91Ó°ÊÓ

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:

$S=NkIn\frac{V}{N{v}_Q}+\frac{5}{2}-\frac{\mathrm{Ï‘}{F}_{int}}{\mathrm{Ï‘}T}$

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 $T$):

$\frac{\mathrm{Ï‘}{F}_{int}}{\mathrm{Ï‘}t}=-NkIn(Z_e{Z}_{rot}0-nk$

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:

$Z_{rot}=\frac{kT}{2∈}$

Here,$∈$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:

$v_Q=\frac{h^2}{2\mathrm{Ï€³¾°ì°Õ}}$

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

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

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:

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

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

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

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:

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

substitute with the values to get:

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