# Entropy (Part 6): Randomness and ensembles

After rolling 2, 3, 4, 10 and Avogadro’s dice, as seen in the entries below, it becomes clear that the most random states (most number of ways of rolling a number) always dominate while those with fewer arrangements occur less frequently:

1 Entropy: Randomness by rolling two dice

2 Entropy: Randomness by rolling three dice

3 Entropy: Randomness by rolling four dice

4 Entropy: Randomness by rolling ten dice

5 Entropy: Randomness by rolling Avogadro’s dice

In this final entry of randomness and entropy, the concept of an ensemble is discussed.

We are using a die to represent a particle that has six states that come up randomly. Hence we have treated systems with 2, 3, 4, 10 and Avogadro’s constant (let’s use 10^{23}) of particles (dice) and have shown that as the number increases, the total number of accessible states, is given by 6* ^{n}*. Clearly the number of states in Avogadro’s case is 6

^{1023 }: an enormous number!! If you start to roll this many dice, every roll gives an outcome in exactly the same way as for 2 or 3 dice: just add up all the rolls (it will take a bit of time!! but it is still a specific number). However when your roll, and if you could add them all up, the answer would be always very close to 3.5×10

^{23 }(see end). This is clearly because the total number of accessible states,

*W*, can be replaced by the total number of random accessible states,

*W*

_{random }that all have outcomes clustered around 3.5×10

^{23 }.

Each time you roll that many dice, it is unlikely they will come up in any arrangement other than the ones that give the value of 3.5×10^{23 }.

We call the collection of all those arrangements that give the most probable outcome an **Ensemble**. In French ensemble means a collection.

The ensemble concept is useful in statistical mechanics. So now let us think of a gas moving around in a container at 300 K. We could take repeated snap shots of the gas, and every time we would see the particles frozen in different position. We could also measure the speed of each particle (in principle). Each snap shot is like a roll of Avogadro’s dice. Below there are three of the many different arrangements and these will arise because each is consistent with a value of 300 K.

If the were not consistent with a temperature of 300 K such a state would be very improbable and can be neglected.

Consider he following is an ordered state:

This state is possible but it is not consistent with 300 K. In fact if the 5 balls are distinguishable, then there is only one way to arrange these. Since Boltzmann’s equation of the entropy is

*S*=*k*ln*W*_{random}

then since in that ordered state* W* = 1, then the entropy of that state is zero (ln(1)=0) which means perfect order.

In contrast, let us calculate the entropy of rolling 6^{1023 }dice (and treat them as particles). In this case the entropy is given by S = *k*ln(6^{1023}) = 1.8x*kx10 ^{23 .}* Let us use the value of Boltzmann’s constant,

*k*=1.3806503×10

^{-23}J K

^{-1 }which gives the entropy of

*S*= 2.48 J K

^{-1 }. This is the value of the randomness of the gas.

Suppose we have a system of two bulbs joined together with a closed stop cock. On one side the gas has 10^{23} particles, each with 6 states giving 6^{1023 }accessible states. Let us suppose that the empty bulb is the same size and is under the same conditions, so when the stop cock is opened, the number of accessible state increases to 6^{1023 } x 6^{1023 }. For this larger system there are many, many more accessible states and the gas will move into the evacuated bulb and occupy those states randomly. The entropy must increase. I will not do that calculation, because we need to take into account that the particles are indistinguishable, but it was from these considerations that Boltzmann determined his equation (Probabilities multiply but the entropy adds, so the only function that has this property is the logarithm.)

When Boltzmann was explaining this, Poincaré pointed out that for a finite system (fixed number of dice) it will eventually return to its original state (Poincaré recurrence theorem or ergodic theory). Apparently to this, Boltzmann replied:

“You should live so long!”

Most probable roll:

- 2 dice: (12-2)/2+2=7
- 3 dice: (18-3)/2+3=10.5 or 10 and 11 because we deal with integers only
- 10 dice: (60-10)/2+10=35
- 10
^{23}dice: (6×10^{23}– 10^{23})/2+10^{23 }= 3.5×10^{23}

The interactive software used in this video is part of the General Chemistry Tutorial and General Physics Tutorial, from MCH Multimedia. These cover most of the topics found in AP (Advanced Programs in High School), and college level Chemistry and Physics courses.