For a mole nitrogen $\left(N_2\right)$gas at room temperature and atmospheric pressure, compute the following $U,H,F,G,S$and $\mu$. (The electronic ground state of nitrogen is not degenerate.)

We need to compute the following $U,H,F,G,S$and$\mu$

Considering one mole of nitrogen at room temperature $T=300k$, the mass of the nitrogen molecule $N_2$is $28u$, where $u=1.66\times {10}^{-27}kg$.The total energy is given by:

$U=\frac{3}{2}NkT+U_{int}$

where,

$U_{int}=\frac{f}{2}NkT$

The rotational motion of the diatomic nitrogen corresponds two degree of freedom $f=2$, hence:

$$\begin{gathered} U=\frac{3}{2}NkT+NkT \\ U=\frac{5}{2}Nkt \end{gathered}$$

but $nk=nR$,so:

$U=\frac{5}{2}nRT$

substitute with the givens to get:

$$\begin{gathered} U=\frac{5}{2}\left(1\right)(8.314J/K(mol\left)\right)\left(300K\right) \\ =6235.5J \\ U=6235.5J \end{gathered}$$

the enthalpy is given by:

$H=U+PV$

from the ideal gas law $PV=nRT$,so:

$H=U+nRT$

substitute given values:

$$\begin{gathered} H=6235.5J+\left(1\right)(8.314J/K(mol\left)\right)\left(300K\right) \\ =8729.7J \\ H=8729.7 \end{gathered}$$

The Helmholtz free energy is given as:

$F=-nRTIn\frac{V{Z}_{int}}{N{v}_Q}+1$

The rotation partition function for a diatomic gas is given as

$Z_{rot.}=\frac{kT}{2\in }$

here,$\in$is the energy constant, which equals $0.00025eV$for nitrogen. So at the room temperature the rotational partition function is:

$$\begin{gathered} Z_{rot}=\frac{\left(8.62\times {10}^{-5}eV\right)\left(300K\right)}{2(0.00025eV)} \\ =51.72 \end{gathered}$$

The quantum volume is given as:

$v_Q={\frac{h^2}{2\mathrm{\pi mkT}}}^{3/2}$

substitute with the givens to get (note that $h=6.626\times {10}^{-34}J\left(s\right)$ and $k=1.38\times {10}^{-23}J/K$):

$$\begin{gathered} v_Q=\frac{(6.626\times {10}^{-34}J{\left(s\right))}^2}{2\mathrm{\pi }(28\times 1.66\times -{27}^{\mathrm{kg}})(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K})\left(300\mathrm{K}\right)} \\ =6.92\times {10}^{-33}{\mathrm{m}}^3 \end{gathered}$$

the volume per particle can be written as (from the ideal gas law):

$\frac{V}{N}=\frac{kT}{P}$

at pressure of and the room temperature we have

$$\begin{gathered} \frac{V}{N}=\frac{(1.38\times {10}^{-23}J/K)\left(300K\right)}{1.01\times {10}^{5}Pa} \\ =4.1\times {10}^{-26}{m}^3 \end{gathered}$$

Putting all number into the natural logarithm in equation $\left(1\right)$to get (note that the internal partition function equals the rotational partition function since the electronic degeneracy of nitrogen is $1$see problem $6.48$):

$In\frac{V{z}_{rot}}{N{v}_Q}=In\frac{(4.1\times {10}^{-26}{m}^3)(51.72)}{6.92\times {10}^{-33}{m}^3}=19.54$

now substitute into to get the Helmholtz free energy, so:

$$\begin{gathered} F=-\left(1\right)(8.314J/K(mol\left)\right)\left(300K\right)\left[19.54+1\right] \\ =-51231J \\ F=-512231J \end{gathered}$$

the Gibbs free energy is given as:

$G=F+PV$

from the ideal gas law $PV=nRT$,so:

$G=F+nRT$

substituting the given values:

$$\begin{gathered} G=-51231J+\left(1\right)(8.314J/K0(mol\left)\right)\left(300K\right) \\ =-48736.8J \\ G=-48736.8J \end{gathered}$$

Calculating the entropy fro the defination of the Helmholtz free energy as:

$$\begin{gathered} F=U-TS \\ S=\frac{U-F}{T} \end{gathered}$$

Substituting from the previous results to get:

$$\begin{gathered} F=U-TS \\ S=\frac{U-F}{T} \end{gathered}$$

The chemical potential can be calculated from the Gibbs free energy as:

$$\begin{gathered} G=\mu N \\ \mu =\frac{G}{N} \end{gathered}$$

where $N$is the number of molecules in one mole which the Avogadro number $N_A=6.022\times {10}^{23}$, substitute with the Gibbs free energy to get the following:

$$\begin{gathered} \mu =\frac{-48736.8J}{6.022\times {10}^{23}} \\ =-8.093\times {10}^{-20}J \\ =-0.506eV \\ \mu =-0.506eV \end{gathered}$$