For a mole of nitrogen (N_2) gas at room temperature and atmospheric pressure, compute the internal energy, the enthalpy, the Helmholtz free energy, the Gibbs free energy, the entropy, and the chemical potential

please see the attached file.

We need to evaluate the partition function for the rotational degrees of freedoms for one molecule in the high temperature limit, Z_{rot}.

We can write the total partition function as:

Z = Z_{rot}*Z_{trans}

where Z_{trans} is the partition function for the translational degrees of freedom. Because the thermodynamical quantities are functions of (derivatives of) logarithms of Z, they can be written as a sum of a purely translational part and a rotational part.
We can write any of the thermodynamic quantity X as:

X = X_{rot} + X_{trans}

The translational part is just:

U_{trans} = 3/2 NkT

S_{trans} = Sackur-Tetrode formula

F_{trans} = U_{trans} - T S_{trans}

H_{trans} = U_{trans} + PV = 5/2 NkT

G_{trans} = F_{trans} + PV = F_{trans} + NkT

The chemical potential mu follows from the fact that G = N mu (let me know if you don't know how this is derived)

So, all we have to do is to calculate the quantities X_{rot} for all the above quantities.

The eigenvalues for the operator L^2 are given by h-bar^2 l(l+1). The energy is:

L^2/(2I) and this can thus take the values:

h-bar^2/(2I) l (l+1)

The degeneracy of the energy levels is (2l+1). If we put h-bar^2/(2I) equal to epsilon, which is given as 0.00025 eV we can write the partition function (for one molecule) as:

Z = sum over l from zero to infinity (2l+1) exp[-beta epsilon l(l+1)].

For high temperatures the term beta epsilon will change only slowly and we can approximate the summation by a integral. A systematic way to expand the summation in powers of beta epsilon is as follows:

One can write a summation:

Sum from n1 to n2 of f(n) as:

Integral over x from n1 to n2 of f(x)dx + 1/2 [f(n1) + f(n2)] +

sum over r from 1 to infinity B_{2r}/(2r)! [f^(2r-1)(n2) - f^(2r-1)(n1)].

This is the Euler-McLaurin summation formula, which I explain in detail in the attached PDF file (I give a heuristic derivation), see formula 10 . The numbers B_{2r} are the Bernoulli numbers. For n2--> infinity and n1 = 0, the first few terms of the above expansions gives you:

Integral over x from 0 to infinity of f(x)dx + 1/2 f(0) -1/12 f'(0) + 1/720 f'''(0)

I our case:

f(l) = (2l+1) exp[-beta epsilon l(l+1)]

The first term is the integral from zero to infinity, which yields:

1/(beta epsilon)

The second term yields:

1/2 f(0) = 1/2

The third term yields:

-1/12 f'(0) = -1/12[2-beta epsilon]

The fourth term yields (to order beta epsilon):

1/720 f'''(0) = 1/720 (-12 beta epsilon)

And you can easily see that the later terms won't contribute to order beta epsilon.

If you sum everything up you get the result:

Z_{rot} = 1/(beta epsilon) + 1/3 + beta epsilon/15

You now use the standard formulae to obtain the thermodynamic quantities from Z_{rot} (remember that this is per molecule, so we must multiply by N):

U_{rot} = -Nd Log(Z)/d beta = N[1/beta - epsilon/3 - beta epsilon^2/45]

F_{rot} = -N/beta Log[Z_{rot}] = 1/beta Log[beta epsilon] - epsilon/3 - beta epsilon^2/90

These expressions are all accurate to order beta epsilon^2, the neglected term is of order beta^2 epsilon^3.

From F_{rot} and U_{rot} you solve for S_{rot}:

F_{rot} = U_{rot} - T S_{rot}

Z_{rot} does not depend on the volume so there is no contribution to the pressure. This means that the enthalpy is just:

H_{rot} = U_{rot} + PV = U_{rot} + NkT

And also:

G_{rot} = F_{rot} + NkT

and the contribution of the rotational degrees of freedom to the chemical potential follows from:

G_{rot} = N mu_{rot}.