Entropy of an ideal gas vs. entropy of a non ideal gas Question: Do you expect the entropy of an interacting gas of N particles in volume V at temperature
T to be smaller or bigger than the entropy for an ideal gas of the same (N, V, T)?
My understanding could be flawed. If so, I am happy to corrected. 
My guess is that the entropy of the nonideal gas should be greater. This is because nonideal processes are irreversible and by the second law of thermodynamics we have to factor in an increase in entropy of the universe. Whereas in the ideal gas situation, we don't have to factor this in.
 A: Consider the definition of entropy in the microcanonical ensemble:
$$S=k_B \log \Omega(N,V,E)$$
where 
$$\Omega(N,V,E)=\frac 1 {N! h^{3N}} \int_{\mathcal H<E} d^N\mathbf p \ d^N \mathbf q \tag{1}\label{1}$$
is the microcanonical partition function. $\mathcal H$ is the hamiltionian of the system, which depends on all the momenta $\mathbf p$ and all the generalized coordinates $\mathbf q$, and the energy of the system is $E$.
The entropy is therefore proportional to the logarithm of the hypervolume delimited by the energy hypersurface $\mathcal H = E$.
The hamiltionian can be written in general as the sum of a kinetic term depending on the momenta and an interaction term:
$$\mathcal H (\{\mathbf p, \mathbf q\})=\sum_{i=1}^N\frac{p_i^2}{2 m}+U(\{\mathbf q\})$$
For an ideal gas, the interaction term is zero, so that the energy hypersurface is an hypersphere:
$$\sum_{i=1}^N p_i^2=2 m E$$
The integral in Eq.\ref{1} becomes in this case trivial and can be easily solved, holding the famous Sackur-Tetrode expression for the entropy of an ideal gas:
$$S_{ig}=Nk_B \left [\log \left(\frac V N u^{3/2}\right) +\frac 3 2 \left( \frac 5 3 + \log \frac{4 \pi m}{3h^2}\right) \right]\tag{2}\label{2}$$
where $u=E/N$.
Your question therefore boils down to the following: is
$$\int_{\sum_{i=1}^N p_i^2/2 m+U(\{\mathbf q\}) <E} d^N\mathbf p \ d^N \mathbf q $$
smaller or bigger than
$$\int_{\sum_{i=1}^N p_i^2/2 m <E} d^N\mathbf p \ d^N \mathbf q \ \ ?$$
At first sight, it would seem that it all depends on the interaction term $U(\{\mathbf q\})$. Let's consider the very simple case of a single particle in 1D confined in a segment of length $L$(*); in this case, for an ideal gas ($U=0$) we simply have to calculate the area of a rectangle:
$$\Omega_{ig} = \frac 1 h \int_0^L dq \int_{|p|<\sqrt{2mE}} dp = \frac 1 h  2L \sqrt{2mE} $$
let's now add a term dependent on $q$ by defining
$$\mathcal H(p,q) = \frac{p^2}{2m} + \frac 1 2 m \omega^2 q^2$$
(you will have noticed that this is the hamiltonian of an harmonic oscillator). The hypersurface is an ellipse:
$$\frac{p^2}{2mE} + \frac 1 {2E} m \omega^2 q^2 = \frac{p^2}{a^2} + \frac{q^2}{b^2} = 1$$
and the axes of the ellipse are $a=\sqrt{2mE}$ and $b=\sqrt{2E/m\omega^2}$. If $2b<L$, the partition function is simply the area of the ellipse divided by $h$:
$$\Omega = \frac 1 h A = \frac 1 h \pi a b =  \frac {2 \pi E}{h \omega}$$
If $2b>L$, however, it will be the area of the intersection between the ellipse and the rectangle of the ideal gas. A graphical representation could help to understand:

You can see that no matter the value of $b$, you will always get an area which is smaller than the area of the rectangle corresponding to $\Omega_{ig}$ (**). You can try to devise different forms for the interaction term, but the result will always be the same: you won't be able to obtain an area larger than the one obtained for an ideal gas.
This is because an interaction term always introduce an ulterior limitation to the domain of the integral of Eq.\ref{1}, resulting always in a smaller hypervolume in phase space. We can therefore conclude that, at given $N,V$ and $E$ the entropy of an ideal gas is the largest possible:
$$S \leq S_{ig}$$
This analysis was done in the microcanonical ensemble. However, if $N$ is large, fixing the temperature (canonical ensemble) is equivalent to fixing the energy (microcanonical ensemble), so the result also holds for the canonical ensemble. You could try in principle to show it with the formalism of the canonical ensemble, but it would be slightly more complicated.

(*)  In this case we should not use definition (1) of the partition function, since that definition is actually an approximation for large $N$ (see K.Huang, Statistical Mechanics). However, I just want to prove a point here, and taking a $1D$ example is the best way to do so.
(**) Except for $b \to \infty$.
A: $ S = k_{B} \ln \Omega \; $ where $\Omega $ is the number of accessible microstates. 
$\Omega$ can be characterized by $(q, p)$ the position $q$ and momentum $p$  of particles.
"Naive" physical intuition:
Since interacting gas (e.g. van der Waal's gas) tends to attract each other, these gas particles cannot be separated too far away, compared with ideal gas.
Hence, I guess $\Omega_{non-ideal} < \Omega_{ideal}  \implies S_{non-ideal} < S_{ideal}$.
