Chapter 13, Problem 13B.10P

Interpretation Introduction

Interpretation:

The explicit summation vibrational partition function of ${\text{I}}_{\text{2}}$ at $100\text{K}$ and $298\text{K}$ and the amount of ${\text{I}}_{\text{2}}$ molecules in ground and first two excited levels at two temperature has to be stated.

Concept introduction:

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

  $q\equiv {\displaystyle \sum _i{g}_i{e}^{-\beta {\in }_i}}$

The vibrational partition function for molecules is represented as,

  $q^{\text{V}}=\frac{1}{\text{1}-e^{-\beta hc\tilde{\nu }}}$

Answer to Problem 13B.10P

The explicit summation vibrational partition function of ${\text{I}}_{\text{2}}$ at $100\text{K}$ is $4.049$.

The explicit summation vibrational partition function of ${\text{I}}_{\text{2}}$ at $298\text{K}$ is $4.737$.

The ratio between ground and first two excited states at $100\text{K}$ is $9.76\times {10}^{-5}$.

The ratio between ground and first two excited states at $298\text{K}$ is $0.045$.

Explanation of Solution

The important quantity in the thermodynamic is partition function that is represented as shown below.

  $q\equiv {\displaystyle \sum _i{g}_i{e}^{-\beta {\in }_i}}$

Where

• $q$ is vibrational partition function.
• $g_i$ is degeneracy.
• $\beta$ is Boltzmann constant.
• $e_i$ is energy of system.

The vibrational partition function for molecules is represented as,

  $q^{\text{V}}=\frac{1}{\text{1}-e^{-\beta hc\tilde{\nu }}}$

The value of $\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}$ for temperature $100\text{K}$ is calculated as shown below.

The value of $\frac{kT}{hc}=69.38{\text{cm}}^{-1}$

Now substitute the value $\tilde{\nu }=0{\text{cm}}^{-1}$. Then the value $\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}$ becomes:

  $\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}=\frac{0}{69.38{\text{cm}}^{-1}}=0$

Now substitute the value $\tilde{\nu }=215.30{\text{cm}}^{-1}$ . Then the value $\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}$ becomes:

  $\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}=\frac{215.30{\text{cm}}^{-1}}{69.38{\text{cm}}^{-1}}=3.103$

Now substitute the value $\tilde{\nu }=425.39{\text{cm}}^{-1}$. Then the value $\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}$ becomes:

  $\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}=\frac{425.39{\text{cm}}^{-1}}{69.38{\text{cm}}^{-1}}=6.131$

Now substitute the value $\tilde{\nu }=636.27{\text{cm}}^{-1}$. Then the value $\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}$ becomes:

  $\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}=\frac{636.27{\text{cm}}^{-1}}{69.38{\text{cm}}^{-1}}=9.17$

Now substitute the value $\tilde{\nu }=845.93{\text{cm}}^{-1}$. Then the value $\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}$ becomes

  $\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}=\frac{845.93{\text{cm}}^{-1}}{69.38{\text{cm}}^{-1}}=12.192$

The partition function of iodine at $100\text{K}$ with $0{\text{cm}}^{-1}$ energy is calculated as,

  $\begin{array}{c}{q}^{\text{V}}=\frac{1}{\text{1}-e^{-\frac{hc\tilde{\nu }}{kT}}}\\ =\frac{1}{\text{1}-e^{-0}}\\ =\infty \end{array}$

The partition function of iodine at $100\text{K}$ with $215.30{\text{cm}}^{-1}$ energy is calculated as shown below.

  $\begin{array}{c}{q}^{\text{V}}=\frac{1}{\text{1}-e^{-\frac{hc\tilde{\nu }}{kT}}}\\ =\frac{1}{\text{1}-e^{-3.103}}\\ =1.047\end{array}$

The partition function of iodine at $100\text{K}$ with $425.39{\text{cm}}^{-1}$ energy is calculated as,

  $\begin{array}{c}{q}^{\text{V}}=\frac{1}{\text{1}-e^{-\frac{hc\tilde{\nu }}{kT}}}\\ =\frac{1}{\text{1}-e^{-6.13}}\\ =1.002\end{array}$

The partition function of iodine at $100\text{K}$ with $636.27{\text{cm}}^{-1}$ energy is calculated as,

  $\begin{array}{c}{q}^{\text{V}}=\frac{1}{\text{1}-e^{-\frac{hc\tilde{\nu }}{kT}}}\\ =\frac{1}{\text{1}-e^{-9.17}}\\ =1.000\end{array}$

The partition function of iodine at $100\text{K}$ with $845.93{\text{cm}}^{-1}$ energy is calculated as,

  $\begin{array}{c}{q}^{\text{V}}=\frac{1}{\text{1}-e^{-\frac{hc\tilde{\nu }}{kT}}}\\ =\frac{1}{\text{1}-e^{-12.192}}\\ =1.00\end{array}$

The summation of theses partition function at different energy levels neglecting $\infty$ value is calculated as shown below.

  $\begin{array}{c}{q}^v={\displaystyle \sum q_i^v}\\ =1.047+1.002+1.00+1.00\\ =4.049\end{array}$

The value of $\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}$ for temperature $298\text{K}$ is calculated as shown below.

The value of $\frac{kT}{hc}=207.244{\text{cm}}^{-1}$

Now substitute the value $\tilde{\nu }=0{\text{cm}}^{-1}$. Then the value $\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}$ becomes:

  $\begin{array}{c}\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}\\ =\frac{0}{207.244{\text{cm}}^{-1}}\\ =0\end{array}$

Now substitute the value $\tilde{\nu }=215.30{\text{cm}}^{-1}$ . Then the value $\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}$ becomes

  $\begin{array}{c}\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}\\ =\frac{215.30{\text{cm}}^{-1}}{207.244{\text{cm}}^{-1}}\\ =1.038\end{array}$

Now substitute the value $\tilde{\nu }=425.39{\text{cm}}^{-1}$. Then the value $\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}$ becomes

  $\begin{array}{c}\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}\\ =\frac{425.39{\text{cm}}^{-1}}{207.244{\text{cm}}^{-1}}\\ =2.052\end{array}$

Now substitute the value $\tilde{\nu }=636.27{\text{cm}}^{-1}$. Then the value $\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}$ becomes:

  $\begin{array}{c}\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}\\ =\frac{636.27{\text{cm}}^{-1}}{207.244{\text{cm}}^{-1}}\\ =3.070\end{array}$

Now substitute the value $\tilde{\nu }=845.93{\text{cm}}^{-1}$. Then the value $\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}$ becomes:

  $\begin{array}{c}\beta hc\tilde{\nu }=\frac{hc\tilde{\nu }}{kT}\\ =\frac{845.93{\text{cm}}^{-1}}{207.244{\text{cm}}^{-1}}\\ =4.081\end{array}$

The partition function of iodine at $298\text{K}$ with $0{\text{cm}}^{-1}$ energy is calculated as shown below.

  $\begin{array}{c}{q}^{\text{V}}=\frac{1}{\text{1}-e^{-\frac{hc\tilde{\nu }}{kT}}}\\ =\frac{1}{\text{1}-e^{-0}}\\ =\infty \end{array}$

The partition function of iodine at $298\text{K}$ with $215.30{\text{cm}}^{-1}$ energy is calculated as shown below.

  $\begin{array}{c}{q}^{\text{V}}=\frac{1}{\text{1}-e^{-\frac{hc\tilde{\nu }}{kT}}}\\ =\frac{1}{\text{1}-e^{-1.038}}\\ =1.54\end{array}$

The partition function of iodine at $298\text{K}$ with $425.39{\text{cm}}^{-1}$ energy is calculated as shown below.

  $\begin{array}{c}{q}^{\text{V}}=\frac{1}{\text{1}-e^{-\frac{hc\tilde{\nu }}{kT}}}\\ =\frac{1}{\text{1}-e^{-2.052}}\\ =1.14\end{array}$

The partition function of iodine at $298\text{K}$ with $636.27{\text{cm}}^{-1}$ energy is calculated as shown below.

  $\begin{array}{c}{q}^{\text{V}}=\frac{1}{\text{1}-e^{-\frac{hc\tilde{\nu }}{kT}}}\\ =\frac{1}{\text{1}-e^{-3.070}}\\ =1.04\end{array}$

The partition function of iodine at $298\text{K}$ with $845.93{\text{cm}}^{-1}$ energy is calculated as shown below.

  $\begin{array}{c}{q}^{\text{V}}=\frac{1}{\text{1}-e^{-\frac{hc\tilde{\nu }}{kT}}}\\ =\frac{1}{\text{1}-e^{-4.081}}\\ =1.017\end{array}$

The summation of theses partition function at different energy levels neglecting $\infty$ value is calculated as shown below.

  $\begin{array}{c}{q}^{\text{V}}={\displaystyle \sum q_i^{\text{V}}}\\ =1.54+1.14+1.04+1.017\\ =4.737\end{array}$

The amount of iodine at ground and first two excited state is calculated by using Boltzmann distribution as,

  $\frac{p_i}{p_j}=e^{-\left(\frac{{\epsilon }_j-{\epsilon }_i}{kT}\right)}$

Where,

• $p_i,p_j$ are probabilities states.
• ${\epsilon }_i,{\epsilon }_j$ are energy sates.
• $k$ is Boltzmann constant
• $T$ is Temperature

The ratio between ground and first two excited states at $100\text{K}$ is calculated as shown below.

  $\begin{array}{c}\frac{p_{\text{ground}}}{p_{\text{excited}}}=e^{-\left(\frac{{\epsilon }_{\text{excited}}-{\epsilon }_{\text{ground}}}{kT}\right)}\\ =e^{-\left(\frac{640.69-0}{kT}\right)}\\ =e^{-\left(\frac{640.69\text{}{\text{cm}}^{-1}\times 6.63\times {10}^{-34}\text{J}\cdot \text{s}\times \text{3}\times {\text{10}}^{\text{10}}\text{cm}\cdot {\text{s}}^{\text{-1}}}{1.38\times {10}^{-23}\text{J}\cdot {\text{K}}^{-1}\times 100\text{K}}\right)}\\ =9.76\times {10}^{-5}\end{array}$

The ratio between ground and first two excited states at $298\text{K}$ is calculated as,

  $\begin{array}{c}\frac{p_{\text{ground}}}{p_{\text{excited}}}=e^{-\left(\frac{{\epsilon }_{\text{excited}}-{\epsilon }_{\text{ground}}}{kT}\right)}\\ =e^{-\left(\frac{640.69-0}{kT}\right)}\\ =e^{-\left(\frac{640.69\text{}{\text{cm}}^{-1}\times 6.63\times {10}^{-34}\text{J}\cdot \text{s}\times \text{3}\times {\text{10}}^{\text{10}}\text{cm}{\text{.s}}^{\text{-1}}}{1.38\times {10}^{-23}\text{J}{\text{.K}}^{-1}\times 298\text{K}}\right)}\\ =0.045\end{array}$