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Physical Chemistry: Thermodynamics, Structure, and Change Tenth Edition Solution Manual

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    Foundations A Matter Answers to discussion questions A.2 Metals conduct electricity, have luster, and they are malleable and ductile. Nonmetals do not conduct electricity and are neither malleable nor ductile. Metalloids typically have the appearance of metals but behave chemically like a nonmetal. 1 IA 2 IIA 3 IIIB 4 IVB 5 VB 6 VIB 7 VIIB 8 VIIIB 9 VIIIB 10 VIIIB 11 IB 12 IIB 13 IIIA 14 IVA 15 VA 16 VIA 17 VIIA 18 VIIIA 1 H 1.008 Periodic Table of the Elements 2 He 4.003 3 Li 6.941 4 Be 9.012 5 B 10.81 6 C 12.01 7 N 14.01 8 O 16.00 9 F 19.00 10 Ne 20.18 11 Na 22.99 12 Mg 24.31 13 Al 26.98 14 Si 28.09 15 P 30.97 16 S 32.07 17 Cl 35.45 18 Ar 39.95 19 K 39.10 20 Ca 40.08 21 Sc 44.96 22 Ti 47.88 23 V 50.94 24 Cr 52.00 25 Mn 54.94 26 Fe 55.85 27 Co 58.93 28 Ni 58.69 29 Cu 63.55 30 Zn 65.38 31 Ga 69.72 32 Ge 72.59 33 As 74.92 34 Se 78.96 35 Br 79.90 36 Kr 83.80 37 Rb 85.47 38 Sr 87.62 39 Y 88.91 40 Zr 91.22 41 Nb 92.91 42 Mo 95.94 43 Tc (98) 44 Ru 101.1 45 Rh 102.9 46 Pd 106.4 47 Ag 107.9 48 Cd 112.4 49 In 114.8 50 Sn 118.7 51 Sb 121.8 52 Te 127.6 53 I 126.9 54 Xe 131.3 55 Cs 132.9 56 Ba 137.3 57 La 138.9 72 Hf 178.5 73 Ta 180.9 74 W 183.9 75 Re 186.2 76 Os 190.2 77 Ir 192.2 78 Pt 195.1 79 Au 197.0 80 Hg 200.6 81 Tl 204.4 82 Pb 207.2 83 Bi 209.0 84 Po (209) 85 At (210) 86 Rn (222) 87 Fr (223) 88 Ra 226 89 Ac (227) 58 Ce 140.1 59 Pr 140.9 60 Nd 144.2 61 Pm 145 62 Sm 150.4 63 Eu 152.0 64 Gd 157.3 65 Tb 158.9 66 Dy 162.5 90 Th 232.0 91 Pa (231) 92 U 238.0 93 Np 237 94 Pu (244) 95 Am (243) 96 Cm (247) 97 Bk (247) 98 Cf (251) Transition metals Lanthanoids Actinoids 1 IA 2 IIA 3 IIIB 4 IVB 5 VB 6 VIB 7 VIIB 8 VIIIB 9 VIIIB 10 VIIIB 11 IB 12 IIB 13 IIIA 14 IVA 15 VA 16 VIA 17 VIIA 18 VIIIA 1 H 1.008 Periodic Table of the Elements 2 He 4.003 3 Li 6.941 4 Be 9.012 5 B 10.81 6 C 12.01 7 N 14.01 8 O 16.00 9 F 19.00 10 Ne 20.18 11 Na 22.99 12 Mg 24.31 13 Al 26.98 14 Si 28.09 15 P 30.97 16 S 32.07 17 Cl 35.45 18 Ar 39.95 19 K 39.10 20 Ca 40.08 21 Sc 44.96 22 Ti 47.88 23 V 50.94 24 Cr 52.00 25 Mn 54.94 26 Fe 55.85 27 Co 58.93 28 Ni 58.69 29 Cu 63.55 30 Zn 65.38 31 Ga 69.72 32 Ge 72.59 33 As 74.92 34 Se 78.96 35 Br 79.90 36 Kr 83.80 37 Rb 85.47 38 Sr 87.62 39 Y 88.91 40 Zr 91.22 41 Nb 92.91 42 Mo 95.94 43 Tc (98) 44 Ru 101.1 45 Rh 102.9 46 Pd 106.4 47 Ag 107.9 48 Cd 112.4 49 In 114.8 50 Sn 118.7 51 Sb 121.8 52 Te 127.6 53 I 126.9 54 Xe 131.3 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

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    N H H H Cs 132.9 Ba 137.3 La 138.9 Hf 178.5 Ta 180.9 W 183.9 Re 186.2 Os 190.2 Ir 192.2 Pt 195.1 Au 197.0 Hg 200.6 Tl 204.4 Pb 207.2 Bi 209.0 Po (209) At (210) Rn (222) 87 Fr (223) 88 Ra 226 89 Ac (227) 58 Ce 140.1 59 Pr 140.9 60 Nd 144.2 61 Pm 145 62 Sm 150.4 63 Eu 152.0 64 Gd 157.3 65 Tb 158.9 66 Dy 162.5 90 Th 232.0 91 Pa (231) 92 U 238.0 93 Np 237 94 Pu (244) 95 Am (243) 96 Cm (247) 97 Bk (247) 98 Cf (251) A.4 Valence-shell electron pair repulsion theory (VSEPR theory) predicts molecular shape with the concept that regions of high electron density (as represented by single bonds, multiple bonds, and lone pair) take up orientations around the central atom that maximize their separation. The resulting positions of attached atoms (not lone pairs) are used to classify the shape of the molecule. When the central atom has two or more lone pair, the molecular geometry must minimize repulsion between the relatively diffuse orbitals of the lone pair. Furthermore, it is assumed that the repulsion between a lone pair and a bonding pair is stronger than the repulsion between two bonding pair, thereby, making bond angles smaller than the idealized bond angles that appear in the absence of a lone pair. Solutions to exercises A.1(b) A.2(b) (i) chemical formula and name: CaH 2 , calcium hydride ions: Ca 2+ and H – oxidation numbers of the elements: calcium, +2; hydrogen, –1 (ii) chemical formula and name: CaC 2 , calcium carbide ions: Ca 2+ and C 2 2– (a polyatomic ion) oxidation numbers of the elements: calcium, +2; carbon, –1 (iii) chemical formula and name: LiN 3 , lithium azide ions: Li + and N 3 – (a polyatomic ion) oxidation numbers of the elements: lithium, +1; nitrogen, – ⅓ A.3(b) (i) Ammonia, NH 3 , illustrates a molecule with one lone pair on the central atom. Example Element Ground-state Electronic Configuration ( i ) Group 3 Sc, scandium [Ar]3d 1 4s 2 ( ii ) Group 5 V, vanadium [Ar]3d 3 4s 2 ( iii ) Group 13 Ga, gallium [Ar]3d 10 4s 2 4p 1

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    H O H H F Xe F F (ii) Water, H 2 O, illustrates a molecule with two lone pairs on the central atom. (iii) The hydrogen fluoride molecule, HF, illustrates a molecule with three lone pairs on the central atom. Xenon difluoride has three lone pairs on both the central atom and the two peripheral atoms. A.4(b) (i) Ozone, O 3 . Formal charges (shown in circles) may be indicated. (ii) ClF 3 + (iii) azide anion, N 3 – A.5(b) The central atoms in XeF 4 , PCl 5 , SF 4 , and SF 6 are hypervalent. A.6(b) Molecular and polyatomic ion shapes are predicted by drawing the Lewis structure and applying the concepts of VSEPR theory. (i) H 2 O 2 Lewis structure: O O O O O O F Cl F F N N N H O O H

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    Orientations caused by repulsions between two lone pair and two bonding pair around each oxygen atom: Molecular shape around each oxygen atom: bent (or angular) with bond angles somewhat smaller than 109.5º (ii) FSO 3 – Lewis structure: (Formal charge is circled.) Orientations around the sulfur are caused by repulsions between one lone pair, one double bond, and two single bonds while orientations around the oxygen to which fluorine is attached are caused by repulsions between two lone pair and two single bonds: Molecular shape around the sulfur atom is trigonal pyramidal with bond angles somewhat smaller than 109.5º while the shape around the oxygen to which fluorine is attached is bent (or angular) with a bond angle somewhat smaller than 109.5º. (iii) KrF 2 Lewis structure: Orientations caused by repulsions between three lone pair and two bonding pair: H O O H S S O O O O O O F F S O F O O F Kr F Kr F F

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    Molecular shape: linear with a 180º bond angle. (iv) PCl 4 + Lewis structure: (Formal charge is shown in a circle.) Orientations caused by repulsions between four bonding pair (no lone pair): Molecular shape: tetrahedral and bond angles of 109.5º A.7(b) (i) Nonpolar or weakly polar toward the slightly more electronegative carbon. (ii) (c) A.8(b) (i) O 3 is a bent molecule that has a small dipole as indicated by consideration of electron densities and formal charge distributions. (ii) XeF 2 is a linear, nonpolar molecule. (iii) NO 2 is a bent, polar molecule. (iv) C 6 H 14 is a nonpolar molecule. A.9(b) In the order of increasing dipole moment: XeF 2 ~ C 6 H 14 , NO 2 , O 3 A.10(b) (i) Pressure is an intensive property. (ii) Specific heat capacity is an intensive property. (iii) Weight is an extensive property. (iv) Molality is an intensive property. A.11(b) (i) 1 mol 5.0 g 0.028 mol [A.3] 180.16 g m n M   = = =     (ii) 23 22 A 6.0221 10 molecules 0.028 mol 1.7 10 molecules mol N nN   × = = = ×     A.12(b) Cl P Cl Cl Cl P Cl Cl Cl Cl C H N Cl δ + δ − P S δ + δ −

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    (i) 78.11 g 10.0 mol 781. g [A.3] mol m n M   = = =     (ii) gravity on Mars Mars weight F m g = = ( ) ( ) 2 2 1 kg 781. g 3.72 m s 2.91 kg m s 2.91 N 1000 g − −   = × × = =     A.13(b) F mg p A A = = ( ) ( ) 2 2 6 2 6 2 4 2 5 60 kg 9.81 m s 1 cm 1 bar 3 10 N m 3 10 Pa 2 cm 10 m 10 Pa 30 bar 10 bar − − − ×     = = × = ×         = ± A.14(b) ( ) 1 atm 30 bar 10 bar 30 atm 10 atm 1.01325 bar   ± = ±     A.15(b) (i) 5 5 1.01325 10 Pa 222 atm 225 10 Pa 1 atm   × = ×     (ii) 1.01325 bar 222 atm 225 bar 1 atm   =     A.16(b) / C / K 273.15 90.18 273.15 182.97 [A.4] T θ ° = − = − = − 182.97 C θ = − ° A.17(b) The absolute zero of temperature is 0 K and 0 ºR. Using the scaling relationship 1 ºF / 1 ºR (given in the exercise) and knowing the scaling ratios 5 ºC / 9 ºF and 1 K / 1 ºC, we find the scaling factor between the Kelvin scale and the Rankine scale to be: 1 F 5 C 1 K 5 K 1 R 9 F 1 C 9 R ° °       × × =       ° ° ° °       The zero values of the absolute zero of temperature on both the Kelvin and Rankine scales and the value of the scaling relationship implies that: ( ) ( ) R R 5 9 / K / R or / R / K 9 5 T T θ θ = × ° ° = × Normal freezing point of water: ( ) ( ) R R 9 9 / R / K 273.15 491.67 5 5 491.67 R T θ θ ° = × = × = = °

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    A.18(b) 1 mol 0.325 g 0.0161 mol 20.18 g n   = × =     ( ) ( ) ( ) 1 1 3 3 3 3 4 0.0161 mol 8.314 J K mol 293.15 K dm [A.5] 2.00 dm 10 m 1.96 10 Pa 19.6 kPa nRT p V − − −   = =     = × = A.19(b) [A.5] mRT pV nRT M = = ( )( ) ( ) 3 1 1 1 1 3 where is the mass density [A.2] 0.6388 kg m 8.314 J K mol 373.15 K 0.124 kg mol 124 g mol 16.0 10 Pa mRT RT M pV p ρ ρ − − − − − = = = = = × The molecular mass is four times as large as the atomic mass of phosphorus (30.97 g mol –1 ) so the molecular formula is P 4 . A.20(b) 1 mol 7.05 g 0.220 mol [A.3] 32.00 g n   = × =     ( ) ( ) ( ) 1 1 3 3 6 3 6 0.220 mol 8.314 J K mol 373.15.15 K cm [A.5] 100. cm 10 m 6.83 10 Pa 6.83 MPa nRT p V − − −   = =     = × = A.21(b) 2 2 O CO 0.25 mole and 0.034 mole n n = = ( ) ( ) ( ) 2 2 1 1 3 O O 3 6 3 0.25 mol 8.314 J K mol 283.15 K cm [A.5] 5.9 MPa 100. cm 10 m n RT p V − − −   = = =     Since the ratio of CO 2 moles to O 2 moles is 0.034/0.25, we may scale the oxygen partial pressure by this ratio to find the partial pressure of CO 2 . ( ) 2 2 2 CO O CO 0.034 5.9 MPa 0.80 MPa [1.6] 6.7 MPa 0.25 p p p p   = × = = + =     B Energy Answers to discussion questions B.2 All objects in motion have the ability to do work during the process of slowing. That is, they have energy, or, more precisely, the energy possessed by a body because of its motion is its kinetic energy , E k . The law of conservation of energy tells us that the kinetic energy of an object equals the work done on the object in order to change its motion from an initial (i) state of v i = 0 to a final (f) state of v f = v . For an object of mass m travelling at a speed v , 2 1 2 k [B.8] E m = v . The potential energy , E p or more commonly V , of an object is the energy it possesses as a result

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    ( ) ( ) 0 0 Mars d d t t t t g t t g t = = = = = ∫ ∫ v v v v of its position. For an object of mass m at an altitude h close to the surface of the Earth, the gravitational potential energy is Eqn B.11 assigns the gravitational potential energy at the surface of the Earth, V (0), a value equal zero and g is called the acceleration of free fall . The Coulomb potential energy describes the particularly important electrostatic interaction between two point charges Q 1 and Q 2 separated by the distance r : ( ) 1 2 0 0 in a vacuum [B.14, is the vacuum permittivity] 4 π Q Q V r r ε ε = and Eqn B.14 assigns the Coulomb potential energy at infinite separation, V ( ∞), a value equal to zero. Convention assigns a negative value to the Coulomb potential energy when the interaction is attractive and a positive value when it is repulsive. The Coulomb potential energy and the force acting on the charges are related by the expression F = −d V /d r . B.4 Quantized energies are certain discrete values that are permitted for particles confined to a region of space. The quantization of energy is most important—in the sense that the allowed energies are widest apart—for particles of small mass confined to small regions of space. Consequently, quantization is very important for electrons in atoms and molecules. Quantization is important for the electronic states of atoms and molecules and for both the rotational and vibrational states of molecules. B.6 The Maxwell distribution of speeds indicates that few molecules have either very low or very high speeds. Furthermore, the distribution peaks at lower speeds when either the temperature is low or the molecular mass is high. The distribution peaks at high speeds when either the temperature is high or the molecular mass is low. Solutions to exercises B.1(b) a = d 𝑣 /d t = g so d 𝑣 = g d t . The acceleration of free fall is constant near the surface of the Mars. (i) ( ) 2 [B.11] where 9.81 m s V h mgh g − = = ( ) 1 2 r r 0 in a medium that has the relative permittivity (formerly, dielectric constant) 4 π Q Q V r r ε ε ε = ( ) ( ) ( ) ( ) ( ) 2 1 2 2 1 1 1 2 2 k 1.0 s 3.72 m s 1.0 s 3.72 m s 0.0010 kg 3.72 m s 6.9 mJ E m − − − = × = = = × = v v

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    (ii) B.2(b) The terminal velocity occurs when there is a balance between the force exerted by the pull of gravity, mg = V particle ρg = 4 / 3 π R 3 ρg , and the force of frictional drag, 6π ηRs . It will be in the direction of the gravitational pull and have the magnitude s terminal . 3 terminal 2 terminal 4 π 6π 3 2 9 R g Rs R g s ρ η ρ η = = B.3(b) The harmonic oscillator solution x ( t ) = A sin( ωt ) has the characteristics that 1/ 2 2 f f d ( ) cos( ) where ( / ) or d x t A t k m m k t ω ω ω ω = = = = v x min = x ( t=n π/ ω , n =0,1,2...) = 0 and x max = x ( t= ( n +½)π/ ω , n =0,1,2...) = A At x min the harmonic oscillator restoration force (Hooke’s law, F x = – k f x , Brief illustration B.2) is zero and, consequently, the harmonic potential energy, V , is a minimum that is taken to equal zero while kinetic energy, E k , is a maximum. As kinetic energy causes movement away from x min , kinetic energy continually converts to potential energy until no kinetic energy remains at x max where the restoration force changes the direction of motion and the conversion process reverses. We may easily find an expression for the total energy E ( A ) by examination of either x min or x max . Analysis using x min : 2 2 2 2 1 1 1 k k,max max f 2 2 2 0 E E V E m mA k A ω = + = + = = = v We begin the analysis that uses x max , by deriving the expression for the harmonic potential energy. f ( ) f 0 0 2 1 f 2 d d [B.10] d d d ( ) x V x x V F x k x x V k x x V x k x = − = = = ∫ ∫ Thus, 2 2 1 1 max max f k max f 2 2 ( ) and 0 V V x k A E E V V k A = = = + = + = B.4(b) 2 1 e 2 where is the displacement from equilibrium [Brief illustration B.3] w kx x R R = = − ( ) ( ) 2 1 12 19 19 1 2 510 N m 20 10 m 1.02 10 N m 1.02 10 J w − − − − = × × = × = × B.5(b) 2 1 k 6 4 where 2 for C H and 76.03 g mol for the major isotopes E ze z M φ + − = ∆ = = 1/ 2 2 1 A 2 2 or where / ze m ze m M N m φ φ ∆   = ∆ = =     v v ( ) ( ) ( ) ( ) ( ) 2 1 2 2 1 1 1 2 2 k 3.0 s 3.72 m s 3.0 s 11.2 m s 0.0010 kg 11.2 m s 63 mJ E m − − − = × = = = × = v v

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    ( ) ( ) ( ) ( ) 1/ 2 1/2 1/2 19 3 5 5 1 23 1 1/2 2 2 5 5 1 2 2 1.6022 10 C 20 10 V C V J 3.2 10 3.2 10 kg kg 0.07603 kg mol / 6.022 10 mol kg m s 3.2 10 3.2 10 m s kg − − − − −   × × × × ×       = = × = ×       ×         = × = ×     v ( ) k 2 20 kV 40 keV E E ze e φ = = ∆ = × = B.6(b) The work needed to separate two ions to infinity is identical to the Coulomb potential drop that occurs when the two ions are brought from an infinite separation, where the interaction potential equals zero, to a separation of r . In a vacuum: ( ) ( ) ( ) ( ) ( ) 2 2 1 2 0 0 0 0 2 19 18 12 1 2 1 12 2 2 4 [B.14] 4 π 4π 4π π 1.6022 10 C 3.69 10 J π 8.85419 10 J C m 250 10 m e e Q Q e e w V r r r r ε ε ε ε − − − − − − × −     = − = − = − = =         × = = × × × × In water: ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 2 0 2 19 20 12 1 2 1 12 2 2 [B.15] where 78 for water at 25 C 4 π 4π π π 1.6022 10 C 4.73 10 J π 78 8.85419 10 J C m 250 10 m r r e e Q Q e e w V r r r r ε ε ε ε ε ε − − − − − − × −     = − = − = − = = = °         × = = × × × × × B.7(b) We will model a solution by assuming that the NaCl pair consists of the two point charge ions Na + and Cl – . The electric potential will be calculated along the line of the ions. ( ) + + + Na Cl 0 0 0 Na Cl Na Cl 1 1 [2.17] [B.16] 4 π 4π 4π e e e r r r r φ φ φ ε ε ε − − −   − = + = + = −       When + Na Cl r r − = , the electric potential equals zero in this model. Likewise, + Na Cl r r − = at every point both on the line perpendicular to the internuclear line and crossing the internuclear line at the mid-point so electric potential equals zero at every point on that perpendicular line.

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    ( ) 1 1 1 1 s m 1 mol / 28.24 J K mol 1.228 J K g 22.99 g C C M − − − −   = = × =     ( ) 1 1 1 1 m s 63.55 g 0.384 J K g 24.4 J K mol mol C C M − − − −   = = × =     B.8(b) ( ) ( ) ( ) ( ) ( ) ( ) ethanol 3 1 ethanol ethanol ethanol ethanol ethanol ethano energy dissipated by the electric circuit [B.20] 1.12 A 12.5 V 172 s 2.41 10 C s V s 2.41 kJ [B.21] / U I t U nC T U U T nC mC M φ − ∆ = = ∆ ∆ = × × = × = ∆ = ∆ ∆ ∆ ∆ = = ( ) ( ) ( ) 3 1 1 1 l 2.41 10 J 150 g 111.5 J K mol / 46.07 g mol 6.64 K = 6.64 C − − − × = × = ° B.9(b) 1 50.0 kJ [B.21] 8.67 K or 8.67 C 5.77 kJ K U T C − ∆ ∆ = = = ° B.10(b) 1 mol 10.0 g 0.555 mol 18.01 g n   = × =     ( ) ( ) ( ) m 1 1 [B.21] 0.555 mol 75.2 J K mol 10.0 K 417 J U C T nC T − − ∆ = ∆ = ∆ = = B.11(b) B.12(b) B.13(b) ( ) ( ) 5 1 6 3 1 m m m 3 3 1.00 10 Pa 18.02 g mol 10 m [B.23] 1.81 J mol 0.997 g cm cm pM H U pV ρ − − − − × ×   − = = = =     B.14(b) 2 2 H O(l) H O(s) S S > B.15(b) 2 2 H O(l, 100 C) H O(l, 0 C) S S ° ° > B.16(b) In a state of static equilibrium there is no net force or torque acting on the system and, therefore, there is no resultant acceleration. Examples: When holding an object in a steady position above the ground, there is a balance between the downward gravitational force pulling on the object downward and the upward force of the hold. Release the object and it falls. A movable, but non-moving, piston within a cylinder may be at equilibrium because of equal pressures on each side of the piston. Increase the pressure on one side of the piston and it moves away from that side. In the Bohr atomic model of 1913 there is a balance between the electrostatic attraction of an electron to the nucleus and the centrifugal force acting on the orbiting electron. Should the

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    electron steadily lose kinetic energy, it spirals into the nucleus. B.17(b) ( ) / / e e [B.25a] i j i j kT kT i j N N ε ε ε − − −∆ = = (i) ( ) ( ) ( ) ( ) { } 19 1 23 1 2.0 eV 1.602 10 J eV / 1.381 10 J K 200 K 51 2 1 e 4.2 10 N N − − − − − × × × × − = = × (ii) ( ) ( ) ( ) ( ) { } 19 1 23 1 2.0 eV 1.602 10 J eV / 1.381 10 J K 2000 K 6 2 1 e 9.2 10 N N − − − − − × × × × − = = × B.18(b) ( ) upper / 0 lower lim lim e [B.25a] e 1 kT T T N N ε −∆ − →∞ →∞   = = =     In the limit of the infinite temperature both the upper and the lower state are equally occupied. B.19(b) ( )( ) 34 9 1 24 upper lower 6.626 10 J s 10.0 10 s 6.63 10 J h ε ε ε ν − − − ∆ = − = = × × = × ( ) ( ) ( ) { } 24 23 1 6.63 10 J / 1.381 10 J K 293 K upper / lower e [B.25a] e 0.998 klT N N ε − − − − × × × −∆ = = = The ratio N upper / N lower indicates that the two states are equally populated. A large fraction of gas- phase molecules will be in excited rotational states. B.20(b) Rates of chemical reaction typically increase with increasing temperature because more molecules have the requisite speed and corresponding kinetic energy to promote excitation and bond breakage during collisions at the high temperatures. B.21(b) ( ) 1/ 2 mean / [B.26] T M ∝ v ( ) ( ) ( ) ( ) ( ) ( ) 1/ 2 1/ 2 mean 2 2 2 1/ 2 mean 1 1 1 1/ 2 mean mean / / 303 K 303 K 1.02 293 K 293 K T T M T T T T M   = =       = =     v v v v B.22(b) ( ) 1/ 2 mean / [2.26] T M ∝ v ( ) ( ) ( ) ( ) ( ) ( ) 1/ 2 1/ 2 mean 2 2 1 1/ 2 mean 1 2 1 1/ 2 1 mean 2 1 mean 2 / / H 401.2 g mol 14.11 Hg 2.016 g mol M T M M M M T M − −   = =       = =     v v v v B.23(b) A gaseous helium atom has three translational degrees of freedom (the components of motion in the x , y , and z directions). Consequently, the equipartition theorem assigns a mean

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    m 3 for water vapour U RT = m 3 for Pb(s) U RT = energy of 3 2 kT to each atom. The molar internal energy is ( ) ( ) ( ) 1 1 1 3 3 3 m A 2 2 2 1 m m = 8.3145 J mol K 303 K 3.78 kJ mol 1 mol 3.78 kJ 10.0 g 9.45 kJ 4.00 g mol U N kT RT U nU mM U − − − − = = =    = = = =        B.24(b) A solid state lead atom has three vibrational quadratic degrees of freedom (the components of vibrational motion in the x , y , and z directions). Its potential energy also has a quadratic form in each direction because V ∝ ( x – x eq ) 2 . There is a total of six quadratic degrees of freedom for the atom because the atoms have no translational or rotational motion. Consequently, the equipartition theorem assigns a mean energy of 6 2 3 kT kT = to each atom. This is the law of Dulong and Petit. The molar internal energy is ( ) ( ) ( ) 1 1 1 m A 1 m m 3 3 = 3 8.3145 J mol K 293 K 7.31 kJ mol 1 mol 7.31 kJ 10.0 g 0.353 kJ 207.2 g mol U N kT RT U nU mM U − − − − = = =    = = = =        B.25(b) See exercise B.23(b) for the description of the molar internal energy of helium. ( ) ( ) 3 2 1 1 1 1 m 3 3 ,m 2 2 8.3145 J mol K 12.47 J mol K V RT U C R T T − − − − ∂ ∂ = = = = = ∂ ∂ B.26(b) (i) Water, being a bent molecule, has three quadratic translational and three quadratic rotational degrees of freedom. So, ( ) ( ) 1 1 1 1 m ,m 3 3 3 8.3145 J mol K 24.94 J mol K V RT U C R T T − − − − ∂ ∂ = = = = = ∂ ∂ (ii) See exercise B.24(b) for the description of the molar internal energy of lead. ( ) ( ) 1 1 1 1 m ,m 3 3 3 8.3145 J mol K 24.94 J mol K V RT U C R T T − − − − ∂ ∂ = = = = = ∂ ∂ C Waves Answers to discussion questions C.2 The sound of a sudden ‘bang’ is generated by sharply slapping two macroscopic objects together. This creates a sound wave of displaced air molecules that propagates away from the collision with intensity , defined to be the power transferred by the wave through a unit area normal to the direction of propagation. Thus, the SI unit of intensity is the watt per meter squared ( 2 W m − ) and ‘loudness’ increases with increasing intensity. The ‘bang’ creates a shell of compressed air molecules that propagates away from the source as a shell of higher pressure and

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    density. This longitudinal impulse propagates when gas molecules escape from the high pressure shell into the adjacent, lower pressure shell. Molecular collisions quickly cause momentum transfer from the high density to the low density shell and the effective propagation of the high density shell. The regions over which pressure and density vary during sound propagation are much wider than the molecular mean free path because sound is immediately dissipated by molecular collisions in the case for which pressure and density variations are of the order of the mean free path. Solutions to exercises C.1(b) 8 1 8 1 benzene r 3.00 10 m s [C.4] 1.97 10 m s 1.52 c c n − − × = = = × C.2(b) 6 1 2 1 1 10 μm [C.5] 2.78 μm 3600 cm 10 cm λ ν −   = = =      8 1 14 1 14 6 3.00 10 m s [C.1] 1.08 10 s 1.08 10 Hz 2.78 10 m c ν λ − − − × = = = × = × × Integrated activities F.2 The plots of Problem F.1 indicate that as temperature increases the peak of the Maxwell– Boltzmann distribution shifts to higher speeds with a decrease in the fraction of molecules that have low speeds and an increase in the fraction that have high speeds. Thus, justifying summary statements like 'temperature is a measure of the average molecular speed and kinetic energy of gas molecules', 'temperature is a positive property because molecular speed is a positive quantity', 'the absolute temperature of 0 K is unobtainable because the area under the plots of Problem F.1 must equal 1'.

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    1 The properties of gases 1A The perfect gas Answers to discussion questions 1A.2 The partial pressure of a gas in a mixture of gases is the pressure the gas would exert if it occupied alone the same container as the mixture at the same temperature. Dalton’s law is a limiting law because it holds exactly only under conditions where the gases have no effect upon each other. This can only be true in the limit of zero pressure where the molecules of the gas are very far apart. Hence, Dalton’s law holds exactly only for a mixture of perfect gases; for real gases, the law is only an approximation. Solutions to exercises 1A.1(b) The perfect gas law [1A.5] is pV = nRT, implying that the pressure would be p = nRT V All quantities on the right are given to us except n , which can be computed from the given mass of Ar. n = 25 g 39 . 95 − 1 g mol = 0 . 626 mol so p = (0 . 626 mol) × (8 . 31 × 10 − 2 dm 3 bar K − 1 mol − 1 ) × (30 + 273) K 1 . 5 dm 3 = 10.5bar So no, the sample would not exert a pressure of 2.0 bar. 1A.2(b) Boyle’s law [1A.4a] applies. pV = constant so p f V f = p i V i Solve for the initial pressure: (i) p i = p f V f V i = (1 . 97 bar) × (2 . 14 dm 3 ) (2 . 14 + 1 . 80) dm 3 = 1.07 bar (ii) The original pressure in Torr is p i = (1.07 bar) × 1 atm 1.013 bar       × 760 Torr 1 atm       = 803 Torr 1A.3(b) The relation between pressure and temperature at constant volume can be derived from the perfect gas law, pV = nRT [1A.5] so p ∝ T and p i T i = p f T f The final pressure, then, ought to be p f = p i T f T i = (125 kPa) × (11 + 273)K (23 + 273)K = 120 kPa 1A.4(b) According to the perfect gas law [1.8], one can compute the amount of gas from pressure, temperature, and volume. pV = nRT so n = pV RT = (1.00 atm) × (1 . 013 × 10 5 Pa atm − 1 ) × (4 . 00 × 10 3 m 3 ) (8.3145 J K − 1 mol − 1 ) × (20 + 273)K = 1 . 66 × 10 5 mol Once this is done, the mass of the gas can be computed from the amount and the molar mass: m = (1 . 66 × 10 5 mol) × (16.04 − 1 g mol ) = 2 . 67 × 10 6 g = 2.67 × 10 3 kg 1A.5(b) The total pressure is the external pressure plus the hydrostatic pressure [1A.1], making the total pressure 1

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