물리화학 7판 / Atkins 솔루션 입니다.

물리화학 7판 / Atkins 솔루션 입니다.

atkins-7ed-instructo.zip

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설명 : 물리화학 7판 / Atkins 솔루션 입니다. 다운 받으시기전에 내용을 한번 읽어 보시고 내용이 맞다면 다운을 받으시기 바랍니다. 혹시나 받았는데 내용이 다를시 환불요청 부탁드리겠습니다. 솔루션 보고 공부 대박나길 바랍니다.

1 The properties of gases Solutions to exercises Discussion questions E1.1(b) 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. It 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. E1.2(b) The critical constants represent the state of a system at which the distinction between the liquid and vapour phases disappears. We usually describe this situation by saying that above the critical temperature the liquid phase cannot be produced by the application of pressure alone. The liquid and vapour phases can no longer coexist, though fluids in the so-called supercritical region have both liquid and vapour characteristics. (See Box 6.1 for a more thorough discussion of the supercritical state.) E1.3(b) The van der Waals equation is a cubic equation in the volume, V . Any cubic equation has certain properties, one of which is that there are some values of the coefficients of the variable where the number of real roots passes from three to one. In fact, any equation of state of odd degree higher than 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(odd) to 1. That is, the multiple values of V converge from n to 1 as T → Tc. 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 result as the critical point is reached. Numerical exercises E1.4(b) Boyle’s law applies. pV = constant so pfVf = piVi pf = piVi Vf = (104 kPa) × (2000 cm3) (250 cm3) = 832 kPa E1.5(b) (a) The perfect gas law 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 g mol−1 = 0.626 mol so p = (0.626 mol) × (8.31 × 10−2 L barK−1 mol−1) × (30 + 273K) 1.5L = 10.5 bar not 2.0 bar. 4 INSTRUCTOR’S MANUAL (b) The van der Waals equation is p = RT Vm − b − a V2m so p = (8.31 × 10−2 L barK−1 mol−1) × (30 + 273)K (1.5L/0.626 mol) − 3.20 × 10−2 L mol−1 − (1.337 L2 atm mol−2) × (1.013 bar atm−1) (1.5L/0.62¯6 mol)2 = 10.4 bar E1.6(b) (a) Boyle’s law applies. pV = constant so pfVf = piVi and pi = pfVf Vi = (1.48 × 103 Torr) × (2.14 dm3) (2.14 + 1.80) dm3 = 8.04 × 102 Torr (b) The original pressure in bar is pi = (8.04 × 102 Torr) × 1 atm 760 Torr × 1.013 bar 1 atm = 1.07 bar E1.7(b) Charles’s law applies. V ∝ T so Vi Ti = Vf Tf and Tf = VfTi Vi = (150 cm3) × (35 + 273)K 500 cm3 = 92.4K E1.8(b) The relation between pressure and temperature at constant volume can be derived from the perfect gas law pV = nRT so p ∝ T and pi Ti = pf Tf The final pressure, then, ought to be pf = piTf Ti = (125 kPa) × (11 + 273)K (23 + 273)K = 120 kPa E1.9(b) According to the perfect gas law, one can compute the amount of gas from pressure, temperature, and volume. Once this is done, the mass of the gas can be computed from the amount and the molar mass using pV = nRT so n = pV RT = (1.00 atm) × (1.013 × 105 Pa atm−1) × (4.00 × 103 m3) (8.3145 JK−1 mol−1) × (20 + 273)K = 1.66 × 105 mol and m = (1.66 × 105 mol) × (16.04 g mol−1 ) = 2.67 × 106 g = 2.67 × 103 kg E1.10(b) All gases are perfect in the limit of zero pressure. Therefore the extrapolated value of pVm/T will give the best value of R. THE PROPERTIES OF GASES 5 The molar mass is obtained from pV = nRT = m M RT which upon rearrangement gives M = m V RT p = ρ RT p The best value of M is obtained from an extrapolation of ρ/p versus p to p = 0; the intercept is M/RT . Draw up the following table p/atm (pVm/T )/(L atmK−1 mol−1) (ρ/p)/(gL−1 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 From Fig. 1.1(a), pVm T p=0 = 0.082 061 5 L atmK−1 mol−1 From Fig. 1.1(b), ρ p p=0 = 1.42755 g L−1 atm−1

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