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Unformatted text preview: Chem 444 B Hour Exam #1 Answer Key 1a)
/ / / So,
/ 1b) Recall that Z>1 when Preal>Pideal and Z<1 for Preal<Pideal. So for Z>1
/ 1 or alternatively / and for Z<1
/ 1 2) For an ideal monotonic gas, all of the ensemble averaged values will be given average molecular values times the number of molecules. That is: <E> =N<ε>. For average energy: and
/ , / So, = For the constant volume heat capacity: For the pressure: So 3) If the system is in thermodynamic equilibrium, the relative populations of the various vibrational levels will be given by the Boltzmann equation for a constant temperature, that is:
/ 0.2 / and solving for kBT=1448 cm1 or T=2083 K Checking another state: 0.0016
/ and again kBT=1448 cm1 or T=2083 K. So to answer part a) Yes, the system is in thermal equilibrium. For part b) the vibrational temperature of the gas is 2083 K. For part c) is this value necessarily the same temperature as the translational temperature? Why or why not? I would argue that the two temperatures are the same, since the vibrational levels for N2 are broadly spaced compared to the translational energy levels. The collisions necessary to bring N2 into thermal equilibrium need to be more energetic and for that reason, less numerous, than those to equilibrate the closely spaced translational levels. Since the translational degrees of freedom would be in equilibrium before vibration, vibrational thermal equilibrium guarantees translational equilibrium. For part d) since we are in thermal equilibrium: and For =2330 cm1,
/ / =3352 K, so 1 / / / + Putting in some numbers: / / =3352 K, T=2083 K
. . 1676 1676 838 The first contribution is from zeropoint energy and independent of temperature. The second comes from the vibrationally excited states, and has a “temperature” of 838 K or 838 4a) 0.695 583 So combining together we get: and 4b) Let’s evaluate each contribution to C6 for the two gases C3H8 and NH3 in units of Jm6: udd 1.06x1086 8.15x1078 uind 1.02x1082 1.00x1078 udisp 5.28x1077 6.07x1078 Molecule C3H8 NH3 For C3H8, C6=5.28x1077 Jm6 and the dominant term is from dispersion. For NH3, C6=1.52x1077 Jm6 and the dominant term is from dipoledipole. 5) This reversible process follows a path of constant P/V. Starting with the ideal gas equation:= , we note that and . Since Thus and 1092 , then and 4 =4 x 273 For an ideal monatomic gas ∆ So, ∆ 10.240 ∆ 12.5 1092 273 10240 , since P/V=const, then P=const x V and dP = const x dV , so 0.176 , 3.41 3.41, 0.008314 13.65 819 17.05 / 34.09 . . . . Since 100 J = 1 bar L, ∆ ∆ ∆ 10.24 ∆ 10.24 ...
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 Spring '08
 JamesLis
 Physical chemistry, pH

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