PChem I     Exam 2     13 October 1999

Answer any 5 of the 6 questions!
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    gem-cut diamond (MS Encarta)
  1. DfG°(diamond) = +2.900 kJ/mol ... from graphite, of course.
    The densities are r[graphite] = 2.260 g/cc and r[diamond] = 3.513 g/cc.
    Ignoring their isothermal compressibilities (because I can find only diamond's, 1.87×10-7 atm-1), at room temperature, what pressure would be required to make diamonds?

    DG = [integral p° to p] DV dp ~ DV Dp

    r = M / V
    V = M / r
    Vgraphite = 12.01 g/mol / 2.260 g/cc = 5.314 cc/mol = 5.314×10-6 m3/mol
    Vdiamond = 12.01 g/mol / 3.513 g/cc = 3.419 cc/mol = 3.419×10-6 m3/mol

    DV = Vdiamond - Vgraphite = - 1.895×10-6 m3/mol
    DG = DV Dp = -2.900×103 J/mol
    Dp = -2.900×103 J/mol / ( - 1.895×10-6 m3/mol ) = 1.530×109 J m-3
    Dp = 1.530×104 105 Pa = 15,300 bar = 15,101 atm

    p = Dp - p° = 15,100 atm

  2. Trouton's Rule is touted as holding for many liquids at their boiling points. There are some spectacular exceptions, like He and H2O. But for all substances it's true that the heat capacity of the liquid is higher than that for its vapor at the same temperature. (The vapor loses those intermolecular interactions.) Since substances vaporize at temperatures other than their normal boiling point, what are the consequences of this non-zero DCp for "Trouton's Rule" at other temperatures even for substances that satisfy it at their normal boiling points? What's the trend?

    Trouton's Rule is that DvapS ~ 85 J/mol at the normal boiling point.

    But dS/dT = Cp/T so dDvapS/dT = DvapCp/T which means DvapS varies with T.

    DvapCp would be Cp(vapor) - Cp(liquid) which is negative, and DvapS would therefore fall with increasing T.

    (This is actually the case. DvapS(H2O) is about 107 J/mol at its normal boiling point but almost 150 J/mol at 25°C.)

  3. In class, we proved that lnf = - ap/(RT)2 at low pressure (high volume) where the Van der Waals attractive term a dominates. You can do it too (in an even more satisfying way than I did) by finding the limit of (dZ/dp)T as V blows up, but I don't want you to take the time to do it here. Trust me. Instead I want you to use the first chapter's critical parameters, pc = a/(27b2), Vc = 3b, and Tc = 8a/(27bR) to show that the non-ideal gas correction to the molar Gibb's Free Energy comes through
    lnf = - 3 pr (3 / 8Tr )2
    where xr are the reduced parameters, pr = p/pc and Tr = T/Tc. Use it to correct mq for benzene whose pc = 48.6 atm and Tc = 562.7 K.

    Notice that an error on the exam had you looking for - pr (3Tr/8)2 which isn't correct. That would have led to a correction to m° of only -2 J/mol and benzene is worse than that. Clearly I'll have to be generous with partial credit. Sorry about that!

    This is a substitution problem. We need only find a in terms of the critical parameters, pc and Tc.

    pc = a/(27b2) means that a = 27 b2 pc
    Tc = 8a/(27bR) means that b = 8a/(27RTc) or b2 = a2 (8/[27RTc])2

    Substituting that last into the first formula gives:

    a = 27 pc a2 (8/[27RTc])2
    a-1 = pc (8/[RTc])2 / 27
    a = 27 (RTc/8)2 / pc = 3 (3RTc/8)2 /pc

    - ap/(RT)2 = - 3 (p/pc) (3 Tc / 8 T)2 = - 3 pr (3 / 8Tr )2

    m = m° + RTln(p/p°) + RTlnf

    So the correction when p=p° is
    RTlnf = - 3 RT pr ( 3 / 8Tr )2
        = -3(8.314 J/mol K × 298.15 K) (1 atm/48.6 atm) (3×562.7 K/[8×298.15 K])2
        = - 76.6 J/mol

  4. Mercury is a constant laboratory hazard. Spills must be cleaned up, and that's hard because its high surface tension makes little Hg marbles that roll away from the spill to hide in cracks. If those aren't found, they continue to vaporize, contaminating you and your colleagues with what vapor pressure at standard conditions?
    Tb = 629.7 K and DvapH = 59.30 kJ/mol (presumed constant).
    (Actually the pathologically high surface tension works in your favor, since it means that Hg coats itself with whatever is around; it's near impossible to keep Hg clean. So sprinkling sulfur gives it a HgS coat which cuts the vapor pressure way down for the droplets you don't find.)

    pT2 = pT1 e - (DvapH/R)(T2-1 - T1-1)

    p298 K = 1 atm e- (59,300 J/mol / 8.315 J/mol K)(298-1 - 629.7-1) K-1
        = 3.35×10- 6 atm = 2.54 mtorr

  5. While the solid-liquid equilibrium line on a phase diagram rarely has a negative slope (water being the common exception), the liquid-vapor line never has a negative slope. Why not?

    (dp/dT)eq = DS/DV and, except at the critical point, DvapS>0 and DvapV>0, so their ratio, the slope, has to be positive.

    But need that liquid-vapor slope be finite? Consider that the molar volumes of the liquid and gas become the same at the critical point! Still the slope of that equilibrium line on the phase diagram doesn't become infinite at the critical point. Why not?

    Because DvapS is disappearing at the critical point as well, there being no longer any difference between the liquid and the gas. So while the denominator DvapV vanishes, so does the numerator, and the ratio remains finite.

  6. Napoleon's cheap white tin buttons spontaneously changed to gray tin in a Russian winter and fell off, freezing his soldiers. Gray tin is brittle and crumbles under stress. Thus did the thermodynamics of phase transitions contribute to modern European history. Napoleon's army retreated, literally in tatters, from Moscow.

    For the white to gray tin transition at 25°C, DH° = - 2.03 kJ/mol and DS° = - 7.1 J/mol°K. For pure Sn, at what T(°C) would this slow process start?

    DG° = DH° - TDS°

    But at the transition point, DG=0, so
    T ~ DH°/DS° = - 2,030 J mol-1 / - 7.1 J mol-1 K-1 = 286 J = 13°C

    Of course, the buttons were a tin alloy (not pure). In what direction would that change your answer?

    The freezing point would be lowered by adulteration of pure tin in some alloy. And a good thing too; Moscow winters are considerably colder than 13°C!

Last modified 14 October 1999