Chapter 13, Problem 13C.7P

(a)

Interpretation Introduction

Interpretation:

The electronic partition function of $\text{NO}$ has to be calculated for the value of temperature from $T=0\text{ K}$ to $T=1000\text{ K}$.  The graph of electronic partition function of $\text{NO}$ and temperature has to be plotted.  The population of levels of $\text{NO}$ molecule has to be calculated.

Concept introduction:

Statistical thermodynamics is used to describe all possible configurations in a system at given physical quantities such as pressure, temperature and number of particles in the system.  An important quantity in thermodynamics is partition function that is represented by the expression as shown below.

  $q={\displaystyle \sum _i{g}_i{e}^{-{\epsilon }_i/kT}}$

Answer to Problem 13C.7P

The electronic partition function of $\text{NO}$ for the value of temperature from $T=0\text{ K}$ to $T=1000\text{ K}$ is shown below in Table.

Temperature $\left(T\right)$	Partition $\left(q\right)$
$0\text{ K}$	$\infty$
$200\text{ K}$	$2.837095$
$400\text{ K}$	$3.293905$
$600\text{ K}$	$3.496046$
$800\text{ K}$	$3.608667$
$1000\text{ K}$	$3.680269$

The graph of electronic partition function of $\text{NO}$ and temperature is shown below.

The population of ground and first excited levels of $\text{NO}$ molecule  is $\underset{\_}{0.6412}$ and $\underset{\_}{0.3588}$ respectively.

Explanation of Solution

The partition function in terms of wavenumber is given by the expression as shown below.

    $q={\displaystyle \sum _j{g}_j{\text{e}}^{-\frac{hc{\tilde{v}}_j}{kT}}}\mathrm{....................}\left(1\right)$

Where,

$g_j$ is the degeneracy of the $i^{\text{th}}$ level.

$h$ is the Plank’s constant with a value $6.62608\times {10}^{-34}\text{J s}$.

$c$ is the speed of light with value $2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}$.

$\tilde{v}$ is the wavenumber of the $i^{\text{th}}$ level.

$k$ is the Boltzmann constant $\left(1.381\times {10}^{-23}{\text{J K}}^{-1}\right)$.

$T$ is the temperature.

The degeneracy of the ground and first excited electronic level of $\text{NO}$ molecule is $2$.

The wavenumber of first excited electronic level of $\text{NO}$ molecule is $121.1{\text{ cm}}^{-1}$.

The wavenumber of ground electronic level of $\text{NO}$ molecule is $0{\text{ cm}}^{-1}$.

The temperature range is from $T=0\text{ K}$ to $T=1000\text{ K}$.

The partition function of $\text{NO}$ molecule is given by the expression as shown below.

    $q=g_0{\text{e}}^{-\frac{hc{\tilde{v}}_0}{kT}}+g_1{\text{e}}^{-\frac{hc{\tilde{v}}_1}{kT}}$

Substitute the values of $h$, $c$, ${\tilde{v}}_0$, ${\tilde{v}}_1$, $g_0$, $g_1$,and $k$ in the above equation.

    $\begin{array}{c}q=\left(\begin{array}{c}\left(2\right){\text{e}}^{-\frac{\left(6.62608\times {10}^{-34}\text{J s}\right)\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(0{\text{ cm}}^{-1}\right)}{\left(1.381\times {10}^{-23}{\text{J K}}^{-1}\right)T}}\\ +\left(2\right){\text{e}}^{-\frac{\left(6.62608\times {10}^{-34}\text{J s}\right)\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(121.1{\text{ cm}}^{-1}\right)}{\left(1.381\times {10}^{-23}{\text{J K}}^{-1}\right)T}}\end{array}\right)\\ q=\left(2\right)\left(1\right)+\left(2\right){\text{e}}^{-\left(\frac{174.193\text{ K}}{T}\right)}\\ q=2+2{\text{e}}^{-\left(\frac{174.193\text{ K}}{T}\right)}\mathrm{..............................}\left(2\right)\end{array}$

Substitute the value $T=0\text{ K}$ in the above expression.

    $\begin{array}{c}q=2+2{\text{e}}^{-\left(\frac{174.193\text{ K}}{0\text{ K}}\right)}\\ =2+\infty \\ =\infty \end{array}$

Similarly, the partition is calculated for the temperature range from $T=0\text{ K}$ to $T=1000\text{ K}$ using the equation (1). The calculated values of partition function is shown below in the table.

Temperature $\left(T\right)$	Partition $\left(q\right)$
$0\text{ K}$	$\infty$
$200\text{ K}$	$2.837095$
$400\text{ K}$	$3.293905$
$600\text{ K}$	$3.496046$
$800\text{ K}$	$3.608667$
$1000\text{ K}$	$3.680269$

Therefore, the plot electronic partition function of $\text{NO}$ and temperature is shown below.

Figure 1

The relative population of ground level of $\text{NO}$ is given by the expression as shown below.

    $\frac{N_0}{N}=\frac{g_0{\text{e}}^{-\frac{hc{\tilde{v}}_0}{kT}}}{2+2{\text{e}}^{-\left(\frac{174.193\text{ K}}{T}\right)}}$

Substitute the values of $h$, $c$, ${\tilde{v}}_0$, $g_0$, $T=300\text{ K}$,and $k$ in the above equation.

    $\begin{array}{c}\frac{N_0}{N}=\frac{\left(2\right){\text{e}}^{-\frac{\left(6.62608\times {10}^{-34}\text{J s}\right)\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(0{\text{ cm}}^{-1}\right)}{\left(1.381\times {10}^{-23}{\text{J K}}^{-1}\right)\left(300\text{ K}\right)}}}{2+2{\text{e}}^{-\left(\frac{174.193\text{ K}}{300\text{ K}}\right)}}\\ \frac{N_0}{N}=\frac{2}{2+1.119}\\ \frac{N_0}{N}=0.6412\end{array}$

When the total population of the system is $1$, the pollution of ground level of $\text{NO}$ is calculated as shown below.

    $\begin{array}{c}\frac{N_0}{\left(1\right)}=0.6412\\ N_0=\underset{\_}{0.6412}\end{array}$

The pollution of first excited level of $\text{NO}$ is calculated as shown below.

    $N_1=N-N_0$

Where,

$N_1$ is the population of first excited level of $\text{NO}$.

$N$ is the total population of $\text{NO}$.

$N_0$ is the population of ground level of $\text{NO}$.

Substitute the value of $N_0$ and $N$ in the above equation.

    $\begin{array}{c}{N}_1=1-0.6412\\ =\underset{\_}{0.3588}\end{array}$

Therefore, the population of ground and first excited levels of $\text{NO}$ molecule  is $\underset{\_}{0.6412}$ and $\underset{\_}{0.3588}$ respectively.

(b)

Interpretation Introduction

Interpretation:

The mean electronic energy of $\text{NO}$ at $300\text{ K}$ has to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 13C.7P

The mean electronic energy of $\text{NO}$ at $300\text{ K}$ is $\underset{\_}{8.756\times 10^{-22}J}$.

Explanation of Solution

The degeneracy of the ground and first excited electronic level of $\text{NO}$ molecule is $2$.

The wavenumber of first excited electronic level of $\text{NO}$ molecule is $121.1{\text{ cm}}^{-1}$.

The wavenumber of ground electronic level of $\text{NO}$ molecule is $0{\text{ cm}}^{-1}$.

The temperature is $300\text{ K}$.

The partition function of $\text{NO}$ molecule is given by the expression as shown below.

    $q=g_0{\text{e}}^{-hc\beta {\tilde{v}}_0}+g_1{\text{e}}^{-hc\beta {\tilde{v}}_1}$

Where,

$g_j$ is the degeneracy of the $i^{\text{th}}$ level.

$h$ is the Plank’s constant with a value $6.62608\times {10}^{-34}\text{J s}$.

$c$ is the speed of light with value $2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}$.

$\tilde{v}$ is the wavenumber of the $i^{\text{th}}$ level.

$\beta$ is the temperature dependent coefficient $\left(\frac{1}{kT}\right)$.

The mean energy, $\langle \epsilon \rangle$, is given by the expression as shown below.

    $\langle \epsilon \rangle =\left(-\frac{1}{q}\frac{\partial q}{\partial \beta }\right)$

Substitute the value of $q$ in the above equation.

    $\begin{array}{c}\langle \epsilon \rangle =\left(-\frac{1}{g_0{\text{e}}^{-hc\beta {\tilde{v}}_0}+g_1{\text{e}}^{-hc\beta {\tilde{v}}_1}}\frac{\partial \left(g_0{\text{e}}^{-hc\beta {\tilde{v}}_0}+g_1{\text{e}}^{-hc\beta {\tilde{v}}_1}\right)}{\partial \beta }\right)\\ =\left(-\frac{1}{g_0{\text{e}}^{-hc\beta {\tilde{v}}_0}+g_1{\text{e}}^{-hc\beta {\tilde{v}}_1}}\left(-hc{\tilde{v}}_0{g}_0{\text{e}}^{-hc\beta {\tilde{v}}_0}-hc{\tilde{v}}_1{g}_1{\text{e}}^{-hc\beta {\tilde{v}}_0}\right)\right)\end{array}$

Substitute the value of $\beta =\frac{1}{kT}$ in the above expression.

    $\langle \epsilon \rangle =\left(-\frac{1}{g_0{\text{e}}^{\frac{-hc{\tilde{v}}_0}{kT}}+g_1{\text{e}}^{\frac{-hc{\tilde{v}}_1}{kT}}}\left(-hc{\tilde{v}}_0{g}_0{\text{e}}^{\frac{-hc{\tilde{v}}_0}{kT}}-hc{\tilde{v}}_1{g}_1{\text{e}}^{\frac{-hc{\tilde{v}}_1}{kT}}\right)\right)$

Where,

$k$ is the Boltzmann constant $\left(1.381\times {10}^{-23}{\text{J K}}^{-1}\right)$.

$T$ is the temperature.

Substitute the values of $h$, $c$, ${\tilde{v}}_0$, ${\tilde{v}}_1$, $g_0$, $g_1$,and $k$ in the above equation.

    $\langle \epsilon \rangle =\left(-\frac{1}{\left(\begin{array}{c}\left(2\right){\text{e}}^{-\frac{\begin{array}{l}\left(6.62608\times {10}^{-34}\text{J s}\right)\\ \left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\\ \left(0{\text{ cm}}^{-1}\right)\end{array}}{\left(1.381\times {10}^{-23}{\text{J K}}^{-1}\right)\left(300\text{ K}\right)}}\\ +\left(2\right){\text{e}}^{-\frac{\begin{array}{l}\left(6.62608\times {10}^{-34}\text{J s}\right)\\ \left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\\ \left(121.1{\text{ cm}}^{-1}\right)\end{array}}{\left(1.381\times {10}^{-23}{\text{J K}}^{-1}\right)\left(300\text{ K}\right)}}\end{array}\right)}\left(\begin{array}{l}\left(\begin{array}{l}-\left(6.62608\times {10}^{-34}\text{J s}\right)\\ \left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\\ \left(0{\text{ cm}}^{-1}\right)\\ \left(\left(2\right){\text{e}}^{-\frac{\begin{array}{l}\left(6.62608\times {10}^{-34}\text{J s}\right)\\ \left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\\ \left(0{\text{ cm}}^{-1}\right)\end{array}}{\left(1.381\times {10}^{-23}{\text{J K}}^{-1}\right)\left(300\text{ K}\right)}}\right)\end{array}\right)\\ -\left(\begin{array}{l}\left(6.62608\times {10}^{-34}\text{J s}\right)\\ \left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\\ \left(121.1{\text{ cm}}^{-1}\right)\\ \left(2\right){\text{e}}^{-\frac{\begin{array}{l}\left(6.62608\times {10}^{-34}\text{J s}\right)\\ \left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\\ \left(121.1{\text{ cm}}^{-1}\right)\end{array}}{\left(1.381\times {10}^{-23}{\text{J K}}^{-1}\right)\left(300\text{ K}\right)}}\end{array}\right)\end{array}\right)\right)$

The above expression can be further simplified as shown below.

    $\begin{array}{c}\langle \epsilon \rangle =\left(-\frac{1}{\left(2+2{\text{e}}^{-\left(0.5595\right)}\right)}\left(0-\left(\text{4}\text{.811}\times {\text{10}}^{-21}\times {\text{e}}^{-\left(0.5595\right)}\right)\right)\right)\text{ J}\\ =\frac{\text{4}\text{.811}\times {\text{10}}^{-21}\left(0.5715\right)}{3.14}\text{ J}\\ =\underset{\_}{8.756\times 10^{-22}J}\end{array}$

Therefore, the mean electronic energy of $\text{NO}$ at $300\text{ K}$ is $\underset{\_}{8.756\times 10^{-22}J}$.