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Unformatted text preview: Chemistry 132, Winter 2008 Solutions to Homework No. 10 Problem 1. See Levine problem 22.54. Solve for N 2 , at T = 300 K with rot = 2 . 86 K and z rot = 51 . 0. (a) The fraction of molecules in the rotational energy level corresponding to quantum number J is given by (Levine eq. 22.71) h N ( J ) i N = g J e J /kT z rot where z rot = X J g J e J /kT (Levine eq. 22.73) , J = rot = h 2 2 I J ( J + 1) = k rot J ( J + 1) (Levine p. 845) , rot = h 2 2 Ik (Levine eq. 22.83) . Also, the degeneracy for a heteronuclear diatomic molecule is g J = 2 J +1 (Levine p. 845), but must be divided by = 2 in the case of a homonuclear diatomic molecule such as N 2 , because a homonuclear diatomic molecule has half as many quantum states as a heteronuclear diatomic molecule (Levine p. 845). Then, the fraction of N 2 molecules for which J 16 is given by P ( J 16) = 16 X J =0 h N ( J ) i N = 1 z rot 16 X J =0 g J e J /kT 1 z rot Z 1 g J e J /kT dJ + Z 2 1 g J e J /kT dJ + ... + Z 17 16 g J e J /kT dJ (Levine p. 844) = 1 z rot Z 17 g J e J /kT dJ = 1 z rot Z 17 2 J + 1 2 e J ( J +1) rot /T dJ. Substituting w = J ( J +1), dw = (2 J +1) dJ (Levine p. 845), and J = 17 w = 17 18 = 306, in the above integral gives P ( J 16) = 1 z rot Z 306 1 2 e ( rot /T ) w dw = 1 2 z rot T rot h e ( rot /T ) w i 306 = T 2 z rot rot 1 e 306( rot /T ) = 300 K 2(2 . 86 K)(51 . 0) 1 exp 306 2 . 86 K 300 K = 0 . 9728 = 97 . 28% . Problem 4. See Levine problems 22.71 and 22.72. Calculate ln Z in both cases. For the case of CH 3 OH( g ), A U = kT ln Z (Levine eq. 22.122) A = G PV (Levine eqs. 4.27, 4.28) = G nRT (ideal gas) , A m = G m RT. For an ideal gas, H m ,T = U m ,T + RT , so at 0 K, H m , = U m , (Levine p. 858). Then, at 1 bar ln Z = A  U kT = A  H kT = n ( A m H m , ) kT = n ( G m RT H m , ) kT = n k R G m H m , T . Therefore, for 1 mole of CH 3 OH( g ) at 1000 K ln Z = (8 . 3145 J / K mol) ( 257 . 1 J / K mol) (1 . 38065 10 23 J / K) = 1 . 92238 10 25 . For the case of diamond, assume that H m ,T H m , U m ,T U m , because, at constant pressure, H = U + P V and P V is negligible for a solid like diamond at 1 bar. Then, for 1 mole of diamod at 298 . 15 K and at 1 bar ln Z = A  U kT = U  TS  U kT = n ( U m ,T U m , TS m ,T ) kT  n ( H m ,T H m , TS m ,T ) kT = n k S m ,T H m ,T H m , T = 1 mol (1 . 38065 10 23 J / K) (2 . 377 J / K mol) 523 J / mol 298 . 15 K = 4 . 51125 10 22 ....
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 Winter '08
 Lindenberg
 Mole

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