Question

# Consider a 3 molecule in a box system, each molecule may be on the right side...

Consider a 3 molecule in a box system, each molecule may be on the right side or the left side. a) What is the number of states? b) What is the total multiplicity? c) What is the probability of all of the molecules being on the right side of the box?

The first answer I know is 4.

a)

There are 3 identical “molecules” and a box with two parts.
The possible states, written (X,Y) where X is the number in left side and Y is the number in the right side are:

(3,0) (2,1) (1,2) (0,3)

Total number of states is 4.

(b)

The multiplicity, W, of a state is the number of different ways in which that state can be achieved. It is equal to W=N!/n1!n2!. So in this case the multiplicities are:

3! / (3! * 0!) , 3! / (2! * 1!) , 3! / (1! * 2!) , 3! / (0! * 3!) ,

= 1, 3, 3, 1. There are 1 + 3 + 3 + 1 = 8 possible states in total.

The total multiplicity is 8.

(c)

Probability of all of the molecules being on the right side of the box = Multiplicity for state (0, 3) / Total Multiplicity

= 1 / 8

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