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Unformatted text preview: uting this expression for Q into the expression above for T2, and algebraically eliminating the time t and the 4π factors, gives Chapter 13 Problems ( 719 ) eσ T 4 − T 4 L
0
T2 = 600.0 °C − 2 r2 kasbestos r1 ( J
4
4 −8
1073.2 K ) − (873.2 K ) 1.0 × 10−2 m
( 0.90 ) 5.67 × 10
2
4 ( s⋅m ⋅K = 600.0 °C − J
2 0.090 s ⋅ m ⋅ C° (10.0 ) = 558 °C ) CHAPTER 14 THE IDEAL GAS LAW AND KINETIC THEORY
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (d) This statement is not true, because the number of molecules that have a mass of ten grams
depends on the molecular mass of the substance, which is different for different substances.
2. 10.2 g nRT
. Increasing n, T, and V together may or may not
V
cause P to increase. Increasing n and T certainly causes P to increase. However, increasing
V causes P to decrease. Therefore, the result of increasing all three variables together
depends on the relative amounts by which they are increased. 3. (b) For an ideal gas, we know that P = 4. 2.25 mol
5. 1.25 kg/m3
6. (a) If the speed of every atom in a monatomic ideal gas were doubled, the rootmeansquare
speed of the atoms would be doubled. According to the kinetic theory of gases, the Kelvin
temperature is proportional to the square of the rootmeansquare speed (see Equation 14.6).
Therefore, the Kelvin temperature would be multiplied by a factor of 22 = 4.
7. (d) According to the kinetic theory of gases, the average translational kinetic energy per
molecule is proportional to the Kelvin temperature (see Equation 14.6). Since each gas has
the same temperature, each has the same average translational kinetic energy. However, this
2
kinetic energy is 1 mvrms , and depends on the mass m. Since each type of molecule has the
2
same kinetic energy, the molecules with the larger masses have the smaller translational rms
speeds vrms.
8. (c) According to the kinetic theory of gases, the internal energy U of a monatomic ideal gas
is U = 3 nRT (Equation 14.7). However, the ideal gas law indicates that PV = nRT, so that
2
U = 3 PV . Since P is doubled and V is reduced by onehalf, the product PV and the internal
2
energy are unchanged. Chapter 14 Answers to Focus on Concepts Questions 721 9. (e) Increasing the crosssectional area of the diffusion channel and increasing the difference
in solute concentrations between its ends certainly would increase the diffusion rate.
However, increasing its length would decrease the rate (see Equation 14.8). Therefore,
increasing all three factors simultaneously could lead to a decrease in the diffusion rate,
depending on the relative amounts of the change in the factors.
10. 6.0 × 10−11 kg/s 722 THE IDEAL GAS LAW AND KINETIC THEORY CHAPTER 14 THE IDEAL GAS LAW AND KINETIC THEORY
PROBLEMS
______________________________________________________________________________
1. SSM REASONING AND SOLUTION Since hemoglobin has a molecular mass of
64 500 u, the mass per mole of hemoglobin is 64 500 g/mol. The number of hemoglobin
molecules per mol is Avogadro's number, or 6.022 × 1023 mol–1. Therefore, one molecule of
hemoglobin has a mass (in kg) of 64 500 g/mol 1 kg = 1.07×10−22 kg 23
−1 1000 g 6.022×10 mol ______________________________________________________________________________
2. REASONING The number of molecules in a known mass of material is the number n of
moles of the material times the number NA of molecules per mole (Avogadro's number).
We can find the number of moles by dividing the known mass m by the mass per mole.
SOLUTION Using the periodic table on the inside of the text’s back cover, we find that the
molecular mass of Tylenol (C8H9NO2) is
Molecular mass
u) 9 (1.00794 u)
of Tylenol = 8 (12.0113 + 14 244
1424
4
3
Mass of 8
carbon atoms Mass of 9
hydrogen atoms + 14.0067 u + 2 (15.9994) = 151.165 u
1 24
4 3 14243
Mass of
nitrogen atom Mass of 2
oxygen atoms The molecular mass of Advil (C13H18O2) is
Molecular mass
= 134 244 + 18 4 244 + 2 (15.9994) = 206.285 u
u)
u)
of Advil
1 (12.0113 1 (1.00794 3 14243
4
4
Mass of 13
carbon atoms Mass of 18
hydrogen atoms Mass of 2
oxygen atoms a. Therefore, the number of molecules of pain reliever in the standard dose of Tylenol is Chapter 14 Problems 723 m Number of molecules = n N A = NA Mass per mole 325 × 10−3 g 23
21
−1
= ( 6.022 ×10 mol ) = 1.29 ×10
151.165 g/mol b. Similarly, the number of molecules of pain reliever in the standard dose of Advil is m Number of molecules = n N A = NA Mass per mole 2.00 × 10−1 g 23
−1
20
= ( 6.022 ×10 mol ) = 5.84 ×10 206.285 g/mol ______________________________________________________________________________
3. REASONING AND SOLUTION
a. The molecular mass of a molecule is the sum of the atomic masses of its atoms. Thus, the
molecular mass of aspartame (C14H18N2O5) is (see the periodic table on the inside of the
text’s back cover) Molecular mass = 14 (12.011 u) + 18 (1.00794 u)
14 244 14 244
4
3
4
3
Mass of 14
carbon atoms Mass of 18
hydrogen atoms + 2 (14.0067 u) + 5 (15.9994) = 294.307 u
14 244 14243
4...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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