Chapter 13, Problem 13C.6AE

Interpretation Introduction

Interpretation:

The mean vibrational energy of ${\text{CCl}}_4$ with the wavenumbers, $\text{459}{\text{cm}}^{-1}$, $\text{217}{\text{cm}}^{-1}$, $\text{776}{\text{cm}}^{-1}$, $\text{314}{\text{cm}}^{-1}$ and has to be calculated at different temperatures.  The graph between mean vibrational energy of ${\text{CCl}}_4$ at different values of temperatures has to be drawn.  The temperature at which the equipartition value within $5\text{\%}$ is accurate 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 given by the expression as shown below.

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

The mean vibrational energy per mode is given by the expression as shown below.

  $\langle {\epsilon }^{\text{V}}\rangle =\frac{hc\tilde{\nu }}{e^{hc\tilde{\nu }/kT}-1}$

Answer to Problem 13C.6AE

The mean vibrational energy of ${\text{CCl}}_4$ is calculated at different temperatures as shown below in the table.

Temperature $\left(\text{K}\right)$	The mean vibrational energy of ${\text{CCl}}_4$ $\left(\text{J}\right)$
$1000$	$8.97\times {10}^{-20}$
$2000$	$2.11\times {10}^{-19}$
$3000$	$3.35\times {10}^{-19}$
$4000$	$4.59\times {10}^{-19}$
$5000$	$5.83\times {10}^{-19}$
$6000$	$7.07\times {10}^{-19}$

The graph between mean vibrational energy of ${\text{CCl}}_4$ at different values of temperatures is shown below.

The temperature at which the equipartition value within $5\text{\%}$ is accurate is $6000\text{ K}$.

Explanation of Solution

The mean vibrational energy per mode is given by the expression as shown below.

  $\langle {\epsilon }^{\text{V}}\rangle =\frac{hc\tilde{\nu }}{e^{hc\tilde{\nu }/kT}-1}$

Where,

• $k$ is the Boltzmann constant($1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}$).
• $T$ is the temperature.
• $h$ is the Planck’s constant. $\left(6.62608\times {10}^{-34}\text{J}\text{s}\right)$
• $c$ is the speed of light. $\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)$
• $\tilde{\nu }$ is the vibrational wavenumber.

The table that represents the wavenumber and degeneracy of the stated of ${\text{CCl}}_4$ is shown below.

Wavenumber	Degeneracy
$\text{459}{\text{cm}}^{-1}$	$1$
$\text{217}{\text{cm}}^{-1}$	$2$
$\text{776}{\text{cm}}^{-1}$	$3$
$\text{314}{\text{cm}}^{-1}$	$3$

The mean vibrational energy for ${\text{CCl}}_4$ is given by the expression as shown below.

  $\langle {\epsilon }^{\text{V}}\rangle =\frac{hc{\tilde{\nu }}_1}{e^{hc{\tilde{\nu }}_1/kT}-1}+\left(2\right)\frac{hc{\tilde{\nu }}_2}{e^{hc{\tilde{\nu }}_2/kT}-1}+\left(3\right)\frac{hc{\tilde{\nu }}_3}{e^{hc{\tilde{\nu }}_3/kT}-1}+\left(3\right)\frac{hc{\tilde{\nu }}_4}{e^{hc{\tilde{\nu }}_4/kT}-1}$        (1)

Substitute the values of wavenumber at different levels, $k$, $c$, and $h$ in the equation (1).

    $\langle {\epsilon }^{\text{V}}\rangle =\left(\begin{array}{l}\frac{\left(6.62608\times {10}^{-34}\text{J}\text{s}\right)\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(\text{459}{\text{cm}}^{-1}\right)}{e^{\left(6.62608\times {10}^{-34}\text{J}\text{s}\right)\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(\text{459}{\text{cm}}^{-1}\right)/\left(1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\right)T}-1}\\ +\left(2\right)\frac{\left(6.62608\times {10}^{-34}\text{J}\text{s}\right)\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(\text{217}{\text{cm}}^{-1}\right)}{e^{\left(6.62608\times {10}^{-34}\text{J}\text{s}\right)\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(\text{217}{\text{cm}}^{-1}\right)/\left(1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\right)T}-1}\\ +\left(3\right)\frac{\left(6.62608\times {10}^{-34}\text{J}\text{s}\right)\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(\text{776}{\text{cm}}^{-1}\right)}{e^{\left(6.62608\times {10}^{-34}\text{J}\text{s}\right)\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(\text{776}{\text{cm}}^{-1}\right)/\left(1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\right)T}-1}\\ +\left(3\right)\frac{\left(6.62608\times {10}^{-34}\text{J}\text{s}\right)\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(\text{314}{\text{cm}}^{-1}\right)}{e^{\left(6.62608\times {10}^{-34}\text{J}\text{s}\right)\left(2.997945\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(\text{314}{\text{cm}}^{-1}\right)/\left(1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\right)T}-1}\end{array}\right)$

The above expression can be simplified as shown below.

    $\begin{array}{c}\langle {\epsilon }^{\text{V}}\rangle =\left(\begin{array}{l}\frac{6.626\times {10}^{-21}}{e^{6.626\times {10}^{-21}/\left(1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\right)T}-1}+\left(2\right)\frac{4.3106\times {10}^{-21}}{e^{4.3106\times {10}^{-21}/\left(1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\right)T}-1}\\ +\left(3\right)\frac{1.5415\times {10}^{-20}}{e^{1.5415\times {10}^{-20}/\left(1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\right)T}-1}+\left(3\right)\frac{6.2375\times {10}^{-21}}{e^{6.2375\times {10}^{-21}/\left(1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\right)T}-1}\end{array}\right)\text{ J}\\ \langle {\epsilon }^{\text{V}}\rangle =\left(\frac{6.626\times {10}^{-21}}{e^{480.14/T}-1}+\frac{8.6212\times {10}^{-21}}{e^{312.36/T}-1}+\frac{4.6245\times {10}^{-20}}{e^{1117.03/T}-1}+\frac{18.8025\times {10}^{-21}}{e^{451.99/T}-1}\right)\text{ J}\end{array}$

Substitute the value of $T=1000\text{ K}$ in the above expression.

    $\begin{array}{c}\langle {\epsilon }^{\text{V}}\rangle =\left(\frac{6.626\times {10}^{-21}}{e^{480.14/1000}-1}+\frac{8.6212\times {10}^{-21}}{e^{312.36/1000}-1}+\frac{4.6245\times {10}^{-20}}{e^{1117.03/1000}-1}+\frac{18.8025\times {10}^{-21}}{e^{451.99/1000}-1}\right)\text{ J}\\ =\left(\frac{6.626\times {10}^{-21}}{1.6163-1}+\frac{8.6212\times {10}^{-21}}{1.3666-1}+\frac{4.6245\times {10}^{-20}}{3.0558-1}+\frac{18.8025\times {10}^{-21}}{1.5714-1}\right)\text{ J}\\ =\left(1.0751\times {10}^{-20}+2.3517\times {10}^{-20}+2.2495\times {10}^{-20}+3.2906\times {10}^{-20}\right)\text{ J}\\ =8.9669\times {10}^{-20}\text{ J}\end{array}$

Similarly, the mean vibrational energy of ${\text{CCl}}_4$ is calculated at different temperatures as shown below in the table.

Temperature $\left(\text{K}\right)$	The mean vibrational energy of ${\text{CCl}}_4$ $\left(\text{J}\right)$
$1000$	$8.97\times {10}^{-20}$
$2000$	$2.11\times {10}^{-19}$
$3000$	$3.35\times {10}^{-19}$
$4000$	$4.59\times {10}^{-19}$
$5000$	$5.83\times {10}^{-19}$
$6000$	$7.07\times {10}^{-19}$

The graph between mean vibrational energy of ${\text{CCl}}_4$ at different values of temperatures is shown below.

Figure 1

The equipartition value of ${\text{CCl}}_4$ is $9kT$.

The equipartition value of ${\text{CCl}}_4$ is calculated at $T=1000\text{ K}$ as shown below.

    $\begin{array}{c}\epsilon =9kT\\ =9\left(1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\right)\left(1000\text{ K}\right)\\ =1.2429\times {10}^{-19}\text{J}\end{array}$

The equipartition value of ${\text{CCl}}_4$ is calculated at different temperatures as shown below in the table.

Temperature $\left(\text{K}\right)$	The equipartition value of ${\text{CI}}_4$ $\left(\text{J}\right)$
$1000$	$1.2429\times {10}^{-19}$
$2000$	$2.4858\times {10}^{-19}$
$3000$	$3.7287\times {10}^{-19}$
$4000$	$4.9716\times {10}^{-19}$
$5000$	$6.2145\times {10}^{-19}$
$6000$	$7.46\times {10}^{-19}$

It is expected that the difference between the equipartition value and the mean vibrational energy of ${\text{CCl}}_4$ at $6000\text{ K}$ is $5\%$.

The percentage error in the value is calculated as shown below.

  $\Delta E\%=\left|\frac{{\epsilon }_{\text{1}}-{\epsilon }_{\text{2}}}{{\epsilon }_{\text{1}}}\right|\times 100$

Where,

• ${\epsilon }_1$ is the equipartition value of ${\text{CCl}}_4$.
• ${\epsilon }_2$  is the mean vibrational energy of ${\text{CCl}}_4$.

Substitute the value of ${\epsilon }_1$ and ${\epsilon }_2$ in the above equation.

    $\begin{array}{c}\Delta E\%=\left|\frac{7.46\times {10}^{-19}\text{ J}-7.07\times {10}^{-19}}{7.46\times {10}^{-19}\text{ J}}\right|\times 100\%\\ =\frac{0.39}{7.46}\times 100\%\\ \approx 5\%\end{array}$