Connecting Heat and Statistical definitions of Entropy

Question

My Question:

Thanks for reading.

I understand that when entropy was first defined, in an attempt to explain which processes happened and which didn't, it was defined as:

$$\Delta S=\frac{Q}{T}$$

Accepting the definition of entropy as the heat added to a system, divided by the temperature of that system, I've seen the following equation...

$$dS=\frac{nC_V *dT}{T}+\frac{nR *dV}{V}$$

...derived from $Q=dU + W$.

Here's a video in which they derive it: https://www.youtube.com/watch?v=JidcDhDXH5I

On the other hand, I've learned that Boltzmann discovered that the entropy of a system could be defined as...

$$S=k_B \mathrm{ln}(W)$$

...where $W$ is the total number of possible micro-states of the system.

However, I don't quite understand how the two equations are connected, and would really appreciate some help relating the first equation with Boltzmann's statistical interpretation!

Thanks!

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Some Thoughts So Far...

I've learned that there are two types of entropy: volume entropy, and thermal entropy.

Therefor, there are two ways in which entropy can increase.

1. The thermal entropy increases (there are more ways to distribute the energy among the particles) because the internal energy of the system increases.

   2. The volume entropy increases (there are more way to distribute the molecules in a larger volume) because...well, the volume increases.

First, on the term on the left:

$$\frac{nC_V *dT}{T}=\frac{dU}{T}$$

That's definitely accounting for an increase in thermal entropy. It's saying that for a tiny change in internal energy, we have to divide it by the current temperature to get the change in entropy.

But, why this is, statistically speaking (like the way Boltzmann would've explained it)?

Now, on the term on the right:

$$\frac{nR *dV}{V}$$

I'm not quite sure what $nR*dV$ is...but, it must be something.

Why is it (again, statistically) that a tiny change in volume (multiplied by $nR$...I'm not sure why) divided by the current volume would be equal to a tiny change in the entropy?

Thanks again.

Answer 1

In statistical physics, you can define entropy as $$ S = k_B \ln Z + \frac{U}{T} $$ where $Z = \sum_i e^{-E_i / k_B T}$ is the partition function. This is the same as $F = U - TS$ where the Helmholtz free energy is $F = -k_B T \ln Z$. Then, with $p_i = \frac{1}{Z} e^{-E_i / k_B T}$ and using $U = \sum_i p_i E_i$, it is a matter of some simple manipulation of logarithms to show that $$ S = -k_B \sum_i p_i \ln p_i. $$ If all states happen to be equally probable, $p_i = 1/W$, this reduces to $S = k_B \ln W$.

For an ideal gas, the first definition will readily give you $$ S = n (R \ln V + C_V \ln T + \text{const}). $$

As for your question about the interpretation of the volume term, note that $n R \frac{dV}{V} = d(nR \ln V)$. Thus, if the volume doubles, say, the entropy increases by $nR \ln 2 = Nk_B \ln 2$, i.e. $k_B \ln 2$ per molecule. This is exactly what you would expect from $k_B \ln W$: you double $W$ once for each molecule because its number of possible states doubles.