# Statistical thermodynamics [closed]

Entropy of indistinguishable particles $$S$$ is

$$S = k\ln W,$$

but in some places notation is different:

$$S = k\ln Q + \frac{U}{T}$$

Can you explain, please?

## closed as off-topic by airhuff, Todd Minehardt, Mithoron, Karsten Theis, Jon CusterJun 5 at 23:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

The only time I've ever seen that second notation is when referring to the substitution of the Helmholtz energy, as A = U-TS and therefore A = -kTlnQ. If you want an in depth explanation of how the thing is derived I suggest looking at Chapter 10 of Biomolecular Thermodynamics: From Theory to Application by Douglas E. Barrick.

Here are some photos of the key parts:  The second formula you quote follows directly from the first. Starting with $$W=N!/(n_1!n_2!\cdots)$$ for $$N$$ distinguishable particles spread over set of $$i$$ energy levels with energy $$\epsilon_i$$, then $$\ln(W)=\ln(N!)-\sum\ln(n_1!)$$. Using the Sterling formula for factorials the entropy becomes, after some fiddling around and simplifying, $$\displaystyle S=-Nk_B\sum \frac{n_i}{N}\ln\left(\frac{n_i}{N} \right)=-Nk_B\sum p_i\ln(p_i)$$ where $$p_i=n_i/N$$ is the chance that the $$i^{th}$$ state is occupied. This fraction is also $$p_i=e^{-\epsilon_i/k_BT}/Z$$ where $$Z$$ is the partition function and $$Z=\sum_ie^{-\epsilon_i/k_BT}$$ then $$\displaystyle \ln(p_i)=-\frac{\epsilon_i}{k_BT}-\ln(Z)$$ and substituting gives

$$S=-Nk_B \left(-\frac{1}{k_BT}\sum(p_i\epsilon_i)-\ln(Z\sum p_i)\right)$$

which becomes $$\displaystyle S=\frac{U}{T}+Nk_B\ln(Z)$$

after evaluating the terms $$\displaystyle \sum p_i=1$$, the probability over all states is one and the total energy $$\sum p_i\epsilon_i$$ is $$\displaystyle \sum p_i\epsilon_i=\frac{U}{N}$$ where $$U$$ is the internal energy.

This approach is only valid for solids because in some way the atoms in a solid can be identified. In a gas this is not possible and the number of microstates must be divided by $$N!$$ because after exchanging atoms/molecules the state of the gas still it indistinguishable. For a perfect gas it is found that $$\displaystyle S=\frac{U}{T}+Nk_B\ln(Z) -Nk_B(\ln(Z)-1)$$