Page 138 - Essentials of physical chemistry
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100                                                  Essentials of Physical Chemistry

            equations and helped define A and G, two new state variables. The third law was defined and
            connected to the Boltzmann equation for absolute entropy, S ¼ k B ln W. Using that equation, we
            derived the equation for the entropy of mixing in binary solutions. We also discussed the empirical
            observation of Trouton’s rule. Now we can proceed, armed with a basic understanding of the first,
            second, and third laws of thermodynamics and in possession of eight equations, which are more
            powerful than we have yet realized.


            TESTING, GRADING, AND LEARNING?

            We now provide some midterm examinations to assist students in preparing for such a test at about
            this point in the course. It may be helpful to the students to give this test a bit past the actual
            midpoint of the course and to give them time to assimilate the material. This author routinely offers
            the students the option to take their final grade based on the cumulative final examination alone or
            their cumulative average over the semester, whichever is higher. That policy tends to make the
            midterm a ‘‘practice run’’ for the final examination, which will include some new questions related
            to lecture material between the midterm and the final examinations. Providing old tests to the
            students does focus their study, but they should expect the questions to be modified for the actual
            examinations. We note here that no student who cut many classes and took only the final
            examination has ever passed this course.
              Now we present an actual examination that was a midterm examination in 2007 for a one-
            semester presentation in CHEM 305 at Virginia Commonwealth University. After that we show a
            slightly more difficult midterm examination from the 2009 Summer course. The time allowed for
            CHEM 303 is longer when a 2 h lecture period is available, but the CHEM 305 time period was only
            1 h in the one-semester course.

            CHEM 305              Midterm examination, Fall 2007         D. Shillady, Professor
            (Points)                  (Attempt all problems)                   55 min


            (15) 1. Calculate the laminar bulk flow rate in gallons=min for blood with h ¼ 0.015 poise through
                  an aorta 5 in. long and 1=4 in. inner diameter due to a pressure difference of 135–80 mmHg,
                  use a duty cycle factor of 0.05 due to pulsation.        (1.218 gallons=min)
            (15) 2. Calculate the temperature of air compressed adiabatically in a one-cylinder diesel engine
                                                              3
                              3
                  from 1030 cm at 228C and 1 atm pressure to 30 cm , given C V ¼ (5=2)R. Compute the
                  moles of air assuming 22.414 L=mol of air, Q, W, DH, and DU for this compression.
                           (T ¼ 1213.78K, Q ¼ 0, n ¼ 0.0425, DU ¼ W ¼þ193.95 cal, DH ¼þ271:54 cal)
                             r ffiffiffiffiffiffiffiffiffi
                               8RT
                                   for He gas molecules in mph at 208C (He ¼ 4.0026 g=mol).
                               pM
            (10) 3. Calculate   v ¼
                                                                                (2795.3 mph)
            (15) 4. Derive the expression for   v of a gas molecule using the Boltzmann principle.
                                                                           (See chapter notes)
            (15) 5. A 0.500 g sample of n-heptane (C 7 H 16 ) burned in a constant volume calorimeter causes
                  DT ¼ 2:934 C.If C V of the calorimeter is 1954 cal=8C, calculate DH comb of n-heptane at

                  2988K (use C ¼ 12.011 g=mol and H ¼ 1.008 g=mol).
                                                                         ( 1,151,505 cal=mol)
            (10) 6. Compute DH 0
                             298  for the reaction C 6 H 6 þ 3H 2(g) ! C 6 H 12 given the data DH comb (C 6 H 6 ) ¼
                   782:3kcal=mol, DH comb (C 6 H 12 ) ¼ 937:8kcal=mol,and DH comb (H 2 ) ¼ 68:3kcal=mol:
                                                                     (DH rxn ¼ 49:4 kcal=mol)
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