Chapter 12, Problem 12.5P

(a)

Interpretation Introduction

Interpretation:

With the given data the total vibrational partition function of ${\text{CO}}_{\text{2}}$ at $\text{500}\text{K}$ has to be found.

Concept Introduction:

Partition function:

The partition function contains all the thermodynamic information about a system. Such information can be used to calculate thermodynamic properties such as the internal energy, entropy, and the Gibbs energy. In general, the molecular partition function gives an indication of the average number of states that are thermally accessible to a molecule at the temperature of the system.

The vibration partition function has the following expression.

$\begin{array}{l}{\text{q}}^{\text{v}}=\frac{\text{kT}}{\text{hυ}}\\ {\text{q}}^{\text{v}}=\text{Vibrational}\text{partition}\text{function}\end{array}$

$\begin{array}{l}\text{h}\text{=}{\text{Plank}}^{\text{'}}\text{s}\text{constant=}\text{6}\text{.626}{\text{×10}}^{\text{-34}}\text{J}\text{s}\\ \text{υ}=\text{frequency=}79.1975\times {10}^7{\text{s}}^{\text{-1}}\Rightarrow \text{Determined}\\ \text{k=}\text{Boltzmann}\text{constant}\text{=}\text{1}\text{.381}{\text{×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\\ \text{T}\text{=}\text{Temperature=}\text{298}\text{K}\Rightarrow \text{Given}\end{array}$

Answer to Problem 12.5P

The total vibrational partition function of ${\text{CO}}_{\text{2}}$ at $\text{500}\text{K}$ has to be found as ${\text{q}}_{\text{total}}^{\text{v}}=1.37665\times {10}^{23}$.

Explanation of Solution

Converting the given wavenumbers from ${\text{cm}}^{\text{-1}}$ to ${\text{m}}^{\text{-1}}$.

For $\overline{\text{υ}}=\text{1388}{\text{cm}}^{\text{-1}}$

$\begin{array}{c}\text{100}\text{cm}\text{=1}\text{m}\\ \text{1388}\text{c}\bcancel{\text{m}}\times \frac{\text{1}\text{m}}{\text{100}\text{c}\bcancel{\text{m}}}=13.88\text{m}\\ \therefore \Delta \text{υ}\text{=}\text{1388}{\text{cm}}^{-1}=13.88{\text{m}}^{-1}\end{array}$

For $\overline{\text{υ}}=\text{2349}{\text{cm}}^{\text{-1}}$

$\begin{array}{c}\text{100}\text{cm}\text{=1}\text{m}\\ \text{2349}\text{c}\bcancel{\text{m}}\times \frac{\text{1}\text{m}}{\text{100}\text{c}\bcancel{\text{m}}}=\text{23}\text{.49}\text{m}\\ \therefore \Delta \text{υ}\text{=}\text{2349}{\text{cm}}^{-1}=\text{23}\text{.49}{\text{m}}^{-1}\end{array}$

For $\overline{\text{υ}}=\text{667}{\text{cm}}^{\text{-1}}$

$\begin{array}{c}\text{100}\text{cm}\text{=1}\text{m}\\ \text{667}\text{c}\bcancel{\text{m}}\times \frac{\text{1}\text{m}}{\text{100}\text{c}\bcancel{\text{m}}}=\text{6}\text{.67}\text{m}\\ \therefore \Delta \text{υ}\text{=}\text{667}{\text{cm}}^{-1}=\text{6}\text{.67}{\text{m}}^{-1}\end{array}$

Converting each wavenumber into their corresponding wavelengths:

$\begin{array}{c}\overline{\text{υ}}\text{=}\frac{\text{1}}{\text{λ}}\\ \overline{\text{υ}}\text{=}\text{Wavenumber}.\\ \lambda =\text{Wavelength}\text{corresponding}\text{to}\text{the}\text{given}\text{wave}\text{number}\text{.}\end{array}$

For $\overline{\text{υ}}=13.88{\text{m}}^{-1}$

$\begin{array}{c}\overline{\text{υ}}\text{=}\frac{\text{1}}{\text{λ}}\\ 13.88{\text{m}}^{-1}\text{=}\frac{\text{1}}{\text{λ}}\\ \text{λ=}\frac{\text{1}}{13.88{\text{m}}^{-1}}\\ \text{=}\text{0}\text{.0720}\text{m}\end{array}$

$\therefore \text{λ}\text{=}\text{0}\text{.0720}\text{m}$

For $\overline{\text{υ}}=\text{23}\text{.49}{\text{m}}^{-1}$

$\begin{array}{c}\overline{\text{υ}}\text{=}\frac{\text{1}}{\text{λ}}\\ \text{23}\text{.49}{\text{m}}^{-1}\text{=}\frac{\text{1}}{\text{λ}}\\ \text{λ=}\frac{\text{1}}{\text{23}\text{.49}{\text{m}}^{-1}}\\ \text{=}\text{0}\text{.0426}\text{m}\end{array}$

$\therefore \text{λ}\text{=}\text{0}\text{.0426}\text{m}$

For $\overline{\text{υ}}=\text{6}\text{.67}{\text{m}}^{-1}$

$\begin{array}{c}\overline{\text{υ}}\text{=}\frac{\text{1}}{\text{λ}}\\ \text{6}\text{.67}{\text{m}}^{-1}\text{=}\frac{\text{1}}{\text{λ}}\\ \text{λ=}\frac{\text{1}}{\text{6}\text{.67}{\text{m}}^{-1}}\\ \text{=}0.\text{1499}\text{m}\end{array}$

$\therefore \text{λ}\text{=}0.\text{1499}\text{m}$

Converting each wavelength into their corresponding frequencies:

$\begin{array}{l}\text{υ}\text{=}\frac{\text{c}}{\text{λ}}\\ \text{υ}\text{=}\text{Frequency}\\ \text{c}\text{=}\text{Velocity}\text{of}\text{light}\text{=}\text{3}\text{×}{\text{10}}^{\text{8}}{\text{ms}}^{\text{-1}}\\ \text{λ}=\text{Wavelength}\text{.}\end{array}$

For $\text{λ}\text{=}\text{0}\text{.0720}\text{m}$

$\begin{array}{c}\text{υ}\text{=}\frac{\text{c}}{\text{λ}}\\ =\frac{\text{3}\text{×}{\text{10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{\text{0}\text{.3788}\text{m}}\\ =41.6667\times {10}^8{\text{s}}^{\text{-1}}\\ \therefore \text{υ}\text{=}41.6667\times {10}^8{\text{s}}^{\text{-1}}\end{array}$

For $\text{λ}\text{=}\text{0}\text{.0426}\text{m}$

$\begin{array}{c}\text{υ}\text{=}\frac{\text{c}}{\text{λ}}\\ =\frac{\text{3}\text{×}{\text{10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{\text{0}\text{.0426}\text{m}}\\ =70.4225\times {10}^8{\text{s}}^{\text{-1}}\\ \therefore \text{υ}\text{=}70.4225\times {10}^8{\text{s}}^{\text{-1}}\end{array}$

For $\text{λ}\text{=}0.\text{1499}\text{m}$

$\begin{array}{c}\text{υ}\text{=}\frac{\text{c}}{\text{λ}}\\ =\frac{\text{3}\text{×}{\text{10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{0.\text{1499}\text{m}}\\ =20.0133\times {10}^8{\text{s}}^{\text{-1}}\\ \therefore \text{υ}\text{=}20.0133\times {10}^8{\text{s}}^{\text{-1}}\end{array}$

Calculating the vibrational partition function of ${\text{CO}}_{\text{2}}$ at $\text{500}\text{K}$ with respect to each of the determined frequencies:

For $\text{υ}\text{=}41.6667\times {10}^8{\text{s}}^{\text{-1}}$

${\text{q}}^{\text{v}}=\frac{\text{kT}}{\text{hυ}}$

$\begin{array}{c}{\text{q}}^{\text{v}}=\frac{\left(\text{1}\text{.381}{\text{×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\right)\times \left(\text{500}\text{K}\right)}{\left(\text{6}\text{.626}{\text{×10}}^{\text{-34}}\text{J}\text{s}\right)\times \left(41.6667\times {10}^8{\text{s}}^{\text{-1}}\right)}\\ =4.3421\times {10}^{22}\\ \therefore {\text{q}}^{\text{v}}=4.3421\times {10}^{22}\Rightarrow \text{Let }\text{it }\text{be}{\text{q}}_{\text{1}}^{\text{v}}\end{array}$

For $\text{υ}\text{=}70.4225\times {10}^8{\text{s}}^{\text{-1}}$

${\text{q}}^{\text{v}}=\frac{\text{kT}}{\text{hυ}}$

$\begin{array}{c}{\text{q}}^{\text{v}}=\frac{\left(\text{1}\text{.381}{\text{×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\right)\times \left(\text{500}\text{K}\right)}{\left(\text{6}\text{.626}{\text{×10}}^{\text{-34}}\text{J}\text{s}\right)\times \left(70.4225\times {10}^8{\text{s}}^{\text{-1}}\right)}\\ =7.3388\times {10}^{22}\\ \therefore {\text{q}}^{\text{v}}=7.3388\times {10}^{22}\Rightarrow \text{Let }\text{it }\text{be}{\text{q}}_2^{\text{v}}\end{array}$

For $\text{υ}\text{=}20.0133\times {10}^8{\text{s}}^{\text{-1}}$

$\begin{array}{c}{\text{q}}^{\text{v}}=\frac{\left(\text{1}\text{.381}{\text{×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\right)\times \left(\text{500}\text{K}\right)}{\left(\text{6}\text{.626}{\text{×10}}^{\text{-34}}\text{J}\text{s}\right)\times \left(20.0133\times {10}^8{\text{s}}^{\text{-1}}\right)}\\ =2.0856\times {10}^{22}\\ \therefore {\text{q}}^{\text{v}}=2.0856\times {10}^{22}\Rightarrow \text{Let }\text{it }\text{be}{\text{q}}_3^{\text{v}}\end{array}$

The total vibrational partition function of ${\text{CO}}_{\text{2}}$ at $\text{500}\text{K}$ can be calculated as shown here:

$\begin{array}{c}{\text{q}}_{\text{total}}^{\text{v}}={\text{q}}_1^{\text{v}}+{\text{q}}_2^{\text{v}}{\text{q}}_3^{\text{v}}\\ =\left(4.3421\times {10}^{22}\right)+\left(7.3388\times {10}^{22}\right)+\left(2.0856\times {10}^{22}\right)\\ =\left(4.3421+7.3388+2.0856\right)\times {10}^{22}\\ =1.37665\times {10}^{23}\end{array}$

${\text{q}}_{\text{total}}^{\text{v}}=1.37665\times {10}^{23}$

Hence, total vibrational partition function of ${\text{CO}}_{\text{2}}$ at $\text{500}\text{K}$ is ${\text{q}}_{\text{total}}^{\text{v}}=1.37665\times {10}^{23}$.

(b)

Interpretation Introduction

Interpretation:

With the given data the total vibrational partition function of ${\text{CO}}_{\text{2}}$ at $\text{1000}\text{K}$ has to be found.

Concept Introduction:

Partition function:

The partition function contains all the thermodynamic information about a system. Such information can be used to calculate thermodynamic properties such as the internal energy, entropy, and the Gibbs energy. In general, the molecular partition function gives an indication of the average number of states that are thermally accessible to a molecule at the temperature of the system.

The vibration partition function has the following expression.

$\begin{array}{l}{\text{q}}^{\text{v}}=\frac{\text{kT}}{\text{hυ}}\\ {\text{q}}^{\text{v}}=\text{Vibrational}\text{partition}\text{function}\end{array}$

$\begin{array}{l}\text{h}\text{=}{\text{Plank}}^{\text{'}}\text{s}\text{constant=}\text{6}\text{.626}{\text{×10}}^{\text{-34}}\text{J}\text{s}\\ \text{υ}=\text{frequency=}79.1975\times {10}^7{\text{s}}^{\text{-1}}\Rightarrow \text{Determined}\\ \text{k=}\text{Boltzmann}\text{constant}\text{=}\text{1}\text{.381}{\text{×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\\ \text{T}\text{=}\text{Temperature=}\text{298}\text{K}\Rightarrow \text{Given}\end{array}$

Answer to Problem 12.5P

The total vibrational partition function of ${\text{CO}}_{\text{2}}$ at $\text{1000}\text{K}$ has to be found as $2.7533\times {10}^{23}$.

Explanation of Solution

Calculating the vibrational partition function of ${\text{CO}}_{\text{2}}$ at $\text{1000}\text{K}$ with respect to each of the determined frequencies:

For $\text{υ}\text{=}41.6667\times {10}^8{\text{s}}^{\text{-1}}$

${\text{q}}^{\text{v}}=\frac{\text{kT}}{\text{hυ}}$

$\begin{array}{c}{\text{q}}^{\text{v}}=\frac{\left(\text{1}\text{.381}{\text{×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\right)\times \left(\text{1000}\text{K}\right)}{\left(\text{6}\text{.626}{\text{×10}}^{\text{-34}}\text{J}\text{s}\right)\times \left(41.6667\times {10}^8{\text{s}}^{\text{-1}}\right)}\\ =8.6842\times {10}^{22}\\ \therefore {\text{q}}^{\text{v}}=8.6842\times {10}^{22}\Rightarrow \text{Let }\text{it }\text{be}{\text{q}}_{\text{1}}^{\text{v}}\end{array}$

For $\text{υ}\text{=}70.4225\times {10}^8{\text{s}}^{\text{-1}}$

${\text{q}}^{\text{v}}=\frac{\text{kT}}{\text{hυ}}$

$\begin{array}{c}{\text{q}}^{\text{v}}=\frac{\left(\text{1}\text{.381}{\text{×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\right)\times \left(\text{1000}\text{K}\right)}{\left(\text{6}\text{.626}{\text{×10}}^{\text{-34}}\text{J}\text{s}\right)\times \left(70.4225\times {10}^8{\text{s}}^{\text{-1}}\right)}\\ =14.6776\times {10}^{22}\\ \therefore {\text{q}}^{\text{v}}=14.6776\times {10}^{22}\Rightarrow \text{Let }\text{it }\text{be}{\text{q}}_2^{\text{v}}\end{array}$

For $\text{υ}\text{=}20.0133\times {10}^8{\text{s}}^{\text{-1}}$

$\begin{array}{c}{\text{q}}^{\text{v}}=\frac{\left(\text{1}\text{.381}{\text{×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\right)\times \left(\text{1000}\text{K}\right)}{\left(\text{6}\text{.626}{\text{×10}}^{\text{-34}}\text{J}\text{s}\right)\times \left(20.0133\times {10}^8{\text{s}}^{\text{-1}}\right)}\\ =4.1712\times {10}^{22}\\ \therefore {\text{q}}^{\text{v}}=4.1712\times {10}^{22}\Rightarrow \text{Let }\text{it }\text{be}{\text{q}}_3^{\text{v}}\end{array}$

The total vibrational partition function of ${\text{CO}}_{\text{2}}$ at $\text{1000}\text{K}$ can be calculated as shown here:

$\begin{array}{c}{\text{q}}_{\text{total}}^{\text{v}}={\text{q}}_1^{\text{v}}+{\text{q}}_2^{\text{v}}{\text{q}}_3^{\text{v}}\\ =\left(8.6842\times {10}^{22}\right)+\left(14.6776\times {10}^{22}\right)+\left(4.1712\times {10}^{22}\right)\\ =\left(4.3421+7.3388+2.0856\right)\times {10}^{22}\\ =2.7533\times {10}^{23}\end{array}$

${\text{q}}_{\text{total}}^{\text{v}}=2.7533\times {10}^{23}$

Hence, total vibrational partition function of ${\text{CO}}_{\text{2}}$ at $\text{1000}\text{K}$ is ${\text{q}}_{\text{total}}^{\text{v}}=2.7533\times {10}^{23}$.