For a diatomic gas near room temperature, the internal partition function is simply the rotational partition function computed in section 6.2, multiplied by the degeneracy $Z_e$of the electronic ground state.

(a) Show that the entropy in this case is

$S=Nk\left[In\left(\frac{V{Z}_e{Z}_{rot}}{N{v}_Q}\right)+\frac{7}{2}\right]$.

Calculate the entropy of a mole of oxygen at room temperature and atmospheric pressure, and compare to the measured value in the table at the back of this book.

(b) Calculate the chemical potential of oxygen in earth's atmosphere near sea level, at room temperature. Express the answer in electron-volts

We have given that a diatomic gas near room temperature, the internal partition function is simply the rotational partition function computed in section 6.2, multiplied by the degeneracy $Z_e$

We need to calculate the entropy of a mole of oxygen at room temperature and atmospheric pressure, and compare to the measured value in the table at the back of this book.

The entropy is given as:

$S=NkIn\frac{V}{N{v}_Q}+\frac{5}{2}-\frac{\vartheta F_{int}}{\vartheta T}$

where $F_{int}$is the internal Helmholtz free energy and it's given in terms of the internal partition as:

$F_{int}=-NktIn\left(Z_{int}\right)$

for a diatomic gas, the internal partition function is simply the rotational partition function multiplied by the degeneracy of the electronic ground state $Z_e$,the at is:

$Z_{int}=Z_e{Z}_{rot}$

substitute into $\left(2\right)$to get:

$F_{int}=-NkTIn\left(Z_e{Z}_{rot}\right)$

the partial derivative with respect to the temperature is, So(note that $Z_{rot}$is proportional to $T$):

$\frac{\vartheta F_{int}}{\vartheta t}=-NkIn(Z_e{Z}_{rot}0-nk$

substitute into $\left(1\right)$to get:

$$\begin{gathered} S=NkIn\frac{V}{N{v}_Q}+\frac{5}{2}+nKIn\left(Z_e{Z}_{rot}\right)+Nk \\ S=NkIn\frac{V}{N{v}_q}+In\left(Z_e{Z}_{rot}\right)+\frac{5}{2} \\ S+NKIn\frac{V{Z}_e{Z}_{rot}}{N{v}_Q}+\frac{7}{2} \end{gathered}$$ $\left(3\right)$

Considering an oxygen molecule at the room temperature and atmospheric pressure where $Z_e=3$. The rotation partition for a diatomic gas is given by:

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

Here,$\in$is the energy constant, which equals $0.00018eV$for oxygen. So at the room temperature, the rotational partition function is:

The quantum volume will be given as:

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

The mass of the oxygen molecule $O_2$is $32u$,here$u=1.{66}^{-27}kg$, substituting (not that $h=6.626\times {10}^{-34}J.s$and $k=1.38\times {10}^{-23}J/K$):

$$\begin{gathered} v_Q=\frac{{(6.626\times {10}^{-34}J.s)}^2}{2\mathrm{\pi }(32\times 1.66\times {10}^{-27}\mathrm{kg})(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K})\left(300\mathrm{K}\right)} \\ =5.66\times {10}^{-33}{\mathrm{m}}^3 \end{gathered}$$

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

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

at a pressure of $1atm$and the room temperature we have:

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

plug all these numbers into the natural logarithm in the equation $\left(3\right)$to get:

$In\frac{V{Z}_e{Z}_{rot}}{N{v}_Q}=In\frac{(4.1\times {10}^{-26}{m}^3)\left(3\right)(71.8)}{5.66\times {10}^{-33}{m}^3}=21.17$

now substitute into the equation $\left(3\right)$to get the entropy as:

$S=Nk21.17+\frac{7}{2}$

but $Nk=nR$,

$S=n(8.314J/K.mol)21.17+\frac{7}{2}$

for one mole $n=1$,

$$\begin{gathered} S=(8.314J/K)21.17+\frac{7}{2}=205.1J/K \\ S=205.1J/K \end{gathered}$$

We have given that a diatomic gas near room temperature, the internal partition function is simply the rotational partition function computed in section 6.2, multiplied by the degeneracy $Z_e$of the electronic ground state.

We need to calculate the chemical potential of oxygen in earth's atmosphere near sea level, at room temperature.

The chemical potential is given by:

$\mu =-kTIn\frac{V{Z}_e{Z}_{rot}}{N{v}_Q}$

substitute with the values to get:

$$\begin{gathered} \mu =-(8.62\times {10}^{-5}eV/K)\left(300K\right)(21.17) \\ =-0.547eV \\ \mu =-0.547eV \end{gathered}$$