Chapter 13, Problem 13E.4AE

(i)

Interpretation Introduction

Interpretation:

The plot of molar heat capacity of a collection of harmonic oscillators as a function of $T/{\theta }^{\text{V}}$ is to be plotted.  Also, the vibrational heat capacity of ethyne at $298\text{K}$ is to be calculated.

Concept introduction:

According to the equipartition theorem, the contribution of each degree of freedom to the heat capacity is $\frac{1}{2}R$.  The molar heat capacity is calculated by using the formula shown below.

    $C_{V,m}/R=f^2$

Answer to Problem 13E.4AE

The plot of molar heat capacity of a collection of harmonic oscillators as a function of $x=T/{\theta }^{\text{V}}$ is shown below.

The vibrational heat capacity of ethyne is $\underset{\_}{14.93J/mol\cdot K}$.

Explanation of Solution

The molar heat capacity is calculated by using the formula shown below.

    $C_{V,m}/R=f^2$        (1)

The value of $f$ is expressed by the formula shown below.

  $f=\left(\frac{{\theta }^{\text{V}}}{T}\right)\times \left(\frac{e^{-{\theta }^{\text{V}}/2T}}{1-e^{-{\theta }^{\text{V}}/T}}\right)$        (2)

Therefore, the value of molar heat capacity is calculated by using the formula shown below.

    $C_{V,m}/R={\left(\left(\frac{{\theta }^{\text{V}}}{T}\right)\times \left(\frac{e^{-{\theta }^{\text{V}}/2T}}{1-e^{-{\theta }^{\text{V}}/T}}\right)\right)}^2$        (3)

Now, first the value of ${\theta }^{\text{V}}$ is calculate by using the formula shown below.

    ${\theta }^{\text{V}}=\frac{hc\overline{\nu }}{k}$        (4)

Where,

• $h$ is the Planck constant $\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)$.
• $c$ is the speed of light $\left(2.998\times {10}^8\text{m}/\text{s}=2.998\times {10}^{10}\text{cm}/\text{s}\right)$.
• $\overline{\nu }$ is the wavenumber.
• $k$ is the Boltzmann constant with value $1.38\times {10}^{-23}\text{J}/\text{K}$.

The values of ${\theta }^{\text{V}}$ for different values of wavenumber for the given data are to be calculated by using equation (4) as shown below.

For $\overline{\nu }=612{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 612{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{1.215\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =880.4\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{880.4\text{K}}{298\text{K}}\\ =2.96\end{array}$

For $\overline{\nu }=729{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 729{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{1.448\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =1049.275\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{1049.275\text{K}}{298\text{K}}\\ =3.52\end{array}$

For $\overline{\nu }=1974{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 1974{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{3.92\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =2840.57\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{2840.57\text{K}}{298\text{K}}\\ =9.53\end{array}$

For $\overline{\nu }=3287{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 3287{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{6.530\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =4731.88\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{4731.88\text{K}}{298\text{K}}\\ =15.88\end{array}$

For $\overline{\nu }=3374{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 3374{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{6.702\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =4856.52\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{4856.52\text{K}}{298\text{K}}\\ =16.30\end{array}$

Now, substitute the value of ${\theta }^{\text{V}}/T$ in equation (3) to calculate the molar heat capacity when $\overline{\nu }=612{\text{cm}}^{-1}$.

    $\begin{array}{c}{C}_{V,m}/R={\left(\left(2.96\right)\times \left(\frac{e^{-1.477}}{1-e^{-2.96}}\right)\right)}^2\\ ={\left(\left(2.96\right)\times \left(\frac{0.228}{1-0.051}\right)\right)}^2\\ ={\left(2.96\times \frac{0.228}{0.949}\right)}^2\\ =0.505\end{array}$

Similarly, substitute the value of ${\theta }^{\text{V}}/T$ in equation (3) to calculate the molar heat capacity when $\overline{\nu }=729{\text{cm}}^{-1}$.

    $\begin{array}{c}{C}_{V,m}/R={\left(\left(3.52\right)\times \left(\frac{e^{-1.760}}{1-e^{-3.52}}\right)\right)}^2\\ ={\left(\left(3.52\right)\times \left(\frac{0.172}{1-0.03}\right)\right)}^2\\ ={\left(3.52\times \frac{0.172}{0.97}\right)}^2\\ =0.389\end{array}$

Similarly, substitute the value of ${\theta }^{\text{V}}/T$ in equation (3) to calculate the molar heat capacity when $\overline{\nu }=1974{\text{cm}}^{-1}$.

    $\begin{array}{c}{C}_{V,m}/R={\left(\left(9.53\right)\times \left(\frac{e^{-4.766}}{1-e^{-9.53}}\right)\right)}^2\\ ={\left(\left(9.53\right)\times \left(\frac{8.514\times {10}^{-3}}{1-\left(7.263\times {10}^{-5}\right)}\right)\right)}^2\\ ={\left(9.53\times \frac{8.514\times {10}^{-3}}{0.999}\right)}^2\\ =0.007\end{array}$

Similarly, substitute the value of ${\theta }^{\text{V}}/T$ in equation (3) to calculate the molar heat capacity when $\overline{\nu }=3287{\text{cm}}^{-1}$.

    $\begin{array}{c}{C}_{V,m}/R={\left(\left(15.88\right)\times \left(\frac{e^{-7.94}}{1-e^{-15.88}}\right)\right)}^2\\ ={\left(\left(15.88\right)\times \left(\frac{3.562\times {10}^{-4}}{1-\left(1.268\times {10}^{-7}\right)}\right)\right)}^2\\ ={\left(15.88\times \frac{3.562\times {10}^{-4}}{0.999}\right)}^2\\ =3.20\times {10}^{-5}\end{array}$

Similarly, substitute the value of ${\theta }^{\text{V}}/T$ in equation (3) to calculate the molar heat capacity when $\overline{\nu }=3374{\text{cm}}^{-1}$.

    $\begin{array}{c}{C}_{V,m}/R={\left(\left(16.30\right)\times \left(\frac{e^{-8.14}}{1-e^{-16.30}}\right)\right)}^2\\ ={\left(\left(16.30\right)\times \left(\frac{3\times {10}^{-4}}{1-\left(8.3368\times {10}^{-8}\right)}\right)\right)}^2\\ ={\left(16.30\times \frac{3\times {10}^{-4}}{0.999}\right)}^2\\ =2.3\times {10}^{-5}\end{array}$

Therefore, the values of $\overline{\nu }\left({\text{cm}}^{-1}\right)$, ${\theta }^{\text{V}}/T$ and $C_{V,m}/R$ at $298\text{K}$ are shown below in the Table.

$\overline{\nu }\left({\text{cm}}^{-1}\right)$	${\theta }^{\text{V}}/T$	$C_{V,m}/R$
$612$	$2.96$	$0.505$
$612$	$2.96$	$0.505$
$729$	$3.52$	$0.389$
$729$	$3.52$	$0.389$
$1974$	$9.53$	$0.007$
$3287$	$15.88$	$3.20\times {10}^{-5}$
$3374$	$16.30$	$2.3\times {10}^{-5}$

Now, the plot of molar heat capacity of a collection of harmonic oscillators as a function of $x=T/{\theta }^{\text{V}}$ is shown below.

Figure 1

The vibrational heat capacity of ethyne is calculated by the formula shown below.

  $C_{V,m}/R$        (5)

The value of gas constant is $8.\text{314 }\text{J}/\text{mol}\cdot \text{K}$.

The sum of all the values of $C_{V,m}/R$ at different wavenumbers is calculated as shown below.

    $\begin{array}{c}{C}_{V,m}/R=0.505+0.505+0.389+0.389+0.007+3.20\times {10}^{-5}+2.3\times {10}^{-5}\\ C_{V,m}/R=1.7960\end{array}$

Therefore, the vibrational heat capacity of ethyne is calculated as shown below.

    $\begin{array}{c}{C}_{V,m}=1.7960\times R\\ =1.7960\times 8.\text{314 }\text{J}/\text{mol}\cdot \text{K}\\ =\underset{\_}{14.93J/mol\cdot K}\end{array}$

The vibrational heat capacity of ethyne is $\underset{\_}{14.93J/mol\cdot K}$.

(ii)

Interpretation Introduction

Interpretation:

The plot of molar heat capacity of a collection of harmonic oscillators as a function of $T/{\theta }^{\text{V}}$ is to be plotted.  Also, the vibrational heat capacity of ethyne at $500\text{K}$ is to be calculated.

Concept introduction:

According to the equipartition theorem, the contribution of each degree of freedom to the heat capacity is $\frac{1}{2}R$.  The molar heat capacity is calculated by using the formula shown below.

    $C_{V,m}/R=f^2$

Answer to Problem 13E.4AE

The plot of molar heat capacity of a collection of harmonic oscillators as a function of $x=T/{\theta }^{\text{V}}$ is shown below.

The vibrational heat capacity of ethyne is $\underset{\_}{25.465J/mol\cdot K}$.

Explanation of Solution

The molar heat capacity is calculated by using the formula shown below.

    $C_{V,m}/R=f^2$        (1)

The value of $f$ is expressed by the formula shown below.

  $f=\left(\frac{{\theta }^{\text{V}}}{T}\right)\times \left(\frac{e^{-{\theta }^{\text{V}}/2T}}{1-e^{-{\theta }^{\text{V}}/T}}\right)$        (2)

Therefore, the value of molar heat capacity is calculated by using the formula shown below.

    $C_{V,m}/R={\left(\left(\frac{{\theta }^{\text{V}}}{T}\right)\times \left(\frac{e^{-{\theta }^{\text{V}}/2T}}{1-e^{-{\theta }^{\text{V}}/T}}\right)\right)}^2$        (3)

Now, first the value of ${\theta }^{\text{V}}$ is calculate by using the formula shown below.

    ${\theta }^{\text{V}}=\frac{hc\overline{\nu }}{k}$        (4)

Where,

• $h$ is the Planck constant $\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)$.
• $c$ is the speed of light $\left(2.998\times {10}^8\text{m}/\text{s}=2.998\times {10}^{10}\text{cm}/\text{s}\right)$.
• $\overline{\nu }$ is the wavenumber.
• $k$ is the Boltzmann constant with value $1.38\times {10}^{-23}\text{J}/\text{K}$.

The values of ${\theta }^{\text{V}}$ for different values of wavenumber for the given data are to be calculated by using equation (4) as shown below.

For $\overline{\nu }=612{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 612{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{1.215\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =880.4\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{880.4\text{K}}{500\text{K}}\\ =1.76\end{array}$

For $\overline{\nu }=729{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 729{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{1.448\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =1049.275\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{1049.275\text{K}}{500\text{K}}\\ =2.09\end{array}$

For $\overline{\nu }=1974{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 1974{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{3.92\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =2840.57\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{2840.57\text{K}}{500\text{K}}\\ =5.68\end{array}$

For $\overline{\nu }=3287{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 3287{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{6.530\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =4731.88\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{4731.88\text{K}}{500\text{K}}\\ =9.46\end{array}$

For $\overline{\nu }=3374{\text{cm}}^{-1}$.

    $\begin{array}{c}{\theta }^{\text{V}}=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 2.998\times {10}^{10}\text{cm}/\text{s}\times 3374{\text{cm}}^{-1}}{1.38\times {10}^{-23}\text{J}/\text{K}}\\ =\frac{6.702\times {10}^{-20}}{1.38\times {10}^{-23}}\text{K}\\ =4856.52\text{K}\end{array}$

Therefore, the value of ${\theta }^{\text{V}}/T$ is calculated as shown below.

    $\begin{array}{c}{\theta }^{\text{V}}/T=\frac{4856.52\text{K}}{500\text{K}}\\ =9.71\end{array}$

Now, substitute the value of ${\theta }^{\text{V}}/T$ in equation (3) to calculate the molar heat capacity when $\overline{\nu }=612{\text{cm}}^{-1}$.

    $\begin{array}{c}{C}_{V,m}/R={\left(\left(1.76\right)\times \left(\frac{e^{-0.8804}}{1-e^{-1.76}}\right)\right)}^2\\ ={\left(\left(1.76\right)\times \left(\frac{0.414}{1-0.1720}\right)\right)}^2\\ ={\left(1.76\times \frac{0.414}{0.828}\right)}^2\\ =0.77\end{array}$

Similarly, substitute the value of ${\theta }^{\text{V}}/T$ in equation (3) to calculate the molar heat capacity when $\overline{\nu }=729{\text{cm}}^{-1}$.

    $\begin{array}{c}{C}_{V,m}/R={\left(\left(2.09\right)\times \left(\frac{e^{-1.049}}{1-e^{-2.09}}\right)\right)}^2\\ ={\left(\left(2.09\right)\times \left(\frac{0.350}{1-0.123}\right)\right)}^2\\ ={\left(2.09\times \frac{0.350}{0.877}\right)}^2\\ =0.70\end{array}$

Similarly, substitute the value of ${\theta }^{\text{V}}/T$ in equation (3) to calculate the molar heat capacity when $\overline{\nu }=1974{\text{cm}}^{-1}$.

    $\begin{array}{c}{C}_{V,m}/R={\left(\left(5.68\right)\times \left(\frac{e^{-2.840}}{1-e^{-5.68}}\right)\right)}^2\\ ={\left(\left(5.68\right)\times \left(\frac{0.058}{1-\left(3.413\times {10}^{-3}\right)}\right)\right)}^2\\ ={\left(5.68\times \frac{0.058}{0.99}\right)}^2\\ =0.110\end{array}$

Similarly, substitute the value of ${\theta }^{\text{V}}/T$ in equation (3) to calculate the molar heat capacity when $\overline{\nu }=3287{\text{cm}}^{-1}$.

    $\begin{array}{c}{C}_{V,m}/R={\left(\left(9.46\right)\times \left(\frac{e^{-4.731}}{1-e^{-9.46}}\right)\right)}^2\\ ={\left(\left(9.46\right)\times \left(\frac{8.817\times {10}^{-3}}{1-\left(7.790\times {10}^{-5}\right)}\right)\right)}^2\\ ={\left(9.46\times \frac{8.817\times {10}^{-3}}{0.999}\right)}^2\\ =0.007\end{array}$

Similarly, substitute the value of ${\theta }^{\text{V}}/T$ in equation (3) to calculate the molar heat capacity when $\overline{\nu }=3374{\text{cm}}^{-1}$.

    $\begin{array}{c}{C}_{V,m}/R={\left(\left(9.71\right)\times \left(\frac{e^{-4.856}}{1-e^{-9.71}}\right)\right)}^2\\ ={\left(\left(9.71\right)\times \left(\frac{7.7815\times {10}^{-3}}{1-\left(6.067\times {10}^{-5}\right)}\right)\right)}^2\\ ={\left(9.71\times \frac{7.7815\times {10}^{-3}}{0.999}\right)}^2\\ =5.7\times {10}^{-3}\end{array}$

    $C_{V,m}/R\approx 0.006$

Therefore, the values of $\overline{\nu }\left({\text{cm}}^{-1}\right)$, ${\theta }^{\text{V}}/T$ and $C_{V,m}/R$ at $500\text{K}$ are shown below in the Table.

$\overline{\nu }\left({\text{cm}}^{-1}\right)$	${\theta }^{\text{V}}/T$	$C_{V,m}/R$
$612$	$1.76$	$0.77$
$612$	$1.76$	$0.77$
$729$	$2.09$	$0.70$
$729$	$2.09$	$0.70$
$1974$	$5.68$	$0.110$
$3287$	$9.46$	$0.007$
$3374$	$9.71$	$0.006$

Now, the plot of molar heat capacity of a collection of harmonic oscillators as a function of $x=T/{\theta }^{\text{V}}$ is shown below.

Figure 2

The vibrational heat capacity of ethyne is calculated by the formula shown below.

  $C_{V,m}/R$        (5)

The value of gas constant is $8.\text{314 }\text{J}/\text{mol}\cdot \text{K}$.

The sum of all the values of $C_{V,m}/R$ at different wavenumbers is calculated as shown below.

    $\begin{array}{c}{C}_{V,m}/R=0.77+0.77+0.70+0.70+0.110+0.007+0.006\\ C_{V,m}/R=3.063\end{array}$

Therefore, the vibrational heat capacity of ethyne is calculated as shown below.

    $\begin{array}{c}{C}_{V,m}=3.063\times R\\ =3.063\times 8.\text{314 }\text{J}/\text{mol}\cdot \text{K}\\ =\underset{\_}{25.465J/mol\cdot K}\end{array}$

The vibrational heat capacity of ethyne is $\underset{\_}{25.465J/mol\cdot K}$.