The Chemistry Maths Book

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1 Solutions for Chapter The Chemistr Maths Book Erich Steiner Universit of Eeter Second Edition 008 Solutions Chapter. Differentiation. Concepts. The process of differentiation. Continuit. Limits.5 Differentiation from first principles.6 Differentiation b rule.7 Implicit functions.8 Logarithmic differentiation.9 Successive differentiation.0 Stationar points. Linear and angular motion. The differential

2 Solutions for Chapter Section.. For, find (i) the change Δ in that corresponds to change Δ in, (ii) Δ Δ. (i) +Δ ( +Δ ) + Δ + ( Δ ) + ( Δ) Δ Δ + ( Δ ) + ( Δ) (ii) Δ + Δ + ( Δ ) Δ. For, find Δ lim Δ 0 Δ From Eercise, Δ + Δ + ( Δ ) Δ Δ lim Δ 0 Δ. For the Langmuir isotherm θ Kp + Kp find (i) the change Δ θ in θ that corresponds to change Δ p in p, (ii) lim Δ p 0 Δθ Δ p (i) Kp ( ), K p +Δ θ θ +Δ θ p + Kp + K( p+δp) Then p+δp p Δ θ K K + K( p+δ p) + Kp KΔ p [ + K( p +Δ p)][ + Kp] p +Δ p+ Kp + KpΔp p Kp KpΔp [ + K( p+δ p)][ + Kp] (ii) Δ θ K Δ p [ + K( p+δ p)][ + Kp] Δθ K lim Δ 0 Δ p ( + Kp)

3 Solutions for Chapter Section. Find the discontinuities of the following functions and state which are essential and which removable. Sketch graphs to demonstrate our answers.. + as from values > (from the right in Figure ) + as from values < (from the left) The function has an essential discontinuit at. Figure 5. if 0, but is not defined at 0. The function has a removable discontinuit at 0. Figure 6. ( ) The function has a removable discontinuit at 0 and an essential discontinuit at. Section. Figure Find the limits 7. lim 0 if 0, so that lim 0 Δ 0 8. lim if 0, so that as 0 ± lim 0 lim is continuous at 0, and lim ( ) if. ( ) ( + ) + lim lim. + Δ

4 Solutions for Chapter.. + lim + + as. + lim + lim. + as. lim 0. lim + as. as +. lim when 0 lim lim 0 e e (+ + + ) when 0 e lim 0 6. lim (ln ln ) 0 ln ln ln ln when 0 lim (ln ln ) ln 0 7. lim ln( ) ln( + ) ln( ) ln( + ) ln ln ln as + lim ln( ) ln( + ) ln

5 Solutions for Chapter 5 Section.5 Differentiate from first principles Let + + Then +Δ ( +Δ ) + ( +Δ ) + ( + + ) + (+ ) Δ + ( Δ) Δ (+ ) Δ + ( Δ ) Δ Δ + + Δ, lim + Δ d Δ 0 Δ 9. +Δ ( +Δ) + Δ + 6 ( Δ ) + ( Δ ) + ( Δ) Δ Δ + 6 ( Δ ) + ( Δ ) + ( Δ ) Δ + 6 ( Δ ) + ( Δ ) + ( Δ ), lim Δ d Δ 0 Δ Δ 0. +Δ ( +Δ) Δ ( Δ) Δ ( ) ( ) +Δ +Δ Δ ( Δ) Δ, lim Δ ( +Δ) 0 d Δ Δ. +Δ ( +Δ ) Δ ( +Δ) ( +Δ) ( +Δ ) + ( +Δ ) + ( +Δ) Δ + ( Δ ) + ( Δ) ( +Δ ) + ( +Δ ) + Δ + Δ + ( Δ) Δ, lim 0 Δ ( +Δ ) + d Δ Δ

6 Solutions for Chapter 6. e ( +Δ) Δ ( Δ) +Δ e e e e ( Δ + ) ( Δ) Δ e ( Δ + ) Δ Δ Δ e + e +, lim e Δ d Δ 0 Δ Section.6 Differentiate b rule / d d d 5 5 d 5 (5) d d / (/) / 6. d d d d 7. d ( + + 5sin 6 cos + 7e 8ln ) d d d d d d d d d () ( ) + ( ) ( ) + 5 (sin ) 6 (cos ) + 7 ( e ) 8 (ln ) d d d d d d d d 0 + ( ) ( ) + 5 cos 6 ( sin ) + 7 e cos + 6sin + 7e 8 8. The virial equation of state of a gas at low pressure is pv nrt nb V. Find dp dv at constant T and n (assume B is also constant). nb nb pv nrt p nrt V V V Then dp d d nrt nb nrt nb dv dv V dv + V V V nrt V nb + V

7 Solutions for Chapter 7 Products and quotients Differentiate 9. ( ) cos Put Then u v ( ) cos d dv du uv u + v d d d d d ( ) cos + (cos ) ( ) d d ( ) ( sin ) + (cos ) ( 8 ) ( ) sin 8cos 0. ( + ) e d d d ( + ) e ( + ) e + e ( + ) d d d ( + ) e + e (5 + e ). e cos d d d e cos e cos + cos e e ( sin ) + (cos ) e d d d e (cos sin ). ln d d d ln ln + ln + ln d d d + ln. ( + + ) ( + ) Put Then uv ( + + ) ( + ) d du dv uv v u v d d d d d ( + ) (+ + ) (+ + ) ( + ) ( + ) d d ( + ) ( + 6 ) ( + + )( ) ( + ) 6+ 8 ( + )

8 Solutions for Chapter 8. ( ) sin d d d ( ) sin sin ( ) ( ) sin sin d d d (sin ) ( 8 ) ( ) (cos ) sin 8sin + ( )cos sin 5. cos sin d d d cos sin sin cos cos sin sin d d d (sin ) ( sin ) (cos ) (cos ) sin (sin + cos ) cosec sin sin 6. (ln ) d d d (ln ) ln ln (ln ) d d d ( ln) Chain rule Differentiate 7. 5 ( + ) Write Then u 5, where u ( + ) du (5 u ) () d du d 5( + ) 8. + u, where u ( + ) u ( ) ( + ) d + 9. u, where u u ( ) ( ) d ( )

9 Solutions for Chapter 9 0. ( ) u, where u ( ) u ( ) ( ) ( ) d. sin sin u, where u cos u () cos d. e u e, where u u e ( ) e d. e + u e, where u + u e ( ) ( ) e d +. ln( + ) 5. cos( + ) ln u, where u + ( ) ( ) d u + cos u, where u + sin u ( ) ( )sin( + ) d 6. sin e u e, where u sin u sin e cos e cos d 7. ln(cos ) ln u, where u cos sin ( sin ) tan d u cos

10 Solutions for Chapter 0 e + cos( ) 8. Let u e, where u cos( + ) cos v, where v + d d d d and d d u u d d u d v. u v d du dv d du dv d u ( e ) (sin v) (6 ) 6sin( + ) e cos( +) 9. + ln ln( ) ln( ) ln ln, where and + u v u + v ( ) d u v ln (sin + sin ) ln u, where u sin + sin and u sin v+ w, where v and w sin Then du d d sin d d d d v+ w u u d d cos v + w cos u cos + sin cos cos + sin sin + sin sin + sin 5. ( ) + ( + ) u v dv du u + v ( + ) + ( + ) 6 d d d 6 ( + ) + ( + )

11 Solutions for Chapter 5. sin cos sin cos u v du dv u sin, cos ; v cos, sin d d dv du u + v sin ( sin ) + cos cos d d d cos cos sin sin 5. tan cos tan cos u v du u tan, sec d dv v cos, (cos ) ( sin ) sin d dv du u + v (tan ) ( sin ) + (cos ) (sec ) d d d cos sec tansin + 5. e + e u v du dv u, ; v e, e d d + + dv du + + u + v e + e d d d (+ ) e ( + ) ( + ) u v du dv u, 6 ; v ( + ), ( + ) d d dv du u + v ( )( + ) + 6 ( + ) d d d 6 ( + ) + ( + ) ( + ) ( + )

12 Solutions for Chapter Inverse functions 56. If +, find. d d +, d d Find dv dp at constant T and n for the following equations of state (assume that B, a and b are constants). nb 57. pv nrt + V 58. p( V nb) nrt 0 nb nb pv nrt + p nrt + V V V dp nb nrt nrt ( V nb dv + ) V V V dv V dp nr( V + nb) T dv V nb nrt pv ( nb) nrt 0 p V nb dp nrt p ( V nb) dv V nb dp p 59. na p + ( V nb) nrt V na nrt na p + ( V nb) nrt p V V nb V dp nrt n a + dv ( V nb) V dv n a nrt dp V ( V nb)

13 Solutions for Chapter Differentiate 60. sin sin sin d cos d cos (or b formula from Table.5 with a ) 6. tan tan du tan u, where u d du d du d du Then, b formula from Table.5, d + u + 6. cos + du cos cos u, where u + + d ( + ) Then, b formula from Table.5, du d du d u ( + ) Now u + ( + ) du ( + ) d du d (+ ) ( + ) 6. sinh B formula from Table.6 with a, d sinh d + + ( )

14 Solutions for Chapter 6. tanh B formula from Table.6, tanh tanh u, where u du d du d u Section.7 Find d d d d d d d d d d + d 67. ln d d ln ln ln d + d d + ln ( ln ) d d d d (+ ln ) d 0 d d d d d d

15 Solutions for Chapter 5 Section.8 Differentiate ln ln( ) ln( + ) + 7 d + ( )( + ) 7 d ( )( + ) ( + )( ) (+ )( + ) ( + )( ) (+ )( + ) ln ln( + ) + ln( ) ln(+ ) ln( + ) d + ( ) + ( + ) 6 + ( + )( ) + d + ( ) + ( + ) (+ )( + ) 7. sin cos ( + ) tan sin cos ( + ) tan ln ln( sin ) + ln( cos( + ) ) + ln(tan ) cos + ( sin( + )) + sec d sin cos( + ) tan cot 6tan( + ) + sin cot 6tan( ) sin cos ( ) tan d sin

16 Solutions for Chapter 6 7. Show that the equations dln p ΔH dp pδh dt RT dt RT vap vap and are equivalent epressions of the Clausius-Claperon equation. dln p dp dt p dt dp ΔH dp pδh pdt RT dt RT vap vap and 7. The decomposition of dinitrogen pentoide in tetrachloromethane at T 5 C has stoichiometr NO NO + O 5 and obes first-order kinetics. From the volumes of ogen liberated after various times t, the following concentrations of NO 5 were obtained [ 5] t s NO mol dm Plot a graph of ln against ts and determine the rate constant. ln t s In first-order kinetics, a plot of ln against t gives a straight line with slope d(ln ) dt k. The tabulated values for the decomposition of NO 5 are plotted in Figure. Figure In the present case, the straight line fit obtained b the method of least squares (Section.0) is almost identical to that obtained b drawing a straight line through the end points. The slope corresponds to rate constant k 6. 0 s

17 Solutions for Chapter 7 Section.9 7. Find all the nonzero derivatives of the function () (5) Find d d d,,, d d d d for the function ln. ln d d d ( ) d d d d ( ) d d d d ( ) 6 6 d d 76. Find a general formula for the nth derivative of e. d d d e e, e e, e e, and so on d d d d ( n) d ( n) e n e

18 Solutions for Chapter Find a general formula for the nth derivative of cos. cos sin cos () + sin () + cos (5) 5 sin and so on Then ( n) n/ n ( ) cos when n is an even integer (or zero), ( n) ( n+ )/ n ( ) sin when n is an odd integer. Section.0 Find the maimum and minimum values and the points of inflection of the following functions. In each case, sketch the graph and show the positions of these points , d 0 when The quadratic has a single stationar value at, when + Figure 5 The second derivative is positive, so that the stationar value is a (local) minimum value. d The sketch of the graph should look like Figure ( )( 5) d 0 when and 5 0 when, a minimum point > 6 0 when 5, a maimum point < d 0 when 7, a point of inflection

19 Solutions for Chapter 9 The graph of the function has minimum point at, when 0 maimum point at 5, when / 7 point of inflection at 7, when 6 / 7 The sketch of the graph should look like Figure Figure ( + ) d 0 when 0 and 0 when 0, a minimum point > + 0 when, 5 a maimum point < d 0 when, a point of inflection The sketch of the graph should look like Figure 7. Figure 7 8. e (see Figure.0) e ( e ) 0 when d < 0 when, e a maimum point ( e ) d 0 when, e a point of inflection The sketch of the graph should look like Figure 8. Figure 8

20 Solutions for Chapter 0 8. Confirm that the cubic , discussed in Eample., has local maimum and minimum values at and ( 8)( ) d 0 when 8 and + > 0 when 8 a minimum point 6 d < 0 when a maimum point 8. Find the maimum and minimum values and the points of inflection of Sketch a graph to show the positions of these points ( ) d 0 when 0, a triple root and > 0 when, a minimum point ( ) 0 when 0,? d 0 when, 57 8 a point of inflection The nature of the point (0, ) is determined b the first non-zero higher derivative ( ) 0 when 0 d 0( ) < 0 when 0, a maimum point d The sketch of the graph should look like Figure 9. Figure 9

21 Solutions for Chapter 8. The Lennard-Jones potential for the interaction of two molecules separated b distance R is A B U( R) 6 R R where A and B are constants. The equilibrium separation R is that value of R at which U( R) is a minimum and the binding energ is De U( R e ). Epress (i) A and B in terms of Re and De, (ii) U( R) in terms of R, R and D. e e e To obtain the minimum value of the function Let Then A B A B U( R), where R 6 R R du du d A B 5 6 A + 6R B 7 6 dr d dr R R 6 0 when R A B 6 6 A B A Re A B, De U( Re). R R R 6 e e e Then (i) A De A D er e Re 6 6 e e e R A B B D R (ii) 6 A B R e Re U( R) D 6 e R R R R 85. The probabilit that a molecule of mass m in a gas at temperature T has speed v is given b the Mawell-Boltzmann distribution m f() v π v e πkt mv kt where k is Boltzmann s constant. Find the most probable speed (for which f () v is a maimum). Let m av f () v π (), v where () v v e, a m kt πkt Then, for a maimum, d () ( ( e av a ) ( ) ( a e v v v) + v v ) dv av ve ( av ) 0 when v a and the most probable speed is v kt m

22 Solutions for Chapter k k 86. The concentration of species B in the rate process A B C, consisting of two consecutive irreversible first-order reactions, is given b (when k k) k kt k t [B] [A] 0 ( e e ) k k (i) Find the time t, in terms of the rate constants and k, at which B has its maimum k concentration, and (ii) show that the maimum concentration is [B] k [A] k ma 0 ( ) k k k k kt kt (i) Let [B] [A] 0 ( e e ) k k Then, for a maimum, k [A] 0 bt ( ), where bt ( ) e e k k kt kt db ke + ke 0 when ke ke dt kt kt kt kt k ( k k) t k e ln ( k k) t k k k t ln (Equation ) k k k kt k t (ii) At the maimum, ke k e. k kt kt kt kt [B] [A] 0 ( e e ) [A] 0 ( ke k e ) k k k k [A] kt 0 e Now, b Equation, k k k kt ln ln k k k k k ( k k) kt k e k k ( k k) and, at maimum concentration of B, kt k [B] ma [A] 0 e [A] 0 k k ( k k)

23 Solutions for Chapter Section. 87. A particle moving along a straight line travels the distance s t t in time t. (i) Find the velocit v and acceleration a at time t. (ii) Sketch graphs of s and v as functions of t in the interval t 0, (iii) find the stationar values, and describe the motion of the particle. (i) ds dv s t t, v t, a dt dt (ii) The sketch of the graphs should look like Figure 0. ds (iii) v t dt 0 when t, s 9 8 Figure 0 The particle moves from s 0 at time t 0 with velocit v, in the negative s-direction (downwards in Figure 0). The acceleration is constant, a +, in the positive s-direction. The particle slows down, and turns at t and s 98, when v 0 It then moves in the positive s-direction (upwards) with increasing speed. 88. A particle moving on the circumference of a circle of radius r travels distance s t t t in time t. (i) Epress the distance travelled in terms of the angle θ subtended at the centre of the circle, (ii) find the angular velocit ω and acceleration ω around the centre of the circle, (iii) Sketch graphs of θ, ω and ω as functions of t in the interval t 0, (iv) find the stationar values, and describe the motion of the particle. (i) (ii) s ( s rθ θ t t t) r dθ ω dt ( ) dω ω (6 t ) t dt t t Figure

24 Solutions for Chapter (iii) The sketch of the graphs should look like Figure. Figure (iv) ω ( t t ) ( t+ )( t ) 0 when t The particle moves from θ 0 at time t 0 in a clockwise direction around the circle with decreasing speed, comes to rest at t when θ, then moves in an anticlockwise direction with increasing speed. Section. Find the differential 89. d d d d (6 + ) d d 9. sin d cos d d 9. The volume of a sphere of radius r is V πr. Derive the differential dv from first principles. Give a geometric interpretation of the result. π π V +Δ V ( r+δ r) r + r Δ r+ r( Δ r) + ( Δr) π ΔV π Δ V r Δ r+ r( Δ r) + ( Δr) r + rδ r+ ( Δr) Δr Then dv dr ΔV lim πr Δr 0 Δr and dv πr dr surface area of sphere differential radius volume of a spherical shell of radius r and thickness dr

1. The following problems are not related: (a) (15 pts, 5 pts ea.) Find the following limits or show that they do not exist: arcsin(x)

1. The following problems are not related: (a) (15 pts, 5 pts ea.) Find the following limits or show that they do not exist: arcsin(x) APPM 5 Final Eam (5 pts) Fall. The following problems are not related: (a) (5 pts, 5 pts ea.) Find the following limits or show that they do not eist: (i) lim e (ii) lim arcsin() (b) (5 pts) Find and classify