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SOLUTIONS MANUAL
, Contents
Preface vii
1 Problem Solving and Numerical Mathematics e1 10 Mathematical Series e79
2 Mathematical Functions e7 11 Functional Series and Integral Transforms e89
3 Problem Solving and Symbolic Mathematics: 12 Differential Equations e97
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 Independent 16 Data Reduction and the
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
Exercise 1.1. Take a few fractions, such as 2 , 4 or 3
and significant digits
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. b. 46.45 ≈ 46.4
2
= 0.66666666 · · ·
3 Exercise 1.5. Find the pressure P of a gas obeying the ideal
4
= 0.4444444 · · · gas equation
9 P V = n RT
3
= 0.428571428571 · · · if the volume V is 0.200 m3, the temperature T is 298.15 K and
7 the amount of gas n is 1.000 mol. Take the smallest and
Exercise 1.2. Express the following in terms of SI base units. largest value of each variable and verify your number of
The electron volt (eV), a unit of energy, equals 1.6022 × significant digits. Note that since you are dividing by V the
10−18 J. smallest value of the quotient will correspond to the
1.6022 × 10 −19 J
a. (13.6 eV) = 2.17896 × 10−19 J largest value of V.
1 eV
≈ 2.18 × 10−18 J
P = n RT
5280 ft 12 in 0.0254m V
b. (24.17 mi) 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
−3
5280 ft 12 in 0.0254 m = 12395 J m = 12395 N m−2 ≈ 1.24 × 104 Pa
c. (55 mi h−1)
1 mi 1 ft 1 in Pmax = n RT
1h = 24.59 m s−1 ≈ 25 m s−1 V
12
3600 s = (1.0005 mol)(8.3145 J K−1 mol−1)(298.155 K)
1m ps
d. (7.53 nm ps−1) 10 0.1995 m3
109 nm 1s 4
= 1.243 × 10 Pa
= 7.53 × 103 m s−1 Pmin = n RT
Exercise 1.3. Convert the following numbers to scientific V −1 −1
notation: (0.9995 mol)(8.3145 J K mol )(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 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 to
numbers of significant digits. 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:
ln (0.000123) = −9.00 7. A U. S. gallon is defined as 231.00 cubic inches.
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 = 39.37 in = 3.785 l
(1.000 m)
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 at
number of miles in 1.000 km, using the definition of the 0.00 ◦C (273.15 K) and 1.000 atm is
inch. 22.414 liters. Express this volume in gallons and in
5280 ft 12 in 0.0254 m cubic feet.
(1.000 mi) 1 mi 1 ft 1 in
= 1609 m 1 m3 1 in 3
1000 m 1 in (22.414 l)
(1.000 km) 1 ft 1000 l 0.0254 m3
1 km 0.0254 m 12 in = 5.9212 gal
1 mi 1 gal 3
× 5280 ft = 0.6214 × 231.00 in3 3
3. Find the speed of light in miles per second. 1m 1 in
(22.414 l)
1000 l 0.0254 m3
1 in 3
(299792458 m s−1) 1 ft 1 ft
0.0254 m 12 in × 12 in = 0.79154 ft3
−1
1 mi = 186282.397 mi s
× 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 was
1in 1 ft
(299792458 m s−1) thought to be unattainable. The best runners now
0.0254 m 12 in run 100 m in 10 seconds or less. Express 100.0 m in
3600 s yards. If a runner runs 100.0 m in 10.00 s, find his
1 mi = 670616629 mi h−1 1 h
× 5280 ft time for 100 yards, assuming a constant speed.
5. A furlong is exactly one-eighth of a mile and a fortnight is 1 in
(100.0 m)
exactly 2 weeks. Find the speed of light in furlongs per 0.0254 m 1 yd
fortnight, using the correct number of significant digits. = 109.4 m 36
100.0 yd in
1 in 1 ft
(299792458 m s−1) (10.00 s) = 9.144 s
0.0254 m 12 in 109.4 m
1 mi 8 9. Find the average length of a century in seconds and in
× 5280 ft furlongs minutes. Use the rule that a year ending in 00 is not a
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.
,CHAPTER | 1 Problem Solving and Numerical Mathematics e3
Number of days in 400 years b. Find the Rankine temperature at 0.00 ◦F.
= (365 d)(400 y) + 97 d = 146097 d
Average number of days in a century
273.15 K − 18.00 K = 255.15 K
146097 d 9 ◦F
= = 36524.25 d (255.15 K) = 459.27 ◦R
4 5K
24 h 60 min
1 century = (36524.25 d)
1d 1h 12. The volume of a sphere is given by
= 5.259492 × 107 min
1 century 4
(5.259492 × 107 min) V = πr 3
1 × 106 microcenturies 3
= 52.59492 min
60 s where V is the volume and r is the radius. If a certain
(52.59492 min) = 3155.695 s
1 min sphere has a radius given as 0.005250 m, find its
volume, specifying it with the correct number of digits.
10. A light year is the distance traveled by light in one year. Calculate the smallest and largest volumes that the
sphere might have with the given information and
a. Express this distance in meters and in kilometers. check your first answer for the volume.
4 4
V = πr 3 = V = π(0.005250 m)3
Use the average length of a year as described in 3 3
−7 3
the previous problem. How many significant digits can = 6.061 × 10−7 m3
4
be given? V = π(0.005245 m) 3
min = 6.044 × 10 m
3
60 s 4
Vmax = π(0.005255 m)3 = 6.079 × 10−7 m3
(299792458 m s−1)
1 min 3
5
×(5.259492 × 10 min) V = 6.06 × 10−7 m3
= (9.46055060 × 1015 m)
1km
(9.46055060 × 1015 m) The rule of thumb gives four significant digits, but the
1000 m calculation shows that only three significant digits can be
= 9.4605506 × 1012 km specified and that the last digit can be wrong by one.
≈ 9.460551 × 1012 km 13. The volume of a right circular cylinder is given by
Since the number of significant digits in the number V = πr 2h,
of days in an average century is seven, we round to
seven significant digits.
where r is the radius and h is the height. If a right
b. Express a light year in miles.
circular cylinder has a radius given as 0.134 m and a
1 in 1 ft height given as 0.318 m, find its volume, specifying it
(9.460551 × 1015 m)
0.0254 m 12 in with the correct number of digits. Calculate the
smallest and largest volumes that the cylinder might
1 mi = 5.878514 × 1012 mi have with the given information and check your first
× 5280 ft answer for the volume.
11. The Rankine temperature scale is defined so that the
Rankine degree is the same size as the Fahrenheit V = π(0.134 m)2(0.318 m) = 0.0179 m3 Vmin =
degree, and absolute zero is 0 ◦R, the same as 0 K.
π(0.1335 m)2(0.3175 m) = 0.01778 m3 Vmax =
a. Find the Rankine temperature at 0.00 ◦C. π(0.1345 m)2(0.3185 m) = 0.0181 m3
◦
14. The value of an angle is given as 31◦. Find the measure
◦ 9 F ◦ of the angle in radians. Find the smallest and largest
0.00 C ↔ (273.15 K) = 491.67 R
5K values that its sine and cosine might have and specify
,e4 Mathematics for Physical Chemistry
the sine and cosine to the appropriate number of digits. V = 10.00 l, and T = 298.15 K. Convert your answer to
2π rad atmospheres and torr.
(31◦) = 0.54 rad
360◦ RT a
P=
sin (30.5◦) = 0.5075 V −b —
m Vm2 −1 −1
sin (31.5◦) = 0.5225 (8.3145 J K mol )(298.15 K)
sin (31◦) = 0.51 =
7.3110×10−2 m3 mol−1 − 4.267×10−5 m3 mol−1
cos (30.5◦) = 0.86163 −2
0.3640 Pa m 6 mol −1
−
cos (31.5◦) = 0.85264 (7.3110 × 10−2 m3 mol )2
cos (31◦) = 0.86 RT a
P= −
V m—b V2 m −1 −1
15. Some elementary chemistry textbooks give (8.3145 J K mol )(298.15 K)
the value of−1 R, −1the ideal gas constant, as = 7.3110×10−2 m63 mol−1−2− 4.267×10−5 m3 mol−1
0.0821 l atm K mol .
0.3640 Pa m mol
a. Using the SI value, 8.3145 J K−1 mol−1, obtain −
(7.3110 × 10−2 m3 mol −1)2
the value in l atm K−1 mol−1 to five significant digits. = 3.3927 × 104 J m−3 − 68.1 Pa
= 3.3927 × 104 Pa − 68.1 Pa = 3.386 Pa
1 Pa m 1 atm (3.3859 Pa) 1 atm = 0.33416 atm
(8.3145 J K−1 mol−1) 3
101325 Pa
1J 101325 Pa The prediction of the ideal gas equation is
1000 l −1 −1
× 1 m3 = 0.082058 l atm K mol −1 −1
(8.3145 J K mol )(298.15 K)
P = 7.3110 × 10−2 m3 mol−1
b. Calculate the pressure in atmospheres and in N
m−2 (Pa) of a sample of an ideal gas with n = 0.13678 = 3.3907 × 104 J m−3 = 3.3907 × 104 Pa
mol, V = 10.000 l and T = 298.15 K. 17. The specific heat capacity (specific heat) of a substance is
n RT crudely defined as the amount of heat required to raise
P= the temperature of unit mass of the substance by
V −1 −1 1 degree Celsius (1 ◦C). The specific heat capacity of
(0.13678 mol)(0.082058 l atm K mol )(298.15 K) water is 4.18 J ◦C−1 g−1. Find the rise in temperature if
= 1.000 l 100.0 J of heat is transferred to 1.000 kg of water.
= 0.33464 atm
n RT
P= 100.0 J 1kg
V −1 −1 T = (4.18 J ◦C−1 g−1)(1.000 kg) 1000 g
(0.13678 mol)(8.3145 J K mol )(298.15 K) = 0.0239 ◦C
= 10.000 × 10−3 m3 18. The volume of a cone is given by
= 3.3907 × 104 J m−3 = 3.3907 × 104 N m−2 1
V = πr 2h
= 3.3907 × 104 Pa 3
where h is the height of the cone and r is the radius of
16. The van der Waals equation of state gives better its base. Find the volume of a cone if its radius is
accuracy than the idealagas equation of state. It is
P+ (Vm — b) = RT given as 0.443 m and its height is given as 0.542 m.
Vm2 1 3 1 2 3
where a and b are parameters that have different V = πr h = π(0.443 m) (0.542 m) = 0.111 m 3
values for different gases and where Vm = 3
V /n, the molar volume. For carbon dioxide, 19. The volume of a sphere is equal to 43 πr 3 where r is the
a = 0.3640 Pa m6 mol−2, b = 4.267 × radius of the sphere. Assume that the earth is spherical with
−5 3 −1
10 m mol . Calculate the pressure of carbon a radius of 3958.89 miles. (This is the radius of a sphere
dioxide in pascals, assuming that n = 0.13678 mol, with the same volume as the earth, which
,CHAPTER | 1 Problem Solving and Numerical Mathematics e5
is flattened at the poles by about 30 miles.) Find the We give two significant digits since the use of 1 as a
volume of the earth in cubic miles and in cubic meters. Use a single digit would specify a possible error of about 50%. It
value of π with at least six digits and give the correct is a fairly common practice to give an extra digit when the
number of significant digits in your answer. last significant digit is 1.
4 4 21. The hectare is a unit of land area defined to equal
V = πr 3 = π(3958.89 mi)3
exactly 10,000 square meters, and the acre is a unit
3 3 of land area defined so that 640 acres equals exactly
= 2.59508 × 1011 mi3 one square mile. Find the number of square meters in
5280 ft 3 1.000 acre, and find the number of acres equivalent to
(2.59508 × 1011 mi3) 12 in
3
1 mi 1 ft 1.000 hectare.
= 1.08168 × 1021 m3 (5280 ft) 2 12 in
0.0254 m 3
1.000 acre =
× 1 in 640
2
1 ft
2
20. Using the radius of the earth in the previous problem and 0.0254 m = 4047 m2
the fact that the surface of the earth is about 70% × 1 in
covered by water, estimate the area of all of the bodies of 1.000 hectare = (1.000 hectare) 10000 m2
water on the earth. The area of a sphere is equal to four 1 hectare
times the area of a great circle, or 4πr 2, where r
is the radius of the sphere. 1 acre = 2.471 acre
× 4047 m2
A ≈ (0.7)4πr 2 = (0.7)4π(3958.89 mi)2
= 1.4 × 108 mi2
,This page intentionally left blank
, ✤ ✜
Chapter 2
✣ ✢
Mathematical Functions
EXERCISES 0.1 = 1/10
Exercise 2.1. Enter a formula into cell D2 that will log (0.1) = − log (10) = −1
compute the mean of the numbers in cells A2,B2, and C2. 0.01 = 1/100
= (A2 + B2 + C2)/3 log (0.01) = − log (100) = −2
Exercise 2.2. Construct a graph representing the function 0.001 = 1/1000
y(x ) = x 3 − 2x 2 + 3x + 4 (2.1) log (0.001) = − log (1000) = −3
Use Excel or Mathematica or some other software to 0.0001 = 1/10000
construct your graph. log (0.001) = − log (10000) = −4
Exercise 2.4. Using a calculator or a spreadsheet, evaluate the
Here is the graph, constructed with Excel:
quantity (1+ 1 )n for several integral values of n ranging from 1 to
1,000,000. Notice how n the value approaches the
value of e as n increases and determine the value of n needed to
provide four significant digits.
Here is a table of values
✬ ✩
x (1 + 1/n)n
1 2
2 2.25
5 2.48832
10 2.59374246
Exercise 2.3. Generate the negative logarithms in the short table
of common logarithms.
✬ ✩
100 2.704813829
1000 2.716923932
x y = log 10 (x) x y = log 10 (x)
10000 2.718145927
100000 2.718268237
1 0 0.1 −1
1000000 2.718280469
10 1 0.01 −2 ✫ ✪
✫
100 2 0.001 −3 ✪
1000 3 0.0001 −4
Mathematics for Physical Chemistry. http://dx.doi.org/10.1016/B978-0-12-415809-2.00026-4
© 2013 Elsevier Inc. All rights reserved. e7
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SOLUTIONS MANUAL
, Contents
Preface vii
1 Problem Solving and Numerical Mathematics e1 10 Mathematical Series e79
2 Mathematical Functions e7 11 Functional Series and Integral Transforms e89
3 Problem Solving and Symbolic Mathematics: 12 Differential Equations e97
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 Independent 16 Data Reduction and the
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
Exercise 1.1. Take a few fractions, such as 2 , 4 or 3
and significant digits
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. b. 46.45 ≈ 46.4
2
= 0.66666666 · · ·
3 Exercise 1.5. Find the pressure P of a gas obeying the ideal
4
= 0.4444444 · · · gas equation
9 P V = n RT
3
= 0.428571428571 · · · if the volume V is 0.200 m3, the temperature T is 298.15 K and
7 the amount of gas n is 1.000 mol. Take the smallest and
Exercise 1.2. Express the following in terms of SI base units. largest value of each variable and verify your number of
The electron volt (eV), a unit of energy, equals 1.6022 × significant digits. Note that since you are dividing by V the
10−18 J. smallest value of the quotient will correspond to the
1.6022 × 10 −19 J
a. (13.6 eV) = 2.17896 × 10−19 J largest value of V.
1 eV
≈ 2.18 × 10−18 J
P = n RT
5280 ft 12 in 0.0254m V
b. (24.17 mi) 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
−3
5280 ft 12 in 0.0254 m = 12395 J m = 12395 N m−2 ≈ 1.24 × 104 Pa
c. (55 mi h−1)
1 mi 1 ft 1 in Pmax = n RT
1h = 24.59 m s−1 ≈ 25 m s−1 V
12
3600 s = (1.0005 mol)(8.3145 J K−1 mol−1)(298.155 K)
1m ps
d. (7.53 nm ps−1) 10 0.1995 m3
109 nm 1s 4
= 1.243 × 10 Pa
= 7.53 × 103 m s−1 Pmin = n RT
Exercise 1.3. Convert the following numbers to scientific V −1 −1
notation: (0.9995 mol)(8.3145 J K mol )(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 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 to
numbers of significant digits. 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:
ln (0.000123) = −9.00 7. A U. S. gallon is defined as 231.00 cubic inches.
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 = 39.37 in = 3.785 l
(1.000 m)
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 at
number of miles in 1.000 km, using the definition of the 0.00 ◦C (273.15 K) and 1.000 atm is
inch. 22.414 liters. Express this volume in gallons and in
5280 ft 12 in 0.0254 m cubic feet.
(1.000 mi) 1 mi 1 ft 1 in
= 1609 m 1 m3 1 in 3
1000 m 1 in (22.414 l)
(1.000 km) 1 ft 1000 l 0.0254 m3
1 km 0.0254 m 12 in = 5.9212 gal
1 mi 1 gal 3
× 5280 ft = 0.6214 × 231.00 in3 3
3. Find the speed of light in miles per second. 1m 1 in
(22.414 l)
1000 l 0.0254 m3
1 in 3
(299792458 m s−1) 1 ft 1 ft
0.0254 m 12 in × 12 in = 0.79154 ft3
−1
1 mi = 186282.397 mi s
× 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 was
1in 1 ft
(299792458 m s−1) thought to be unattainable. The best runners now
0.0254 m 12 in run 100 m in 10 seconds or less. Express 100.0 m in
3600 s yards. If a runner runs 100.0 m in 10.00 s, find his
1 mi = 670616629 mi h−1 1 h
× 5280 ft time for 100 yards, assuming a constant speed.
5. A furlong is exactly one-eighth of a mile and a fortnight is 1 in
(100.0 m)
exactly 2 weeks. Find the speed of light in furlongs per 0.0254 m 1 yd
fortnight, using the correct number of significant digits. = 109.4 m 36
100.0 yd in
1 in 1 ft
(299792458 m s−1) (10.00 s) = 9.144 s
0.0254 m 12 in 109.4 m
1 mi 8 9. Find the average length of a century in seconds and in
× 5280 ft furlongs minutes. Use the rule that a year ending in 00 is not a
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.
,CHAPTER | 1 Problem Solving and Numerical Mathematics e3
Number of days in 400 years b. Find the Rankine temperature at 0.00 ◦F.
= (365 d)(400 y) + 97 d = 146097 d
Average number of days in a century
273.15 K − 18.00 K = 255.15 K
146097 d 9 ◦F
= = 36524.25 d (255.15 K) = 459.27 ◦R
4 5K
24 h 60 min
1 century = (36524.25 d)
1d 1h 12. The volume of a sphere is given by
= 5.259492 × 107 min
1 century 4
(5.259492 × 107 min) V = πr 3
1 × 106 microcenturies 3
= 52.59492 min
60 s where V is the volume and r is the radius. If a certain
(52.59492 min) = 3155.695 s
1 min sphere has a radius given as 0.005250 m, find its
volume, specifying it with the correct number of digits.
10. A light year is the distance traveled by light in one year. Calculate the smallest and largest volumes that the
sphere might have with the given information and
a. Express this distance in meters and in kilometers. check your first answer for the volume.
4 4
V = πr 3 = V = π(0.005250 m)3
Use the average length of a year as described in 3 3
−7 3
the previous problem. How many significant digits can = 6.061 × 10−7 m3
4
be given? V = π(0.005245 m) 3
min = 6.044 × 10 m
3
60 s 4
Vmax = π(0.005255 m)3 = 6.079 × 10−7 m3
(299792458 m s−1)
1 min 3
5
×(5.259492 × 10 min) V = 6.06 × 10−7 m3
= (9.46055060 × 1015 m)
1km
(9.46055060 × 1015 m) The rule of thumb gives four significant digits, but the
1000 m calculation shows that only three significant digits can be
= 9.4605506 × 1012 km specified and that the last digit can be wrong by one.
≈ 9.460551 × 1012 km 13. The volume of a right circular cylinder is given by
Since the number of significant digits in the number V = πr 2h,
of days in an average century is seven, we round to
seven significant digits.
where r is the radius and h is the height. If a right
b. Express a light year in miles.
circular cylinder has a radius given as 0.134 m and a
1 in 1 ft height given as 0.318 m, find its volume, specifying it
(9.460551 × 1015 m)
0.0254 m 12 in with the correct number of digits. Calculate the
smallest and largest volumes that the cylinder might
1 mi = 5.878514 × 1012 mi have with the given information and check your first
× 5280 ft answer for the volume.
11. The Rankine temperature scale is defined so that the
Rankine degree is the same size as the Fahrenheit V = π(0.134 m)2(0.318 m) = 0.0179 m3 Vmin =
degree, and absolute zero is 0 ◦R, the same as 0 K.
π(0.1335 m)2(0.3175 m) = 0.01778 m3 Vmax =
a. Find the Rankine temperature at 0.00 ◦C. π(0.1345 m)2(0.3185 m) = 0.0181 m3
◦
14. The value of an angle is given as 31◦. Find the measure
◦ 9 F ◦ of the angle in radians. Find the smallest and largest
0.00 C ↔ (273.15 K) = 491.67 R
5K values that its sine and cosine might have and specify
,e4 Mathematics for Physical Chemistry
the sine and cosine to the appropriate number of digits. V = 10.00 l, and T = 298.15 K. Convert your answer to
2π rad atmospheres and torr.
(31◦) = 0.54 rad
360◦ RT a
P=
sin (30.5◦) = 0.5075 V −b —
m Vm2 −1 −1
sin (31.5◦) = 0.5225 (8.3145 J K mol )(298.15 K)
sin (31◦) = 0.51 =
7.3110×10−2 m3 mol−1 − 4.267×10−5 m3 mol−1
cos (30.5◦) = 0.86163 −2
0.3640 Pa m 6 mol −1
−
cos (31.5◦) = 0.85264 (7.3110 × 10−2 m3 mol )2
cos (31◦) = 0.86 RT a
P= −
V m—b V2 m −1 −1
15. Some elementary chemistry textbooks give (8.3145 J K mol )(298.15 K)
the value of−1 R, −1the ideal gas constant, as = 7.3110×10−2 m63 mol−1−2− 4.267×10−5 m3 mol−1
0.0821 l atm K mol .
0.3640 Pa m mol
a. Using the SI value, 8.3145 J K−1 mol−1, obtain −
(7.3110 × 10−2 m3 mol −1)2
the value in l atm K−1 mol−1 to five significant digits. = 3.3927 × 104 J m−3 − 68.1 Pa
= 3.3927 × 104 Pa − 68.1 Pa = 3.386 Pa
1 Pa m 1 atm (3.3859 Pa) 1 atm = 0.33416 atm
(8.3145 J K−1 mol−1) 3
101325 Pa
1J 101325 Pa The prediction of the ideal gas equation is
1000 l −1 −1
× 1 m3 = 0.082058 l atm K mol −1 −1
(8.3145 J K mol )(298.15 K)
P = 7.3110 × 10−2 m3 mol−1
b. Calculate the pressure in atmospheres and in N
m−2 (Pa) of a sample of an ideal gas with n = 0.13678 = 3.3907 × 104 J m−3 = 3.3907 × 104 Pa
mol, V = 10.000 l and T = 298.15 K. 17. The specific heat capacity (specific heat) of a substance is
n RT crudely defined as the amount of heat required to raise
P= the temperature of unit mass of the substance by
V −1 −1 1 degree Celsius (1 ◦C). The specific heat capacity of
(0.13678 mol)(0.082058 l atm K mol )(298.15 K) water is 4.18 J ◦C−1 g−1. Find the rise in temperature if
= 1.000 l 100.0 J of heat is transferred to 1.000 kg of water.
= 0.33464 atm
n RT
P= 100.0 J 1kg
V −1 −1 T = (4.18 J ◦C−1 g−1)(1.000 kg) 1000 g
(0.13678 mol)(8.3145 J K mol )(298.15 K) = 0.0239 ◦C
= 10.000 × 10−3 m3 18. The volume of a cone is given by
= 3.3907 × 104 J m−3 = 3.3907 × 104 N m−2 1
V = πr 2h
= 3.3907 × 104 Pa 3
where h is the height of the cone and r is the radius of
16. The van der Waals equation of state gives better its base. Find the volume of a cone if its radius is
accuracy than the idealagas equation of state. It is
P+ (Vm — b) = RT given as 0.443 m and its height is given as 0.542 m.
Vm2 1 3 1 2 3
where a and b are parameters that have different V = πr h = π(0.443 m) (0.542 m) = 0.111 m 3
values for different gases and where Vm = 3
V /n, the molar volume. For carbon dioxide, 19. The volume of a sphere is equal to 43 πr 3 where r is the
a = 0.3640 Pa m6 mol−2, b = 4.267 × radius of the sphere. Assume that the earth is spherical with
−5 3 −1
10 m mol . Calculate the pressure of carbon a radius of 3958.89 miles. (This is the radius of a sphere
dioxide in pascals, assuming that n = 0.13678 mol, with the same volume as the earth, which
,CHAPTER | 1 Problem Solving and Numerical Mathematics e5
is flattened at the poles by about 30 miles.) Find the We give two significant digits since the use of 1 as a
volume of the earth in cubic miles and in cubic meters. Use a single digit would specify a possible error of about 50%. It
value of π with at least six digits and give the correct is a fairly common practice to give an extra digit when the
number of significant digits in your answer. last significant digit is 1.
4 4 21. The hectare is a unit of land area defined to equal
V = πr 3 = π(3958.89 mi)3
exactly 10,000 square meters, and the acre is a unit
3 3 of land area defined so that 640 acres equals exactly
= 2.59508 × 1011 mi3 one square mile. Find the number of square meters in
5280 ft 3 1.000 acre, and find the number of acres equivalent to
(2.59508 × 1011 mi3) 12 in
3
1 mi 1 ft 1.000 hectare.
= 1.08168 × 1021 m3 (5280 ft) 2 12 in
0.0254 m 3
1.000 acre =
× 1 in 640
2
1 ft
2
20. Using the radius of the earth in the previous problem and 0.0254 m = 4047 m2
the fact that the surface of the earth is about 70% × 1 in
covered by water, estimate the area of all of the bodies of 1.000 hectare = (1.000 hectare) 10000 m2
water on the earth. The area of a sphere is equal to four 1 hectare
times the area of a great circle, or 4πr 2, where r
is the radius of the sphere. 1 acre = 2.471 acre
× 4047 m2
A ≈ (0.7)4πr 2 = (0.7)4π(3958.89 mi)2
= 1.4 × 108 mi2
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, ✤ ✜
Chapter 2
✣ ✢
Mathematical Functions
EXERCISES 0.1 = 1/10
Exercise 2.1. Enter a formula into cell D2 that will log (0.1) = − log (10) = −1
compute the mean of the numbers in cells A2,B2, and C2. 0.01 = 1/100
= (A2 + B2 + C2)/3 log (0.01) = − log (100) = −2
Exercise 2.2. Construct a graph representing the function 0.001 = 1/1000
y(x ) = x 3 − 2x 2 + 3x + 4 (2.1) log (0.001) = − log (1000) = −3
Use Excel or Mathematica or some other software to 0.0001 = 1/10000
construct your graph. log (0.001) = − log (10000) = −4
Exercise 2.4. Using a calculator or a spreadsheet, evaluate the
Here is the graph, constructed with Excel:
quantity (1+ 1 )n for several integral values of n ranging from 1 to
1,000,000. Notice how n the value approaches the
value of e as n increases and determine the value of n needed to
provide four significant digits.
Here is a table of values
✬ ✩
x (1 + 1/n)n
1 2
2 2.25
5 2.48832
10 2.59374246
Exercise 2.3. Generate the negative logarithms in the short table
of common logarithms.
✬ ✩
100 2.704813829
1000 2.716923932
x y = log 10 (x) x y = log 10 (x)
10000 2.718145927
100000 2.718268237
1 0 0.1 −1
1000000 2.718280469
10 1 0.01 −2 ✫ ✪
✫
100 2 0.001 −3 ✪
1000 3 0.0001 −4
Mathematics for Physical Chemistry. http://dx.doi.org/10.1016/B978-0-12-415809-2.00026-4
© 2013 Elsevier Inc. All rights reserved. e7
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