Chm 1316 Honors Freshman Chemistry II
Spring 1999
Exam 1     15 February 1999

  1. OK, let's see how close my guess was about the true "thickness" of the atmosphere. On each m2 of Earth, the atmosphere presses down with a Force of 101,325 Newtons (Pa=N/m2). That comes from the total atmospheric mass, M, in the column above that 1 m2. By Newton's Law, F=Mg, where g is the acceleration of gravity (9.80665 m s-2). Since everything is given in SI here, that M will end up in kg. Given the average MW of (dry) air to be 0.02898 kg/mol, calculate the "thickness" (height above that 1 m2) of an atmosphere of constant pressure and temperature (0°C). Naturally, we have to turn off gravity and capture the atmosphere inside a crystal sphere of the proper height to make the pressure and temperature constant throughout. Consider that done. J

    F=1.01325×105 N = Mg = M×9.80665 m s-2 (where N = J/m = kg m s-2)
    M = 1.033×104 kg (!)

    n = M/MW = 1.033×104 kg / 0.02898 kg mol-1
    n = 3.5653×105 mol

    V = nRT/P
       = 3.5654×105 mol (8.3145 J/mol K)(273.15 K) / 1.01325×105 Pa
       = 7991 m3   (since J/Pa = kg m2 s-2 / kg m-1 s-2 = m3)

    V = Ah = 1 m2 h = 7991 m3
    h = 7.991 km   (8 km, eh? I'd guessed 4-6 km. Pretty close.)

  2. A soap bubble filled with warm, humid air (1% Ar, 5% H2O, 20% O2, 74% N2) drifts into a room filled with dry He at the same T. While it's sinking, explain what happens to it based on your calculated ratio of effusion rates of He and (take the average MW) air.

    MW = 0.01(39.95) + 0.05(18.016) + 0.20(32.00) + 0.74(28.02)
            = 28.44 g/mol

    vHe / vair = [ MWair / MWHe ]½ = [28.44/4.003]½ = 2.665

    Since vHe > vair by a long shot, the He effuses into the bubble 2.7× faster than the air effuses out. So the bubble initially inflates.

  3. If T for the He in question #2 is body temperature, 98.6°F, and the soap film doesn't weigh anything, how hot (Kelvin) would we have to make the bubble's air so that it didn't sink in the He? What do you think of the bubble's chances of surviving? L

    THe = (98.6-32)×(5/9) + 273.16 = 310 K

    r = m/V = (n/V)MW = (P/RT)MW = (P/R)(MW/T)
    implies that we can counter increased MW by scaling up T in proportion.

    (MWair / MWHe) = 28.44/4.003 = 7.1 so
    (Tair / THe) = 7.1 will compensate (by bloating the volume of the bubble to 7.1× it's original size).

    Tair = 7.1×310 K = 2200 K (!)     The bubble can't survive.

    [Bonus: at that new T for the super-bubble, what would the new effusion rate ratio (warm He vs. hot air) become?]

    ½mv2 = KE = (3/2)RT
    v = [(3R/2)(T/m)]½
    (vHe / vair) = [(THe/Tair)×(MWair/MWHe)]½

    But the ratio of T's is the same as the ratio of MW's. So (vHe/vair)=1.

  4. You're a scuba diver who discovers too late that you forgot to refill your air tank after your last dive. While the tank had enough pressure to let you swim down 100 ft in Loch Ness, you realize that it is now about to run out. Clearly, you have to ascend immediately. Fortunately, you've not been down so long that you need to worry about The Bends, so you cast off your weights and start a rocket ride to the surface. It occurs to you that you might conserve what little air you have left by holding your breath, but just then the Spirit of Robert Boyle appears before your astonished eyes (he's Nessie?) and tells you what? Why?

    "Breathe!"

    Because if you don't breathe, your lungs will rupture just after your eardrums do. At 100 feet down, there's not 1 atm of air in your lungs but rather four atmospheres, the extra 3 due to 3×33 feet, each one the equivalent of an atmosphere since 33 feet of 1 g/cc density is equivalent in mass to 760 mm of 13.6 g/cc (Hg) density. So as you ascend, your lungs will expand under the reducing pressure, and you must let that extra air out to keep your lungs at their same volume!

  5. The density of Al is 2.702 g/cc. You've solved the transportable hydrogen problem by developing a mossy aluminum with only 80% of that density. At saturation, each Al atom in your discovery captures a single hydrogen atom. What would have been the H2 pressure (at 25°C) in a gas cylinder of the same volume as your new hydrogen sponge? [Hint: solve it using a mole of aluminum just for convenience.] Collect your Nobel Prize.

    rmossy Al = 0.8×2.702 = 2.16 g/cc = m/V = MW/Vmolar    nAl=1
    Vmolar = MW/r = 26.98 g/mol / 2.16 g/cc = 12.48 cc/mol = 1.248×10-2 L

    nH2 = ½ nAl = ½
    P = nRT/V = ½ (0.08206 atm L/mol K)(298 K) / 0.01248 L = 980 atm !

  6. A perspicacious assistant of yours on the project in question 5 applies your techinques to Mg (density 1.74 g/cc) in hopes of making an even lighter hydrogen sponge. However, he never makes it to market because of the incredible flammability of magnesium. What volume (L) of dry air at STP (1 atm, 0°C) is required for the reaction with 10 kg of magnesium? (Assume that it reacts only with the oxygen, but that isn't really true. Mg is so reactive that it will "burn" in pure nitrogen gas to make Mg3N2 if heated hot enough! It used to be used in vacuum tubes to scavenge up the last traces of air. It prefers the O2, of course. And we prefer transisitors.)

    Mg + ½ O2 ---> MgO

    nO2 = ½ nMg = ½ mMg/MWMg
          = ½ × 10×103 g / 24.30 g/mol = 205.8 mol O2

    VO2 = nRT/P
          = 205.8 mol (0.08206 atm L/mol K)(273 K) / 1 atm = 4610 LO2

    Vair = VO2 / XO2 = 4610 L / 0.21 = 21,950 L

Comments to Chris Parr Return to CHM 1316 HomePage Last modified 15 February 1999.