(Solution Manual) Atkins’ physical chemistry

Atkins, 10

Chemistry

Solution Manual

10th Edition

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Solution Manual for Atkins’ physical chemistry, The downloadable file consists of 551 pages, and the total size of the file is 10.52 MB. Purchase now only for highly discounted price.

Author(s): Atkins

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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 wo uld exert if it occupied alone the same container as the mixture at the same temperature. Dalto n’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 mo lecules of the gas are very far apart. Hence, Dalton’s law holds exactl y 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 −1g mol=0.626 mol so p=(0.626 mol) ×(8.31×10−2dm3 bar K−1 mol−1)×(30+273) K 1.5 dm3=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 pfVf = piVi Solve for the initial pressure: (i) pi=pfVf Vi=(1.97 bar) ×(2.14dm3) (2.14+1.80)dm3=1.07 bar (ii) The original pressure in Torr is pi=(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 f rom the perfect gas law, pV = nRT [1A.5] so p∝Tandpi Ti=pf Tf The final pressure, then, ought to be pf=piTf Ti=(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) ×(1.013×105Pa atm−1)×(4.00×103m3) (8.3145 J K−1mol−1)×(20+273)K=1.66×105mol Once this is done, the mass of the gas can be computed from the a mount and the molar mass: m=(1.66×105mol) ×(16.04 −1g mol)=2.67×106g=2.67 ×103 kg 1A.5(b) The total pressure is the external pressure plus the hydrostatic pressu re [1A.1], making t he total pressure 1 p = pex + ρgh . Let pex be the pressure at the top of the straw and p the pressure on the surface of the liquid (atmospheric pressure). Thus the pressure difference is p−pex=ρgh=(1.0−3gcm)×1 kg 103 g×1 cm 10−2 m  3 ×(9.81 m s−2)×(0.15m) =1.5×103Pa=1.5×10−2 atm 1A.6(b) The pressure in the apparatus is given by p = pex + ρgh [1A.1] where pex = 760 Torr = 1 atm = 1.013 ×105 Pa, and ρgh=13.55 g cm−3×1 kg 103 g  ×1 cm 10−2 m  3 ×0.100 m ×9.806 m s−2=1.33×104 Pa p=1.013 ×105 Pa+1.33 ×104 Pa=1.146 ×105 Pa = 115 kPa 1A.7(b) Rearrange the perfect gas equation [1A.5] to give R=pV nT=pVm T All gases are perfect in the limit of zero pressure. Therefore the value of pVm/T extrapolated to zero pressure will give the best val ue of R. The molar mass can be introduced through pV=nRT =m MRT which upon rearrangement gives M=m VRT p=ρRT p The best value of M is obtained from an extrapolation of ρ/p versus p to zero pressure; the intercept is M/RT. Draw up the following table : From Figure 1 A.1(a), R=lim p→0pVm T  =0.082 062 dm3 atm K−1 mol−1 Figure 1A.1 (a) p/atm (pVm/T)/(dm3 atm K–1 mol–1) (ρ/p)/(g dm–3 atm–1) 0.750 000 0.082 0014 1.428 59 0.500 000 0.082 0227 1.428 22 0.250 000 0.082 0414 1.427 90 2 (b) From Figure 1 A.1(b), lim p→0ρ p  =1.427 55 g dm-3 atm−1 M=lim p→0RTρ p  =(0.082062 dm3 atm K−1 mol−1)×(273.15 K) ×(1.42755 g dm-3 atm−1) =31.9988 g mol−1 The value obtained for R deviates from the accepted valu e by 0.005 per cent, better than can be expected from a linear extrapolation from three data points. 1A.8(b) The mass density ρ is related to the molar volume Vm by Vm=V n=V m×m n=M ρ where M is the molar mass. Putting this relation into the perfect gas law [1A.5] yields pVm = RT so pM ρ=RT Rearranging this result gives an expression for M; once we know the molar mass, we can divide by the molar mass of phosphorus atoms to determine the number of a toms per gas molecule. M=RTρ p=(8.3145 Pa m3 −1mol)×[(100 +273)K] ×(0.6388kg m−3) 1.60 ×104 Pa = 0.124 kg mol−1=124 g mol−1 The number of atoms per molecule is 124−1gmol 31.0−1gmol=4.00 suggesting a f ormula of P4. 1A.9(b) Use the perfect gas equation [1A.5] to compute the amount; then convert to mass. pV = nRT so n=pV RT We need the partial pressure of water, which is 53 per cent of the equilibr ium vapour pressure at the given temperature and standard pres sure. (We must look it up in a handbook like the CRC or other resource such as the NIST Chemistry WebBook.) p=(0.53)×(2.81 ×103Pa)=1.49×103Pa 3 so n=(1.49×103 Pa)×(250 m3) (8.3145 J K−1 mol−1)×(23+273)K=151 mol and m=(151 mol) ×(18.0 −1gmol)=2.72×103 g=2.72 kg 1A.10(b) (i) The volume occupied by each gas is the same, since each completely fills the contai ner. Thus solving for V we have (assuming a perfect gas, eqn. 1A.5) V=nJ RT pJ We have the pressure of neon, so we focus on it nNe=0.225 g 20.18 g mol−1=1.115 ×10−2 mol Thus V=1.115×10−2 mol ×8.3145 Pa m3 K−1 mol−1×300 K 8.87 ×103 Pa=3.14 ×10−3 m3=3.14 dm3 (ii) The total pressure is determined from the total amount of gas, n=nCH4+nAr+nNe. nCH4=0.320 g 16.04 g mol−1=1.995×10−2 molnAr=0.175 g 39.95 g mol−1=4.38 ×10−3 mol n=1.995+0.438 +1.115 ( ) ×10−2mol=3.55 ×10−2 mol and p=nRT V=3.55 ×10−2 mol ×8.3145 Pa m3 K−1 mol−1×300 K 3.14 ×10−3 m3 =2.82 ×104 Pa=28.2 kPa 1A.11(b) This exercise uses the formula, M=ρRT p, which was developed and used in Exercis e 1A.8(b). First the density must first be calculated. ρ=33.5×10−3g 250cm3×103 cm3 dm3  =0.134g dm−3 M=(0.134g dm−3)×(62.36−1dm3torrK−1mol)×(298K) 152torr=16.4 g mol−1 1A.12(b) This exercise is similar to Exercise 1.12(a) in that it uses the definiti on of absolute zero as that temperature at which the volume of a sample of gas would become zero if t he substanc e remained a gas at low temperatures. The solution uses the experimen tal fact that the volume is a linear function of the Celsius temperature: V = V0 + αθ where V0 = 20.00 dm3 and α = 0.0741 dm3 °C–1 . At absolute zero, V = 0 = V0 + αθ so θ(abs.zero) = −V0 α= −20.00 dm3 0.0741 dm3 ¡C−1=Ğ270°C which is close to the accepted value of –273C. 1A.13(b) (i) Mole fractions are xN=nN ntotal[1A.9] =2.5 mol (2.5+1.5) mol=0.63 Similarly, xH=0.37 According to the perfect gas law ptotV = ntotRT 4 so ptot=ntotRT V=(4.0 mol) ×(0.08206 dm3 atm mol−1 K−1)×(273.15 K) 22.4 dm3=4.0 atm (ii) The partial pressures are pN=xNptot=(0.63)×(4.0 atm) =2.5 atm and pH=(0.37)×(4.0 atm) =1.5 atm (iii) p=pH+pN[1A.10]=(2.5+1.5) atm =4.0 atm Solutions to problems 1A.2 Solving for n from the perfect gas equat ion [1A.5] yields pVnRT=. From the definition of molar mass mnM=, hence ρ=m V=Mp RT. Rearrangement yields the desired relation, namely RTpMρ=. Therefore, for ideal gases p ρ=RT M and M=RT p/ρ. For real gases, find the zero -pressure limit of p ρby plotting it against p. Draw up the following tab le. p/(kPa) 12.223 25.20 36.97 60.37 85.23 101.3 ρ/(kg m–3) 0.225 0.456 0.664 1.062 1.468 1.734 p/ρ 103 m2 s−2 54.3 55.3 55.7 56.8 58.1 58.4 Bear in mind that 1 kPa = 103 kg m–1 s–2. p ρ is plotted in Figure 1 A.2. A straight line fits the data rather well. The extrapolation to p = 0 yields an intercept of 54.0 ×103 m2 s–2 . Then M=RT 5.40×104 m2 s−2=(8.3145 J K−1 mol−1)×(298.15K) 5.40×104m2 s−2 =0.0459 kg mol−1=45.9−1gmol Figure 1A.2 5 Comment. This method of the determination of the molar masses of gaseous com pounds is due to Cannizarro who presented it at the Karlsruhe Congress of 1860. That conference ha d been called to resolve the problem of the determination of the molar masses of at oms and molecules and the molecular formulas of compounds. 1A.4 The mass of displaced gas is ρV, where V is the volume of the bulb and ρ is the density of the displaced gas. The balance condition for the two gases is m(bulb) = ρV(bulb) and m(bulb) = ρ′V(bulb) which implies that ρ = ρ′. Because [Problem 1.2] ρ=pM RT the balance condition is pM = p′M′ , which implies that ′M=p ′p×M This relation is valid in the limit of zero pressure (for a gas behaving perfectly). In experiment 1, p = 423.22 Torr, p′ = 327.10 Torr; hence ′M=423.22Torr 327.10Torr×70.014gmol−1=90.59gmol−1 In experiment 2, p = 427.22 Torr, p′ = 293.22 Torr; hence ′M=427.22Torr 293.22Torr×70.014gmol−1=102.0gmol−1 In a proper series of experiments one should reduce the pressure (e.g. by adjus ting the balanced weight). Ex periment 2 is closer to zero pressure than experiment 1, so it is more likely to be close to the true value: ′M≈102gmol−1 The molecules CH 2FCF3 and CHF 2CHF2 have molar mass of 102 g mol–1. Comment . The substantial difference in molar mass between the two experi ments ought to make us wary of confidently accepting the result of Experiment 2, even if i t is the more likely estimate. 1A.6 We assume that no H 2 remains after the reaction has gone to completion. The balanced equation is N 2 + 3 H2 → 2 NH3 . We can draw up t he following table N2 H2 NH3 Total Initial amount n n′ 0 n + n′ Final amount n−1 3′n 0 2 3′n n+1 3′n Specifically 0.33 mol 0 1.33 mol 1.66 mol Mole fraction 0.20 0 0.80 1.00 p=nRT V=(1.66mol) ×(0.08206 dm3 atm mol−1 K−1)×(273.15K) 22.4dm3  =1.66 atm p(H2) = x(H2)p = 0 p(N2) = x(N2)p = 0.20 × 1.66 atm = 0.33 atm p(NH3) = x(NH3)p = 0.80 × 1.66 atm = 1.33 atm 1A.8 The perfect gas law is pV = nRT so n=pV RT At mid-latitudes n=(1.00atm) ×{(1.00dm2)×(250 ×10−3 cm)/10 cm dm−1} (0.08206dm3atmK−1mol−1)×(273K)=1.12 ×10−3 mol In the ozone hole n=(1.00atm) ×{(1.00dm2)×(100 ×10−3 cm)/10cm dm−1} (0.08206dm3atm K−1mol−1)×(273 K)=4.46 ×10−4 mol The corresponding concentrations are 6 n V=1.12 ×10−3 mol (1.00dm2)×(40×103 m)×(10dm m−1)=2.8×10−9 moldm−3 and n V=4.46 ×10−4 mol (1.00dm2)×(40×103 m)×(10dm m−1)=1.1×10−9 moldm−3 respectively. 1A.10 The perfect gas law [1A.5] can be rearranged to n=pV RT The volume of the balloon is V=4π 3r3=4π 3×(3.0 m)3=113 m3 (a) n=(1.0atm) ×(113×103dm3) (0.08206 dm3 atm mol−1 K−1)×(298K)=4.62×103mol (b) The mass that the balloon can lift is the difference between the mass of displaced air and the mass of the balloon. We assume that the mass of the balloon is essenti ally that of the gas it encloses: m=m(H2)=nM(H2)=(4.62×103mol) ×(2.02−1gmol)=9.33×103g Mass of displ aced air =(113m3)×(1.22−3kgm)=1.38×102kg Therefore, the mass of the maximum payload is 138kg −9.33kg =1.3×102kg (c) For helium, m=nM(He) =(4.62×103mol) ×(4.00g mol−1)=18kg The maximum payload is now 138kg −18kg =1.2×102kg 1A.12 Avogadro’s principle states that equal volumes of gases contain equa l amounts (moles) of the gases, so the volume mixing ratio is equal to the mole fraction. The definition of partial pressures i s pJ = xJp . The perfect gas law is pV=nRTsonJ V=pJ RT=xJp RT (a) n(CCl3F) V=(261 ×10−12)×(1.0atm) (0.08206dm3atmK−1mol−1)×(10+273)K=1.1×10−11moldm-3 and n(CCl2F2) V=(509 ×10−12)×(1.0atm) (0.08206dm3atmK−1mol−1)×(10+273)K=2.2×10−11moldm-3 (b) n(CCl3F) V=(261 ×10−12)×(0.050atm) (0.08206dm3 atm K−1mol−1)×(200K)=8.0×10−13moldm-3 and n(CCl2F2) V=(509 ×10−12)×(0.050atm) (0.08206dm3atmK−1mol−1)×(200K)=1.6×10−12moldm-3 1B The kinetic model Answers to discussion questions 1B.2 The formula for the mean free path [eqn 1B.13] is λ=kT σp In a container of constant volume, the mean free path is directly proporti onal to temperature and inversely proportional to pressure. The former dependence can be rationali zed by noting that the faster the molecules travel, the farther on average t hey go between collisions. The latter also makes sense in that the lower the pressure, the less frequent are collisions, 7 and therefore the further the average distance between collisions. Perhaps m ore fundamental than either of these considerations are dependences on size. As pointed out i n the text, the ratio T/p is directly proportional t o volume for a perfect gas, so the average distance between collisions is directly proportional to the size of the container hold ing a set number of gas molecules. Finally, the mean free path is inversely proportion al to the size of the molecules as given by the collision cross section (and therefore inversely proportional t o the square of the molecules’ radius). Solutions to exercises 1B.1(b) The mean speed is [1B.8] vmean=8RT πM  1/2 The mean translational kinetic energy is Ek=1 2mv2=1 2m v2=1 2mvrms2=m 23RT M  [1B.3] =3kT 2 The ratios of species 1 to specie s 2 at the same temperature are vmean,1 vmean,2=M2 M1  1/2 and Ek1 Ek2=1 (i) vmean,H2 vmean,Hg=200.6 4.003  1/2 =7.079 (ii) The mean translation kinetic energy is independent of molecular mass and depends upon temperature alone! Consequently, because the mean transla tional kinetic energy for a gas is proportional to T, the ratio of mean tr anslational kinetic energies for gases at the same temperature always equals 1. 1B.2(b) The root mean square speed [1B. 3] is vrms=3RT M  1/2 For CO 2 the molar mass is M = (12.011 + 2 ×15.9994) ×10–3 kg mol–1 = 44.010 ×10–3 kg mol–1 so vrms=3(8.3145 J K−1 mol−1)(20+273) K 44.01 ×10−3 kg mol−1  1/2 =408 m s−1 For He vrms=3(8.3145 J K−1 mol−1)(20+273) K 4.003 ×10−3 kg mol−1  1/2 =1.35 ×103 m s−1=1.35 km s−1 1B.3(b) The Maxwell-Boltzmann distribution of speeds [1B.4] is f(v)=4πM 2πRT  3/2 v2e−Mv2/2RT and the fraction of molecules that have a speed between v and v+dv is f(v)dv. The fraction of molecules to have a speed in the range between v1 and v2 is, therefore, f(v)dv v1v2∫. If the range is relatively small, however, such that f(v) is nearly constant over that range, the integral may be approximated by f(v)∆v, where f(v) is evaluated anywhere within the range and ∆v = v2 – v1 . Thus, we have, with M = 44.010 ×10–3 kg mol–1 [Exericse 1B.2(b)], 2 13/231 12 1144.010 10 kg mol( )d ( ) 4 (402.5 m s )2 (8.3145 J K mol )(400 K)v vfv v fv v ππ−− − −− ×≈ ∆=  ∫ 8 ×exp −(44.010 ×10−3 kg mol−1)(402.5 m s−1)2 2(8.3145 J K−1 mol−1)(400 K)  ×(405 −400) m s−1 =0.0107, just over 1% 1B.4(b) The most probable, mean, and mean relative speeds are, respectively vmp=2RT M  1/2 [1B.9] vmean=8RT πM  1/2 [1B.8] vrel=8RT πµ  1/2 [1B.10b] The temperature is T = (20+273) K = 293 K. so vmp=2(8.3145 J K−1 mol−1)(293 K) 2×1.008 ×10−3 kg mol−1  1/2 =1.55 ×103 m s−1 and vmean=8(8.3145 J K−1 mol−1)(293 K) π(2×1.008 ×10−3 kg mol−1)  1/2 =1.75 ×103 m s−1 For many purposes, air can be considered as a gas with an average molar mas s of 29.0 g mol–1 . In that case, the reduced molar mass [1B.10b] is µ=MAMB MA+MB=(29.0 g mol−1)(2×1.008 g mol−1) (29.0 +2×1.008) g mol−1=1.88 g mol−1 and vrel=8(8.3145 J K−1 mol−1)(293 K) π(1.88 ×10−3 kg mol−1)  1/2 =1.81×103 m s−1 Comment . One computes the average molar mass of air just as one computes the avera ge molar mass of an isotopically mixed element, namely by taking an average of the species that have diffe rent masses weighted by their abundances. Comment . Note that vrel and vmean are very nearly equal. This is because the reduced mass between two very dissimilar species is nearly equal to the mass of the lig hter species (in this case, H 2). 1B.5(b) (i) vmean=8RT πM  1/2 [1B.8] =8(8.3145 J K−1 mol−1)(298 K) π(2×14.007 ×10−3 kg mol−1)  1/2 =475 m s−1 (ii) The mean free path [1B.13] is λ=kT σp=kT πd2p=(1.381 ×10−23 J K−1)(298 K) π(395 ×10−12 m)2(1×10−9 Torr)×1 Torr 133.3 Pa =6.3×104 m=63 km The mean free path is much larger than the dimensions of the pumping appara tus used to generate the very low pressure. (iii) The collision frequency is related to the mean free path and relat ive mean speed by [1B.12] λ=vrel z so z=vrel λ=21/2vmean λ [1B.10a] z=21/2(475 m s−1) 6.3×104 m=1.1×10−2 s−1 1B.6(b) The collision diameter is related to the collision cross section by σ = πd2 so d = (σ/π)1/2 = (0.36 nm2/π)1/2 = 0.34 nm . The mean free path [1B.13] is λ=kT σp Solve this expression for the pressure and set λ equal to 10 d: 9 p=kT σλ=(1.381×10−23 J K−1)(293 K) 0.36×(10−9 m)2(10×0.34 ×10−9 m)=3.3×106 J m−3=3.3 MPa Comment . This pressure works out to 33 bar (about 33 atm), conditions under which the assumption of perfect gas behavior and kinetic model applicability at leas t begins to come into question. 1B.7(b) The mean free path [1B.13] is λ=kT σp=(1.381 ×10−23 J K−1)(217 K) 0.43×(10−9 m)2(12.1 ×103 Pa atm−1)=5.8×10−7 m Solutions to problems 1B.2 The number of molecules that escape in unit time is the number per unit time t hat would have collided with a wall section of area A equal to the area of the small hole. This quantity is readily expressed in terms of ZW, the collision flux (collisions per uni t time with a unit area), given in eqn 19A.6. That is, dN dt= −ZWA=−Ap (2πmkT)1/2 where p is the (constant) vapour pressure of the solid. The change in the number o f molecules inside the cell in an interval t∆ is therefore WN ZAt∆= − ∆ , and so the mass loss is ∆w=m∆N= −Apm 2πkT  1/2 ∆t= −ApM 2πRT  1/2 ∆t Therefore, the vapour pressure of the substance in the cell is 1/22RT wpAt Mπ −∆  = ×   ∆ For the vapour pressure of germanium p=43×10−9 kg π(0.50 ×10−3 m)(7200 s)  ×2π(8.3145 J K−1 mol−1)(1273 K) 72.64 ×10−3 kg mol−1  1/2 =7.3×10−3 Pa=7.3 mPa 1B.4 We proceed as in Justification 1B.2 except that, instead of taking a product of three one - dimensional distributions in order to get the three -dimensional distribution, we m ake a product of two one -dimensional distributions. f(vx,vy)dvxdvy=f(vx2)f(vy2)dvxdvy=m 2πkT  e−mv2/2kTdvxdvy where v2=vx2+vy2. The probability f(v)dv that the molecules have a two -dimensional speed, v, in the range v to v + dv is the sum of the probabilities that it is in any of the area elements dvxdvy in the circular shell of radius v. The sum of the area elements is the area of the circular shell of radius v and thickness dv which is π(ν+dν)2 – πν2 = 2πνdν . Therefore , f(v)=m kT  ve−mv2/2kT=M RT  ve−Mv2/2RT M R=m k   The mean speed is determined as vmean=vf(v)dv 0∞∫=m kT  v2e−mv2/2kTdv 0∞∫ Using integral G.3 from the Resource Section yields vmean=m kT  ×π1/2 4  ×2kT m  3/2 =πkT 2m  1/2 =πRT 2M  1/2 1B.6 The distribution [1B.4] is 10 f(v)=4π3/2M 2πRT  v2e−Mv2/2RT. The proportion of molecules with speeds less than vrms is P=f(v)dv 0vrms∫=4π3/2M 2πRT  v2e−Mv2/2RTdv 0vrms∫ Defining a≡R/2RT, P=4π3/2a π  v2e−av2dv 0vrms∫= −4π3/2a π  d dae−av2dv 0vrms∫ Defining 2 2 1/2. Then, d d andav v a χχ−≡= P= −4π3/2a π  d da1 a1/2e−χ2dχ 0vrmsa1/2 ∫{ } = −4π3/2a π  −1 23/21 a()e−χ2dχ 0vrmsa1/2 ∫+1/21 a()d dae−χ2dχ 0vrmsa1/2 ∫     Then we use the error function [Integral G.6]: e−χ2dχ 0vrmsa1/2 ∫=π1/2/2( )erf(vrmsa1/2). d dae−χ2dχ 0vrmsa1/2 ∫=dvrmsa1/2 da  ×(e−avrms2)=1 2c a1/2  e−avrms2 where we have used d dzf(y)dy 0z∫=f(z) Substituting and cancelling we obtain P=erf(vrmsa1/2)−2vrmsa1/2/π1/2( ) e−avrms2 Now vrms=3RT M  1/2 so vrmsa1/2=3RT M  1/2 ×M 2RT  1/2 1/2 =3 2  1/2 and P=erf3 2      −1/26 π  e−3/2=0.92−0.31=0.61 Therefore , (a) 1 – P = 39% have a speed greater than the root mean square speed. (b) P = 61% of the molecules have a speed less than the root mean square speed . (c) For the proportions in terms of the mean speed vmean, replace vrms by vmean=8kT/πm ( )1/2=8/3π( )1/2vrms so vmeana1/2 = 2/π1/2 . Then P=erf(vmeana1/2)−2vmeana1/2/π1/2( ) ×(e−av2 mean) =erf 2/ π1/2( ) −4/π( )e−4/π=0.889−0.356=0.533 That is, 53% of the molecules have a speed less than the mean, and 47% have a speed greater than the mean. 1B.8 The average is obtained by substituting the distribution (eqn 1B.4) into eqn 1B.7: vn=vnf(v)dv 0∞∫=4πM 2πRT  3/2 vn+2e−Mv2/2RTdv 0∞∫ For even values of n, use Integral G.8: vn=4πM 2πRT  3/2(n+1)!! 2n+4 2  2RT M  n+2 2  2πRT M  1/2 =(n+1)!!RT M  n 2   where (n+1)!! = 1 × 3 × 5 ... × (n+1) Thus vn1/n=(n+1)!!RT M      1/2 even n 11 For odd values of n, use Integral G.7: vn=4πM 2πRT  3/2n+1 2  ! 22RT M  n+3 2   =2 π1/22RT M  n/2 Thus vn1/n=2 π1/22RT M  n/2    1/n =21/n π1/2n2RT M  1/2 odd n Question . Show that these expressions reduce to vmean and vrms for n = 1 and 2 respectively. 1B.10 Dry atmospheric air is 78.08% N 2, 20.95% O 2, 0.93% Ar, 0.04% CO 2, plus traces of other gases. Nitrogen, oxygen, and carbon dioxide contribute 99.06% of the molecules in a volume with each molecule contributing an average rotational energy equal to kT. (Linear molecules can rotate in two dimensions, contributing two “quadratic terms” of r otational energy, or kT by the equipartition theorem [Topic B.3(b)]. The rotational energy density is given by ρR=ER V=0.9906NεR V=0.9906NkT V=0.9906p =0.9906(1.013 ×105 Pa)=1.004 ×105 J m−3=0.1004 J cm−3 The total energy density is translational plus rotational (vibrational energy contributing negligibly): ρtot=ρT+ρR=0.15 J cm−3+0.10 J cm−3=0.25 J cm−3 1B.12 The fraction of molecules (call it F) between speeds a and b is given by F(a,b)=f(v)dv ab∫ where f(v) is given by eqn 1B.4. This integral can be approximated by a sum over a discrete set of velocity values. For convenience, let the velocities vi be evenly spaced within the interval such that vi+1 = vi + ∆v: (,) () Δi Fab fv v ≈∑ On a spreadsheet or other mathematical software, make a column of velocity values and then a column for f(v) [1B.4] at 300 K and at 1000 K. Figure 1 B.1 shows f(v) plotted against v for these two temperatures. Each curve is labeled with the numerical value of T/K, and each is shaded under the curve between the speeds of 100 and 200 m s–1. F(a,b) is simply the area under the curve between v = a and v = b. One should take some care to avoid double counting at the edges of the interval, that is, not including both endpoints of the i nterval with full weight. example, beginning the sum with the area under the curve at those spe eds. Using a spreadsheet that evaluates f(v) at 5-m s–1 intervals, and including points at both 100 and 200 m s–1 with half weight, F(100 m s–1, 200 m s–1) ≈ 0.281 at 300 K and 0.066 at 1000 K. Figure 1B.1 12 1C Real gases Answers to discussion questions 1C.2 The critical constants represent the state of a system at which the distinctio n between the liquid and vapour phases disappears. We usually describe this situat ion by saying that above the critical temperature the liquid phase cannot be produced by the applicati on of pressure alone. The liquid and vapour phases can no longer coexist, though supercrit ical fluids have both liquid and vapour characteristics. 1C.4 The van der Waa ls equation is a cubic equation in the volume, V. Every cubic equation has some values of the coefficients for which the number of real roots pas ses from three to one. In fact, any equation of state of odd degree n > 1 can in principle account for critical behavior because for equations of odd degree in V there are necessarily some values of temperature and pressure for which the number of real roots of V passes from n to 1. That is, the multiple values of V converge from n to 1 as the temperature approache s the critical temperature. This mathematical result is consistent with passing from a two phase region (more than one volume for a given T and p) to a one phase region (only one V for a given T and p), and this corresponds to the observed experimental res ult as the critical point is reached. Solutions to exercises 1C.1(b) The van der Waals equation [1C.5a] is p=nRT V−nb−an2 V2 From Table 1C.3 for H 2S, a = 4.484 dm6 atm mol–1 and b = 0.0434 dm3 mol–1. (i) p=(1.0 mol) ×(0.08206 dm3 atm mol−1 K−1)×(273.15 K) 22.414 dm3−(1.0 mol) ×(4.34×10−2 dm3 mol−1) −(4.484 dm6 atm mol−2)×(1.0 mol)2 (22.414 dm3)2=0.99 atm (ii) p=(1.0mol) ×(0.08206 dm3 atm mol−1 K−1)×(500K) 0.150dm3−(1.0mol) ×(4.34×10−2dm3 mol−1) −(4.484 dm3 atm mol−1)×(1.0 mol)2 (0.150 dm3)2=190 atm (2 sig. figures) 1C.2(b) The conversions needed are as follows: 1 atm = 1.013 ×105 Pa, 1 Pa = 1 kg m–1 s–2, 1 dm6 = (10–1 m)6 = 10–6 m6, 1 dm3 = 10–3 m3. Therefore, a=1.32 atm dm6 mol−2×1.013 ×105 kg m−1 s−2 1 atm×10−6 m6 dm6 =1.34 ×10−1 kg m5 s−2 mol−2 and b=0.0426 dm3 mol−1×10−3 m3 dm3=4.26 ×10−5 m3 mol−1 1C.3(b) The compression factor Z is [1C.1] Z=Vm Vm°=pVm RT (i) Because Vm=Vmο+0.12 Vmο=(1.12)Vmο, we have Z=1.12 (ii) The molar volume is 13 Vm=(1.12)Vmο=(1.12)×RT p   =(1.12)×(0.08206 dm3 atm mol−1 K−1)×(350K) 12atm  =2.7 dm3 mol−1 Since Vm>Vmo repulsive forces dominate. 1C.4(b) (i) According to the perfect gas law Vmo=RT p=(8.3145 J K−1 mol−1)×(298.15K) (200bar) ×(105 −1Pabar)×1 dm 10−1 m  3 =0.124 dm3 mol−1 (ii) The van der Waals e quation [1C.5b] is a cubic equation in Vm. Cubic equations can be solved analytically. However, this approach is cumbersome, so we proceed as in Example 1C.1. The van der Waals equation is rearranged to the cubic form Vm3−b+RT p  Vm2+a p  Vm−ab p=0 or x3−b+RT p  x2+a p  x−ab p=0 with x = Vm/(dm3 mol–1) . It will be convenient to have the pressure in atm: 200 bar ×1 atm 1.013 bar=197.4 atm The coefficients in the equation are b+RT p=(3.183 ×10−2 dm3 mol−1)+(0.08206 dm3 atm mol−1 K−1)×(298.15 K) 197.4 atm =(3.183 ×10−2+0.1239) dm3 mol−1=0.1558 dm3 mol−1 a p=1.360 dm6 atm mol−2 197.4 atm=6.89 ×10−3 dm6 mol−2 ab p=(1.360 dm6 atmmol−2)×(3.183 ×10−2 dm3 mol−1) 197.4 atm=2.193×10−4 dm9 mol−3 Thus, the equation to be solved is x3−0.1558x2+(6.89 ×10−3)x−(2.193×10−4)=0. Calculators and computer software for the solution of polynomials are readi ly available. In this case we find x = 0.112 and Vm = 0.112 dm3 mol–1 . The perfect -gas value is about 15 percent greater than the van der Waals result. 1C.5(b) The molar volu me is obtained by solving Z=pVm RT[1C.2], for Vm , which yields Vm=ZRT p=(0.86)×(0.08206 dm3 atm mol−1 K−1)×(300K) 20atm=1.06 dm3 mol−1 (i) Then, V=nVm=(8.2×10−3mol) ×(1.06 dm3 mol−1)=8.7×10−3dm3=8.7cm3 (ii) An approximate value of B can be obtained from eqn 1C.3b by truncation of the series expansion after the second term, B/Vm, in the series. Then, B=VmpVm RT−1  =Vm×(Z−1) =(1.06 dm3 mol−1)×(0.86−1)= −0.15 dm3 mol−1 1C.6(b) Equations 1C.6are solv ed for b and a, respectively, and yield b = Vc/3 and a = 27b2pc = 3Vc2pc . Substituting the critical constants b=148 cm3 mol−1 3=49.3 cm3 mol−1=0.0493 dm3 mol−1 14 and a=3×(0.148 dm3 mol−1)2×(48.20atm) =3.17 dm6 −2atmmol But this problem is overdetermined. We have another piece of information Tc=8a 27Rb If we use Tc along with Vc as above, we would arrive at the same value of b along with a=27RbTc 8=9RVcTc 8 =9(0.08206 dm3 atm mol−1 K−1)(0.148 dm3 mol−1)(305.4 K) 8 =4.17 dm6 atmmol−2 Or we could use Tc along with pc. In that case, we can solve the pair of equations for a and b by first setting the two exp ressions for a equal to each other: a=27b2pc=27RbTc 8 Solving the resulting equation for b yields b=RTc 8pc=(0.08206 dm3 atm mol−1 K−1)(305.4 K) 8(48.20 atm)=0.06499 dm3 mol−1 and then a = 27(0.06499 dm3 mol–1)2(48.20 atm) = 5.497 dm6 atm mol–2 These results are summarized in the following tab le Using a/dm6 atm mol–2 b/dm3 mol–1 Vc & pc 3.17 0.0493 Vc & Tc 4.17 0.0493 pc & Tc 5.497 0.06499 One way of selecting best values for these parameters would be to take the mean of the three determinations, namely a = 4.28 dm6 atm mol–2 and b = 0.0546 dm3 mol–1. By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular size. The centres of spherical particles are excluded from a sphere whose radius is the diameter of those spherical particles (i.e., twice thei r radius); that volume times the Avogadro constant is the molar excluded volume b b=NA4π(2r)3 3  sor=1 21/3 3b 4πNA   r=1 21/33(0.0546 dm3 mol−1) 4π(6.022 ×1023 mol−1)  =1.39×10−9dm=0.139 nm 1C.7(b) The Boyle temperature, TB, is the temperature at which the virial coefficient B = 0. In order to express TB in terms of a and b, the van der Waals equation [1C.5b] must be recast into the form of the virial equation. p=RT Vm−b−a Vm2 Factoring out RT Vm yields p=RT Vm1 1−b/Vm−a RTVm     So long as b/Vm < 1, the first term inside the brackets can be expanded using (1–x)–1 = 1 + x + x2 + ... , which gives ฀ p=RT Vm1+b−a RT  ×1 Vm  +L     We can now identify the second virial coefficient as B=b−a RT 15 At the Boyle temperature B=0=b−a RTB so TB=a bR=27Tc 8 (i) From Table 1C.3, a = 4.484 dm6 atm mol–2 and b = 0.0434 dm3 mol–1. Therefore, TB=(4.484 dm6 atm mol−2) (0.08206 L atm mol−1 K−1)×(0.0434 dm3 mol−1)=1259K (ii) As in Exercise 1C.6(b), b=NA4π(2r)3 3  sor=1 21/3 3b 4πNA   r=1 21/33(0.0434 dm3 mol−1) 4π(6.022 ×1023 mol−1)  =1.29×10−9dm=1.29×10−10m=0.129nm 1C.8(b) States that have the same reduced pressure, temperature, and volume [1 C.8] are said to correspond. The reduced pressure and temperature for N 2 at 1.0 atm and 25°C are [Table 1C.2] pr=p pc=1.0atm 33.54atm=0.030 and Tr=T Tc=(25+273)K 126.3K=2.36 The corresponding states are (i) For H2S (critical constants obtained from NIST Chemistry WebBook ) T = 2.36(373.3 K) = 881 K p = 0.030(89.7 atm) = 2.67 atm (ii) For CO 2 T = 2.36(304.2 K) = 718 K p = 0.030(72.9 atm) = 2.2 atm (iii) For Ar T = 2.36(150.7 K) = 356 K p = 0.030(48.0 atm) = 1.4 atm 1C.9(b) The van der Waals equation [1C.5b] is p=RT Vm−b−a Vm2 which can be solved for b b=Vm−RT p+a Vm2=4.00×10−4m3mol−1−(8.3145 J K−1 mol−1)×(288K) 4.0×106Pa+0.76m6Pamol−2 (4.00×10−4m3mol−1)2   =1.3×10−4 m3 mol−1 The compression factor is Z=pVm RT[1C.2] =(4.0×106Pa)×(4.00×10−4m3 mol−1) (8.3145 J K−1 mol−1)×(288K)=0.67 Solutions to problems 1C.2 From the definition of Z [1C.1] and the virial equation [1C.3b], Z may be expressed in virial form as ฀ Z=1+B1 Vm  +C2 1 Vm  +L 16 Since Vm=RT p (by assumption of approximate perfec t gas behavior), 1 Vm=p RT; hence upon substitution, and dropping terms beyond the second power of 1 Vm   Z=1+Bp RT  +Cp RT  2 =1+(−21.7×10−3dm3mol−1)×100atm (0.08206 dm3 atm mol−1 K−1)×(273K)   +(1.200×10−3dm6mol−2)×2100atm (0.08206 dm3 atm mol−1 K−1)×(273K)   =1−0.0968 +0.0239 =0.927 Vm=(0.927)RT p   =(0.927)(0.08206 dm3 atm mol−1 K−1)(273 K) 100 atm  =0.208 dm3 Question . What is the value of Z obtained from the nex t approximation using the value of Vm just calculated? Which value of Z is likely to be more accurate? 1C.4 Since B′(TB) = 0 at the Boyle temperature [Topic 1.3b]: ′B(TB)=a+be−c/TB2=0 Solving for TB: ()1/2 1/2 2 2 B 1 1(1131K )5.0 10 K ( 01993bar ) lnln(0 2002bar )cTa b− − −−  = = = ×  − −−.      .  1C.6 From Table 1C.4 Tc=2 3  ×2a 3bR  1/2 , pc=1 12  ×2aR 3b3  1/2 ()122 3a bR/ may be solved for from the expression for pc and yields 12bpc R  . Thus Tc=2 3  ×12pcb R  =8 3  ×pcVc R   =8 3  ×(40 atm) ×(160 ×10−3 dm3 mol−1) 0.08206 dm3 atm mol−1 K−1  =210K By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular size. The centres of spherical particles are excluded from a sph ere whose radius is the diameter of those spherical particles (i.e., twice their radius); that volume times the Avogadro constant is the molar excluded volume b b=NA4π(2r)3 3  sor=1 21/3 3b 4πNA   [Exercise 1C.6(b)] =1 21/3Vc 4πNA   r=1 21/3160 cm3 mol−1 4π(6.022 ×1023 mol−1)  =1.38×10−8cm=0.138 nm 1C.8 Substitute the van der Waals equation [1C.5b] into the defin ition of the compression factor [1C.2] 17 Z=pVm RT=1 1−b Vm  −a RTVm [Exercise 1C.7(a)] which upon expansion of 12 m mm11b bb V VV−  − = ++ +     yields 2 2 mm111aZb bRT V V  = +− × + +      We note that all terms beyond the second are necessarily positive, so only if 2 mm ma bb RTV V V>+ +  can Z be less than one. If we ignore terms beyond b Vm, the conditions are simply stated as Z<1 whena RT>b Z >1 whena RT<b Thus Z < 1 when attractive forces predominate and when there is insufficient therma l energy to disrupt those forces, and Z > 1 when size effects (short -range repulsions) predominate. 1C.10 The Dieterici equation is p=RTe−a/RTVm Vm−b [Table 1C.4] At the critical point the derivatives of p with respect to Vm equal zero along the isotherm defined by T = Tc . This means that (∂p/ ∂Vm)T=0 and (∂2p/ ∂Vm2)T=0 at the critical point. T∂p ∂Vm  =paVm−ab−RTVm2 Vm2(Vm−b)(RT)     and ( ) { }2 322 2 mm mmm 22 32 m m mm mm24 2 ( )( ) ( )( )T TaV V ab RTV ab aV ab RTV pppV V V V b RT V V b RT−+ + −     −− ∂∂= +   ∂ ∂−  −      Setting the Dieterici equation equal to the critical pressure and making th e two derivatives vanish at the critical point yields three equations: and pc=RTce−a/RTcVc Vc−baVc−ab−RTcVc2=0 −2aVc2+4Vcab+RTcVc3−2ab2=0 Solving the middle equation for Tc, substitution of the result into the last equation, and solving for Vc yields the result Vc = 2b or b = Vc / 2 (The solution Vc = b is rejected because there is a singularity in the Dieterici equation at the point Vm = b.) Substitution of Vc = 2b into the mi ddle equation and solving for Tc gives the result Tc = a / 4bR or a = 2RTcVc Substitution of Vc = 2b and Tc = a / 4bR into the first equation gives pc=ae−2 4b2=2RTce−2 Vc The equations for Vc, Tc, pc are substituted into the equation for the critical compression factor [1C.7] to give Zc=pcVc RTc=2e−2=0.2707. This is significantly lower than the critical compression factor that is pre dicted by the van der Waals equation: Zc(vdW) =pcVc/RTc=3/8=0.3750. Experimental values for Zc are 18 summarized in Table 1C.2 where it is seen that the Dieterici equation prediction is often better. 1C.12 2 m1pVBp CpRT′′ = ++ +  [1C.3a] m 2 mm1pV CB RT V V= +++  [1C.3b] Thus 2 2 mmCBBp CpVV′′+ += + +  Multiply through by Vm, replace pVm by RT{1+(B/Vm) + ...}, and equate coefficients of powers of 1/ Vm: 22 mmBBRT CR T CBRT BVV′′+′+ += + +  Hence, B′RT = B, implying that ′B=B RT Also BB′RT + C′R2T2 = C = B2 + C′R2T2, implying that ′C=C−B2 R2T2 1C.14 Write Vm = f(T, p); then dVm= p∂Vm ∂T  dT+ T∂Vm ∂p  dp Restricting the variations of T and p to those which leave Vm constant, that is d Vm = 0, we obtain p∂Vm ∂T  = − T∂Vm ∂p  × Vm∂p ∂T  = − T−1 ∂p ∂Vm  × Vm∂p ∂T   From the equation of state T∂p ∂Vm  = −RT Vm2−2(a+bT) Vm3= −VmRT+2(a+bT) Vm3 and Vm∂p ∂T  =R Vm+b Vm2=RVm+b Vm2 Substituting P∂Vm ∂T  =Vm3 VmRT+2(a+bT)  RVm+b Vm2  =RVm2+bVm VmRT+2(a+bT) From the equation of state, a + bT = pVm2 – RTVm Then P∂Vm ∂T  =RVm2+bVm VmRT+2pVm2−2RTVm=RVm+bm 2pVm−RT 1C.16 Z=Vm Vmo [1C.1], where Vm° = the molar volume of a perfect gas From the given equation of state Vm=b+RT p=b+Vmo For Vm =10b, we have 1 0b = b + Vm°, so Vm° = 9b . Then Z=10b 9b=10 9=1.11 1C.18 The virial equation is m 2 mm1BCpV RTVV= +++  [1C.3b] 19 or m 2 mm1pV BC RT V V= +++  (a) If we assume that the series may be truncated after the B term, then a plot of pVm RT vs 1 Vm will have B as its slope and 1 as its y-intercept. Transforming the data gives p/MPa Vm/(dm3 mol–1) (1/Vm)/(mol dm–3) pVm/RT 0.4000 6.2208 0.1608 0.9976 0.5000 4.9736 0.2011 0.9970 0.6000 4.1423 0.2414 0.9964 0.8000 3.1031 0.3223 0.9952 1.000 2.4795 0.4033 0.9941 1.500 1.6483 0.6067 0.9912 2.000 1.2328 0.8112 0.9885 2.500 0.98357 1.017 0.9858 3.000 0.81746 1.223 0.9832 4.000 0.60998 1.639 0.9782 Figure 1C.1(a) The data are plotted in Figure 1 C.1(a). The data fit a straight line reasonably well, and the y- intercept is very close to 1. The regression yields B = –1.324 ×10–2 dm3 mol–1. (b) A quadratic function fits the data somewhat better (Figure 1 C.1(b)) with a slightly better correlation coefficient and a y-intercept closer to 1. This fit implies that truncation of the virial series after the term with C is more accurate than after just the B term. The regression then yields 20 Figure 1C.1(b) B = –1.503 ×10–2 dm3 mol–1 and C = –1.06×10–3 dm6 mol–2 1C.20 The perfect gas equation [1A.5] gives Vm=RT p=(8.3145 J K−1 mol−1)(250 K) 150×103 Pa=0.0139 m3=13.9 dm3 The van der Waals equation [1C.5b] is a cubic equation in Vm. Cubic equations can be solved analytically. However, this approach is cumbersome, so we proceed as in Exa mple 1C.1. The van der Waals equation is rearranged to the cubic form Vm3−b+RT p  Vm2+a p  Vm−ab p=0 or x3−b+RT p  x2+a p  x−ab p=0 with x = Vm/(dm3 mol–1) . It will be convenient to have the pressure in atm: 150 kPa ×1 atm 101.3 kPa=1.481 atm The coefficients in the equation are b+RT p=(5.42 ×10−2 dm3 mol−1)+(0.08206 dm3 atm mol−1 K−1)×(250 K) 1.481 atm =(5.42 ×10−2+13.85) dm3 mol−1=13.91 dm3 mol−1 a p=6.260 dm6 atm mol−2 1.481 atm=4.23 dm6 mol−2 ab p=(6.260 dm6 atmmol−2)×(5.42 ×10−2 dm3 mol−1) 1.481 atm=2.291×10−2 dm9 mol−3 Thus, the equation to be solved is x3−13.91x2+4.23x−(2.291×10−2)=0. Calculators and computer software for the solution of polynomi als are readily available. In this case we find x = 13.6 and Vm = 13.6 dm3 mol–1 . Taking the van der Waals result to be more accurate, the error in the perfect -gas value is 13.9 −13.6 13.6×100% =2% 21 Assume all gases are perfect unless stated otherwise. Unless otherwise stated, thermochemical data are for 298.15 K. 2 The First L aw 2A Internal energy Answers to d iscussion questions 2A.2 Work is a precisely defined mechanical concept. It is produced from the applicati on of a force through a distance. The technical definition is based on the realization that both force an d displacement are vector quantities and it is the component of the force acting in the direc tion of the displacement that is used in the calculation of the amount of work, that is, work is the sca lar product of the two vectors. In vector notation cos w fd θ = −⋅ = − dF , where θ is the angle between the force and the displacement. The negative sign is inserted to conform to the standard thermodynamic convent ion. Heat is associated with a non -adiabatic process and is defined as the difference between the adiabatic work and the non -adiabatic work associated with the same change in state of the system. T his is the formal (and best) definition of heat and is based on the definition of wor k. A less precise definition of heat is the statement that heat is the form of energy that is transferred b etween bo dies in thermal contact with each other by virtue of a difference in temperature. The interpretations of heat and work in terms of energy levels and popula tions is based upon the change in the total energy of a system that arises from a change in the molecul ar energy levels of a system and from a change in the populat ions of those levels as explained more fully in Chapter 15 of this text. The statistical thermo dynamics of Chapter 15 allows us to express the change in total energy of a system in the followi ng form: dddi i ii iiN NNεε ε〈 〉= +∑∑ The work done by the system in a reversible, isothermal expansion can be i dentified with the second term on the right of this expression, since there is no chan ge in the populations of the levels which depend only on temperature; hence, the first term on the right is zero. Because the influx of energy as heat does not change the energy levels of a system, but does resul t in a change in temperature, the second term on the right of the above equation is zero and the heat associated with the process (a constant volume process, with no additional work) can be identified with the first term. The change in populations is due to the change in temperature, which redistribute s the molecules over the fixed energy levels. Solutions to exercises 2A.1(b) See the solution to Exercise 2A .1(a) where we introduced the following equation based on the material of Chapter 15. CV,m=1 2(3+vR*+2vV*)R with a mode active if T>θM (where M is T, R, or V). (i) O3:CV,m=1 2(3+3+0)R=3R [experimental = 3.7 R] 11 13 3 8.314 J K mol 298.15 K= 7.436 kJ molE RT−− −= =×× (ii) C2H6:CV,m=1 2(3+3+2×1)R=4R [experimental = 6.3 R] 11 14 4 8.314 J K mol 298.15 K= 9.915 kJ molRTE−− −=×× = 2:1 (iii) 1 2 ,m 2SO : (3 3 0) 3VC RR = ++ = [experimental = 3.8 R] 11 13 3 8.314 J K mol 298.15 K= 7.436 kJ molE RT−− −= =×× Consultation of Herzberg references, G. Herzberg, Molecular spectra and Molecular structure, II, Chapters 13 and 14, Van Nostrand, 1945, turns up only one vibrational mode among these molecules whose frequency is low enough to have a vibrational temperature near r oom temperature. That mode was in C 2H6, corresponding to the “internal rotation” of CH 3 groups. The discrepancies between the estimates and the experimental values suggest that there are vibrational mo des in each molecule that contribute to the heat capacity —albeit not to the full equipartition value —that our estimates have classified as inactive. 2A.2(b) (i) volume, (iii) internal energy, and (iv) density are state functions. 2A.3(b) This is an expansion against a constant external pressure; hence ex [2A.6] w pV= −∆ The change in volume is the cross -sectional area times the linear displacement: 3 2 33 1m(75.0cm ) (25.0cm) 1.87 10 m100cmV− ∆= × × = ×   so 3 33(150 10 Pa) (1.87 10 m ) 281Jw−= −× × × = − as 1 Pa m3 = 1 J 2A.4(b) For all cases ∆U=0, since the internal energy of a perfect gas depends only on temperature. From th e definition of enthalpy, H = U + pV, so ∆H= ∆U+ ∆(pV)= ∆U+ ∆(nRT) (perfect gas). ∆H=0 as well, at constant temperature for all processes in a perfect g as. (i) ∆U= ∆H=0 f i 3 11 3 3 3ln [2A.9] 20.0dm(2 00mol) (8 3145JK mol ) 273K ln 6.29 10 J5.0dm 6.29 10 JVw nRTV qw−−= −   = −. ×. × × = − × = −= × (ii) ∆U= ∆H=0 w= −pex∆V [2A.6] where pex in this case can be computed from the perfect gas law pV=nRT so 11 13 5 3(2 00mol) (8 3145JK mol ) 273K(10dmm ) 2.22 10 Pa20.0dmp−− − . ×. ×= ×= × and 53 3 13(2.22 10 Pa) (20.0 5.0)dm3.34 10 J(10dmm )w−−× × −= = −× 33.34 10 J qw= −= × (iii) ∆U= ∆H=0 w=0 [free expansion] q= ∆U−w=0−0=0 Comment. An isothermal free expansion of a perfect gas is also adiabatic. 2:2 2A.5(b) The perfect gas law leads to p1V p2V=nRT1 nRT2orp2=p1T2 T1=(111k Pa) ×(356K) 277K=143k Pa There is no change in volume, so w = 0. The heat flow is q=CVdT≈CV∆T=(2.5)×(8.3145JK−1mol−1)×(2.00mol) ×(356 −277)K ∫ =3.28×103J ∆U=q+w=3.28 ×103J 2A.6(b) (i) 33 ex 13(7 7 10 Pa) (2 5dm )19J(10dmm )w pV−− .× × .= − ∆= = − (ii) f iln [2A.9]Vw nRTV= −   ( ) ( )( )3 11 132.5 18.5 dm 6.56g8.3145J K mol 305 K ln39.95gmol 18.5dm 52.8 Jw−− −+ =− × ×× = − Solutions to p roblems 2A.2 ex f i f ex[2A.6] sonRTw pV V V VVp= − ∆ = >> ; ∆ ≈ Hence w≈(−pex)×nRT pex  = −nRT =(−1.0mol) ×(8.314JK−1mol−1)×(1073K) w ≈ –8.9 kJ Even if there is no physical piston, the gas drives back the atmosphere, so the work is also w ≈ –8.9 kJ 2A.4 By multiplying and dividing the value of each variable by its critical value we obtain w= −nR×T Tc  Tc×lnV2 Vc−nb Vc V1 Vc−nb Vc        −n2a Vc  ×Vc V2−Vc V1   w= − V1V2∫pdV= −nRT V1V2∫dV V−nb+n2a V1V2∫dV V2 = −nRTlnV2−nb V1−nb  −n2a1 V2−1 V1   2:3 rr c c cc83 [Table1C.4]27TV aT V T V nbT V Rb= ,= , = ,= w= −8na 27b  ×(Tr)×lnVr,2−1 3 Vr,1−1 3      −na 3b  ×1 Vr,2−1 Vr,1   The van der Waals constants can be eliminated by defining wr=3bw a, then w=awr 3band wr= −8 9nTrlnVr,2−1/3 Vr,1−1/3  −n1 Vr,2−1 Vr,1   Along the critical isotherm, Tr = 1, Vr,1 = 1, and Vr,2 = x. Hence wr n= −8 9ln3x−1 2  −1 x+1 2A.6 One obvious limitation is that the model treats only displacements along t he chain, not displacements that take an end away from the chain. (See Fig. 2A.2 in the Student’s Solutions Manual ) (a) The displacement is twice the persistence length, so x = 2l, n = 2, ν = n/N = 2/200 = 1/100 and F=kT 2l ln1+ν 1−ν  =(1.381 ×10−23 J K−1)(298 K) 2×45×10−9 m ln1.01 0.99  =9.1×10−16 N Figure 2A.1 (b) Fig. 2A.1 displays a plot of force vs. displacement for Hooke’s law and for the one -dimensional freely jointed chain. For small displacements the plots very nearly coincide. However, for large displacements, the magnitude of the force in the one -dimensional model grows much faster. In fact, in the one-dimensional model, the magnitude of the force approaches infinity for a finite displacement, -5-4-3-2-1012345 -1 -0.8 -0.6 -0.4 -0.2 00.2 0.4 0.6 0.8 1 displacement Hooke 1-D model 2:4