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Expert-verified Found in: Page 255 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # 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

(a) The statement is proved below and the entropy of a mole of oxygen at room temperature and atmospheric pressure is$S=205.1J/k$.

(b) The chemical potential of oxygen in earth's atmosphere near sea level, at room temperature is$\mu =-0.547eV$.

See the step by step solution

## Part (a) step 1: Given information

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.

## Part (a) step 2: Simplify

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 $T$ ):

$\frac{\vartheta {F}_{int}}{\vartheta t}=-NkIn\left({Z}_{e}{Z}_{rot}0-nk$

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

$S=NkIn\frac{V}{N{v}_{Q}}+\frac{5}{2}+nKIn\left({Z}_{e}{Z}_{rot}\right)+Nk\phantom{\rule{0ex}{0ex}}S=NkIn\frac{V}{N{v}_{q}}+In\left({Z}_{e}{Z}_{rot}\right)+\frac{5}{2}\phantom{\rule{0ex}{0ex}}S+NKIn\frac{V{Z}_{e}{Z}_{rot}}{N{v}_{Q}}+\frac{7}{2}$ $\left(3\right)$

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 $h=6.626×{10}^{-34}J.s$ and $k=1.38×{10}^{-23}J/K$ ):

${v}_{Q}=\frac{{\left(6.626×{10}^{-34}J.s\right)}^{2}}{2\mathrm{\pi }\left(32×1.66×{10}^{-27}\mathrm{kg}\right)\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(300\mathrm{K}\right)}\phantom{\rule{0ex}{0ex}}=5.66×{10}^{-33}{\mathrm{m}}^{3}$

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:

$\frac{v}{N}=\frac{\left(1.38×{10}^{-23}J/K\right)\left(300K\right)}{1.01×{10}^{5}Pa}\phantom{\rule{0ex}{0ex}}=4.1×{10}^{-26}{m}^{3}$

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

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

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

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

but $Nk=nR$ ,

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

for one mole $n=1$ ,

$S=\left(8.314J/K\right)21.17+\frac{7}{2}=205.1J/K\phantom{\rule{0ex}{0ex}}S=205.1J/K$

## Part (b) step 1: Given information

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.

## Part (b) step 2: Simplify

The chemical potential is given by:

$\mu =-kTIn\frac{V{Z}_{e}{Z}_{rot}}{N{v}_{Q}}$

substitute with the values to get:

$\mu =-\left(8.62×{10}^{-5}eV/K\right)\left(300K\right)\left(21.17\right)\phantom{\rule{0ex}{0ex}}=-0.547eV\phantom{\rule{0ex}{0ex}}\mu =-0.547eV$

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