Working Content > Thermodynamics and statistical physics > Boltzmann distribution

Prerequisites:

• Gibbs free energy
• Boltzmann distribution

The Boltzmann factor, e^-E/k_BT,tells us the relative probability of a particular arrangement (with a given energy).  And if we're dealing with a system at constant pressure, we can use enthalpy as the relevant energy, and call it e^-H/k_BT.

But as we discussed at the end of the Boltzmann distribution page, that's not the whole story: we need to multiply it by a weighting factor. While the Boltzmann distribution gives the probability of a given degree of freedom having a given energy, there may be different numbers of states that have the same energy. We have to multiply our probability by the number of different arrangements there are that have a given energy. Let's do the math and see what happens.

We know S = k_B ln W (where S is the entropy and W is the number of arrangements), so solving for W we get W = e^S/k_B

So to find the overall probability that a system is in a given state, we multiply these together:

W * e^-H/k_BT = e^S/k_B * e^-H/k_BT = e^TS/k_BT* e^-H/k_BT = e^-(H-TS)/k_BT

What's H - TS?  Hey, it's Gibbs free energy G!    G=H-TS

So the overall probability - based on both the energy of the microstates and the number of possible microstates with that energy -  is actually proportional to e^-G/k_BT.

This tells us two things:

1) Our reasoning above is meant to illustrate that Gibbs free energy combines the effects of energy (here, that's the original Boltzmann factor) and of entropy (the number of possible arrangements)

2) For systems at constant pressure and temperature, the Gibbs free energy is what really tells us which states are more probable, and therefore what a system is actually going to do.  That's why Gibbs free energy gets so much attention in chemistry and biology.

Ben Dreyfus 1/9/12

Wolfgang Losert 2/8/2013