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

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FoundationsA MatterAnswers to discussion questionsA.2Metals 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.1IA2IIA3IIIB4IVB5VB6VIB7VIIB8VIIIB9VIIIB10VIIIB11IB12IIB13IIIA14IVA15VA16VIA17VIIA18VIIIA1H1.008Periodic Table of the Elements2He4.0033Li6.9414Be9.0125B10.816C12.017N14.018O16.009F19.0010Ne20.1811Na22.9912Mg24.3113Al26.9814Si28.0915P30.9716S32.0717Cl35.4518Ar39.9519K39.1020Ca40.0821Sc44.9622Ti47.8823V50.9424Cr52.0025Mn54.9426Fe55.8527Co58.9328Ni58.6929Cu63.5530Zn65.3831Ga69.7232Ge72.5933As74.9234Se78.9635Br79.9036Kr83.8037Rb85.4738Sr87.6239Y88.9140Zr91.2241Nb92.9142Mo95.9443Tc(98)44Ru101.145Rh102.946Pd106.447Ag107.948Cd112.449In114.850Sn118.751Sb121.852Te127.653I126.954Xe131.355Cs132.956Ba137.357La138.972Hf178.573Ta180.974W183.975Re186.276Os190.277Ir192.278Pt195.179Au197.080Hg200.681Tl204.482Pb207.283Bi209.084Po(209)85At(210)86Rn(222)87Fr(223)88Ra22689Ac(227)58Ce140.159Pr140.960Nd144.261Pm14562Sm150.463Eu152.064Gd157.365Tb158.966Dy162.590Th232.091Pa(231)92U238.093Np23794Pu(244)95Am(243)96Cm(247)97Bk(247)98Cf(251)Transition metalsLanthanoidsActinoids1IA2IIA3IIIB4IVB5VB6VIB7VIIB8VIIIB9VIIIB10VIIIB11IB12IIB13IIIA14IVA15VA16VIA17VIIA18VIIIA1H1.008Periodic Table of the Elements2He4.0033Li6.9414Be9.0125B10.816C12.017N14.018O16.009F19.0010Ne20.1811Na22.9912Mg24.3113Al26.9814Si28.0915P30.9716S32.0717Cl35.4518Ar39.9519K39.1020Ca40.0821Sc44.9622Ti47.8823V50.9424Cr52.0025Mn54.9426Fe55.8527Co58.9328Ni58.6929Cu63.5530Zn65.3831Ga69.7232Ge72.5933As74.9234Se78.9635Br79.9036Kr83.8037Rb85.4738Sr87.6239Y88.9140Zr91.2241Nb92.9142Mo95.9443Tc(98)44Ru101.145Rh102.946Pd106.447Ag107.948Cd112.449In114.850Sn118.751Sb121.852Te127.653I126.954Xe131.3555657727374757677787980818283848586

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NHHHCs132.9Ba137.3La138.9Hf178.5Ta180.9W183.9Re186.2Os190.2Ir192.2Pt195.1Au197.0Hg200.6Tl204.4Pb207.2Bi209.0Po(209)At(210)Rn(222)87Fr(223)88Ra22689Ac(227)58Ce140.159Pr140.960Nd144.261Pm14562Sm150.463Eu152.064Gd157.365Tb158.966Dy162.590Th232.091Pa(231)92U238.093Np23794Pu(244)95Am(243)96Cm(247)97Bk(247)98Cf(251)A.4Valence-shell electron pair repulsion theory (VSEPR theory)predicts molecularshape 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 theirseparation. The resulting positions of attached atoms (not lone pairs) are used to classify theshape of the molecule. When the central atom has two or more lone pair, the molecular geometrymust minimize repulsion between the relatively diffuse orbitals of the lone pair. Furthermore, itis assumed that the repulsion between a lone pair and a bonding pair is stronger than therepulsion between two bonding pair, thereby, making bond angles smaller than the idealizedbond angles that appear in the absence of a lone pair.Solutions to exercisesA.1(b)A.2(b)(i)chemical formula and name: CaH2, calcium hydrideions: Ca2+and H–oxidation numbers of the elements: calcium, +2; hydrogen, –1(ii)chemical formula and name: CaC2, calcium carbideions: Ca2+and C22–(a polyatomic ion)oxidation numbers of the elements: calcium, +2; carbon, –1(iii)chemical formula and name: LiN3, lithium azideions: Li+and N3–(a polyatomic ion)oxidation numbers of the elements: lithium, +1; nitrogen, –⅓A.3(b)(i)Ammonia, NH3, illustrates a molecule with one lone pair on the central atom.ExampleElementGround-state Electronic Configuration(i)Group 3Sc, scandium[Ar]3d14s2(ii)Group 5V, vanadium[Ar]3d34s2(iii)Group 13Ga, gallium[Ar]3d104s24p1

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HOHHFXeFF(ii)Water, H2O, 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 thecentral atom. Xenon difluoride has three lone pairs on both the central atom and the twoperipheral atoms.A.4(b)(i)Ozone, O3. Formal charges (shown in circles) may be indicated.(ii)ClF3+(iii)azide anion, N3–A.5(b)The central atoms in XeF4, PCl5, SF4, and SF6are hypervalent.A.6(b)Molecular and polyatomic ion shapes are predicted by drawing the Lewis structure andapplying the concepts of VSEPR theory.(i)H2O2Lewis structure:OOOOOOFClFFNNNHOOH

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Orientations caused by repulsions between two lone pair and two bonding pair aroundeach oxygen atom:Molecular shape around each oxygen atom: bent (or angular) with bond angles somewhatsmaller than 109.5º(ii)FSO3–Lewis structure:(Formal charge is circled.)Orientations around the sulfur are caused by repulsions between one lone pair, onedouble bond, and two single bonds while orientations around the oxygen to whichfluorine 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 somewhatsmaller than 109.5º while the shape around the oxygen to which fluorine is attached isbent (or angular) with a bond angle somewhat smaller than 109.5º.(iii)KrF2Lewis structure:Orientations caused by repulsions between three lone pair and two bonding pair:HOOHSSOOOOOOFFSOFOOFKrFKrFF

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Molecular shape: linear with a 180º bond angle.(iv)PCl4+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 109.5ºA.7(b)(i)Nonpolar or weakly polar toward the slightly more electronegative carbon.(ii)(c)A.8(b)(i)O3is a bent molecule that has a small dipole as indicated by consideration of electrondensities and formal charge distributions.(ii)XeF2is a linear, nonpolar molecule.(iii)NO2is a bent, polar molecule.(iv)C6H14is a nonpolar molecule.A.9(b)In the order of increasing dipole moment: XeF2~ C6H14, NO2, O3A.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 mol5.0 g0.028 mol[A.3]180.16 gmnM===(ii)2322A6.022110molecules0.028 mol1.710moleculesmolNnN×===×A.12(b)ClPClClClPClClClClCHNClδ+δ−PSδ+δ−

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(i)78.11 g10.0 mol781. g[A.3]molmn M===(ii)gravity on MarsMarsweightFm g==()()221 kg781. g3.72 m s2.91 kg m s2.91 N1000 g−−=××==A.13(b)FmgpAA==()()22626242560 kg9.81 m s1 cm1 bar310N m310Pa2 cm10m10Pa30 bar10 bar−−−×==×=×=±A.14(b)()1 atm30 bar10 bar30 atm10 atm1.01325 bar±=±A.15(b)(i)551.0132510Pa222 atm22510Pa1 atm×=×(ii)1.01325 bar222 atm225 bar1 atm =A.16(b)/ C/ K273.1590.18273.15182.97[A.4]Tθ°=−=−= −182.97Cθ = −°A.17(b)The absolute zero of temperature is 0 K and 0 ºR. Using the scaling relationship 1 ºF / 1ºR (given in the exercise) and knowing the scaling ratios 5 ºC / 9 ºF and 1 K / 1 ºC, we find thescaling factor between the Kelvin scale and the Rankine scale to be:1F5C1 K5 K1R9F1C9R°°××=°°°°The zero values of the absolute zero of temperature on both the Kelvin and Rankine scales andthe value of the scaling relationship implies that:()()RR59/ K/ Ror/ R/ K95TTθθ=×°°=×Normal freezing point of water:()()RR99/ R/ K273.15491.6755491.67RTθθ°=×=×==°

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A.18(b)1 mol0.325 g0.0161 mol20.18 gn=×=()()()11333340.0161 mol8.314 J Kmol293.15 Kdm[A.5]2.00 dm10m1.9610Pa19.6 kPanRTpV−−−===×=A.19(b)[A.5]mRTpVnRTM==()()()311113whereis the mass density [A.2]0.6388 kg m8.314 J Kmol373.15 K0.124 kg mol124 g mol16.010PamRTRTMpVpρρ−−−−−=====×The molecular mass is four times as large as the atomic mass of phosphorus (30.97 g mol–1) sothe molecular formula is P4.A.20(b)1 mol7.05 g0.220 mol [A.3]32.00 gn=×=()()()11336360.220 mol8.314 J Kmol373.15.15 Kcm[A.5]100. cm10m6.8310Pa6.83 MPanRTpV−−−===×=A.21(b)22OCO0.25 moleand0.034 molenn==()()()22113OO3630.25 mol8.314 J Kmol283.15 Kcm[A.5]5.9 MPa100. cm10mnRTpV−−−===Since the ratio of CO2moles to O2moles is 0.034/0.25, we may scale the oxygen partial pressureby this ratio to find the partial pressure of CO2.()222COOCO0.0345.9 MPa0.80 MPa[1.6]6.7 MPa0.25pppp=×==+=BEnergyAnswers to discussion questionsB.2All 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 itskinetic energy,Ek. The law of conservation of energy tells us that the kinetic energy of an objectequals the work done on the object in order to change its motion from an initial (i) state ofvi= 0to a final (f) state ofvf=v. For an object of massmtravelling at a speedv,212k[B.8]Em=v.Thepotential energy,Epor more commonlyV, of an object is the energy it possesses as a result

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( )( )00Marsddttttgttgt=====∫∫vvvvof its position. For an object of massmat an altitudehclose to the surface of the Earth, thegravitational potential energy isEqn B.11 assigns the gravitational potential energy at the surface of the Earth,V(0), a valueequal zero andgis called theacceleration of free fall.The Coulomb potential energy describes the particularly important electrostatic interactionbetween two point chargesQ1andQ2separated by the distancer:( )1200in a vacuum [B.14,is the vacuum permittivity]4πQ QVrrεε=andEqn B.14 assigns the Coulomb potential energy at infinite separation,V(∞), a value equal tozero. Convention assigns a negative value to the Coulomb potential energy when the interactionis attractive and a positive value when it is repulsive. The Coulomb potential energy and theforce acting on the charges are related by the expressionF= −dV/dr.B.4Quantized energies are certain discrete values that are permitted for particles confined toa region of space.The quantization of energy is most important—in the sense that the allowed energies are widestapart—for particles of small mass confined to small regions of space. Consequently, quantizationis very important for electrons in atoms and molecules. Quantization is important for theelectronic states of atoms and molecules and for both the rotational and vibrational states ofmolecules.B.6The Maxwell distribution of speeds indicates that few molecules have either very low orvery high speeds. Furthermore, the distribution peaks at lower speeds when either thetemperature is low or the molecular mass is high. The distribution peaks at high speeds wheneither the temperature is high or the molecular mass is low.Solutions to exercisesB.1(b)a= d𝑣/dt=gsod𝑣=gdt. The acceleration of free fall is constant near the surfaceof the Mars.(i)()2[B.11]where9.81 m sVhmghg−==( )12rr0in a medium that has the relative permittivity(formerly, dielectric constant)4πQ QVrrεε ε=()()()()()212211122k1.0 s3.72 m s1.0 s3.72 m s0.0010 kg3.72 m s6.9 mJEm−−−=×===×=vv

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(ii)B.2(b)The terminal velocity occurs when there is a balance between the force exerted by thepull of gravity,mg=Vparticleρg=4/3πR3ρg, and the force of frictional drag, 6πηRs. It will be in thedirection of the gravitational pull and have the magnitudesterminal.3terminal2terminal4π6π329RgRsRgsρηρη==B.3(b)The harmonic oscillator solutionx(t) =Asin(ωt) has the characteristics that1/ 22ffd( )cos()where(/)ordxtAtkmmktωωωω====vxmin=x(t=nπ/ω,n=0,1,2...) = 0andxmax=x(t=(n+½)π/ω,n=0,1,2...) =AAtxminthe harmonic oscillator restoration force (Hooke’s law,Fx= –kfx, Brief illustration B.2)is zero and, consequently, the harmonic potential energy,V, is a minimum that is taken to equalzero while kinetic energy,Ek, is a maximum. As kinetic energy causes movement away fromxmin, kinetic energy continually converts to potential energy until no kinetic energy remains atxmaxwhere the restoration force changes the direction of motion and the conversion processreverses. We may easily find an expression for the total energyE(A) by examination of eitherxminorxmax.Analysis usingxmin:2222111kk,maxmaxf2220EEVEmmAk Aω=+=+===vWe begin the analysis that usesxmax, by deriving the expression for the harmonic potentialenergy.f()f0021f2dd[B.10]ddd( )xVxxVFxk xxVk xxV xk x= −===∫∫Thus,2211maxmaxfkmaxf22()and0VV xk AEEVVk A===+=+=B.4(b)21e2whereis the displacement from equilibrium [Brief illustration B.3]wkxxRR==−()()2112191912510 N m2010m1.0210N m1.0210Jw−−−−=××=×=×B.5(b)21k64where2 for C Hand76.03 g molfor the major isotopesEzezMφ+−=∆==1/ 221A22orwhere/zemzemMNmφφ∆=∆==vv()()()()()212211122k3.0 s3.72 m s3.0 s11.2 m s0.0010 kg11.2 m s63 mJEm−−−=×===×=vv

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()()()()1/ 21/21/21935512311/222551221.602210C2010VC VJ3.2103.210kgkg0.07603 kg mol/ 6.02210molkg ms3.2103.210m skg−−−−−×××××==×=××=×=×v()k220 kV40 keVEEzeeφ==∆=×=B.6(b)The work needed to separate two ions to infinity is identical to the Coulomb potentialdrop that occurs when the two ions are brought from an infinite separation, where the interactionpotential equals zero, to a separation ofr.In a vacuum:()()()()()22120000219181212112224[B.14]4π4π4ππ1.602210C3.6910Jπ 8.8541910JCm25010meeQ QeewVrrrrεεεε−−−−−−× −= −= −= −==×==××××In water:()()()()()()2212021920121211222[B.15] where78 for water at 25 C4π4πππ1.602210C4.7310Jπ 788.8541910JCm25010mrreeQ QeewVrrrrεεεεε ε−−−−−−× −= −= −= −===°×==×××××B.7(b)We will model a solution by assuming that the NaCl pair consists of the two point chargeions Na+and Cl–. The electric potential will be calculated along the line of the ions.()+++NaCl000NaClNaCl11[2.17][B.16]4π4π4πeeerrrrφφφεεε−−−−=+=+=−When+NaClrr−=, the electric potential equals zero in this model. Likewise,+NaClrr−=at every pointboth on the line perpendicular to the internuclear line and crossing the internuclear line at themid-point so electric potential equalszeroat every point on that perpendicular line.

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()1111sm1 mol/28.24 J Kmol1.228 J Kg22.99 gCCM−−−−==×=()1111ms63.55 g0.384 J Kg24.4 J KmolmolCC M−−−−==×=B.8(b)()()()()()()ethanol31ethanolethanolethanolethanolethanolethanoenergy dissipated by the electric circuit[B.20]1.12 A12.5 V172 s2.4110C sV s2.41 kJ[B.21]/UItUnCTUUTnCmCMφ−∆==∆∆=××=×=∆=∆∆∆∆==()()()3111l2.4110J150 g111.5 J Kmol/ 46.07 g mol6.64 K = 6.64C−−−×=×=°B.9(b)150.0 kJ[B.21]8.67 K or 8.67C5.77 kJ KUTC−∆∆===°B.10(b)1 mol10.0 g0.555 mol18.01 gn=×=()()()m11[B.21]0.555 mol75.2 J Kmol10.0 K417 JUCTnCT−−∆=∆=∆==B.11(b)B.12(b)B.13(b)()()51631mmm331.0010Pa18.02 g mol10m[B.23]1.81 J mol0.997 g cmcmpMHUpVρ−−−−××−====B.14(b)22H O(l)H O(s)SS>B.15(b)22H O(l, 100C)H O(l, 0C)SS°°>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 thedownward 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 equalpressures on each side of the piston. Increase the pressure on one side of the piston and it movesaway from that side.In the Bohr atomic model of 1913 there is a balance between the electrostatic attraction of anelectron 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)()//ee[B.25a]iji jkTkTijNNεεε−−−∆==(i)()()()(){}1912312.0 eV1.602 10J eV/1.381 10J K200 K5121e4.210NN−−−−−××××−==×(ii)()()()(){}1912312.0 eV1.602 10J eV/1.381 10J K2000 K621e9.210NN−−−−−××××−==×B.18(b)()upper/0lowerlimlim e[B.25a]e1kTTTNNε−∆−→∞→∞ ===In the limit of the infinite temperature both the upper and the lower state are equally occupied.B.19(b)()()349124upperlower6.62610J s10.010s6.6310Jhεεεν−−−∆=−==××=×()()(){}242316.63 10J /1.381 10J K293 Kupper/lowere[B.25a]e0.998klTNNε−−−−×××−∆===The ratioNupper/Nlowerindicates 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 moremolecules have the requisite speed and corresponding kinetic energy to promote excitation andbond breakage during collisions at the high temperatures.B.21(b)()1/ 2mean/[B.26]TM∝v()()()()()()1/ 21/ 2mean2221/ 2mean1111/ 2meanmean//303 K303 K1.02293 K293 KTTMTTTTM== ==vvvvB.22(b)()1/ 2mean/[2.26]TM∝v()()()()()()1/ 21/ 2mean2211/ 2mean1211/ 21mean21mean2//H401.2 g mol14.11Hg2.016 g molMTMMMMTM−−== ==vvvvB.23(b)A gaseous helium atom has three translational degrees of freedom (the components ofmotion in thex,y, andzdirections). Consequently, the equipartition theorem assigns a mean

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m3for water vapourURT=m3for Pb(s)URT=energy of3 2kTto each atom. The molar internal energy is()()()111333mA2221mm=8.3145 J molK303 K3.78 kJ mol1 mol3.78 kJ10.0 g9.45 kJ4.00 gmolUN kTRTUnUmMU−−−−=======B.24(b)A solid state lead atom has three vibrational quadratic degrees of freedom (thecomponents of vibrational motion in thex,y, andzdirections). Its potential energy also has aquadratic form in each direction becauseV∝(x–xeq)2. There is a total of six quadratic degreesof freedom for the atom because the atoms have no translational or rotational motion.Consequently, the equipartition theorem assigns a mean energy of6 23kTkT=to each atom.This is the law of Dulong and Petit. The molar internal energy is()()()111mA1mm33= 3 8.3145 J molK293 K7.31 kJ mol1 mol7.31 kJ10.0 g0.353 kJ207.2 gmolUN kTRTUnUmMU−−−−=======B.25(b)See exercise B.23(b) for the description of the molar internal energy of helium.()()321111m33,m228.3145 J molK12.47 J molKVRTUCRTT−−−−∂∂=====∂∂B.26(b)(i)Water, being a bent molecule, has three quadratic translational and three quadraticrotational degrees of freedom. So,()()1111m,m333 8.3145 J molK24.94 J molKVRTUCRTT−−−−∂∂=====∂∂(ii)See exercise B.24(b) for the description of the molar internal energy of lead.()()1111m,m333 8.3145 J molK24.94 J molKVRTUCRTT−−−−∂∂=====∂∂CWavesAnswers to discussion questionsC.2The sound of a sudden ‘bang’ is generated by sharply slapping two macroscopic objectstogether. This creates a sound wave of displaced air molecules that propagates away from thecollision withintensity, defined to be the power transferred by the wave through a unit areanormal to the direction of propagation. Thus, the SI unit of intensity is the watt per meter squared(2W m−) and ‘loudness’ increases with increasing intensity. The ‘bang’ creates a shell ofcompressed 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 pressureshell into the adjacent, lower pressure shell. Molecular collisions quickly cause momentumtransfer from the high density to the low density shell and the effective propagation of the highdensity shell. The regions over which pressure and density vary during sound propagation aremuch wider than the molecular mean free path because sound is immediately dissipated bymolecular collisions in the case for which pressure and density variations are of the order of themean free path.Solutions to exercisesC.1(b)8181benzener3.0010m s[C.4]1.9710m s1.52ccn−−×===×C.2(b)6121110μm[C.5]2.78μm3600 cm10cmλν−===811411463.0010m s[C.1]1.0810s1.0810Hz2.7810mcνλ−−−×===×=××Integrated activitiesF.2The plots of Problem F.1 indicate that as temperature increases the peak of theMaxwell–Boltzmann distribution shifts to higher speeds with a decrease in the fraction of molecules thathave low speeds and an increase in the fraction that have high speeds. Thus, justifying summarystatements like 'temperature is a measure of the average molecular speed and kinetic energy ofgas molecules', 'temperature is a positive property because molecular speed is a positivequantity', 'the absolute temperature of 0 K is unobtainable because the area under the plots ofProblem F.1 must equal 1'.

Physical Chemistry: Thermodynamics, Structure, and Change Tenth Edition Solution Manual - Page 16 preview image

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1The properties of gases1AThe perfect gasAnswers to discussion questions1A.2The partial pressure of a gas in a mixture of gases is the pressure the gas would exert if itoccupied alone the same container as the mixture at the same temperature. Dalton’s law is alimiting law because it holds exactly only under conditions where the gases have no effectupon each other. This can only be true in the limit of zero pressure where the molecules ofthe gas are very far apart. Hence, Dalton’s law holds exactly only for a mixture of perfectgases; for real gases, the law is only an approximation.Solutions to exercises1A.1(b)The perfect gas law [1A.5] ispV=nRT,implying that the pressure would bep=nRTVAll quantities on the right are given to us exceptn, which can be computed from the givenmass of Ar.n=25 g39.95−1g mol=0.626 molsop=(0.626 mol)×(8.31×10−2dm3bar K−1mol−1)×(30+273) K1.5 dm3=10.5barSo no, the sample would not exert a pressure of 2.0 bar.1A.2(b)Boyle’s law [1A.4a] applies.pV= constantsopfVf=piViSolve for the initial pressure:(i)pi=pfVfVi=(1.97 bar)×(2.14 dm3)(2.14+1.80) dm3=1.07 bar(ii)The original pressure in Torr ispi=(1.07 bar)×1 atm1.013 bar ×760 Torr1 atm =803 Torr1A.3(b)The relation between pressure and temperature at constant volume can be derived from theperfect gas law,pV=nRT[1A.5]sop∝TandpiTi=pfTfThe final pressure, then, ought to bepf=piTfTi=(125 kPa)×(11+273)K(23+273)K=120 kPa1A.4(b)According to the perfect gas law [1.8], one can compute the amount of gas from pressure,temperature, and volume.pV=nRTson=pVRT=(1.00 atm)×(1.013×105Pa atm−1)×(4.00×103m3)(8.3145 J K−1mol−1)×(20+273)K=1.66×105molOnce this is done, the mass of the gas can be computed from the amount and the molarmass:m=(1.66×105mol)×(16.04−1g mol)=2.67×106g=2.67×103kg1A.5(b)The total pressure is the external pressure plus the hydrostatic pressure [1A.1], making thetotal pressure1

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