(Solution Manual) Mathematics for Physical Chemistry

Mortimer, 4

Chemistry

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Solutions Manual for Mathematics for Physical Chemistry Solutions Manual for Mathematics for Physical Chemistry Fourth Edition Robert G. Mortimer Professor Emeritus Rhodes College Memphis, Tennessee AMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORD • PARIS SAN DIEGO • SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO Academic Press is an Imprint of Elsevier Preface vii 1 Problem Solving and Numerical Mathematics e1 2 Mathematical Functions e7 3 Problem Solving and Symbolic Mathematics: Algebra e13 4 Vectors and Vector Algebra e19 5 Problem Solving and the Solution of Algebraic Equations e23 6 Differential Calculus e35 7 Integral Calculus e51 8 Differential Calculus with Several Independent Variables e59 9 Integral Calculus with Several Independent Variables e69 10 Mathematical Series e79 11 Functional Series and Integral Transforms e89 12 Differential Equations e97 13 Operators, Matrices, and Group Theory e113 14 The Solution of Simultaneous Algebraic Equations with More Than Two Unknowns e125 15 Probability, Statistics, and Experimental Errors e135 16 Data Reduction and the Propagation of Errors e145 v Contents This book provides solutions to nearly of the exercises and problems in Mathematics for Physical Chemistry , fourth edition, by Robert G. Mortimer. This edition is a revision of a third edition published by Elsevier/Academic Press in 2005. Some of exercises and problems are carried over from earlier editions, but some have been modified, and some new ones have been added. I am pleased to acknowledge the cooperation and help of Linda Versteeg-Buschman, Beth Campbell, Jill Cetel, and their collaborators at Elsevier. It is also a pleasure to acknowledge the assistance of all those who helped with all editions of the book for which this is the solutions manual, and especially to thank my wife, Ann, for her patience, love, and forbearance. There are certain errors in the solutions in this manual, and I would appreciate learning of them through the publisher. Robert G. Mortimer vii Preface = = ≈ = Problem Solving and Numerical Mathematics , Chapter 1 \ J EXERCISES Exercise 1.1. Take a few fractions, such as 2 , 4 or 3 and Exercise 1.4. Round the following numbers to three significant digits 3 9 7 represent them as decimal numbers, finding either all of the nonzero digits or the repeating pattern of digits. 2 3 = 0.66666666 ··· 4 9 = 0.4444444 ··· 3 7 = 0.428571428571 ·· · Exercise 1.2. Express the following in terms of SI base units. The electron volt (eV), a unit of energy, equals 1.6022 × 10−18 J. a. 123456789123 ≈ 123,000,000,000 b. 46.45 ≈ 46.4 Exercise 1.5. Find the pressure P of a gas obeying the ideal gas equation PV = nRT if the volume V is 0.200 m3, the temperature T is 298.15 K and the amount of gas n is 1.000 mol. Take the smallest and largest value of each variable and verify your number of significant digits. Note that since you are dividing by a. (13.6 eV) 1.6022 × 10−19 J 1 eV 2.17896 × 10−19 J V the smallest value of the quotient will correspond to the largest value of V. ≈ 2.18 × 10−18 J b. (24.17 mi) 5280 ft 12 in 0.0254m P nRT V 1 mi = 3.890 × 104 m 1 ft 1 in (1.000 mol)(8.3145 J K−1 mol−1)(298.15 K) = 0.200 m3 c. −1 5280 ft 12 in 0.0254 m = 12395 J m−3 = 12395 N m−2 ≈ 1.24 × 104 Pa (55 mi h ) 1 h 1 mi 1 ft 1 in Pmax = nRT V 24.59 m s−1 25 m s−1 3600 s (1.0005 mol)(8.3145 J K−1 mol−1 )(298.155 K) d. (7.53 nm ps−1 1 m 1012 ps ) 109 nm 1 s = 0.1995 m3 = 1.243 × 104 Pa = 7.53 × 103 m s−1 Exercise 1.3. Convert the following numbers to scientific notation: a. 0.00000234 = 2,34 × 10−6 Pmin = = nRT V (0.9995 mol)(8.3145 J K−1 mol−1)(298.145 K) 0.2005 m3 b. 32.150 = 3.2150 × 101 Mathematics for Physical Chemistry. http://dx.doi.org/10.1016/B978 -0-12-415809-2.00025-2 = 1.236 × 104 Pa © 2013 Elsevier Inc. All rights reserved. e1 ≈ + + + − = = = 1 mi 1 ft 1 in 231.00 in3 0.0254 m 12 in 5280 ft 1 h 0.0254 m 36 in 1 h 1 d 1 fortnight Mathematics for Physical Chemistry Exercise 1.6. Calculate the following to the proper numbers of significant digits. a. 17.13 14.6751 3.123 7.654 8.123 34.359 34.36 b. ln (0.000123) 6. The distance by road from Memphis, Tennessee to Nashville, Tennessee is 206 miles. Express this distance in meters and in kilometers. (206 mi) 5380 ft 12 in 0.0254 m 1 mi 1 ft 1 in ln (0.0001235) = −8.99927 ln (0.0001225) = −9.00740 The answer should have three significant digits: ln (0.000123) = −9.00 = 3.32 × 105 m = 332 km 7. A U. S. gallon is defined as 231.00 cubic inches. a. Find the number of liters in 1.000 gallon. PROBLEMS 231.00 in3 0.0254 m 3 1000 l 1. Find the number of inches in 1.000 meter. (1.000 m) 1 in 39.37 in 0.0254 m (1 gal) = 3.785 l 1 gal 1 in 1 m3 2. Find the number of meters in 1.000 mile and the number of miles in 1.000 km, using the definition of the inch. (1.000 mi) 5280 ft 12 in 0.0254 m b. The volume of 1.0000 mol of an ideal gas at 0.00 ◦C (273.15 K) and 1.000 atm is 22.414 liters. Express this volume in gallons and in cubic feet. = 1609 m 1 m3 1 in 3 (1.000 km) 1000 m 1 in 1 ft (22.414 l) 1000 l 0.0254 m3 1 km 1 mi 0.0254 m 12 in × 1 gal = 5.9212 gal 3. Find the speed of light in miles per second. 1 m3 1 in 3 (299792458 m s−1) 1 in 1 ft 1 ft 3 1 mi × 5280 ft 0.0254 m = 186282.397 mi s 12 in −1 × 12 in = 0.79154 ft3 4. Find the speed of light in miles per hour. (299792458 m s−1) 1 in 1 ft 8. In the USA, footraces were once measured in yards and at one time, a time of 10 .00 seconds for this distance was thought to be unattainable. The best runners now × 1 mi 3600 s = 670616629 mi h−1 yards. If a runner runs 100.0 m in 10.00 s, find his 5. A furlong is exactly one-eighth of a mile and a fortnight is exactly 2 weeks. Find the speed of light (100.0 m) 1 in 1 yd = 109.4 m significant digits. (299792458 m s−1) 1 in 1 ft (10.00 s) 100.0 yd 9.144 s 109.4 m 0.0254 m × 1 mi 8 furlongs 12 in 9. Find the average length of a century in seconds and in 5280 ft 1 mi minutes. Use the rule that a year ending in 00 is not a 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 = 1.80261750 × 1012 furlongs fortnight−1 microcentury. will be 97 leap years. Find the number of minutes in a in furlongs per fortnight, using the correct number of time for 100 yards, assuming a constant speed. run 100 m in 10 seconds or less. Express 100.0 m in 5280 ft e2 × = 0.6214 (22.414 l) 1000 l 0.0254 m3 ◦9 F = = = 3 3 × 5 K × = × × CHAPTER | 1 Problem Solving and Numerical Mathematics Number of days in 400 years = (365 d)(400 y) + 97 d = 146097 d Average number of days in a century 146097 d = 4 = 36524.25 d 1 century = (36524.25 d) 24 h 60 min b. Find the Rankine temperature at 0.00 ◦F. 273.15 K − 18.00 K = 255.15 K (255.15 K) 459.27 ◦R 5 K 1 d 1 h = 5.259492 × 107 min 12. The volume of a sphere is given by (5.259492 × 107 min) 1 century 1 × 106 microcenturies V 4 πr 3 3 = 52.59492 min (52.59492 min) 60 s 3155.695 s 1 min 10. A light year is the distance traveled by light in one year. a. Express this distance in meters and in kilometers. where V is the volume and r is the radius. If a certain sphere has a radius given as 0 .005250 m, find its volume, specifying it with the correct number of digits. Calculate the smallest and largest volumes that the sphere might have with the given information and check your first answer for the volume. V = 4 πr 3 = V = 4 π(0.005250 m)3 Use the average length of a year as described in 3 3 the previous problem. How many significant digits can be given? = 6.061 × 10−7 m3 Vmin = 4 π(0.005245 m)3 = 6.044 × 10−7 m3 (299792458 m s−1) 60 s 1 min Vmax = 4 π(0.005255 m)3 = 6.079 × 10−7 m3 ×(5.259492 × 105 min) = (9.46055060 × 1015 m) (9.46055060 1015 m) 1 km 1000 m = 9.4605506 × 1012 km ≈ 9.460551 × 1012 km Since the number of significant digits in the number of days in an average century is seven, we round to seven significant digits. b. Express a light year in miles. (9.460551 1015 m) 1 in 1 ft 0.0254 m 12 in 1 mi 5.878514 1012 mi 5280 ft 11. The Rankine temperature scale is defined so that the Rankine degree is the same size as the Fahrenheit degree, and absolute zero is 0 ◦R, the same as0 K. a. Find the Rankine temperature at 0.00 ◦C. 0.00 ◦C ↔ (273.15 K) 9 ◦F = 491.67 ◦R V = 6.06 × 10−7 m3 The rule of thumb gives four significant digits, but the calculation shows that only three significant digits can be specified and that the last digit can be wrong by one. 13. The volume of a right circular cylinder is given by V = πr 2h, where r is the radius and h is the height. If a right circular cylinder has a radius given as 0 .134 m and a height given as 0 .318 m, find its volume, specifying it with the correct number of digits. Calculate the smallest and largest volumes that the cylinder might have with the given information and check your first answer for the volume. V = π(0.134 m)2(0.318 m) = 0.0179 m3 Vmin = π(0.1335 m)2(0.3175 m) = 0.01778 m3 Vmax = π(0.1345 m)2(0.3185 m) = 0.0181 m3 14. The value of an angle is given as 31◦. Find the measure of the angle in radians. Find the smallest and largest values that its sine and cosine might have and specify e3 = = ◦ 3 6 0 V = m = = = m V 2 = = = × m = Mathematics for Physical Chemistry the sine and cosine to the appropriate number of digits. (31◦) 2π rad = 0.54 rad sin (30.5◦) = 0.5075 V 10.00 l, and T 298.15 K. Convert your answer to atmospheres and torr. RT a = Vm − b − 2 sin (31.5◦) = 0.5225 sin (31◦) = 0.51 cos (30.5◦) = 0.86163 cos (31.5◦) = 0.85264 (8.3145 J K−1 mol−1)(298.15 K) = 7.3110×10−2 m3 mol−1 − 4.267×10−5 m3 mol−1 0.3640 Pa m6 mol−2 − (7.3110 × 10−2 m3 mol−1)2 cos (31◦) = 0.86 15. Some elementary chemistry textbooks give P RT Vm − b a — V 2 (8.3145 J K−1 mol−1)(298.15 K) the value of R, the ideal gas constant, as 0.0821 l atm K−1 mol−1. a. Using the SI value, 8 3145 J K−1 mol−1, obtain = 7.3110×10−2 m3 mol−1 − 4.267×10−5 m3 mol−1 0.3640 Pa m6 mol−2 − (7.3110 × 10−2 m3 mol−1)2 . the value in l atm K−1 digits. mol−1 to five significant = 3.3927 × 104 J m−3 − 68.1 Pa = 3.3927 × 104 Pa − 68.1 Pa = 3.386 Pa (8.3145 J K−1 mol−1) 1 Pa m3 1 atm (3.3859 Pa) 1 atm 0.33416 atm 101325 Pa 1 J 101325 Pa 1000 l × 1 m3 = 0.082058 l atm K−1 mol−1 The prediction of the ideal gas equation is −1 −1 P = (8.3145 J K mol )(298.15 K) b. Calculate the pressure in atmospheres and in N m−2 (Pa) of a sample of an ideal gas with n = 0.13678 mol, V = 10.000 l and T = 298.15 K. P nRT V (0.13678 mol)(0.082058 l atm K−1 mol−1)(298.15 K) 7.3110 × 10−2 m3 mol−1 = 3.3907 × 104 J m−3 = 3.3907 × 104 Pa 17. The specific heat capacity (specific heat) of a substance is crudely defined as the amount of heat required to raise the temperature of unit mass of the substance by 1 degree Celsius (1 ◦C). The specific heat capacity of = 0.33464 atm P nRT V 1.000 l water is 4.18 J ◦C−1 g−1. Find the rise in temperature if 100.0 J of heat is transferred to 1.000 kg of water. ∆T = 100.0 J 1 kg −1 −1 (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 = 3.3907 × 104 J m−3 = 3.3907 × 104 N m−2 = 3.3907 × 104 Pa 16. The van der Waals equation of state gives better accuracy than the ideal gas equation of state. It is P + a (Vm − b) = RT where a and b are parameters that have different 18. The volume of a cone is given by V 1 πr 2h 3 where h is the height of the cone and r is the radius of its base. Find the volume of a cone if its radius is given as 0.443 m and its height is given as 0.542 m. V = 1 πr 3h = 1 π(0.443 m)2(0.542 m) = 0.111 m3 3 3 values for different gases and where Vm = 4 3 V /n, the molar volume. For carbon dioxide, 19. The volume of a sphere is equal to 3 πr where r is the a 0.3640 Pa m6 mol−2, b 4.267 10−5 m3 mol−1. Calculate the pressure of carbon dioxide in pascals, assuming that n = 0.13678 mol, radius of the sphere. Assume that the earth is spherical with a radius of 3958.89 miles. (This is the radius of a sphere with the same volume as the earth, which e4 P × = . m CHAPTER | 1 Problem Solving and Numerical Mathematics is flattened at the poles by about 30 miles.) Find the volume of the earth in cubic miles and in cubic meters. Use a value of π with at least six digits and give the correct number of significant digits in your answer. V = 4 πr 3 = 4 π(3958.89 mi)3 We give two significant digits since the use of 1 as a single digit would specify a possible error of about 50%. It is a fairly common practice to give an extra digit when the last significant digit is 1. 21. The hectare is a unit of land area defined to equal 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 (2.59508 × 1011 mi3) 0.0254 m 3 5280 ft 3 12 in 3 1 mi 1 ft 21 3 1.000 acre, and find the number of acres equivalent to 1.000 hectare. (5280 ft)2 12 in 2 20. Using the radius of the earth in the previous problem and the fact that the surface of the earth is about 70% covered by water, estimate the area of all of the bodies of water on the earth. The area of a sphere is equal to four times the area of a great circle, or 4πr 2, where r 0.0254 m 2 × 1 in 1.000 hectare = (1.000 hectare) = 4047 m2 10000 m2 1 hectare is the radius of the sphere. A ≈ (0.7)4πr 2 = (0.7)4π(3958.89 mi)2 = 1.4 × 108 mi2 1 acre 2 471 acre 4047 m2 1 ft 640 1 in e5 × = 1.08168 × 10 1.000 acre = n + Mathematical Functions , Chapter 2 \ J EXERCISES Exercise 2.1. Enter a formula into cell D2 that will compute the mean of the numbers in cells A2,B2, and C 2. = ( A2 + B2 + C 2)/3 Exercise 2.2. Construct a graph representing the function y(x) = x 3 − 2x 2 + 3x + 4 (2.1) Use Excel or Mathematica or some other software to construct your graph. Here is the graph, constructed with Excel: 0.1 = 1/10 log (0.1) = − log (10) = −1 0.01 = 1/100 log (0.01) = − log (100) = −2 0.001 = 1/1000 log (0.001) = − log (1000) = −3 0.0001 = 1/10000 log (0.001) = − log (10000) = −4 Exercise 2.4. Using a calculator or a spreadsheet, evaluate the quantity (1 1 )n for several integral values of n ranging from 1 to 1,000,000. Notice how 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 , Exercise 2.3. Generate the negative logarithms in the short table of common logarithms. , 100 2.704813829 1000 3 0.0001 −4 \ J \ J Mathematics for Physical Chemistry. http://dx.doi.org/10.1016/B978 -0-12-415809-2.00026-4 © 2013 Elsevier Inc. All rights reserved. e7 x (1 + 1/n)n 1 2 2 2.25 5 2.48832 10 2.59374246 x y = log10 (x) x y = log10 (x) 1000 2.716923932 1 0 0.1 −1 10000 2.718145927 10 1 0.01 −2 100000 2.718268237 100 2 0.001 −3 1000000 2.718280469 = = [ ] = [ ] ≺ ≺ r y r + b b − = ≈ Mathematics for Physical Chemistry To twelve significant digits, the value of e is 2.71828182846. The value for n 1000000 is accurate to six significant digits. Four significant digits are obtained with n = 10000. Exercise 2.5. Without using a calculator or a table of logarithms, find the following: a. ln (100.000) = ln (10) log10 (100.000) = (2.30258509 ·· ·)(2.0000) = 4.60517 There is no round-off error to 11 digits in the calculator that was used. Exercise 2.9. Using a calculator and displaying as many digits as possible, find the values of the sine and cosine of 49.500◦. Square the two values and add the results. See if there is any round-off error in your calculator. sin (49.500◦) = 0.7604059656 cos (49.500◦) = 0.64944804833 b. ln (0.0010000) = ln (10) log10 (0.0010000) = (2.30258509 ·· ·)(−3.0000) = −6.90776 (0.7604059656 )2 + (0.64944804833 )2 = 1.00000000000 c. log10 ln (e) (e) = ln (10) 1 = 2.30258509 ··· = 0.43429 ··· Exercise 2.10. Construct an accurate graph of sin (x) and tan (x) on the same graph for values of x from 0 to 0 .4 rad Exercise 2.6. For a positive value of b find an expression in terms of b for the change in x required for the function ebx to double in size. and find the maximum value of x for which the two functions differ by less than 1%. f (x + ∆x) f (x) = 2 = eb(x +∆x) ebx = e b∆x ∆x = ln (2) = 0.69315 ·· · Exercise 2.7. A reactant in a first-order chemical reaction without back reaction has a concentration governed by the same formula as radioactive decay, [A]t = [A]0e−kt , where A 0 is the concentration at time t 0, A t is the concentration at time t, and k is a function of temperature called the rate constant. If k 0.123 s−1 find the time required for the concentration to drop to 21.0% of its initial value. t = 1 ln [A]0 = 1 ln 100.0 The two functions differ by less than 1% at 0 .14 rad. Notice that at 0 .4 rad, sin (x) x tan (x) and that the three quantities differ by less than 10%. Exercise 2.11. For an angle that is nearly as large as π/2, find an approximate equality similar to Eq. (2.36) involving (π/2) − α, cos (α), and cot (α). k = 12.7 s [A]t 0.123 s−1 21.0 Construct a right triangle with angle with the angle (π/2) − α, where α is small. The triangle is tall, with a Exercise 2.8. Using a calculator, find the value of the cosine of 15.5◦ and the value of the cosine of 375.5◦. small value of x (the horizontal leg) and a larger value of y (the vertical leg). Let r be the hypotenuse, which is nearly Display as many digits as your calculator is able to display. Check to see if your calculator produces any round-off error equal to y. cos ((π/2) − α) = x in the last digit. Choose another pair of angles that differ by 360◦ and repeat the calculation. Set your calculator to use angles measured in radians. Find the value of sin (0.3000). Find the value of sin (0.3000 2π ). See if there is any round-off error in the last digit. cos (15.5◦) = 0.96363045321 cos (375.5◦) = 0.96363045321 sin (0.3000) = 0.29552020666 sin (0.3000 + 2π) = sin (6.58318530718 ) = 0.29552020666 cot ((π/2) α) x x . The measure of the angle in radians is equal to the arc length subtending the angle α divided by r and is very nearly equal to x/r . Therefore cos ((π/2) − α) ≈ α cot ((π/2) − α) ≈ α cos ((π/2) − α) ≈ cot ((π/2) − α) Exercise 2.12. Sketch graphs of the arcsine function, the arccosine function, and the arctangent function. Include only the principal values. e8 CHAPTER | 2 Mathematical Functions Here are accurate graphs: Exercise 2.13. Make a graph of tanh (x) and coth (x) on the same graph for values of x ranging from 0.1 to 3.0. Exercise 2.14. Determine the number of significant digits in sin (95.5◦). We calculate sin (95.45◦) and sin (95.45◦). Using a calculator that displays 8 digits, we obtain sin (95.45◦) = 0.99547946 sin (95.55◦) = 0.99531218 We report the sine of 95 .5◦ as 0.9954, specifying four significant digits, although the argument of the sine was given with three significant digits. We have followed the common policy of reporting a digit as significant if it might be incorrect by one unit. Exercise 2.15. Sketch rough graphs of the following functions. Verify your graphs using Excel or Mathematica. a. e−x/5 sin (x). Following is a graph representing each of the factors and their product: b. sin2 (x) = [sin (x)]2 Following is a graph representing sin (x) and sin2 (x). PROBLEMS 1. The following is a set of data for the vapor pressure of ethanol taken by a physical chemistry student. Plot these points by hand on graph paper, with the temperature on the horizontal axis (the abscissa) and e9 = [ ] = [ ] = = k 0.232 s−1 Mathematics for Physical Chemistry the vapor pressure on the vertical axis (the ordinate). Decide if there are any bad data points. Draw a smooth curve nearly through the points, disregarding any bad points. Use Excel to construct another graph and notice how much work the spreadsheet saves you. , a function of the reciprocal of the Kelvin temperature. Why might this graph be more useful than the graph in the previous problem? \ J This graph might be more useful than the first graph Here is a graph constructed with Excel: The third data point might be suspect. Here is a graph omitting that data point: because the function is nearly linear. However, the third data point still lies off the curve. Here is a graph with that data point omitted. Thermodynamic theory implies that it should be nearly linear if there were no experimental error. 3. A reactant in a first -order chemical reaction without back reaction has a concentration governed by the same formula as radioactive decay, [A]t = [A]0e−kt , where A 0 is the concentration at time t 0, A t is the concentration at time t, and k is a function of temperature called the rate constant. If k 0.232 s−1 at 298.15 K find the time required for the concentration to drop to 33.3% of its initial value at a constant temperature of 298.15 K. t = ln [A]0/[A]t = ln (1/0.333) = 4.74 s 2. Using the data from the previous problem, construct a graph of the natural logarithm of the vapor pressure as 4. Find the value of each of the hyperbolic trigonometric functions for x 0 and x π/2. Compare these values with the values of the ordinary (circular) trigonometric functions for the same values of the independent variable. e10 Temperature/◦C Vapor pressure/torr 25.00 55.9 30.00 70.0 35.00 97.0 40.00 117.5 45.00 154.1 50.00 190.7 55.00 241.9 J J = − = = × CHAPTER | 2 Mathematical Functions Here are two table of values: , x sinh(x) cosh( x) tanh( x) csch( x) sech( x) coth( x) 0 0 1 0 ∞ 1 ∞ π /2 2.3013 2.5092 0.91715 0.43454 0.39854 1.09033 \ , x sin(x) cos(x) tan(x) csc(x) sec(x) cot(x) 0 0 1 0 ∞ 0 ∞ π /2 1 0 ∞ 1 ∞ 0 \ 5. Express the following with the correct number of significant digits. Use the arguments in radians: a. tan (0.600) a. x 2e−x/2 Following is a graph of the two factors and their product. b. 1/x 2 Following is the graph: tan (0.600) = 0.684137 tan (0.5995) = 0.683403 tan (0.60005) = 0.684210 We report tan (0.600) 0.684. If a digit is probably incorrect by 1, we still treat it as significant. b. sin (0.100) c. (1 x)e−x Following is a graph showing each factor and their product. sin (0.100) = 0.099833 sin (0.1005) = 0.100331 sin (0.0995) = 0.099336 We report sin (0.100) 0.100. c. cosh (12.0) cosh (12.0) = 81377 cosh (12.05) = 85550 cosh (11.95) = 77409 We report cosh (12.0) 8 104. There is only one significant digit. d. sinh (10.0) d. xe−x2 Following is a graph showing the two factors and their product. sinh (10.0) = 11013 sinh (10.01) = 11578 sinh (9.995) = 10476 We report sinh (10.0) = 11000 = 1.1 × 104 6. Sketch rough graphs of the following functions. Verify your graphs using Excel or Mathematica: 7. Tell where each of the following functions is discontinuous. Specify the type of discontinuity: e11 a − = = = = = = Mathematics for Physical Chemistry a. tan (x) Infinite discontinuities at x = π/2, x = 3π/2, x = 5π/2, · ·· b. csc (x) Infinite discontinuities at x = 0, x = π , x = 2π, · ·· c. |x | Continuous everywhere, although there is a sharp change of direction at x = 0. 8. Tell where each of the following functions is discontinuous. Specify the type of discontinuity: a. cot (x) Infinite discontinuities at x = 0,x = π , x = 2π, · ·· b. sec (x) Infinite discontinuities at x = π/2, x = 3π/2, x = 5π/2, · ·· c. ln (x −1) Infinite discontinuity at x = 1, function not defined for x ≺ 1. 9. If the two ends of a completely flexible chain (one that requires no force to bend it) are suspended at the same height near the surface of the earth, the curve representing the shape of the chain is called a catenary. It can be shown1 that the catenary is represented by y = a cosh x For this graph, we have plotted y 11.4538 such that this quantity vanishes at the ends of the chain. 10. For the chain in the previous problem, find the force necessary so that the center of the chain is no more than 0.500 m lower than the ends of the chain. By trial and error, we found that the center of the chain is 0 .499 m below the ends when a 25.5 m. This corresponds to T = gρa = (9.80 m s−2)(0.500 kg m−1)(25.5 m) = 125 N 11. Construct a graph of the two functions: 2 cosh (x) and ex for values of x from 0 to 3. At what minimum value of x do the two functions differ by less than 1%? where a T gρ and where ρ is the mass per unit length, g is the acceleration due to gravity, and T is the tension force on the chain. The variable x is equal to zero at the center of the chain. Construct a graph of this function such that the distance between the two points of support is 10.0 m and the mass per unit length is 0 .500 kg m−1, and the tension force is 50.0 N. T 50.0 kg m s2 a = gρ = (9.80 m2 s−2)(0.500 kg m−1) = 10.20 m y = (10.20 m) cosh (x/10.20 m) 1 G. Polya, Mathematical Methods in Science, The Mathematical Associa - tion of America, 1977, pp. 178ff. By inspection in a column of values of the difference, the two functions differ by less than 1% at x 2.4. 12. Verify the trigonometric identity sin (x + y) = sin (x) cos (y) + cos (x) sin (y) for the angles x 1.00000 rad, y 2.00000 rad. Use as many digits as your calculator will display and check for round-off error. sin (3.00000) = 0.14112000806 sin (1.00000) cos (2.00000) + cos (1.00000) × sin (2.00000) = 0.14112000806 There was no round -off error in the calculator that was used. 13. Verify the trigonometric identity cos (2x) = 1 − 2 sin2 (x) for x 0.50000 rad. Use as many digits as your calculator will display and check for round-off error. cos (1.00000) = 0.54030230587 1 − 2 sin2 (0.50000) = 1 − 0.45969769413 = 0.54030230587 There was no round -off error to 11 significant digits in the calculator that was used. e12 3 4 . √ m V 2 m — − = = , = − + − = m = − Problem Solving and Symbolic Mathematics: Algebra , Chapter 3 \ J EXERCISES Exercise 3.1. Write the following expression in a simpler form: b. Find ρ and θ if x = 5.00 and y = 10.00. ρ = , x 2 + y2 = √ 125.0 = 11.18 (x 2 + 2x)2 − x 2(x − 2)2 + 12x 4 6x + 12x x 2(x 2 + 4x + 4) − x 2(x 2 − 4x + 4) + 12x 4 θ = arctan (y/x) = arctan (2.00) = 1.107 rad = 63.43◦ Exercise 3.4. Find the spherical polar coordinates of the point whose Cartesian coordinates are (2.00, 3.00, 4.00). B 6x 3 + 12x 4 r = , (2.00)2 + (3.00)2 − (4.00)2 = 29.00 = 5.39 x 4 + 4x 3 + 4x 2 − x 4 + 4x 3 − 4x 2 + 12x 4 6x 3 + 12x 4 θ = arctan 3.00 2.00 = 0.98279 rad = 56.3◦ = 12x 4 + 8x 3 = 12x + 8 = 6x + 4 ζ = arccos 4.00 = 0.733 rad = 42.0◦ 6x 3 + 12x 4 12x + 6 6x + 3 Exercise 3.2. Manipulate the van der Waals equation into an expression for P in terms of T and Vm. Since the pressure is independent of the size of the system (it is an intensive variable), thermodynamic theory implies that it can depend on only two independent intensive variables. P + a (Vm − b) = RT 5.39 Exercise 3.5. Find the Cartesian coordinates of the point whose cylindrical polar coordinates are ρ = 25.00,θ = 60.0◦, z = 17.50 x = ρ cos (θ) = 25.00 cos (60.0◦) = 25.00 × 0.500 = 12.50 y = ρ sin (θ) = 25.00 sin (60.0◦) = 25.00 × 0.86603 = 21.65 a RT 2 = Vm − b P RT a Vm − b V 2 Exercise 3.3. a. Find x and y if ρ 6.00 and θ π/6 radians z = 17.50 Exercise 3.6. Find the cylindrical polar coordinates of the point whose Cartesian coordinates are ( 2.000, 2.000, 3.000). ρ ( 2.00)2 ( 2.00)2 2.828 −2.00 ◦ x = (6.00) cos (π/6) = (6.00)(0.866025) = 5.20 y = ρ sin (θ) = (6.00)(0.500) = 3.00 Mathematics for Physical Chemistry. http://dx.doi.org/10.1016/B978 -0-12-415809-2.00027-6 θ = arctan z = 3.000 −2.00 = 0.7854 rad = 45.0 © 2013 Elsevier Inc. All rights reserved. e13 = + V P B = = = = + = = = + = + 2 5.657e − = Mathematics for Physical Chemistry Exercise 3.7. Find the cylindrical polar coordinates of the point whose spherical polar coordinates are r = 3.00, ζ = 30.00◦, θ = 45.00◦. b. B = (3 + 7i)3 − (7i)2. B∗ = (3 − 7i)3 − (−7i)2 = (3 − 7i)3 — (7)2 z = r cos (ζ ) = (3.00) cos (30.00◦) = 3.00 × 0.86603 = 2.60 ρ = r sin (ζ ) = (3.00) sin (30.00◦) = (3.00)(0.500) = 1.50 θ = 45.00◦ Exercise 3.8. Simplify the expression (4 + 6i )(3 + 2i) + 4i (4 + 6i )(3 + 2i) + 4i = 12 + 18i + 8i − 12 + 4i = 30i Exercise 3.9. Express the following complex numbers in the form reiθ: a. z = 4.00 + 4.00i r = √ 32.00 = 5.66 θ arctan 4.00 0.785 4.00 z = 4.00 + 4.00i = 5.66e0.785i b. z = −1.00 z = −1 = eπi Exercise 3.10. Express the following complex numbers in the form x + iy a. z = 3eπi/2 Exercise 3.12. Write a complex number in the form x iy and show that the product of the number with its complex conjugate is real and nonnegative (x + iy)(x − iy) = x 2 + ixy − ixy + y2 = x 2 + y2 The square of a real number is real and nonnegative, and x and y are real. Exercise 3.13. If z (3.00 2.00i)2, find R(z), I (z), r , and θ. z = 9.00 + 6.00i − 4.00 = 5.00 + 6.00i R(z) = 5.00 I (z) 6.00 r = √ 25.00 + 36.00 = 7.781 θ = arctan (6.00/5.00) = 0.876 rad = 50.2◦ Exercise 3.14. Find the square roots of z 4.00 4.00i . Sketch an Argand diagram and locate the roots on it. z reiθ r = √ 32.00 = 5.657 θ = arctan (1.00) = 0.785398 rad = 45.00◦ x = r cos (θ) = 3 cos (π/2) = 3 × 0 = 0 y = r sin (θ) = 3 sin (π/2) = 3 × 1 = 3 √z = 5.657 exp √ 5.657e0.3927i 0.785398 +2π i = √ 3.534i z = 3i b. z = e3πi/2 x = r cos (θ) = cos (3π/2) = 0 y = r sin (θ) = sin (3π/2) = −1 z = −i Exercise 3.11. Find the complex conjugates of a. A = (x + iy)2 − 4eixy A = x 2 + 2ixy + y2 − 4eixy A∗ = x 2 − 2ixy + y2 − 4e−ixy = (x − iy)2 − 4e−ixy Otherwise by changing the sign in front of every i: A∗ = (x − iy)2 − 4e−ixy To sketch the Argand diagram, we require the real and imaginary parts. For the first possibility √z = √ 5.657( cos (22.50◦)) + i sin (22.50◦) = √ 5.657(0.92388 + i(0.38268) = 2.1973 + 0.91019i For the second possibility √z = √ 5.657( cos (202.50◦)) + i sin (202.50◦) = √ 5.657(−0.92388 + i(−0.38268) = −2.1973 − 0.91019i Exercise 3.15. Find the four fourth roots of −1. 1 eπi , e3πi , e5πi , e7πi ,4 eπi = eπi/4, e3πi/4, e5πi/4, e7πi/4 e14 √ = b + + = = + [ − ] = + m √ m = = = = √ 3 = × − 3 m 2 m m m m + 2 − |− | 3 + − | − | + − + | − | + − | − | CHAPTER | 3 Problem Solving and Symbolic Mathematics: Algebra Exercise 3.16. Estimate the number of house painters in Chicago. The 2010 census lists a population of 2,695,598 for the city of Chicago, excluding surrounding areas. Assume that about 20% of Chicagoans live in single -family houses or duplexes that would need exterior painting. With an average family size of four persons for house -dwellers, this would give about 135,000 houses. Each house wo uld be painted about once in six or eight years, giving roughly 20,000 house-painting jobs per year. A crew of two painters might paint a house in one week, so that a crew of two painters could paint about 50 houses in a year. This gives about 400 two-painter crews, or 800 house painters in Chicago. The zero in 30 m is significant, which we indicated with a bar over it. 4. The equation x 2 y2 z2 c2 where c is a constant, represents a surface in three dimensions. Express the equation in spherical polar coordinates. What is the shape of the surface? x 2 + y2 + z2 = r 2 = c2 This represents a sphere with radius c. 5. Express the equation y b, where b is a constant, in plane polar coordinates. y = ρ sin (θ) = b ρ = sin (θ) = b csc (θ) PROBLEMS 1. Manipulate the van der Waals equation into a cubic equation in Vm. That is, make a polynomial with terms proportional to powers of Vm up to V 3 on one side of the equation. 6. Express the equation y mx b, where m and b are constants, in plane polar coordinates. ρ sin (θ) = mρ cos (θ) + b ρ tan (θ) = mρ + b sec (θ) ρ tan (θ) m b sec (θ) b sec (θ) P a V 2 (Vm − b) = RT ρ [tan (θ) − m] m Multiply by V 2 (PV 2 + a)(Vm − b) = RT V 2 7. Find the values of the plane polar coordinates that correspond to x = 3.00, y = 4.00. ρ = 9.00 + 16.00 = 5.00 4.00 P Vm + aVm − bP Vm − ab = RT V 2 PV 3 − (b + RT )V 2 + aVm − ab = 0 2. Find the value of the expression 3(2 4)2 6(7 17 )3 √37 1 3 (1 + 22)4 − (|− 7|+ 63)2 + √ 12 + | − 4| 3(2 4) 6(7 17 ) 37 1 (1 + 22)4 − (|− 7|+ 63)2 + √12 +|− 4| 3(6)2 − 6(24)3 + √ 36 3 = (5)4 − (223)2 + √ 16 3.00 8. Find the values of the Cartesian coordinates that correspond to r 5.00,ζ 45.0◦,θ 135.0◦. 9. A surface is represented in cylindrical polar coordinates by the equation z ρ2. Describe the shape of the surface. This equation represents a paraboloid of revolution, produced by revolving a parabola around the z axis. 10. The solutions to the Schrödinger equation for the electron in a hydrogen atom have three quantum numbers associated with them, called n, l, and m, and these solutions are denoted by ψnlm. a. The ψ210 function is given by 72 − 82944 + 216 −82656 1 1 3/2 r −r/2a = 625 − 49729 + 4 = −49100 = 1.683 ψ210 2π a0 a0 e 0 cos (ζ ) 3. A Boy Scout finds a tall tree while hiking and wants to estimate its height. He walks away from the tree and finds that when he is 45 m from the tree, he must look upward at an angle of 32◦ to look at the top of the tree. His eye is 1.40 m from the ground, which is perfectly where a0 0.529 10−10 m is called the Bohr radius. Write this function in terms of Cartesian coordinates. cos (ζ ) = z 1 1 3/2 level. How tall is the tree? h = (45 m) tan (32◦) + 1.40 m = 28.1 m + 1.40 m = 29.5 m ≈ 30 m ψ210 = 4√ 2π z × a0 exp a0 (x 2 + y2 + z2)1/2 2a0 √ = 4 e15 θ = arctan = 53.1◦ = 0.927 rad = √ ⎧⎪⎨ 1.710e0.30910i ⎪ 2i |ψ211| = (ψ211ψ211) = 8√π ⎪⎩ √4 √π Mathematics for Physical Chemistry b. The ψ211 function is given by 12. Find the sum of 4.00e3.00i and 5.00e2.00i . 3.00i 1 1 3/2 r ψ211 = 8√π a0 a0 e −r/2a0 sin (ζ )eiθ 4.00e = (4.00)[cos (3.00) + i sin (3.00)] = (4.00)( − 0.98999 + 0.14112i) = −3.95997 + 0.56448i Write an expression for the magnitude of the ψ211 function. 5.00e2.00i = (5.00) ∗∗∗ ∗[cos (2.00]+ i sin (2.00) = (5.00)( − 0.41615 + 0.90930i) ∗ 1/2 1 1 3/2 = −2.08075 + 4.54650i × r e−r/2a0 sin (ζ )(ei θe−iθ)1/2 = −6.04 + 5.11i a0 1 = 8√π 1 3/2 r a0 a0 e−r/2a 0 sin (ζ ) 13. Find the difference 3.00eπi − 2.00e2πi . 3.00eπi − 2.00e2πi = −3.00 − 2.00 = 5.00 c. The ψ211 function is sometimes called ψ2 p1 . Write 14. Find the three cube roots of z = 3.000 + 4.000i . r = √ 9.000 + 16.000 = 5.000 expressions for the real and imaginary parts of the function, which are proportional to the related θ = arctan (4.000/3.000) = 0.92730 rad functions called ψ2px and ψ2py. 1 z = ⎧ ⎪⎨ 5.000eiθ = 5.000e0.92730i 5.000e(2π +θ) = 5.000e7.21048i R(ψ211) = = 2 ψ211 + ψ2∗11 1 1 3/2 r e−r/2a 0 sin (ζ ) z1/3 = ⎩⎪ 5.000e(4π +θ) = 5.000e13.49367 1.710e2.40349i × 1 eiθ + e−iθ 1.710e4.49789i 2 1 1 3/2 r −r/2a 15. Find the four fourth roots of 3.000i . ⎧ 3 000eπi/2 = 8√ a a e 0 ⎪ . π 0 0 3 000i = ⎨ 3.000e5πi/2 × sin (ζ ) cos (θ) . 3.000e9πi/2 ⎪⎩ 3.000e13π i/2 I (ψ211) = 2i ψ211 − ψ2∗ 11 √4 ⎧ ⎪⎪⎨ √4 3.000eπi/8 = 1.316eπi/8 √4 3.000e5πi/8 = 1.316e5πi/8 1 1 3/2 r 3.000i = ⎪ √4 3.000e9πi/8 = 1.316e9πi/8 × 1 (eiθ − e−iθ) 16. Find the real and imaginary parts of 1 1 3/2 r z = (3.00 + i)3 + (6.00 + 5.00i)2 × sin (ζ ) sin (θ) 11. Find the complex conjugate of the quantity e2.00i + 3eiπ e2.00i + 3eiπ = e2.00i − 3 = cos (2.00) + i sin (2.00) − 3 (e2.00i + 3eiπ )∗ = cos (2.00) − i sin (2.00) − 3 = e−2.00i − 3 (3.00 + i)3 = (3.00 + i )(9.00 + 6.00i − 1) = 27.00 + 18.00i − 3.00 + 9.00i −6.00 − i = 18.00 + 26.00i (6.00 + 5.00i)2 = 36.00 + 60.00i − 25.00 = 11.00 + 60.00i (3.00 + i)3 + (6.00 + 5.00i)2 = 29.00 + 86.00i z∗ = 29.00 − 86.00i Find z∗. a0 a0 8 e−r/2a0 sin (ζ ) a0 a0 π 8 4.00e3.00i + 5.00e2.00i 0 a e16 1 √ = 8 π a0 a0 3.000e 13 πi/8 = 1.316e 13 πi/8 = e−r/2a0 41 7 2 × × = = − − = + 41 + i + 41 × × 5 + = = CHAPTER | 3 Problem Solving and Symbolic Mathematics: Algebra 17. If z = 3 + 2i 2 4 + 5i , find R(z), I (z), r, and θ. 19. Estimate the number of grains of sand on the beaches of the major continents of the earth. Exclude islands and inland bodies of water. (4 5i)−1 4 − 5i 4 − 5i 16 + 25 41 Assume that the earth has seven continents with an average radius of 2000 km. Since the coastlines are 3 + 2i 4 + 5i = 4 − 5i (3 + 2i) somewhat irregular, assume that each continent has a coastline of roughly 10000 km = 1 × 107 m for a 12 −15 + 8 10 total coastline of 7 × 10 m. Assume that the average 22 7i = − of 1.75 × 1010 m3. Assume that the average grain of 22 7i 2 41 − 41 41 41 22 2 = 41 2 22 7 — (41)2 i + 41 sane is roughly 0 .3 mm in diameter, so that each cubic millimeter contains roughly 30 grains of sand. This is equivalent to 3 1010 grains per cubic meter, so that we have roughly 5 1020 grains of sand. If we were to include islands and inland bodies of water, we would likely have a number of grains of sand nearly equal to = 0.43378 − 0.18322i R(z) = 0.43378 I (z) = −0.18322 r = , (0.43378)2 + (0.18322)2 = 0.47079 Avogadro’s constant. 20. A gas has a molar volume of 20 liters. Estimate the average distance between nearest-neighbor molecules. Assume that each molecule is found in a cube such that 20 liters is divided into a number of cubes equal to Avogadro’s constant: 1 m3 θ arctan −0.18322 arctan ( 0.42238) 0.43378 volume of a cube = (20 l) 1000 l = 3×10−26 m3 = −0.39965 The principal value of the arctangent is in the fourth 6 × 10 The length of the side of the cube is roughly the average distance between molecules: quadrant, equal to 0.39965 rad. Since θ ranges from 0 to 2π , we subtract 0.39965 from 2π to get average distance = 3×10−26 m3 1/3 =3 × 10−9 m θ = 5.8835 rad 18. Obtain the famous formulas = 30 Å This is a reasonable value, since it is roughly ten times as large as a molecular diameter. 21. Estimate the number of blades of grass in a lawn with an area of 1000 square meters. Assume approximately 10 blades per square centimeter. If z = eiθ The real part is obtained from number = (10 cm−2) 100 cm 2 1 m (1000 m2) ∗ i −i = 1 × 108 R(z) = z + z 2 e θ + e θ 2 22. Since in its early history the earth was too hot for and the imaginary part is obtained from liquid water to exist on it, it has been hypothesized that all of the water on the earth came from collisions I (z) = z − z∗ 2i eiθ − e−iθ 2i of comets with the earth. Assume an average diameter for the head of a comet and assume that it is completely composed of water ice. Estimate the volume of water in agreement with the formula eiθ cos (θ) i sin (θ) on the earth and estimate how many comets would have collided with the earth to supply this much water. cos (θ) = eiθ + e−iθ 2 = R(e ) iθ 41 e17 sin (θ) = eiθ − e−iθ 2i = I (e ) iθ 23 = = = stretch of coastline has sand roughly 5 m deep and 50 m wide. This gives a total volume of beach sand 3 × Mathematics for Physical Chemistry Assume that an average comet has a spherical nucleus 50 miles (80 kilometers) in radius. This corresponds to a volume V (comet) = 4 π(8 × 104 m)3 = 2 × 1015 m3 The radius of the earth is roughly 6.4 106 m, so its surface area is Roughly 70% of the earth’s surface is covered by water. Assume an average depth of 1.0 kilometer for all bodies of water. V (wat er) = (0.70)(5.1 × 1014 m2)(1000 m) = 3.6 × 1017 m3 3.6 × 1017 m3 A = 4π(6.4 × 106 m) 2 = 5.1 × 1014 m2 Number of comets ≈ 2 × 1015 m3 ≈ 200 e18 √ = − = − + AB (3.606)(4.123) Vectors and Vector Algebra , Chapter 4 \ J EXERCISES Exercise 4.1. Draw vector diagrams and convince your - self that the two schemes presented for the construction of D = A − B give the same result. Exercise 4.2. Find A − B if A = (2.50,1.50) and B = (1.00,−7.50) A − B = (1.50,9.00) = 1.50i + 9.00j Exercise 4.3. Let |A|= 4.00,|B|= 2.00, and let the angle between them equal 45.0◦. Find A · B. A · B = (4.00)(2.00) cos (45◦) = 8.00 × 0.70711 = 5.66. Exercise 4.4. If A = (3.00)i − (4.00)j and B = (1.00)i + (2.00)j. a. Draw a vector diagram of the two vectors. b. Find A · B and (2.00A) · (3.00B). A · B = 3.00 × 1.00 +(−4.00)(2.00) = −5.00 (2A) · (3B) = 6(−5.00) = −30.00 Exercise 4.5. If A 2.00i 3.00 j and B 1.00i j a. Find |A| and |B|. |A| = A = √ 4.00 + 9.00 = 3.606 |B| = B = √ 1.00 + 16.00 = 4.123 b. Find the components and the magnitude of 2.00A − B. 2.00A − B = i(4.00 + 1.00)+j(−6.00 + 4.00) = i(5.00)+j(−2.00) |2.00A − B| = 25.00 + 4.00 = 5.385 c. Find A · B. A · B = (2.00)(−1.00) + (−3.00)(4.00) = −14.00 d. Find the angle between A and B. cos (α) = A · B = −14.00 = −0.94176 α = arccos (−0.94176) = 2.799 rad = 160.3◦ Exercise 4.6. Find the magnitude of the vector A = (−3.00, 4.00, −5.00). A = √ 9.00 + 16.00 + 25.00 = √ 50.00 = 7.07 Exercise 4.7. a. Find the Cartesian components of the position vector whose spherical polar coordinates are r = 2.00, ζ = 90◦, θ = 0◦. Call this vector A. x = 2.00 y = 0.00 z = r cos (ζ ) = 0.00 A = (2.00)i b. Find the scalar product of the vector A from part a and the vector B whose Cartesian components are (1.00, 2.00, 3.00). A · B = 2.00 + 0 + 0 = 2.00 Mathematics for Physical Chemistry. http://dx.doi.org/10.1016/B978 -0-12-415809-2.00028-8 © 2013 Elsevier Inc. All rights reserved. e19 = = = = · = = × = = Mathematics for Physical Chemistry c. Find the angle between these two vectors. We must first find the magnitude of B. |B| = B = √ 1.00 + 4.00 + 9.00 = √ 14.00 = 3.742 By the right-hand rule, the angular momentum is vertically upward. PROBLEMS cos (α) 2.00 (2.00)(3.742) = 0.26726 1. Find A − B if A = 2.00i + 3.00 j and B = 1.00i + 3.00 j − 1.00k. α = arccos (0.26726) = 74.5◦ = 1.300 rad Exercise 4.8. From the definition, show that . This result follows immediately from the screw -thread rule or the right-hand rule, since reversing the order of the factors reverses the roles of the thumb and the index finger. Exercise 4.9. Show that the vector C is perpendicular to B. We do this by showing that B · C = 0. B · C = Bx Cx + ByCy + BzCz = −1 + 2 − 1 = 0. Exercise 4.10. The magnitude of the earth’s magnetic field ranges from 0.25 to 0 .65 G (gauss). Assume that the average magnitude is equal to 0 .45 G, which is equivalent to 0.000045 T. Find the magnitude of the force on the electron in the previous example due to the eart h’s magnetic field, assuming that the velocity is perpendicular to the magnetic field. |F|= F = (1.602 × 10−19 C) ×(1.000 × 105 m s−1)(0.000045 T) = 7.210 × 10−19 A s m s−1 kg s−2 A−1 = 7.210 × 10−19 kg m s−2 = 7.210 × 10−19 N Exercise 4.11. A boy is swinging a weight around his head on a rope. Assume that the weight has a mass of 0.650 kg, that the rope plus the effective length of the boy’s arm has a length of 1.45 m and that the weight makes a complete circuit in 1.34 s. Find the magnitude of the angular momentum, excluding the mass of the rope and that of the boy’s arm. If the mass is moving counterclockwise in a horizontal circle, what is the direction of the angular momentum? v 2π(1.45 m) 6.80 m s−1 1.34 s L = mvr = (0.650 kg)(6.80 m s−1) ×(1.45 m) = 6.41 kg m2 s−1 A − B = −1.00i + 2.00k. 2. An object of mass m 10.0 kg near the surface of the earth has a horizontal force of 98 .0 N acting on it in the eastward direction in addition to the gravitational force. Find the vector sum of the two forces (the resultant force). Let the gravitational force be denoted by W and the eastward force be denoted by F = (98.0 N)i . Denote the resultant force by R. R = F + W = (98.0 N)i − mgk = (98.0 N)i − (10.0 kg)(9.80 m s−2)k = (98.0 N)i − (98.0 kg m s−2k = (98.0 N)i − (98.0 N)k The direction of this force is 45◦ from the vertical in the eastward direction. 3. Find A · B if A = (0,2) and B = (2,0). A · B = 0 + 0 = 0 4. Find |A| if A = 3.00i + 4.00 j − 5.00k. |A|= √ 9.00 + 16.00 + 25.00 = 7.07 5. Find A · B if A = (1.00)i + (2.00) j + (3.00)k and B = (1.00)i + (3.00) j − (2.00)k. A · B = 1.00 + 6.00 − 6.00 = 1.00 6. Find A B if A (1.00,1.00,1.00) and B (2.00,2.00,2.00). A · B = 2.00 + 2.00 + 2.00 = 6.00 7. Find A B if A (0.00,1.00,2.00) and B (2.00,1.00,0.00). A × B = A × B = i( Ay Bz − Az By) + j( Az Bx − Ax Bz) + k( Ax By − Ay Bx ) = i(0.00 − 2.00) + j(4.00 − 0.00) + k(0.00 − 2.00) = −2.00i + 4.00j − 2.00k 8. Find A × B if A = (1,1,1) and B = (2,2,2). A × B = 0 e20 A × B = −(B × A) √ √ √ √ , √ = = − √ √ = √ + √ + = = − = , = V CHAPTER | 4 Vectors and Vector Algebra 9. Find the angle between A and B if A = 1.00i +2.00 j + 1.00k and B = 1.00i − 1.00k. A · B = 1.00 + 0 − 2.00 = −1.00 A = 1.00 + 4.00 + 1.00 = 6.00 = 2.4495 B = 1.00 + 1.00 = 2.00 = 1.4142 cos (α) −1.00 0.28868 (2.4495)(1.4142) 12. An object has a force on it given by F = (4.75 N)i + (7.00 N) j + (3.50 N)k. a. Find the magnitude of the force. F = (4.75 N)2 + (7.00 N)2 + (3.50 N)2 = 83.8125 = 915 N b. Find the projection of the force in the x-y plane. α = arccos (−0.28868) = 107◦ = 1.86 rad 10. Find the angle between A and B if A = 3.00i +2.00 j + 1.00k and B = 1.00i + 2.00 j + 3.00k. A · B = 3.00 + 4.00 + 3.00 = 10.00 A = 9.00 + 4.00 + 1.00 = 14.00 = 3.7417 B 1.00 4.00 9.00 14.00 3.7417 10 That is, find the vector in the x-y plane whose head is reached from the head of the force vector by moving in a direction perpendicular to the x-y plane. Fprojection = (4.75 N)i + (7.00 N)j 13. An object of mass 12 .000 kg is moving in the x direction. It has a gravitational force acting on it equal to mgk, where m is the mass of the object and g is the acceleration due to gravity, equal to 9.80 m s−1. cos (α) = .00 (3.7417)(3.7417) = 0.71429 There is a frictional force equal to (0.240 N)i . What is the magnitude and direction of the resultant force α = arccos (0.71429) = 44.4◦ = 0.775 rad 11. A spherical object falling in a fluid has three forces acting on it: (1) The gravitational force, whose magnitude is Fg mg, where m is the mass of the object and g is the acceleration due to gravity, equal to 9.80 m s−2; (2) The buoyant force, whose magnitude is Fb = mfg, where mf is the mass of the displaced (the vector sum of the forces on the object)? Ftotal = −(12.000 kg)(9.80 m s−1)k + (0.240 N)i = −(117.60 N)k + (0.240 N)i Ftotal = (117.6 N)2 + (0.240 N)2 = 118 N The angle between this vector and the negative z axis is 0.240 fluid, and whose direction is upward; (3) The frictional force, which is given by Ff = −6π εr v, where r α = arctan 117.6 = 0.117◦ = 0.00294 rad is the radius of the object, v its velocity, and ε the coefficient of viscosity of the fluid. This formula for the frictional forces applies only if the flow around the object is laminar (flow in layers). The object is falling at a constant speed in glycerol, which has a viscosity of 1490 kg m−1 s−1. The object has a mass of 0.00381 kg, has a radius of 0.00432 m, a mass of 14. The potential energy of a magnetic dipole in a magnetic field is given by the scalar product V = −µ · B, where B is the magnetic induction (magnetic field) and μ is the magnetic dipole. Make a graph of | as a 0.00381 kg, and displaces a mass of fluid equal to 0.000337 kg. Find the speed of the object. Assume that the object has attained a steady speed, so that the net force vanishes. Fz,total = 0 = −(0.00381 kg)(9.80 m s−2) + (0.000337 kg)(9.80 m s−2) + 6π(1490 kg m−1 s−1)(0.00432 m)v .−(0.00381 kg)(9.80 m s−2)+(0.000337 kg)(9.80 m s−2). function of the angle between μ and B the angle from 0◦ to 180◦. V |µ||B| = − cos (α) Here is the graph µ B for values of v = . 0.18 m s−1 6π(1490 kg m−1 s−1)(0.00432 m) . || | e21 = Mathematics for Physical Chemistry 15. According to the Bohr theory of the hydrogen atom, the electron in the atom moves around the nucleus in one of various circular orbits with radius r = a0n2 For n = 1: L (6.626 × 10−34 J s) where a0 is a distance equal to 0.529 × 10−10 m, v = mr = 2π(9.109 × 10−31 kg)(0.529 × 10−10 m) called the Bohr radius and n is a positive integer. The mass of the electron is 9.109 × 10−31 kg. According to the theory, L = nh/2π , where h is Planck’s constant, equal to 6.626 × 10−34 J s. Find the speed of the electron for n = 1 and = 2.188 × 106 m s−1 For n = 2 L 2(6.626 × 10−34 Js) for n = 2. Since the orbit is circular, the position vector and the velocity are perpendicular to each other, and L = mvr . v = mr = 2π(9.109 × 10−31 kg)22(0.529 × 10−10 m) = 1.094 × 106 m s−1 Notice that the speed for n 1 is nearly 1% of the speed of light. e22