Group Exercise 1--Moles for Noles--Solution Begin by calculating the volume of the atmosphere in one of two ways. One way, calculate volume of the earth as (4/3)pi(6.37 x 10^3 km)^3, which comes to 1.08269 x 10^12 km^3. Then, calculate volume of earth plus atmosphere, so that the radius becomes 6.37 x 10^3 km = 20 km, or 6.39 x 10 ^3 km. (Do you now see where sig. fig. starts to be a problem, that is you'd get the same answer if I had said assume 17 km or 23 km height). Also, this of course neglects the fact that the earth is not a perfect sphere. Either of these heights gives a volume of (4/3)pi(6.39 x 10^3 km)^3, or 1.09292 x 10^12 km^3. The two volumes are good to only 3 sig. fig., so when you subtract you get 0.01 x 10^12, or 1 x 10^10 km^3. (A more accurate radius would give you a better figure, but distortions of the earth would come into play) The alternative calculation, getting surface area as 4pir^2 gives 4pi(6.37 x 10^3 km)^2 or 5.09 x 10^8 km^2, and multiplying this by a height of 20 km gives 1.02 x 10^10 km^3. Not an exact calculation either, but note how consistent the two are. Now if the significant figures bother you on the first method, and the approximation of the earth's surface as flat bother you on the second, consider a slightly more complete calculation in which you keep the radius as a variable, "r", until after the subtraction: volume of earth + atmosphere: 4/3(pi)(r+20)^3 = 4/3(pi)(r^3+60r^2+1200r+800) volume of earth alone: 4/3(pi)r^3 Subtracting the second from the first gives the volume as: 4/3(pi)60r^2 + 4/3(pi)1200r + 4/3(pi)800 From which you see that the first term is the same you get in the second approximation above (4(pi)20r^2), and the second and third terms are much smaller and negligible when you are considering even 3 significant figures in the answer, i.e. the three terms are: 1.0198x10^10 km^3 + 0.00032019x10^10 km^3 + 0.0000000335x10^10km^3 which is 1.02x10^10 km^3 to three significant figures. Lets take the value 1.0 x 10^10 km3, which can be converted to 1.0 x 10^22 dm^3 or 1.0 x 10^22 Liters. Now, lets assume Caesar takes a breath which contains 0.5 L x 0.044 moles/L or 0.022 moles of air (at atmospheric pressure, remember). With Argon present at 0.94 mole%, this is 0.022 moles x 0.0094 x 6.022 x 10 ^23 atoms/mole = 1.245 x 10^20 atoms of Ar from just one breath. Now lets take those atoms and disperse them over the whole atmosphere: 1.245 x 10^20 atoms/1.02 x 10^22 Liters = 1.22 x 10^-2 atoms/Liter. This is the average distribution, and remember we are assuming the average is about 1/2 the concentration at the surface of the earth (0.022 mole of air per Liter versus 0.044 moles), so at the surface there should be about 2.44 x 10^-2 atoms (i.e., the "marked atoms" from one breath of Caesar) per Liter, or you should find one of these atoms in every 40.9 liters of air. Therefore after about 80 breaths, you should encounter one of the atoms. If you now want to consider that Caesar undoubtedly took more than one breath during his life-time, and even though each breath does not contain all new Ar atoms, it is easy to see that every breath you take must have a number of atoms once breathed by Caesar (or by anyone who lived long enough ago for their breaths to become dispersed throughout the atmosphere). Of course, these calculations are not "exact". One could get a better value for the size of the atmosphere and the number of moles in it from the pressure at sea level and the surface area of the earth (essentially "weighing" the atmosphere). And one would also have to develop some model for the rate of dispersion of the atoms throughout the atmosphere to take into account how many atoms in each breath are "new", and how many had been previously breathed. But you will encounter many situations is science where "order of magnitude" estimates such as were done here still give insights into a problem. Next group problem coming up soon. Be on the lookout. Dr. Light