06 - Integration, Function calculations.pdf
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Department
Chemistry
Course
CHEM 5
Professor
Douglas Tobias
Semester
Fall

Description
Sabrina Lugo Section B 11-9-13 Homework #6 #1) The Debye theory of solids provides an accurate description of the temperature dependance of the heat capacity of monatomic solids over a wide temperature range. According to the Debye theory, the molar heat capacity at constant volume, C for an absolute temperature T, is given by the following expres- v sion C = 9 R J T N * HQDêTL x *e „x v QD Ÿ0 He -1L J R= gas constant = 8.314 K*mol QD= the Debye temperature Given that QD= 309 K for copper, Calc. the molar heat capacity, which will come out J in units of K*molfor copper at 103 K using the eq. above. R = 8.314 J ; K * mol Q = 309 K; T = 103 K; 4 x T 3 HQêTL x * E Cv = 9 * R * * ‡ „x; Q 0 HE - 1L 2 Cv 16.5305 J K mol #2) According to the kinetic theory of gases, the distribution of speeds of atoms or moleucles in a ideal gas is given by the Maxwell-Boltzmann Distribution: M 3ê2 2 -Mu ë2 RT f HuL = 4 p I 2 pRT u e Here, f(u)du is hte probability that an atom/molecule in the gas has a speed between u and u+du, M = the molar mass of the atom/molecule R = the gas constant T = the absolute temperature a) Derive an expression for the most-probable speed by solving the equaiton df(u)/d = 0 for u. Printed by Wolfram Mathematica Student Edition 2 sclugo_hw6.nb M 3ê2 2 [email protected]_D := 4 * Pi * * u * E -IM*u MëH2*R*T; H2 * Pi * R * TL [email protected]; [email protected]; eq = [email protected]@[email protected], uD ã 0, uD; [email protected]@3DD 2 R T :u Ø > M b) Calculate an exprssion for the most-probable value of the speed for N 2 (M = 0.028 kg/mol) at T=300K, using J R=8.314 so your answer will be in m/s K*mol [email protected], T, RD; value = [email protected]@[email protected], uD ã 0, uD; 0.028 kg 8.314 J [email protected]@3DD ê. :M Ø , T Ø 300 K, R Ø > mol K * mol J 422.087 K K mol :u Ø > kg mol 2 2 K * 2g*m 2 H*Recall J= kg*m \ s *K*mol= m = m*L s2 Kg s2 s mol #3) The equation constant for the reaction: 3Al 2l 6HgL 2Al C3 9HgL -4 is K p 1.04x 10 at 454 K. A reaction vessel is charged with an initial pressure of Al Cl of P 0 .If -x is the change in the pressure of Al Cl that occurs as the reac- 2 6 A2 C6 2 6 tion reaches equilibrium, then the law of mass action is: 2 x J3 N K p 0 3 IP A2 6lxM a) Calc. the equilibrium partial pressures of Al 2l 6 and Al3Cl 9 using the usual 0 0 assumption that x is small compared to P A2 C6, using P A2 C6 = 2atm. Printed by Wolfram Mathematica Student Edition sclugo_hw6.nb 3 Kp = 1.04 * 10 ; PAl2Cl6 = 2; 2 J H2*xLN 3 equation = : 3 ã Kp>; HPAl2Cl6L roots = [email protected], xD 88x Ø -0.0432666; 3 HPAl2Cl6 - xL root = [email protected], xD 88x Ø -4267.5
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