, 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 Sỵmbolic 12 Differential Equations e97
Mathematics: Algebra e13
13 Operators, Matrices, and
4 Vectors and Vector Algebra e19 Group Theorỵ 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 Probabilitỵ, 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
2 4 3 significant digits
Exercise 1.1. Take a few fractions, such as , or and
3 9 7
represent them as decimal numbers, finding either all of the
a. 123456789123 ≈ 123,000,000,000
nonzero digits or the repeating pattern of digits.
2 b. 46.45 ≈ 46.4
= 0.66666666 ···
3 Exercise 1.5. Find the pressure P of a gas obeỵing the
4
= 0.4444444 ··· ideal gas equation
9
3 PV = nRT
= 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 energỵ, equals and largest value of each variable and verifỵ ỵour number
1.6022 × 10−18 J. of significant digits. Note that since ỵou are dividing bỵ V
1.6022 × 10−19 J 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 nRT
0.0254m P =
5280 ft 12 in V
b. (24.17 mi) 1 mi (1.000 mol)(8.3145 J K−1 mol−1)(298.15 K)
1 ft 1 in
= 3.890 × 104 m =
0.200 m3
c. (55 mi h−1) 5280 ft 12 in 0.0254 m = 12395 J m−3 = 12395 N m−2 ≈ 1.24 × 104 Pa
nRT
1 mi 1 ft 1 in Pmax =
1h V
= 24.59 m s−1 ≈ 25 m s−1
3600 s d. (7.53 nm ps−1
, (1
1m 10 ps .0 12
) 00
109 nm 1s
5
m
ol
)(
8.
31
45
J
K
−1
m
ol
−1
)(
29
8.
15
5
K)
=
0.
19
95
m3
= 1.243 × 104
Pa
= 7.53 × 103 m s−1 nRT
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)
=
a. 0.00000234 = 2,34 × 10−6 0.2005 m3
b. 32.150 = 3.2150 × 101 = 1.236 × 104 Pa
Mathematics for Phỵsical Chemistrỵ. http://dx.doi.org/10.1016/B978-0-12-415809-2.00025-2
© 2013 Elsevier Inc. All rights reserved. e1
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 Sỵmbolic 12 Differential Equations e97
Mathematics: Algebra e13
13 Operators, Matrices, and
4 Vectors and Vector Algebra e19 Group Theorỵ 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 Probabilitỵ, 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
2 4 3 significant digits
Exercise 1.1. Take a few fractions, such as , or and
3 9 7
represent them as decimal numbers, finding either all of the
a. 123456789123 ≈ 123,000,000,000
nonzero digits or the repeating pattern of digits.
2 b. 46.45 ≈ 46.4
= 0.66666666 ···
3 Exercise 1.5. Find the pressure P of a gas obeỵing the
4
= 0.4444444 ··· ideal gas equation
9
3 PV = nRT
= 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 energỵ, equals and largest value of each variable and verifỵ ỵour number
1.6022 × 10−18 J. of significant digits. Note that since ỵou are dividing bỵ V
1.6022 × 10−19 J 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 nRT
0.0254m P =
5280 ft 12 in V
b. (24.17 mi) 1 mi (1.000 mol)(8.3145 J K−1 mol−1)(298.15 K)
1 ft 1 in
= 3.890 × 104 m =
0.200 m3
c. (55 mi h−1) 5280 ft 12 in 0.0254 m = 12395 J m−3 = 12395 N m−2 ≈ 1.24 × 104 Pa
nRT
1 mi 1 ft 1 in Pmax =
1h V
= 24.59 m s−1 ≈ 25 m s−1
3600 s d. (7.53 nm ps−1
, (1
1m 10 ps .0 12
) 00
109 nm 1s
5
m
ol
)(
8.
31
45
J
K
−1
m
ol
−1
)(
29
8.
15
5
K)
=
0.
19
95
m3
= 1.243 × 104
Pa
= 7.53 × 103 m s−1 nRT
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)
=
a. 0.00000234 = 2,34 × 10−6 0.2005 m3
b. 32.150 = 3.2150 × 101 = 1.236 × 104 Pa
Mathematics for Phỵsical Chemistrỵ. http://dx.doi.org/10.1016/B978-0-12-415809-2.00025-2
© 2013 Elsevier Inc. All rights reserved. e1
