We're teaching this course in a classroom which is 30'x15'x8' or 97,200 L (ignoring the volume
of your bodies). At 293 K, that corresponds to 4043 moles of (ideal) air or 2.4x1027 molecules.
Those molecules (if ideal) are bouncing back and forth at random between the walls at roughly
the speed of sound (760 mph). We feel no such wind since they're not all going in the same direction.
But what's to prevent that if they're truly bouncing at random?
In fact, what's to prevent a random build-up of more than half the molecules in less than half of
the room, resulting in a pressure fluctuation? Nothing. In fact, pressure fluctuations are to be
expected from such random motion. But they're never manifest. Why not?
Let's take an egregious case: all the molecules "choosing" at once to be in the left half of the room.
Sinister, no? Half of you would asphyxiate while the other half imploded. At least briefly, at any rate
since we'd expect the compressed gas to expand in a thunderclap which would take out the wall!
No one's running for the door because you've all done the mental math.
If the probability of one molecule being on the left is (1/2) then the probability of two on the left is (1/2)(1/2).
Probability of independent events (ideal gas molecules are independent) multiply.
That means that the likelihood of all 4,043 moles being on the left is (1/2) raised to the power
2.4x1027. A little logarithmetic shows that to be 1 in 10^(7.2x1026)...that's
roughly 1 followed by 1200 times Avogadro's Number of zeroes!! Such numbers are inconceivable...except
by mathematicians who call 10100 a googol and 10 googol a googolplex.
So we're something in between.
So it's NOT going to happen. Not in the lifetime of this universe or in a googol of universes!
While such a gross fluctuation ISN'T going to occur, what magnitude of fluctuation will?
It's clear that the probability calculation above was just like flipping (2.4x1027) coins.
Heads you're left; tails you're right. In preparation for this class, I flipped over 1,000,000 coins
(on the computer) to simulate this problem. And here are the results:
10 trials with 10 coins.
7 5 5 8 4
3 3 5 8 2
Average was 5 ± 2 or 0.5 ± 0.2
10 trials with 100 coins.
42 45 50 47 53
48 49 41 55 48
Average was 47.8 ± 4.166536 or 0.478 ± 0.0417
10 trials with 1000 coins.
499 502 519 483 499
487 504 498 473 523
Average was 498.7 ± 14.44281 or 0.4987 ± 0.0144
10 trials with 10000 coins.
4926 4925 5000 5012 5050
5019 5016 5094 5073 5037
Average was 5015.2 ± 52.50715 or 0.5015 ±- 0.0053
10 trials with 100000 coins.
49827 49961 50091 50021 50001
49851 49874 50064 49842 50054
Average was 49958.6 ± 95.42494 or 0.4996 ± 0.00095
Two things stand out from these trials. The first is that the deviations from 50:50 grow (2 to 95 on average).
The second is that the relative deviations diminish (from 0.2 to 0.00095 on average).
In other words, 50:50 is a better bet the more coins you toss. The deviation grows more slowly than
the average. In fact, in the limit of large numbers, the deviation grows on the order of the square
root of the number.
# of Heads (# of Heads)1/2 Deviation Above
5 2.2 2
48 7.0 4.1
499 22.4 14.4
5015 70.2 52.5
49959 223.6 95.4
So if we had the expected number of molecules in the left half of the room (exactly half the total or
1.2x1027), we'd expect a deviation of roughly square root of that or 3.5x1013 extra
molecules on one side or the other. Nonchemists might be concerned about "so many," but we realize that
the trivial overpressure implied by that is only (3.5x1013) / (1.2x1027) = 2.9x10 -14
atm! Unmeasurable.
And so it is with virtually any variation from thermodynamic equilibrium of any sort. There are variations, but they're down below the 12th decimal place! Thermodynamic tables are secure. The statistics of large numbers makes for very trustworthy averages.
Were we to deal with molecules a few at a time, all bets would be off. For example, water droplets grow easily in saturated air in spite of the increased number of surface molecules which (we'll see near the end of the course) are under surface tension which raises the average energy of the droplet with their addition. This is because the bulk molecules (which, unaffected by surface tension, diminish the average energy) grow faster (as radius cubed) than surface molecules (as only radius squared). But when a droplet is so small that it's all surface and no interior, it lowers its energy by evaporating even into saturated air!
The breakeven point (thermodynamically) between growing and shrinking is on the order of 10 molecules! We see from the table above that the statistics of large numbers isn't going to ensure small deviations from the average under such circumstances! So equilibrium arguments should be made with caution unless one has a comforting Avogadro's Number of molecules to back one up.
Return to the CHM 5414 Lecture Notes or Go To Next or Previous Lectures.
Chris Parr
University of Texas at Dallas
Programs in Chemistry, Room BE3.506
P.O. Box 830688 M/S BE2.6 (for snailmail)
Richardson, TX 75083-0688
Voice: (972) 883-2485
Fax: (972) 883-2925
BBS: (972) 883-2168 (HST) or -2932 (V.32bis)
Internet: parr@utdallas.edu (Click on that address to send Dr. Parr e-mail.)
Last modified 18 September 1997.