v1 / v2 = [ MW2 / MW1 ]½ MWX / MWO2 = [ vO2 / vX ]2 = ¼2 = 1/16 MWX = (1/16) MWO2 = (1/16) 32 g/mol = 2 g/mol X = H2

One clever scheme takes advantage of such high compression of the gas that it condenses! That scheme releases CO₂ deep under the ocean, relying on the added pressure to liquify it. Indeed, this has recently been shown to work!

Suppose it is released at the average depth of the ocean (3.8 km) and the into average seawater density of 1.04 g/cm³. What is the pressure (in atm) at that depth that keeps the CO₂ liquid? (The density of mercury is 13.55 g/cm³.)

This is just a definition of pressure question. It comes down to "what is the gravitational force of the water (and atmosphere) bearing down on each square meter of ocean floor?" Weight force is just Mg where M is the total mass. M = rV and V is the volume of seawater above that one square meter. (Don't confuse that Greek "rho," r with pressure, p.) That volume is 3800 m3 since the height is 3.8 km. So the water pressure is rVg per m2 or 1.04 g/cm3 × 10-3 kg/g × 106 cm3/m3 × 3800 m3 × 9.806 m s-2 / 1 m2 = 3.875×107 Pa × 1 atm / 105 Pa = 387.5 atm of seawater Of course the total pressure counts the atmosphere pressing on the surface of the sea as well. So P = 388.5 atm at the average ocean depth.

1. What's the corresponding daily change in moles per Liter?

   The decadal change is 368-354=14 ppm. Each decade has 10 years of 365.25 days for a total of 3652.5 days. Therefore the change per day was 14 ppm / 3652.5 = 3.83×10-3 ppm = 3.83×10-9 mole fraction daily. That's also 3.83×10-9 atm/day increase. Doesn't sound like much. But we were asked for mol/L, so n/V = p/RT is how the ideal gas equation helps us: d[CO2]/dt = 3.83×10-9 atm/day / (0.08206 atm L/mol K × 273 K)     = 1.71×10-10 mol/L day-1

2. Given our previous estimate of the depth of a constant pressure atmosphere as about 8 km (1 atm at 273 K) and the radius of the Earth as R = 6378 km, calculate the number of moles of CO₂ added to the atmosphere daily.

   Moles/day = mol/L day-1 × V(in L) V ~ 4pR2 h = 4 × 3.14 × (6.378×106 m)2 × 8×103 m V ~ 4.10×1018 m3 (103 L / m3) V ~ 4.10×1021 L in the atmosphere Accumulated moles/day = 1.71×10-10 mol/L day-1 × 4.10×1021 L = 7×1011 mol/day CO2 accumulation.

3. How many tons of gasoline would we have to stop burning daily to halt the rise?

   Clearly, each mole of C8H18 produces 8 moles of CO2 by stoichiometry. So the excess CO2 per day, when translated into gasoline, is Moles octane = (1/8) × 7×1011 mol/day = 9×1010 mol/day excess octane = 0.114 kg / mole octane × 9×1010 mol/day = 1010 kg/day (1 lb/0.453 kg) (1 ton/2000 lb) = 11 megatons/day (That works out to about 2 billion gallons of gas a day or about 2 quarts of fuel for every man, woman, and child on the planet. Of course, it's not just vehicles that burn fossil fuels. You need to count power plants, water heaters, iron works, etc., etc.)

4. Why would (c) be an insufficient response to Global Warming?

   Even if that huge change stopped the CO2 from accumulating, it's already at far too high a level to maintain our historical climate. We need it not to level off at too high a value but rather to return to its pre-industrial value.

But hydrogen is a dangerous gas. No one wants to see freeways full of exploding cars (except, possibly, Hollywood). So we can't simply store H₂ in (breakable) pressure tanks! Another possibility is mossy metals which absorb and trap hydrogen, releasing it in controlled ways and not exploding if broken.

But metals are too heavy and grow brittle with hydrogen absorption. So a new method has just arisen (in Shenyang, China): using carbon nanotubes! You're already heard about Buckminsterfullerene, C₆₀, as the wonder allotrope (proposed as the Texas State Molecule because it was studied at Rice). These nanotubes are likewise composed entirely of carbon in sort of chicken-wire cylinders.

The critical aspect of the storage is that one can easily cram hydrogen in such tubes at a density equivalent to one hydrogen atom per two carbon atoms. Assuming that the density of nanotubes is that of graphite (2.25 g/cm³), what would the equivalent gas pressure of the captured H₂ be to attain its stored density (at 25°C, say)?

p = (n / V)RT implies that knowledge of gas concentration in moles per L and the temperature T permits one to calculate the resultant pressure. Since there are two C per H in these nanotubes, the [H] is half [C]. So we need to convert g/cc into mol/L: [C] = 2.25 g/cc × (1 mol/12 g) × (1000 cc/L) = 188 mol/L [H] = ½ × 188 mol/L = 94 M     [H2] = ½ 94 M = 47 M pH2 = 47 mol/L × 0.08206 atm L/mol K (298 K) = 1150 atm but with near perfect safety.

I told you M was the averaged molecular weight of air, about 29 g/mol. But in a still atmosphere, each molecule sets up its own altitude distribution via that formula at its own molecular weight. So first let us assume a weather-free troposphere (oxymoron) so that gases don't get stirred up (as they actually do). Then methane (16 g/mol) will fall off in pressure more slowly with altitude than air (29 g/mol). Given P₀(CH₄) = 2×10^-6 atm and (of course) P₀(air) = 1 atm at sea level, calculate the apparent enrichment of methane at the tropopause (8 km) for a constant T = 0°C.

In other words, at sea level, methane is 2 ppm. What is it at the top of the troposphere? (Ignoring weather-driven stirring.)

ppm (parts per million) scale directly with partial pressures. After all, 1,000,000 ppm is 1 atm at sea level. ½ atm = 500,000 ppm, and so on. So the Barometric Formula can work directly with ppm just at is can with atm as long as we compare ppm to ppm. But we need to know not only how methane falls off from sea level to 8 km but also how air does so. The ratio will be the new ppm of methane at that altitude if we keep the scaling right. pCH4 = 2 m atm e-0.016 kg/mol × 9.8 m s-1 × 8×103 m / (8.314 J/mol K × 273 K) = 1.15 m atm pair = 1 atm e-0.029 kg/mol × 9.8 m s-1 × 8×103 m / (8.314 J/mol K × 273 K) = 0.367 atm Methane/air ratio = 1.15×10-6 / 0.367 = 3.13×10-6 = 3 ppm, an apparent 50% enrichment. In reality, the mixing that weather makes in the troposphere ensures that the proportion of gases in the atmosphere (save water which freezes out) is pretty much the same throughout. However, the absence of mixing in the stratosphere means that such calculations would be valid there.

So assume you weigh 160 lbs (not trying to insult anyone, mind you). How big (radius in meters) would your (weightless) balloon have to be in order to make you neutral buoyant with only 1°C temperature difference with the surrounding 25°C air? It wouldn't be a problem living in that balloon...which might (scaled up) be a floating city! (Shades of Flash Gordon.) Remember that M_air = 29 g/cc.

We need the DM, the difference in mass between some unknown volume of cold vs. hot air, is 160 lbs (0.453 kg/1 lb) or 72.5 kg. DM = V Dr r = m/V = (n/V)MW = MW (p/RT) Dr = (MW p/R) D(1/T) V = DM / Dr = (DM / D[1/T]) (R/p MW) = { 72.5 kg / [ (1/298 K) - (1/299 K) ] } × (8.314 J/mol K / [101,325 Pa × 0.029 kg/mol]) = 18,300 m3 V = (4/3)pR3 R = (0.75 V / p)1/3 = (0.75×1.83×104/3.14)1/3 = 16.4 m In other words, a (weightless) balloon only about 100 feet across, if gently warmed by only one degree will lift you!

Similarly, nitric acid, another acid rain problem, has the anhydride that Rehal discovered in class. But nitric too has a lower oxidation state brother, nitrous acid, HNO₂, a really weak acid. What is its anhydride gas? And balance the formula for the production of nitrous acid from what is indeed called nitrous anhydride.

The nitrogen atom in the oxide, NxOy must have the same oxidation number as it does in HNO2, that is, +3. Since oxygen has an oxidation number of -2, that suggests N2O3 for nitrous anhydride. N2O3 + H2O ® 2 HNO2

Deja vu! This is the nanotubes question made dull. p = (n / V)RT = 1 g/cc (1 mol/18 g) (1000 cc/L) 0.08206 atm L/mol K (373 K) p = 1700 atm

The problem gave V=22.45 L, a typo. pideal = nRT / V = 1 mol (0.08206 atm L/mol K × 298 K) / 22.45 L = 1.089 atm ?! Oops...should've used 24.45 for one mole of an ideal gas...sorry for the typo. pideal = 1 mol (0.08206 × 298) / 24.45 = 1 atm ! (p + an2 / V2) V = nRT (ignoring the volume correction) pnonideal = (nRT / V) - an2 / V2 = 1 atm - 1 mol2 ×3.59 L2 atm/mol2/(24.45 L)2 = 1 - 0.006 = 0.994 atm A ½% difference would have been really tough to observe in our apparatus. We need a worse gas, like decane (C10H22, a=49 L2 atm/mol2), but its boiling point is 174°C!

With the thumbhole open, the pressure inside the thief is the same as that outside. So when it is lowered into water, the water fills the gourd to the same level inside as outside. But when the thumbhole is closed, the air in the gourd (above the water level) is trapped. As the thief is lifted out of the water, some of its internal water escapes, but this increases the volume the trapped air must then fill! Since n and T are fixed, Boyle tells us that the increasing volume must betoken decreased pressure. When the pressure difference inside and outside the thief is about equal to the pressure exerted by the small height of water (remember that 1 atm = 33 feet of water!) left in the thief, the water stops leaking! The thief has successfully stolen the water. The holes in the bottom have to be small so that water's natural surface tension prevents air from bubbling back into the thief as it would were the holes large.