Classical and Semi-classical treatments of the ideal gas In the semi-classical treatment of the ideal gas, we write the partition function for the system as $$Z = \frac{Z(1)^N}{N!}$$ where $Z(1)$ is the single particle partition function and $N$ is the number of particles.  It is semi-classical in the sense that we consider the indistinguishability of the particles, so we divide by $N!$.
The resulting expression for the entropy of the system is $$S = Nk \left(\ln \left[\left(\frac{V}{N}\right) \left(\frac{2\pi mkT}{h^2}\right)^{3/2} \right] + \frac{5}{2}\right)$$
Now consider a fully classical analysis. There $Z(1) = \sum_{E} \exp (-\frac{E}{kT})$, where $E = p^2/2m$ (assuming no interaction potentials).  The problem can be mapped to an integral over the phase space of the Hamiltonian to give $$Z(1) \rightarrow \int \exp \left(-\frac{1}{2mkT} (p_x^2 + p_y^2 + p_z^2) \right)\text{d}^3 \underline{p} \,\text{d}^3 \underline{x}$$ This can then be rewritten like $$ \int \exp \left(-\frac{p_x^2}{2mkT}\right) \text{d}p_x \int \exp \left(-\frac{p_y^2}{2mkT}\right) \text{d}p_y \int \exp \left(-\frac{p_z^2}{2mkT} \right) \text{d}p_z \cdot V$$ where $V$ is the volume of the container.  Those are Gaussian integrals and so evaluation is immediate. The result is that $Z(1) = (2\pi mkT)^{3/2} V$  The corresponding entropy can be calculated and the result is that $$S = Nk \left(\frac{3}{2} + \ln\left(\frac{(2\pi mkT)^{3/2}}{V}\right)\right)$$.
My question is: What is the significance of the factors $5/2$ in the semi-classical treatment and the factor $3/2$ in the classical treatment and why are they different?  They look like the number of degrees of freedom a monatomic and diatomic molecule would have at room temperature, but I think this is a coincidence.
 A: Why in the heck are you worrying about the difference between 3/2 and 5/2?  Your formulae differ not just by 3/2 and 5/2 but also by $\ln(N)$.  For the thermodynamic limit, $\ln(N)$ is infinity, and so a lot bigger than 1.  
The number 3/2 has in this context no meaning, except that it is 5/2-1.  The number 5/2 gives the correct entropy correct for the ideal, non-relativistic gas, composed of indistinguishable particles with a single spin state and no degrees of freedom other than the position and momentum.  However, you should understand that the entropy is defined up to a constant only by the third law of thermodynamics.  This third law, in turn, would not be valid if there were not an underlying quantum mechanics.  So, this is an intrinsically QUANTUM notion, and can not really be understood without quantum mechanics.
A: Semi-classically, from Zemansky:
$$ S = k \ln \Omega \quad , \quad \Omega = \prod_i \left( \frac{g_i^{N_i}}{N_i!} \right) \quad , \quad N_i = \frac{N}{Z} g_i e^{-\epsilon_i / kT} $$
one finds, applying Stirling's approximation: $\ln(x!) \approx x \ln x - x $:  
$$ \ln \Omega = \sum_i N_i \ln \frac{g_i}{N_i} + N  =  \sum_i N_i \left( \ln \frac{Z}{N} + \frac{\epsilon_i}{kT} \right) + N = N \left(\ln \frac{Z}{N} + \frac{U}{NkT} + 1  \right) $$
$$ S = Nk \left( \ln \frac{Z}{N} + \frac{U}{NkT} + 1 \right)  $$
The $U$ term for a monatomic gas contributes $3/2$. 
[By the way, your classical $Z$ needs some kind of constant with the dimensions of momentum x distance in the denominator to make the result unit-less.  Huang calls it "h"...  ]
