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Unformatted text preview: 14 STUDENT'S SOLUTIONS MANUAL: PHYSICAL CHEMISTRY Solutions to exercises E1.1(a) (a) The perfect gas equation [1.8] is pV = nRT Solving for the pressure gives p = - ART V The amount of xenon is n = 131g 131 g mol-1 7 = 1.00 mol p = (1.00 mol) x (0.0821 dm3 atm K-1 mol-1) x (298.15 K) 1.0 dm3 24 atm That is, the sample would exert a pressure of 24 atm if it were a perfect gas, not 20 atm. (b) The van der Waals equation [1.21a] for the pressure of a gas is p = nRT an2 V - nb V2 For xenon, Table 1.6 gives a = 4.137 dm atm mol-2 and b = 5.16 x 10-2 dm3 mol-1. Inserting these constants, the terms in the equation for p become nRT (1.00 mol) x (0.08206 dm' atm K-1 mol-) x (298.15 K) V - nb 1.0 dm' - {(1.00 mol) x (5.16 x 10-2 dm' mol-1)} 2 =25.8 atm an? _(4.137 dm atm mol-2) x (1.00 mol)2 V 2 (1.0 dm3)2 - = 4.137 atm Therefore, p = 25.8 atm - 4.137 atm = 22 atm E1.2(a) Boyle's law [1.5] applies. pV = constant so pfVr = p;Vi This equation can be solved for either initial or final pressure, hence Pi = V. V. = 4.65 dm3, V; = (4.65 + 2.20) dm' = 6.85 dm', pf = 5.04 bar Therefore, (a) pi = 4.65 dm3 6.85 dm3 x (5.04 bar) = 3.42 bar (b) Since 1 atm = 1.013 bar, p; = 3.42 bar x 1 atm 1.013 bar 3.38 atm E1.3(a) The perfect gas law, PV = nRT [1.8], can be rearranged to P = nk constant. Hence, ~ Pr = Pi or, solving for Po Pr= T. It xpi V = constant, if n and V are...
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