0 c for the path abc work is done only during the

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Unformatted text preview: ame material as the container. The volume of water that was lost is V = Vwater Vcontainer = V0water T V0container T = V0 (water container) T; (0.35 g)/(0.98324 g/mL) = (55.50 mL)[210 106 (C)1 container](60C 20C), which gives container = 50 106 (C)1. (b) From Table 171, copper is the most likely material. 17.35 If we assume argon is an ideal gas, we have PV = nRT = (m/M)RT; P(35.0 103 m3) = [(105.0 kg)(103 g/kg)/(40 g/mol)](8.315 J/mol K)(293 K), which gives P = 1.83 108 Pa = 1.80 103 atm. 17.42 The pressure at the bottom of the lake is Pbottom = Ptop + gh = 1.013 105 Pa + (1000 kg/m3)(9.80 m/s2)(37.0 m) = 4.64 105 Pa. For the two states of the gas we can write PbottomVbottom = nRTbottom , and PtopVtop = nRTtop , which can be combined to give (Pbottom/Ptop)(Vbottom/Vtop) = Tbottom/Ttop ; (4.64 105 Pa/1.013 105 Pa)(1.00 cm3/Vtop) = (278.7 K/294.2 K), which gives Vtop = 4.83 cm3. 17.46 (a) We find the number of moles from n = +V/M = +4R2d/M = (1000 kg/m3)3(6.4 106 m)2(3 103 m)(103 g/kg)/(18 g/mol) = 6 1022 mol. (b) For the number of molecules we have N = nNA = (6 1022 mol)(6.02 1023 molecules/mol) = 4 1046 molecules. 18.8 The average kinetic energy depends on the temperature: mvrms2 = 8kT, or kT/m = @vrms2. With M the mass of the gas and m the mass of a molecule, we write the ideal gas law as PV = NkT = (M/m)kT, or P = (M/V)(kT/m) = (@vrms2) = @vrms2, or vrms = (3P/)1/2. 18.11 (a) The average kinetic energy depends on the temperature: !mvrms2 = 8kT, which gives vrms = (3kT/m)1/2 = [3(1.38 1023 J/K)(273 K)/(32 u)(1.66 1027 kg/u)]1/2 = 461 m/s. (b) The molecule, on the average, will have a component in one direction less than the average speed. If we take the rms speed as the average speed, from the analysis of the molecular motion, we know that (vx2)av = @vrms2. Thus the time to go back and forth is t = 2/(vx)av 23/vrms . The frequency of collisions with one wall is N = 1/t = vrms/23 = (461 m/s)/2(7.0 m) 3 = 19 s1. 18.31 (a) We write the Van der Waals equation as P = (an2/V2) + nRT/(V nb). When we differentiate, with the temperature constant, we get dP/dV = (2an2/V3)...
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