For a diatomic gas near room temperature, the internal partition function is simply the rotational partition function multiplied by the degeneracy Ze of the electronic ground state

We can calculate the entropy from the partition function as follows. The partition function for an ideal gas is:

Z = Z1^(N)/N!

where Z1 is the partition function for a single molecule. Z1 factorizes:

Z1 = Z_{trans}*Z_{rot}*Z_{vib}*Z_{elec}*etc.

The translational part of the partition function is:

Z_{trans} = V*(2 pi m kT/h^2)^(3/2)

Since a partition function must be dimensionless the factor

(2 pi m kT/h^2)^(3/2)

must have the dimensions of an inverse volume. You can easily check this. This is the volume at which quantum effects become important. We define the quantity:

Vq = (2 pi m kT/h^2)^(-3/2)

to simplify expressions, but note that Vq depends on the temperature.

We ignore the vibrational part of the partition function and the electronic part is given by the degeneracy of the ground state, which is 3 in this problem.

The rotational part of the partition function was calculated in a previous problem and is approximately:

Z_{rot} = 1/(beta epsilon) = kT/epsilon

We have obtained all the factors of the partition function. To calculate the entropy, we can use the relations:

F = -kT Log[Z] (1)

E = -d Log[Z]/d beta (2)

S = [E - F]/T (3)

You can see from this that we can consider all the factors of Z separately.

Let's look at how a constant (i.e. temperature independent) factor contributes to S. From (2) you see that it doesn't affect E, but it does affect F as you can see from (1). So, you can see that a factor C contributes a term:

k Log[C] (4)

to the entropy.

How does a factor T^n affect the entropy? From Eq. (1), (2) and (3) you easily find that the contribution is:

nk + k Log[T^n] (5)

From (4) and (5) you see that the rule by which you can calculate the entropy if the partition function only contains factors that are powers of T and constant factors is very simple. You just have to add up the logarithms of the factors multiplied by k and for each nth power of T there is a term nk. In our case the entire partition function has this structure, therefore:

S = k Log[Z] + Nk*5/2 (6)

The last term comes from the fact that Z_{trans} is proportional to T^(3/2) and that Z_{rot} is proportional to T. Both these terms appear to the Nth power in the full partition function, so in the full partition function Z there is a factor T^(5/2N).

Let's simplify Log[Z].

The 1/N! factor yields:

-Log[N!] = -N log[N] + N

And we write Z as the Nth power of the internal partition functions and the translational part divided by N! we get:

S = N k Log[N^(-1)V/Vq Z_{rot}Z_{elec}] + Nk*7/2

Z_{elec} = 3, because we can ignore the electronic excitations, but the ground state is three fold degenerate.

I don't know exactly what epsilon is, but if I use the same figure as for nitrogen, which is 2.5*10^(-4) eV, I find that at 298.15 K and 10^5 Pa the entropy is 208 J K^(-1) for one mole.

Codata gives:

http://www.codata.org/resources/databases/key1.html

S(298.15 K) = 205.152 ± 0.005 J K^(-1) mol^(-1)

To calculate the chemical potential you can first calculate the Gibbs energy:

G = E - TS + PV = 7/2NkT - TS

And then use that G = N mu --->

- k T Log[N^(-1)V/Vq Z_{rot}Z_{elec}]

You must evaluate N by using the ideal gas law and the fact that the partial pressure is about 0.2 bar, So, N is lower by a factor of 5 than at 1 bar