Entropy of a two-level system Consider a two-level system with energies and degeneracies $\epsilon_0 = 0, g_0=1$ and $\epsilon_1 = \epsilon, g_1=4$. I can show that the temperature at which both levels are equally populated is given by $\exp\left(\frac{\epsilon}{kT_0}\right) = 4$, at which the partition function, $q=2$ and the molar internal energy $U(T_0) = \frac{1}{2}N_A\epsilon$.
This means that the entropy at $T_0$ is $S(T_0) = R\ln 2 + \frac{N_A\epsilon}{2T_0}$, right? Is there any special interpretation of this value of the entropy? Can the two terms here be described separately by something physically meaningful?
 A: Take a similar system of $N_A$ coins, where each can be either heads or tails with equal probability. The entropy of the system? $N_A k_B \ln 2$. This is to say that $k_B \ln 2$ is a natural "unit" of entropy (the information in one bit).
Your system is slightly different (although both can be described as canonical ensembles with the fixed number of 1 mole of distinguishable particles): Each of the subsystems (each particle) can be in one of 5 microstates (which correspond to 2 macrostates at the particle level). When the probability of each macrostate is equal, we get the probabilities 1/2, 1/8, 1/8, 1/8, 1/8 for the microstates. Entropy being additive, we may write $S = \frac{k_B}{2}N_A(\ln2 + 3\ln2) = 2R\ln2$. This is indeed the same thing as your result, because $\frac{\varepsilon}{T_0} = k_B \ln 4$. 
To try to build some physical intuition, although this is not straightforward, we could say that the entropy from not knowing which macrostate the particles are in is, like in the coin flipping case, $R\ln2$. The rest of the entropy can then be attributed to the entropy of internal states (degeneracy) of the macrostate 1 (i.e. not knowing which microstate the particle is in). As you probably noticed, if one of the states does not have degeneracy, the second term in the entropy will always drop out, so this physical picture does make sense.
