## Derivation of Boltzwann equation for n moles of gas

Boltzmann Equation for Entropy:

Cameron Sasmor 1G
Posts: 11
Joined: Fri Sep 26, 2014 2:02 pm

### Derivation of Boltzwann equation for n moles of gas

I don't understand how kB is changed into nR in the Boltzwann equation for n moles of gas. Can someone explain this please?

Neil DSilva 1L
Posts: 70
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Derivation of Boltzwann equation for n moles of gas

All of this assumes two states.
So we have as our equation.
For 1 mole (NA),

We use the log rule to get
, so we get
This is for one mole, so if we have n moles, we just multiple that by n to get

If you have n moles, that's , and plugging that into the equations above and following all the same rules will get you

Let's assume
We have
Then, we use the log rule to get:
This simplifies to , where , so we have

Chem_Mod
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### Re: Derivation of Boltzwann equation for n moles of gas

I assume you are probably more comfortable with the gas constant R rather than the Boltzmann constant k. These two are really just the same constant, but on different scales.

R has units of J/mol*K and relates to the entropy of 1 mole of particles. R is "macroscopic"
k has units of J/K and relates to the entropy of just 1 particle, so we often say k is "microscopic"

S = k*ln(W) gives the entropy per particle of a system with W states. So Nk*ln(W) is the total entropy of N particles. Dividing N by avogadro's number gives n, the number of moles. Multiplying k by avogadro's number gives R. Since we divided and multiplied by the same number, nothing has changed, and Nk = nR.

Justin Le 2I
Posts: 142
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Derivation of Boltzwann equation for n moles of gas

I would recommend remembering the log rule where the exponent can be turned into a coefficient. If the equation is left as it is, your calculator might overflow like with one of the homework problems.

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