Unformatted text preview: n 14.6 ( KE = 2 mv2 ), v rms = 2(KE)/ m . Before we can use the last
rms
expression for the translation rms speed v rms , we must determine the mass m of a water
molecule and the average translational kinetic energy KE .
1 Using the periodic table on the inside of the text’s back cover, we find that the molecular
mass of a water molecule is
2(1.00794 u) + 15.99943 = 18.0153 u
24u
14 244
4
3 14
Mass of one
oxygen atom Mass of two
hydrogen atoms The mass of a single molecule is m= 18.0153 × 10 –3 kg/mol
= 2.99 × 10 –26 kg
23
–1
6.022 × 10 mol The average translational kinetic energy of water molecules at 293 K is, according to
Equation 14.6,
KE = 2 kT = 2 (1.38 × 10 –23 J/K) ( 293 K ) = 6.07 × 10 –21 J
3 3 Therefore, the translational rms speed of water molecules is
v rms = 2(KE )
m = ( 2 6.07 × 10 –21 J
2.99 × 10 –26 kg ) = 637 m/s Thus, the time t required for a water molecule to travel the distance L = 0.010 m at this
speed is
L
0.010 m
t=
=
= 1.6 × 10 –5 s
v rms 637 m/s c. In part (a), when a water molecule diffuses through air, it makes millions of collisions
each second with air molecules. The speed and direction change abruptly as a result of each
collision. Between collisions, the water molecules move in a straight line at constant speed.
Although a water molecule does move very quickly between collisions, it wanders only very
slowly in a zigzag path from one end of the channel to the other. In contrast, a water
molecule traveling unobstructed at its translational rms speed [as in part (b)], will have a
larger displacement over a much shorter time. Therefore, the answer to part (a) is much
longer than the answer to part (b).
______________________________________________________________________________ Chapter 14 Problems 755 50. 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
–13 mass flow rate of CCl4 at point A is 5.00 × 10 kg/s . b. Solving Fick's law for ∆ C , we obtain
∆C = mL (m / t ) L
=
DAt
DA
= (5.00×10−13 kg/s)(5.00×10−3 m)
(20.0×10 –10 2 m /s)(3.00×10 –4 2 = 4.2×10 −3 kg/m 3 m) Then,
CA = Cleft end − ∆C = (1.00 × 10 –2 kg/m3 ) − (4.2 ×10 –3 kg/m3 )= 5.8 ×10 –3 kg/m3 ______________________________________________________________________________
51. SSM WWW REASONING AND SOLUTION
a. The average concentration is Cav = (1/2) (C1 + C2) = (1/2)C2 = m/V = m/(AL), so that
C2 = 2m/(AL). Fick's law then becomes m = DAC2t/L = DA(2m/AL)t/L = 2Dmt/L2. Solving
for t yields
t = L2 / ( 2 D ) b. Substituting the given data into this expression yields
t = (2.5 × 10–2 m)2/[2(1.0 × 10–5 m2/s)] = 31 s
______________________________________________________________________________
52. REASONING AND SOLUTION Equation 14.8,
DA ∆C t
m=
. Solving for the time t gives
L
mL
t=
DA ∆C gives Fick's law of diffusion: (1) 756 THE IDEAL GAS LAW AND KINETIC THEORY The time required for the water to evaporate is equal to the time it takes for 2.0 grams of
water vapor to traverse the tube and can be calculated from Equation (1) above. Since air in
the tube is completely dry at the right end, the concentration of water vapor is zero,
C1 = 0 kg/m3, and ∆C = C2 – C1 = C2. The concentration at the left end of the tube, C2, is
equal to the density of the water vapor above the water. This can be found from the ideal
gas law:
ρ RT
PV = nRT
⇒
P=
M
where M = 0.0180152 kg/mol is the mass per mole for water (H2O). From the figure that
accompanies Problem 75 in Chapter 12, the equilibrium vapor pressure of water at 20 °C is
2.4 × 103 Pa. Therefore, C2 = ρ = PM ( 2.4 ×103 Pa ) ( 0.0180 kg/mol )
=
= 1.8 ×10−2 kg/m3
RT
[8.31 J/(mol ⋅ K)] (293 K) Substituting values into Equation (1) gives
t= ( 2.0×10−3 kg ) (0.15 m)
=
( 2.4 ×10−5 m 2 /s )( 3.0 ×10−4 m 2 ) (1.8 ×10−2 kg/m3 ) 2.3 × 106 s This is about 27 days!
______________________________________________________________________________
53. REASONING AND SOLUTION To find the temperature T2, use the ideal gas law with n
and V constant. Thus, P1/T1 = P2/T2. Then, 3.01× 105 Pa P T2 = T1 2 = ( 284 K ) = 304 K 2.81× 105 Pa P 1
______________________________________________________________________________ 54. REASONING AND SOLUTION According to the ideal gas law (Equation 14.1), the total
number of moles n of fresh air in a normal breath is
n= PV (1.0 × 105 Pa)(5.0 × 10 –4 m 3 )
–2
=
= 1.94 × 10 mol
RT
[8.31 J/(mole ⋅ K)] (310 K) The total number of molecules in a normal breath is nN A , where NA is Avogadro's number.
Since f...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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