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SOLUTIONS MANUAL
, Contents
Preface vii
1 Problem Solving and Numerical 10 Mathematical Series e79
Mathematics e1
11 Functional Series and Integral
2 Mathematical Functions e7 Transforms e89
3 Problem Solving and Symbolic 12 Differential Equations e97
Mathematics: Algebra e13
13 Operators, Matrices, and
4 Vectors and Vector Algebra e19 Group Theory e113
5 Problem Solving and the Solution 14 The Solution of Simultaneous
of Algebraic Equations e23 Algebraic Equations with More
Than Two Unknowns e125
6 Differential Calculus e35
15 Probability, Statistics, and
7 Integral Calculus e51 Experimental Errors e135
8 Differential Calculus with Several 16 Data Reduction and the
Independent Variables e59 Propagation of Errors e145
9 Integral Calculus with Several
Independent Variables e69
v
, ✤ ✜
Chapter 1
✣ ✢
Problem Solving and Numerical
Mathematics
EXERCISES Exercise 1.4. Round the following numbers to three
significant digits
Exercise 1.1. Take a few fractions, such as 23 , 49 or 37 and
represent them as decimal numbers, finding either all of the
a. 123456789123 ≈ 123,000,000,000
nonzero digits or the repeating pattern of digits.
b. 46.45 ≈ 46.4
2
= 0.66666666 · · ·
3 Exercise 1.5. Find the pressure P of a gas obeying the
4
= 0.4444444 · · · ideal gas equation
9
3 P V = n RT
= 0.428571428571 · · ·
7
if the volume V is 0.200 m3 , the temperature T is 298.15 K
Exercise 1.2. Express the following in terms of SI base and the amount of gas n is 1.000 mol. Take the smallest
units. The electron volt (eV), a unit of energy, equals and largest value of each variable and verify your number
1.6022 × 10−18 J. of significant digits. Note that since you are dividing by
1.6022 × 10−19 J
V the smallest value of the quotient will correspond to the
a. (13.6 eV) = 2.17896 × 10−19 J largest value of V.
1 eV
≈ 2.18 × 10−18 J
n RT
5280 ft
12 in
0.0254m
P =
b. (24.17 mi) V
1 mi 1 ft 1 in (1.000 mol)(8.3145 J K−1 mol−1 )(298.15 K)
= 3.890 × 104 m =
0.200 m3
12395 J m = 12395 N m−2 ≈ 1.24 × 104 Pa
−3
5280 ft 12 in 0.0254 m =
c. (55 mi h−1 )
1 mi 1 ft 1 in n RT
Pmax =
1h −1 −1 V
= 24.59 m s ≈ 25 m s
3600 s (1.0005 mol)(8.3145 J K−1 mol−1 )(298.155 K)
1m
12
10 ps =
d. (7.53 nm ps )−1 0.1995 m3
109 nm 1s = 4
1.243 × 10 Pa
= 7.53 × 103 m s−1 n RT
Pmin =
Exercise 1.3. Convert the following numbers to scientific V
notation: (0.9995 mol)(8.3145 J K−1 mol−1 )(298.145 K)
=
0.2005 m3
a. 0.00000234 = 2,34 × 10−6 4
b. 32.150 = 3.2150 × 101 = 1.236 × 10 Pa
Mathematics for Physical Chemistry. http://dx.doi.org/10.1016/B978-0-12-415809-2.00025-2
© 2013 Elsevier Inc. All rights reserved. e1
, e2 Mathematics for Physical Chemistry
Exercise 1.6. Calculate the following to the proper 6. The distance by road from Memphis, Tennessee
numbers of significant digits. to Nashville, Tennessee is 206 miles. Express this
a. 17.13 + 14.6751 + 3.123 + 7.654 − 8.123 = 34.359 distance in meters and in kilometers.
≈ 34.36
b. ln (0.000123) 5380 ft 12 in 0.0254 m
(206 mi)
1 mi 1 ft 1 in
ln (0.0001235) = −8.99927 = 3.32 × 105 m = 332 km
ln (0.0001225) = −9.00740
The answer should have three significant digits: 7. A U. S. gallon is defined as 231.00 cubic inches.
ln (0.000123) = −9.00
a. Find the number of liters in 1.000 gallon.
PROBLEMS 3
231.00 in3
0.0254 m 1000 l
1. Find the number of inches in 1.000 meter. (1 gal)
1 gal 1 in 1 m3
1 in
(1.000 m) = 39.37 in = 3.785 l
0.0254 m
2. Find the number of meters in 1.000 mile and the
b. The volume of 1.0000 mol of an ideal gas
number of miles in 1.000 km, using the definition of
at 0.00 ◦ C (273.15 K) and 1.000 atm is
the inch.
22.414 liters. Express this volume in gallons and
5280 ft 12 in 0.0254 m in cubic feet.
(1.000 mi)
1 mi 1 ft 1 in
= 1609 m 3
1 m3
1 in
1000 m 1 in 1 ft (22.414 l)
(1.000 km) 1000 l 0.0254 m3
1 km 0.0254 m 12 in
1 gal
1 mi × = 5.9212 gal
× = 0.6214 231.00 in3
5280 ft 3
1 m3
1 in
3. Find the speed of light in miles per second. (22.414 l)
1000 l 0.0254 m3
−1 1 in 1 ft 3
(299792458 m s )
1 ft
0.0254 m 12 in × = 0.79154 ft3
12 in
1 mi
× = 186282.397 mi s−1
5280 ft
8. In the USA, footraces were once measured in yards and
4. Find the speed of light in miles per hour. at one time, a time of 10.00 seconds for this distance
−1 1 in 1 ft was thought to be unattainable. The best runners now
(299792458 m s )
0.0254 m 12 in run 100 m in 10 seconds or less. Express 100.0 m in
1 mi
3600 s
yards. If a runner runs 100.0 m in 10.00 s, find his
× = 670616629 mi h−1 time for 100 yards, assuming a constant speed.
5280 ft 1h
5. A furlong is exactly one-eighth of a mile and a
fortnight is exactly 2 weeks. Find the speed of light 1 in 1 yd
(100.0 m) = 109.4 m
in furlongs per fortnight, using the correct number of 0.0254 m 36 in
significant digits. 100.0 yd
(10.00 s) = 9.144 s
109.4 m
−1 1 in 1 ft
(299792458 m s )
0.0254 m 12 in
1 mi
8 furlongs
9. Find the average length of a century in seconds and in
× minutes. Use the rule that a year ending in 00 is not a
5280 ft 1 mi
leap year unless the year is divisible by 400, in which
3600 s 24 h 14 d
× case it is a leap year. Therefore, in four centuries there
1h 1d 1 fortnight will be 97 leap years. Find the number of minutes in a
= 1.80261750 × 1012 furlongs fortnight−1 microcentury.
SOLUTIONS MANUAL
, Contents
Preface vii
1 Problem Solving and Numerical 10 Mathematical Series e79
Mathematics e1
11 Functional Series and Integral
2 Mathematical Functions e7 Transforms e89
3 Problem Solving and Symbolic 12 Differential Equations e97
Mathematics: Algebra e13
13 Operators, Matrices, and
4 Vectors and Vector Algebra e19 Group Theory e113
5 Problem Solving and the Solution 14 The Solution of Simultaneous
of Algebraic Equations e23 Algebraic Equations with More
Than Two Unknowns e125
6 Differential Calculus e35
15 Probability, Statistics, and
7 Integral Calculus e51 Experimental Errors e135
8 Differential Calculus with Several 16 Data Reduction and the
Independent Variables e59 Propagation of Errors e145
9 Integral Calculus with Several
Independent Variables e69
v
, ✤ ✜
Chapter 1
✣ ✢
Problem Solving and Numerical
Mathematics
EXERCISES Exercise 1.4. Round the following numbers to three
significant digits
Exercise 1.1. Take a few fractions, such as 23 , 49 or 37 and
represent them as decimal numbers, finding either all of the
a. 123456789123 ≈ 123,000,000,000
nonzero digits or the repeating pattern of digits.
b. 46.45 ≈ 46.4
2
= 0.66666666 · · ·
3 Exercise 1.5. Find the pressure P of a gas obeying the
4
= 0.4444444 · · · ideal gas equation
9
3 P V = n RT
= 0.428571428571 · · ·
7
if the volume V is 0.200 m3 , the temperature T is 298.15 K
Exercise 1.2. Express the following in terms of SI base and the amount of gas n is 1.000 mol. Take the smallest
units. The electron volt (eV), a unit of energy, equals and largest value of each variable and verify your number
1.6022 × 10−18 J. of significant digits. Note that since you are dividing by
1.6022 × 10−19 J
V the smallest value of the quotient will correspond to the
a. (13.6 eV) = 2.17896 × 10−19 J largest value of V.
1 eV
≈ 2.18 × 10−18 J
n RT
5280 ft
12 in
0.0254m
P =
b. (24.17 mi) V
1 mi 1 ft 1 in (1.000 mol)(8.3145 J K−1 mol−1 )(298.15 K)
= 3.890 × 104 m =
0.200 m3
12395 J m = 12395 N m−2 ≈ 1.24 × 104 Pa
−3
5280 ft 12 in 0.0254 m =
c. (55 mi h−1 )
1 mi 1 ft 1 in n RT
Pmax =
1h −1 −1 V
= 24.59 m s ≈ 25 m s
3600 s (1.0005 mol)(8.3145 J K−1 mol−1 )(298.155 K)
1m
12
10 ps =
d. (7.53 nm ps )−1 0.1995 m3
109 nm 1s = 4
1.243 × 10 Pa
= 7.53 × 103 m s−1 n RT
Pmin =
Exercise 1.3. Convert the following numbers to scientific V
notation: (0.9995 mol)(8.3145 J K−1 mol−1 )(298.145 K)
=
0.2005 m3
a. 0.00000234 = 2,34 × 10−6 4
b. 32.150 = 3.2150 × 101 = 1.236 × 10 Pa
Mathematics for Physical Chemistry. http://dx.doi.org/10.1016/B978-0-12-415809-2.00025-2
© 2013 Elsevier Inc. All rights reserved. e1
, e2 Mathematics for Physical Chemistry
Exercise 1.6. Calculate the following to the proper 6. The distance by road from Memphis, Tennessee
numbers of significant digits. to Nashville, Tennessee is 206 miles. Express this
a. 17.13 + 14.6751 + 3.123 + 7.654 − 8.123 = 34.359 distance in meters and in kilometers.
≈ 34.36
b. ln (0.000123) 5380 ft 12 in 0.0254 m
(206 mi)
1 mi 1 ft 1 in
ln (0.0001235) = −8.99927 = 3.32 × 105 m = 332 km
ln (0.0001225) = −9.00740
The answer should have three significant digits: 7. A U. S. gallon is defined as 231.00 cubic inches.
ln (0.000123) = −9.00
a. Find the number of liters in 1.000 gallon.
PROBLEMS 3
231.00 in3
0.0254 m 1000 l
1. Find the number of inches in 1.000 meter. (1 gal)
1 gal 1 in 1 m3
1 in
(1.000 m) = 39.37 in = 3.785 l
0.0254 m
2. Find the number of meters in 1.000 mile and the
b. The volume of 1.0000 mol of an ideal gas
number of miles in 1.000 km, using the definition of
at 0.00 ◦ C (273.15 K) and 1.000 atm is
the inch.
22.414 liters. Express this volume in gallons and
5280 ft 12 in 0.0254 m in cubic feet.
(1.000 mi)
1 mi 1 ft 1 in
= 1609 m 3
1 m3
1 in
1000 m 1 in 1 ft (22.414 l)
(1.000 km) 1000 l 0.0254 m3
1 km 0.0254 m 12 in
1 gal
1 mi × = 5.9212 gal
× = 0.6214 231.00 in3
5280 ft 3
1 m3
1 in
3. Find the speed of light in miles per second. (22.414 l)
1000 l 0.0254 m3
−1 1 in 1 ft 3
(299792458 m s )
1 ft
0.0254 m 12 in × = 0.79154 ft3
12 in
1 mi
× = 186282.397 mi s−1
5280 ft
8. In the USA, footraces were once measured in yards and
4. Find the speed of light in miles per hour. at one time, a time of 10.00 seconds for this distance
−1 1 in 1 ft was thought to be unattainable. The best runners now
(299792458 m s )
0.0254 m 12 in run 100 m in 10 seconds or less. Express 100.0 m in
1 mi
3600 s
yards. If a runner runs 100.0 m in 10.00 s, find his
× = 670616629 mi h−1 time for 100 yards, assuming a constant speed.
5280 ft 1h
5. A furlong is exactly one-eighth of a mile and a
fortnight is exactly 2 weeks. Find the speed of light 1 in 1 yd
(100.0 m) = 109.4 m
in furlongs per fortnight, using the correct number of 0.0254 m 36 in
significant digits. 100.0 yd
(10.00 s) = 9.144 s
109.4 m
−1 1 in 1 ft
(299792458 m s )
0.0254 m 12 in
1 mi
8 furlongs
9. Find the average length of a century in seconds and in
× minutes. Use the rule that a year ending in 00 is not a
5280 ft 1 mi
leap year unless the year is divisible by 400, in which
3600 s 24 h 14 d
× case it is a leap year. Therefore, in four centuries there
1h 1d 1 fortnight will be 97 leap years. Find the number of minutes in a
= 1.80261750 × 1012 furlongs fortnight−1 microcentury.
