Chm 1316 Honors Freshman Chemistry II
Spring 1999
Sample Exam 1

  1. You inflate your bicycle tires to the manufacturer's recommended 55 psi on a cold winter's day when the ambient temperature is 32ºF. This being Texas, the next day you're off in 75ºF weather on a bike ride, but the flexing of the tires renders them and the enclosed air at 100ºF. Given no leakage, what gauge pressure must those tires stand if they don't fail you on your ride?

    Of course, when you fill up the tire, you're increasing n, the number of moles of gas inside that volume; hence the pressure rises. But we'll assume that as soon as you cap it, the tire loses no gas. Furthermore, we'll assume that V is fixed which isn't a bad assumption if the tire is fully inflated.

    So this involves P rising with T at fixed n,V. That's not quite any of the named laws, but it follows easily enough from the general PV=nRT. If we just put P and T on the same side of the equation, P/T=nR/V, everything on the right side is constant in this case, which means the rule we want to produce is P1/T1=P2/T2 since they're both equal to nR/V.

    We have values for everything except P2 so P2=P1(T2/T1) will do the trick. All we have to do is convert the temperatures to absolute. (Since the occur as a ratio, we could even use degrees Rankin, the absolute Fahrenheit scale, but I won't do that to you.) T absolute = 273+°C and °C=(°F-32)×(5/9). So T1 = 273 K and T2 = 311 K.

    P2 = 55 psi (311 K/273 K) = 63 psi

  2. The dominant non-ideality in most gases is attractive. If you were to use such a non-ideal gas to estimate the temperature of absolute zero, would your estimate be low, high, or accurate? Why?

    Because of the attractions, the measured volumes at a given pressure are too small. As T is lowered, the effect of the attractions becomes even more acute. So the V vs. T line falls too swiftly toward V = 0 even if we stay above any condensation temperature. That will make the T intercept from this data too high.

  3. Find the density of saturated air at 1 atm and 33ºC where its composition is 75% N2, 20% O2, and 5% H2O. Compare it to dry air (79% N2 and 21% O2) at the same conditions. See? It doesn't even take a temperature difference to cause humid air to rise.

    r = m/V = (n/V)MW = (P/RT)MW where P = 1 atm, T = 306 K, and R = 0.08206 atm L/mol K.

    r = MW×(1 atm / [0.08206 atm L/mol K×306 K]) = 0.0398×MW mol/L

    MW = 0.75×MWN2 + 0.20×MWO2 + 0.05×MWH2O = 28.3 g/mol on average. Which makes humid air r = 1.13 g/L

    But MW = 0.79×MWN2 + 0.21×MWO2 = 28.8 g/mol for dry air. That makes r = 1.15 g/L, and dry air is heavier than humid air.

  4. Fluorine can make gases of the most unlikely materials. For example, fissionable 235U is separated from boring 238U by making hexafluorides (b.pt. 56ºC) of them both, UF6, and counting on their relative rates of diffusion to harvest the faster from the slower molecule. Unfortunately, that difference is so small that the uranium diffusion plants consume the entire hydroelectric power of the Savannah River (for heaters) in the process! What is the ratio of their diffusion rates? (They actually use uranium hexacarbonyl, but I couldn't find its boiling point. Must be classified. J )

    Come to think of it, they SHOULDN'T use U(CO)6 because carbon comes in several isotopes (12C, 13C, etc.) while fluorine has only the 19F isotope. (Be is the earliest element in the periodic table with only a single isotope.)

    We need that isotopic purity (no choice) to ensure that UF6 differs in molecular weight ONLY due to the uranium. Those six fluorines will weigh 6×19.00 = 114 g/mol regardless of its 235U or 238U partners, which weigh 235.0439 u and 238.051 u where u is the atomic mass unit, (1/12)m12C = 1.66054×10-24 g/atom × 6.02214×1023 atom/mol = 1 g/mol (surprise surprise).

    So MW(235UF6) = 349 g/mol and MW(238UF6) = 352 g/mol, and by the Equipartition theorem, ½m235v2235 = ½m238v2238 or (v235/v238) = (m238/m235)½ = (352/349)½ = 1.004! Or the light one's diffusing only 0.4% faster than the heavy one. Can't be an efficient process.

  5. We justified Boyle's Law with an argument about expanding a cube (at constant T and n). Why doesn't it make a difference what shape the container was?

    Because we used the linear transit distances which scale directly with the (duhh) linear scaling factor, the areas which always scale as the square of the dimensions (regardless of shape), and the volume which always scales as the cube of the dimensions. Imagine doubling the dimensions of Michaelangelo's David. How much more would it weigh? Eight times more, and David's no cube.

  6. NOx, the generic name for mixed nitrogen oxides, is a radical scavenger; it attacks other molecules with odd numbers of electrons, like ozone. The automotive industry actually testified before Congress that it should be permitted to increase NOx emissions in engines as a public health benefit since ozone concentrations near freeways was known to be lower than normal due to that cause. Were you a scientifically literate Congressman in those hearings, how would you have responded to Detroit and why?

    Simply put, after NOX has eaten up all the ozone at the freeway, it goes on to greater mischief throughout the city. After all, the NO + O3 reaction produces NO2! Not something benign like N2! Detroit was simply blowing smoke; how appropriate.

    As Mr. Rogers might have asked, "Can you say disingenuous?"

  7. You put CokesTM in the freezer for a party soon to start and only remember they're there late in the party. You rush to the freezer and shake a can. Hearing a muffled slosh, you are relieved to find that it hasn't yet frozen. But when you open that can, out sprays some of the CokeTM and the rest promptly freezes! What has happened and why? (contributed by Dr. Hrncir)

    Of course the spray is the expanding CO2 pushing out the CokeTM. Ah, but if it is expanding, it's also pushing against the atmosphere, doing work. Where will it get the energy to do that? Steal it from the soft drink, cooling it. Since it was almost frozen anyway, it starts to freeze due to that heat loss of gaseous expansion.

    In truth, the energy required for the CO2 to do its expansion is many times less than that given up by freezing all the CokeTM. So it won't all freeze, but some will.

  8. We may have two jet streams (the other's called the Polar or Northern Jet Stream), but Jupiter has lots of them! (Recall all those bands you see in Voyager photos.) Why should there be more than one in any event?

    The first Hadley Cell comes down about 30° N where Earth's surface speed is only cos(30°)=0.866 that at the equator. So it still has a long way to fall to get to zero ground speed at the poles! Hence air rising in thunderstorms north of 30° will also hit the tropopause barrier (where the warmer stratosphere above prevents further rise) and slide still further north along it. Eventually, the velocity difference between it and the ground air is too high to support smooth flow, and it too becomes turbulent, tumbles down, leaving another, but weaker, jet stream in its wake.

    Jupiter and Europa If you calculate where the first Hadley Cell instability of that kind should occur on Jupiter (bigger equator and colder gas), it turns out to be 15° north. Next time you see Jupiter, check out where the first colored band is north of its equator. J

Comments to Chris Parr Return to CHM 1316 HomePage Last modified 4 February 1999.