MT1-cheatsheet

P p1 1 t1 t3 3 t2 2 a 0 v c for the adiabatic process

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Unformatted text preview: nRT/(V nb)2. At the critical point, we have dP/dV = 0 = (2an2/Vcr3) nRTcr/(Vcr nb)2 = [2an2(Vcr nb)2 nRTcrVcr3]/Vcr3(Vcr nb)2, so 2an(Vcr nb)2 = RTcrVcr3. For the second derivative we get d2P/dV2 = (6an2/V4) + 2nRT/(V nb)3. Because the critical point is an inflection point, we have d2P/dV2 = 0 = (6an2/Vcr4) + 2nRTcr/(Vcr nb)3 = [6an2(Vcr nb)3 2nRTcrVcr4]/Vcr4(Vcr nb)3, so 3an(Vcr nb)3 = RTcrVcr4. When we divide the two equations, we get 3(Vcr nb) = 2Vcr , or Vcr = 3nb. If we use this result in one of the equations, we get 2an(3nb nb)2 = RTcr(3nb)3, which gives Tcr = 8a/27bR. For the pressure at the critical point, we have Pcr = (an2/Vcr2) + nRTcr/(Vcr nb) = [an2/(3nb)2] + nR(8a/27bR)/(3nb nb) = a/27b2. (b) We find b from Tcr/Pcr = (8a/27bR)/(a/27b2) = 8b/R; (304 K)/(72.8 atm)(1.013 105 N/m2 atm) = 8b/(8.315 J/mol K), which gives b = 4.28 105 m3/mol. We find a from Tcr2/Pcr = (8a/27bR)2/(a/27b2) = 64a/27R2; (304 K)2/(72.8 atm)(1.013 105 N/m2 atm) = 64a/27(8.315 J/mol K)2, which gives a = 0.365 N m4/mol2. 18.34 (a) If we use the ideal gas law, PV = NkT, in the expression for the mean free path, we find the radius from M = 1/42r2(N/V) = kT/42r2P; 5.6 108 m = (1.38 1023 J/K)(273 K)/42r2(1.013 105 Pa), which gives r = 1.93 1010 m, so d = 3.9 1010 m. (b) For helium we have M = 1/42r2(N/V) = kT/42r2P; 25 108 m = (1.38 1023 J/K)(273 K)/42r2(1.013 105 Pa), which gives r = 9.15 1011 m, so d = 1.8 1010 m. 18.62 If we use the ideal gas law, PV = NkT, in the expression for the mean free path, we have M = 1/42r2(N/V) = kT/42r2P. For the given data we have M = (1.38 1023 J/K)(300 K)/42(1.5 1010 m)2(10 atm)(1.013 105 Pa/atm) = 1.0 108 m....
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