / t ) L
∆C =       =
DAt   DA
(5.00×10−13 kg/s)(5.00×10−3 m)
=           –10   2           –4  2
= 4.2×10−3 kg/m3
(20.0×10 m /s)(3.00×10 m )
Then,

CA = Cleft end − ∆C = (1.00 × 10 –2 kg/m3 ) − (4.2 × 10 –3 kg/m3 )= 5.8 × 10 –3 kg/m3
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