Advanced Thermodynamics   Exam #3

(Open handouts & note sheet)

  1. Balancing surface and pressure-volume work led to the Kelvin equation:
    RT ln (Pr / P0) = 2 gamma M / r rho
    which show that the vapor pressure at equilibrium with a droplet of radius r only reduces to the normal vapor pressure, P0, as r approaches infinity .

    When water vapor is cooled rapidly to 25°C to find the degree of supersaturation required to nucleate water droplets out of dust-free air spontaneously, it is found that the vapor pressure must be FOUR TIMES its equilibrium vapor pressure. [ rho = water density =1 g/cc at 25°C.]

    1. Calculate the radius of a water droplet first formed at this degree of supersaturation. [The surface tension of water at 25°C is 0.07197 N m-1.]

As long as we convert everything to SI (Systeme International, MKS) we ought not to have any problem with the units. gamma is already in SI (N=J/m so N/m=J/m² as you'd expect from a surface energy). So convert rho to 1×10-3 kg / 10-6 m3 or 1×10+3 kg/m3.

r = 2 gamma M / [ rho RT ln(Pr / P0) ]
r = 2×0.07197 Jm-2×0.018 kg/mol / [ 10+3 kg/m3×8.314 J/mol K×298 K×ln(4) ]

r = 7.54×10-10 m = 7.54 Å (not surprising)

  • How many water molecules are there in the droplet? [Vsphere = (4/3) pi r 3]

    • Hardly seems right to characterize 7.54 Å as a droplet, but for want of a better term...(a moat mote maybe?)

    N = NAv rho Vsphere / M
    N = 6.022×10 23 × 10+3 kg/m3 × (4/3) × 3.1416 × (7.54×10-10)3 / 0.018 kg/mol

    N = 60.0 water molecules.

  • Assuming no significant errors in that 4P0 or any of the other data or constants in this problem, what are the standard deviations associated with the answers to (b) and (a)? ( Remember stat. mech.? )

    • We're not in Kansas anymore, Toto. N isn't NAv or even close; so we can't hide behind the Law of Large Numbers. In other words, whereas the uncertainty in thermo properties, (NAv)½ is completely trivial compared to NAv , that's not the case here.

    Even though none of the numbers going into the calculation is in any significant error (being macroscopic measurements), this microscopic result is untrustworthy to the following extent statistically:

    N = 60.0 +/- (60.0)½ = 60.0 +/- 7.7

    And that range from 52.3 to 67.7 will scale the droplet volume by the same, but the radius will scale by the cube root of those, being proportional to the range (52.3)1/3 to (67.7)1/3 = 3.74 to 4.08 which has an average at about 3.91 +/- 0.16 or (dividing both by 3.91 to get a relative scaling) 1.00 +/- 0.04 for the radius. We now scale it back to the answer from (a) to get:

    r = 7.54 × (1.00 +/- 0.04) = 7.54 +/- 0.31 Å

    Now you might've calculated the radius directly from both ends of the N range, and that would be fine. But you should know the better way to propagate errors through a calculation. In fact, this may be the single most useful tool you take away from this course! And here it is...free!

    Let the independent variable(s) all have known expected values and standard deviations, Xi +/- sigma Xi, and the dependent variable, Y, be a known function, Y=f(Xi). Then the square of the standard deviation of the function, sigma Y, assuming Gaussian uncertainties in all the Xi,

    sigma ²Y = Summation i ( Partial f / Partial Xi)² sigma ²Xi

    Applying it in our case with only N as the independent variable:

    r = [ M×N / (NAv rho (4/3) pi ) ]1/3
    r = 1.93×10-10 m × N1/3
    dr/dN = (1/3) 1.93×10-10 N-2/3 = 4.2×10-12 m

    Since there's only one variable, sigma r = | (dr/dN)× sigma N | or

    sigma r = 4.2×10-12 m × 7.7 = 0.32×10-10 m = 0.32 Å

    which appears to justify the cheat we used above. But since you can use the chain rule to propagate through many calculations to find dY/dX, the general formula will save you time in real-world calculations!

    You're welcome.

  • Which of the following tools (a-e) would be the one of choice to attack the problems (i-iv) below?

    1. STANJAN (including JANFILE as needed)

    • permits equilibrium calculations with User-specified species and thermodynamic tables

  • DIPPR (the full-blown program)

    • a compendium of (a mind-numbing variety of) industrially-relevant chemical and physical properties of many important compounds

  • NIST (the industrial strength version)

    • estimates thermodynamics properties of any organic compound

  • Handbook of Chemistry and Physics (student edition's OK)

    • tables of a few variables on a vast number of compounds and detailed values on a few compounds

  • None of the above

    1. Compounds with autoignition temperatures within 40 degrees of the autoignition temperature (Fahrenheit 451) of book paper.

    • (b) DIPPR permits range of properties to identify lists of compounds

  • Solubility of calcium carbonate in boiling water.

    • (d) HPC gives solubilities for all its listed inorganic compounds

  • Overpressures of methane explosions in nitrogen monoxide.

    • (a) STANJAN calculates equilibrium for the constant E and V (appropriate to an explosion) just as easily as it does for constant H and P (appropriate to a flame). By the way, if you work out the stoichiometry of CH4 in NO, you see it has unimpressive overpressures since there is no change in the number of moles of gas between reactants and products!

  • CP of 3-chlorobutanoic acid at 517 K.

    • (c) NIST can easily estimate CP for almost any organic compound anywhere throughout a large temperature range

    Last modified 5 December 1997.