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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 Θ 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 Θ rot J ( J + 1) (Levine p. 845) , Θ 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 σ = 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)Θ 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 (Θ rot /T ) w dw = 1 2 z rot T Θ rot h e (Θ rot /T ) w i 306 = T 2 z rot Θ rot 1 e 306(Θ 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, Δ H = Δ U + P Δ V and P Δ 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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This note was uploaded on 05/22/2008 for the course CHEM 132 taught by Professor Lindenberg during the Winter '08 term at UCSD.
 Winter '08
 Lindenberg
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