# 17.6: Partition Functions of Systems of Distinguishable Molecules are Products of Molecular Partition Functions

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A system such as a gas can consist of a large number of subsystem. How is the partition function of the system built up from those of the subsystems? This depends on whether the subsystems are distinguishable or indistinguishable. Energy is additive so that:

\[E_{tot}(N,V) = ε_1(V) + ε_2(V) + …. \nonumber \]

Each of the molecules can have their energy distributed over their respective energy states (e.g., its vibrations). This means that each ε_{1} is already a summation over the states of the molecule. Let us assume that we can somehow distinguish all the molecules as: a,b,c,d… and denote the energy state they are in by i,j,k:

\[El(N,V) = ε_ia (V) + ε_jb+ \nonumber \]

A good example would be the molecules in a molecular crystal. They only move around a fixed site and so we can distinguish by how far molecule 'a' is from a given corner of the crystal.

The systems partition function becomes:

\[Q(N,V,\beta) = \sum_i e^{\beta ε_i} = \sum_i e^{\beta (Z_1 + )} \nonumber \]

If the particles are independent and * in*distinguishable we can split up the summation into a product of molecular partition functions:

\[Q(N,V,β) = \prod_i^N q_i(V,T)= q_a(V,T) q_b(V,T) q_c(V,T)…. \nonumber \]

Each of this is a sum over the energy states of the molecule:

\[Q(N,V,T) = \prod_i^N q_i(V,T)= \prod_i^N \sum_j e^{-\beta ε_j} \nonumber \]

So far I have done little effort to distinguish between the partition function of a molecular system q as opposed to the whole ensemble Q (the gas e.g.). If the entities that we called 'systems' are **distinguishable** we can simply multiply their partition functions, just like probabilities are multipliable. So we get:

\[Q_{distinguishable}= q^N \nonumber \]

for \(N\) distinguishable systems

If the molecules sit embedded in a crystal we can do that because we can e.g. distinguish them by their location. In a gas that is not possible. This means that if we swap two molecules this should not add any energy levels. There are a total of N! permutations possible and to avoid counting realizations double we get:

\[Q_{indistinguishable} = \dfrac{q^N}{N!} \nonumber \]

for * in*distinguishable systems.

If all the molecules in the crystal are the same, say methane we simply get:

\[Q(N,V,β) = q(V,β)^N \nonumber \]

In a gas the situation is different because we have no way to distinguish one molecule from the other. If we have N molecules we can perform N! permutations (switches) that should not affect the outcome. It can be shown that for such a system we must correct for this effect (make sure we do not count levels twice) and the partition function becomes:

\[Q(N,V,β) = \dfrac{q(V,β)^N}{N!} \nonumber \]