Multiplicity vs Partition function I'm a little confused between all the different notations for the multiplicity and partition function. They're not the same thing, are they?
I know that entropy can be expressed as
$ S = k \ln\Omega $
or
$ S = k\ln Q + kT \frac{\partial \ln(Q)}{\partial T} $
in terms of multiplicity and partition function respectively. They look like they could be related. What is the relationship?
 A: 
They look like they could be related. What is the relationship?

From your two equations, we have
$$k\ln \Omega = k \ln Q + kT\frac{\partial}{\partial T} \ln Q = k \ln Q + \frac{kT}{Q}\frac{\partial}{\partial T}Q$$
but
$$Q = \sum_ie^{-\frac{E_i}{kT}}$$
and so
$$k\ln \Omega = k \ln Q + \frac{kT}{Q}\frac{1}{kT^2}\sum_i E_ie^{-\frac{E_i}{kT}} = k\left(\ln Q + \frac{1}{kT} \langle E \rangle \right)$$
Dividing through by $k$ and exponentiating both sides yields:
$$\Omega = Qe^{\frac{\langle E \rangle}{kT}}$$
A: In the limit that $T\rightarrow\infty$, the partition function and the multiplicity of states are equal.
Why? Well, we have that $Q=\sum_{i} e^{-E_i/kT}$, where $i$ indexes all possible microstates. If $T\rightarrow\infty$, these Boltzman factors all approach one, and we have $Q=\sum_i 1=\Omega$.
You might think that in the limit $T\rightarrow\infty$ the two formulas you gave above would badly disagree, since there's an extra term proportional to $T$ in your second formula. But if you do the derivative explicitly, you'll find that the $\frac{\partial \ln(Q)}{\partial T}$ is proportional to $\frac{1}{T^2}$, so that the term $kT\frac{\partial \ln(Q)}{\partial T}$ goes to zero as $T\rightarrow\infty$.
A: There is a little known fact that is relevant here.  The energies of microstates are only specified up to an arbitrary additive constant (as with all energies, which are always referenced to a zero of the energy).  For example, it is common in statistical mechanics to reference all microstate energies relative to the energy of the lowest energy microstate.  A constant energy can be added or substracted from all microstate energies without changing the probabilities of the microstates:
\begin{align}
\require{cancel}
   E_i' &= E_i - C \\
    p_i &= \frac{e^{-\beta E_i'}}{\sum_s e^{-\beta E_i'}}
        =  \frac{e^{-\beta (E_i-C)}}{\sum_s e^{-\beta (E_i-C)}}
        =  \frac{e^{\beta C} e^{-\beta E_i}}{\sum_s e^{\beta C} e^{-\beta E_i}}
        =  \frac{\cancel{e^{\beta C}} e^{-\beta E_i}}{\cancel{e^{\beta C}} \sum_s e^{-\beta E_i}}
\end{align}
Now, as Alfred Centauri states, the multiplicity and partition function are related by:
$$\Omega = Qe^{\beta \langle E \rangle}$$
If, however, one normalizes the microstate energies so that the average energy $\langle E' \rangle = 0$ ($E_i' = E_i - \langle E \rangle$), we have:
$$\Omega = Q$$
exactly.  
