Fall 2008 CH301 Practice Quiz 6 Answer Key 1. Statistical thermodynamics theory When calculating positional entropy using the Boltzmann formula (S = k·lnW), which of the following statements is/are true? I. W is a theoretical value and its actual value must be experimentally determined. II. The actual value of W is always an integer. III. The equation exactly as written above could describe one atom's or molecule's entropy. 1. I only 2. II only 3. III only 4. I and II only 5. I and III only correct 6. II and III only 7. I, II and III 8. None Explanation: Statement I is true and II is false because an integer value of W assumes that all of the possible orientations of a molecule are equally likely (i.e. there is no bias produced by IMF), which is not the case, and this W must be experimentally determined and is typically smaller than the theoretical value. Statement III is true because using k (as opposed to R) signifies individual particles. 2. Statistical thermodynamics internal energy calculation (E=0.5kT) Which of the following atoms or molecules is incorrectly paired with its total rotational energy? 1. He, 0 2. CH4, 1.5kT 3. I3-, 1.5kT correct 4. N2, 1kT 5. O3, 1.5kT Explanation: The moleculeI3- is linear and therefore has only two rotational modes. 3. Statistical thermodynamics positional entropy calculation (S = klnW) Which of the following systems is incorrectly paired with its positional entropy at absolute zero? 1. 5 molecules of BFH2, 7.58x10-23 J·K-1 2. 100 molecules of O2, 0 J·K-1 3. 2 molecules of CFH3, 3.87x10-23 J·K-1 4. 20 molecules of HF, 1.91x10-22 J·K-1 5. 10 molecules of NH3, 1.52x10-22 J·K-1 correct Explanation: Ammonia has 4 possible orientations as a result of its non-bonding electron pair, the provided number assumes only 3. 4. Internal energy theory Which of the following statements concerning internal energy is/are true? I. Change in internal energy is never zero. II. Change in internal energy is equal to heat when volume is held constant III. Change in internal energy is equal to heat when pressure is held constant 1. I only 2. II only correct 3. III only 4. I and II only 5. I and II only 6. II and III only 7. I, II and III8. None Explanation: Statement I is false because q and w can be equal and opposite in sign, which would result in no change in internal energy. Statement II is true; change in internal energy is equal to heat plus work, but the latter is zero when volume is held constant. Statement III is false; change in enthalpy is equal to heat when pressure is held constant. 5. Internal energy calculation (ΔU = q + w) If an expanding balloon absorbs 100 kJ of heat and exerts 300 J of work, what is its change in internal energy? 1. -99700 J 2. -200 J 3. 400 J 4. 100300 J 5. 99700 J correct Explanation: ΔU = q + w = 100,000 J + (-300 J) = 99700 J. 6. Ranking system entropies Rank the following systems in terms of increasing entropy: I. 1 mole of solid carbon dioxide at 300 K II. 1 mole of solid carbon dioxide at 100 K III. 1 mole of gaseous carbon dioxide at 30 K IV. 1 mole of gaseous carbon dioxide and 1 mole of gaseous oxygen at 500 K V. 1 mole of gaseous carbon dioxide at 500 K 1. III < II < I < V< IV correct 2. III < I < III < IV < V 3. V < IV < III < II < I 4. II < III < IV < V < I 5. I < III < II < V < IV Explanation: Entropy increases as systems go through endothermic phase transitions, when the temperature rises (thus kinetic energy increases), and when there is more matter or more dispersed matter present. 7. Calculation of ΔS from heat transfer If a given system absorbs 1000 J of heat, what will ΔSsurroundings be if this takes place at 127 °C? 1. 3.5 J·K-1 2. 3.0 J·K-1 3. 2.5 J·K-1 correct 4. 4.5 J·K-1 5. 5.2 J·K-1 Explanation: Heat released by the system will increase the entropy of the surroundings by an amount equal to q/T, in this case 1000 J/400 K = 2.5 J·K-1. 8. Calculation involving the second law equation Consider the evaporation of water at standard pressure at 101 °C. Even without knowing the exact values in ΔSvap and ΔHvap, what can you say about the value of ΔSsystem for this process. 1. ΔSsystem would be a small positive number. correct 2. ΔSsystem would be a small negative number. 3. ΔSsystem would be a large negative number. 4. ΔSsystem would be a large positive number. Explanation: The process would be spontaneous (water evaporates at standard pressure and 101 °C), so ΔG would be negative. Thus, given the positive ΔH (endothermic process) and ΔG = ΔH – T ΔS, ΔSsystem must be positive. We are near the phase change temperature of water, so it would also be
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