the improbable tail of two gases. gas molecules occupy the left bulb of the two-equal-bulb assembly...

The Improbable Tail The Improbable Tail of Two Gasesof Two Gases

Gas molecules occupy the left bulb of the two-equal-bulb assembly shown below; the other bulb is closed off by a partition and maintains a perfect vacuum. Let’s call this Condition 1.

FactFact: When the partition is broken, gas escapes from the left bulb and fills the right bulb until equal amounts of gas occupy both bulbs (Condition 2)*.

A similiar and related fact is: a gas occupies uniformly the entire volume of any container.

What would your students answer?

*The experimental evidence that the same amount of gas is in each bulb is that the pressure in each bulb is the same.

Question: Why?Question: Why?






“Pressure is like a force, so the pressure on the left forces gas into the right bulb until the left-right forces equalize.”






“The vacuum on the right sucks the gas out of the left bulb until there is no more vacuum to suck.”

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At T > 0 K, gas molecules are in constant and random motion.

To answer the question, we only need to know one thing about gases.

Suppose that three molecules are confined to the two-equal-bulb assembly. After the partition is broken, the molecules can move rapidlyrapidly and randomlyrandomly between the two bulbs, so the number of the number of molecules in either bulb changes constantlymolecules in either bulb changes constantly.

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At T > 0 K, gas molecules are in constant and random motion.

At time t = 0, the three molecules are in the initial state.

But as time proceeds, other states appear.

In how many other possible states can the three gas molecules be arranged?

There are three possible ways (“microstates”) in which two molecules could be in the left-hand bulb and one in the right-hand bulb).

The number of microstates in a given state is a binomial coefficient:

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23

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32

nk

n!(n-k)!k!

=

This is the number of ways n (=3) molecules can be arranged in which k (=2) of them are chosen to be in the left-hand bulb (“n choose k”).

Note that the previous arrangement is also a binomial coefficient with n = 3, k = 3, and 0! = 1 by definition.

where “!” means factorial: 3! = 321.

There are also three possible arrangements in which one molecule is in the left-hand bulb (n = 3, k = 1), and there is only one possible way to have no molecules in the left-hand bulb (n = 3, k = 0).

The number of states is n+1 since k varies discretely from 0 to n.

The total number of possible arrangments of three molecules in the two bulbs is the total number of microstates (W).

It is easy to show that the total number of ways n molecules can be arranged at random in the two bulbs is:

W = 2n

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molecu

les = 3 states =

4 microstates =

8

As time progresses and the molecules circulate at random between the two bulbs, the probability that any one of the microstates will be visited is 1/8. That is, each microstate has an equal probability of existing.

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The probability that any one of the states will exist is the sum of probabilities of its microstates. Thus, the probability that there would be two molecules in the left bulb (microstates 2 or 3 or 4) is 3/8. The probability that two molecules would be in the right bulb is also 3/8 = 0.375 or 37.5%.

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Note that the probability of an approximately equal distribution of molecules (two molecules in the left or the right bulb) is 0.75; it is less likely (0.25) that all of the molecules will be found in either the left or the right bulb.

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For 30 molecules, there are 31 states and more than 1 billion microstates. Again, the most probable state is the one with an equal (50-50) distribution of molecules between the bulbs. The probability The probability of any one state is equal to its number of microstates divided by the of any one state is equal to its number of microstates divided by the total number of microstatestotal number of microstates. Thus ...

n = 30w = 230 = 1,073,741,824

P(k

)

k = molecules in left bulb

n = 30w = 230 = 1,073,741,824

P(n,k) = n!

2n k!(n-k)!

... the binomial distributionbinomial distribution for this two-bulb model is shown below.

Note that the probability that the distribution is exactly equal decreases as the number of microstates increases; for example:

P(4,2) = 0.375 but P(30,15) = 0.144

(plotted as a continuous function, but it is really a discrete function)

P(k

)

k = molecules in left bulb

n = 30w = 230 = 1,073,741,824

P(n,k) = n!

2n k!(n-k)!

... the binomial distributionbinomial distribution for this two-bulb model is shown below.

But the probability of having a 30-molecule arrangement which deviates from a 15-15 distribution by as much as 5 molecules is only

P(30,10) = P(30,20) = 0.028.

P(k

/n)

In summary, the 50-50 arrangment is the most probablemost probable arrangement for any number n of gas molecules in this 2-bulb apparatus.

The maximum liklihood (the probability that the distribution is exactly 50-50) is Pmax and is impossible to calculate for large n ...

k/n

n = 100

n = 200

n = 400

P(n,k) = n!

2n k!(n-k)!

P(k

/n)

...but easy to estimate using Sterling’s approximation:

k/n

xxex xx 2!

222

2222

2

nne

nnennn

nn

n

2P(k/n = 0.5) ≈ =

n = 100

n = 200

n = 400

P(n,k) = n!

2n k!(n-k)!

k/n

Of more interest is the relative dispersion of the binomial distribution function.

P(k

/n)/

P(0

.5)

n = 100

n = 200

n = 400

P(n,k) = n!

2n k!(n-k)!

One measure of this is the half width at half height.

This number also decreases with n, so the probability of a significant excursion from 50-50 becomes increasingly less likely.

k/n

A more rigorous measure of dispersion is the variance:

where the quantities in brackets are the moments of the distribution:

P(k

/n)/

P(0

.5)

222 kk

n

0k

rr k)P(n,kk or for large n: n

0

rr dkk)P(n,kk

n = 100

n = 200

n = 400

P(n,k) = n!

2n k!(n-k)!

k/n

The most commonly used measure of dispersion is the standard deviation , the square root of the variance. It can be shown that

P(k

/n)/

P(0

.5)

100 = 0.050

200 = 0.035

400 = 0.025 n2

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n

σ

P(n,k) = n!

2n k!(n-k)!

P(k

/n)/

P(0

.5)

For n = NA = 6.02×1023 (Avogadro’s number) of molecules, it is virtually impossible to have an exact 50-50 distribution (Pmax = 1.45×10-12)

k/n

P(n,k) = n!

2n k!(n-k)!

P(k

/n)/

P(0

.5)

For n = NA = 6.02×1023 (Avogadro’s number) of molecules, it is virtually impossible to have an exact 50-50 distribution (Pmax = 1.45×10-12), but it is also highly unlikely to find significant excursions from an approximately equal distribution.

k/n

NA

= 6.44×10-13

P(k

/n)/

P(0

.5)

k/n

NA

= 6.44×10-13

Thus, the gas would always seem to be evenly distributed between the two bulbs

P(k

/n)/

P(0

.5)

k/n

NA

= 6.44×10-13

Thus, the gas would always seem to be evenly distributed between the two bulbs even though the exact numbers of molecules in the two bulbs fluctuate rapidly

P(k

/n)/

P(0

.5)

k/n

NA

= 6.44×10-13

Thus, the gas would always seem to be evenly distributed between the two bulbs even though the exact numbers of molecules in the two bulbs fluctuate rapidly and they are almost never exactly equal.

Now consider a mole of gas molecules in a container of any size or shape.

This is just another form of the “two-bulb” model; thus, the gas appears to fill all parts of the container equally and uniformly.

Divide the container into two equal halves using a partition in any orientation, but allow molecules to pass through the partition.

That is, during the system’s “thermal fluctuations”, the microstates close to a 50-50 distribution are visited often, while those microstates far from an equal distribution are virtually never visited.

Is this result an artifact of using a “two-bulb” model? What if the container is partitioned into more than halves?

Suppose the gas container is divided into ss equal partitions, and again the gas molecules are free to move at random into any one of the equal partitions. At one instant in time ...

s

1p

and the probability of not finding that molecule in that partition is s

1-sq

It is easy to show that the probability of finding kk of the nn gas molecules in a given partition is

the probability of finding a molecule in one of the partitions is

knkqpk)!(nk!

n!s)k,P(n,

and the most probable number of molecules in each partition is s

n

Again, the gas molecules are distributed uniformly throughout the container, because the probability of a non-uniform distribution is vanishingly small!

One mole of gas molecules occupies the left bulb of the two-equal-bulb assembly shown below; the other bulb is closed off by a partition and maintains a perfect vacuum. Let’s call this Condition 1.


Because this is the most likely arrangement.


Other arrangements of the gas in these two bulbs are not impossiblebut their probabilities lie in the tails of the binomial probability distribution function and are just highly unlikely!

the improbable tail of two gases. gas molecules occupy the left bulb of the two-equal-bulb assembly...

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