Re: Thermodynamics

["Carl E. Mungan" <mungan@USNA.EDU>]

> Could I get a few comments on the following phrase which I believe I heard
> from a professor some years back in a stat. mech/thermo class:
> (loosely quoted)
> "In a container filled with a gas there is no reason that all of the air
> molecules couldn't rush to one side of the container, filling half of it and
> leaving the other half in a vacuum.  We just never see this happen (it is
> statistically too improbable)."
>
> and then
>
> "This is similar to saying that in a game of pool, after the 'break' all of
> the balls could end up in the exact same place as they started.  We never see
> this happen as it is too improbable."

You could perhaps make a crude but quick estimate as follows:

POOL CASE:
1. Count the number n of balls. Say n = 20. (Sorry, I haven't played
pool in a while so I don't remember the correct figure!)

2. Estimate the number of available positions N that any one ball or
molecule could occupy. This would be something like (area of
table)/(pi*R^2) where R = radius of ball. Say area of table = 4 m^2
and R = 3 cm.

3. Estimate the probability of getting any specific desired
configuration as (1/N)^n for the pool case. (I am assuming you not
only have to get the balls back into a triangle, but in the same
order as you began with. Otherwise, multiply by n!.)

The result is prob. = 10^-63.

GAS CASE:
1. Estimate the number n of molecules. Say n = 1 mol = Avogadro's number.

2. Estimate the probability of getting half the molecules in one side
as (1/2)^n. (Classical statistics should be good enough for say a
noble gas at STP.)

The result is prob. = 2^-6e-23 = 10^-100,000,000,000,000,000,000,000.

Doesn't look too similar to me - the pool case is WAAAAAAAY more likely.
--
Carl E. Mungan, Asst. Prof. of Physics  410-293-6680 (O) -3729 (F)
      U.S. Naval Academy, Stop 9C, Annapolis, MD 21402-5026
mungan@usna.edu    http://physics.usna.edu/physics/faculty/mungan/