Chapter 2, Problem 2.6PR

(a)

Interpretation Introduction

Interpretation:

For chlorine the temperature at which the equipartition theory becomes valid has to be calculated.

Concept introduction:

Equipartition theorem:

According to the equipartition theorem, the total energy of a molecule is divided equally amongst the various degrees of freedom of the molecules.

Answer to Problem 2.6PR

The temperature is $\text{805}\text{.87}\text{K}$.

Explanation of Solution

According to the equipartition theorem,

Energy contribution by each of the translational degrees of freedom is $\frac{1}{2}{\text{k}}_{\text{B}}\text{T}$ per molecule.

Energy contribution by each of the rotational degrees of freedom is $\frac{1}{2}{\text{k}}_{\text{B}}\text{T}$ per molecule.

For vibrational degrees of freedom, due to collisions between the molecules each vibrational degrees of freedom possess both kinetic energy and potential energy i.e. each vibration involves two degrees of freedom and hence the energy contribution by each of the vibrational degrees of freedom is given by,

  $\begin{array}{l}{\text{E}}_{\text{v}}\text{=}\text{P}\text{.E}\text{+}\text{K}\text{.E}\text{.}\\ \\ {\text{E}}_{\text{v}}\text{=}\frac{\text{1}}{\text{2}}{\text{k}}_{\text{B}}\text{T}\text{+}\frac{\text{1}}{\text{2}}{\text{k}}_{\text{B}}\text{T}\\ \\ {\text{E}}_{\text{v}}\text{=}{\text{k}}_{\text{B}}\text{T}\end{array}$

Hence each vibrational degrees of freedom contributes ${\text{k}}_{\text{B}}\text{T}$ energy per molecule.

Now, the equipartition theorem is valid only if ${\text{k}}_{\text{B}}\text{T}$ is very much greater than the separation between the energy levels of the mode of the motion $\text{Δε}$.  Translational and rotational energy levels are sufficiently close together for most of the molecules at room temperature.

But the separation between the vibrational energy levels is much greater than ${\text{k}}_{\text{B}}\text{T}$ and as a result for vibrational degrees of freedom the contribution in total energy has to be calculated at higher temperature only.

It is given in the question that the energy from vibration is given by $\frac{\text{Δε}}{\text{1}\text{-}{\text{e}}^{\frac{\text{-}\text{Δε}}{\text{kT}}}}$.

According to the question for chlorine the separation between the vibrational energy levels is,

  $\text{Δε}\text{=}\text{6}\text{.70}\text{kJ}{\text{mol}}^{\text{-1}}$

Now the temperature at which equipartition theorem for chlorine is valid that has to be calculated.

Hence,

  $\begin{array}{l}\frac{\text{Δε}}{\text{1}\text{-}{\text{e}}^{\frac{\text{-}\text{Δε}}{\text{kT}}}}\text{=}\text{kT}\\ \\ \frac{\text{Δε}}{\text{kT}}=\text{1}\text{-}{\text{e}}^{\frac{\text{-}\text{Δε}}{\text{kT}}}\end{array}$

Chlorine is linear diatomic molecule and hence it will have degrees of freedom $\text{(3×2}\text{-}\text{5)}\text{=}\text{1}$

Now at higher temperature only vibrational degrees of freedom of chlorine can show equipartition

Now by expansion, ${\text{e}}^{\frac{\text{-}\text{Δε}}{\text{kT}}}=1-\frac{\text{Δε}}{\text{kT}}+{\left(\frac{\text{Δε}}{\text{kT}}\right)}^2-{\left(\frac{\text{Δε}}{\text{kT}}\right)}^3+\mathrm{....}$

But higher temperature all the higher terms can be neglected.

Hence the temperature is, $\text{T}=\frac{{\text{E}}_{\text{v}}}{\text{k}}$

Vibrational energy per mole, ${\text{E}}_{\text{v}}\text{=}\text{RT}$

Hence temperature is,

  $\begin{array}{l}{\text{E}}_{\text{v}}\text{=}\text{RT}\\ \\ \text{T}\text{=}\frac{\text{6}{\text{.70×10}}^{\text{3}}\text{J}{\text{mol}}^{\text{-1}}}{\text{8}\text{.314}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}}\\ \\ \text{T}\text{=}\text{805}\text{.87}\text{K}\end{array}$

Hence the temperature is $\text{805}\text{.87}\text{K}$.

(b)

Interpretation Introduction

Interpretation:

For a given exponential ${\text{e}}^{\text{x}}\text{=}\text{1}\text{+}\text{x}\text{+}\frac{{\text{x}}^{\text{2}}}{\text{2!}}\text{+}\frac{{\text{x}}^{\text{3}}}{\text{3!}}\text{+}\mathrm{....}$ it has to be shown that the exact expression reduced to the result obtained from the equipartition theorem.

Concept introduction:

Equipartition theorem:

According to the equipartition theorem, the total energy of a molecule is divided equally amongst the various degrees of freedom of the molecules.

Explanation of Solution

According to the equipartition theorem,

Energy contribution by each of the translational degrees of freedom is $\frac{1}{2}{\text{k}}_{\text{B}}\text{T}$ per molecule.

Energy contribution by each of the rotational degrees of freedom is $\frac{1}{2}{\text{k}}_{\text{B}}\text{T}$ per molecule.

For vibrational degrees of freedom, due to collisions between the molecules each vibrational degrees of freedom possess both kinetic energy and potential energy i.e. each vibration involves two degrees of freedom and hence the energy contribution by each of the vibrational degrees of freedom is given by,

  $\begin{array}{l}{\text{E}}_{\text{v}}\text{=}\text{P}\text{.E}\text{+}\text{K}\text{.E}\text{.}\\ \\ {\text{E}}_{\text{v}}\text{=}\frac{\text{1}}{\text{2}}{\text{k}}_{\text{B}}\text{T}\text{+}\frac{\text{1}}{\text{2}}{\text{k}}_{\text{B}}\text{T}\\ \\ {\text{E}}_{\text{v}}\text{=}{\text{k}}_{\text{B}}\text{T}\end{array}$

Hence each vibrational degrees of freedom contributes ${\text{k}}_{\text{B}}\text{T}$ energy per molecule.

Now, the equipartition theorem is valid only if ${\text{k}}_{\text{B}}\text{T}$ is very much greater than the separation between the energy levels of the mode of the motion $\text{Δε}$.  Translational and rotational energy levels are sufficiently close together for most of the molecules at room temperature.

But the separation between the vibrational energy levels is much greater than ${\text{k}}_{\text{B}}\text{T}$ and as a result for vibrational degrees of freedom the contribution in total energy has to be calculated at higher temperature only.

It is given in the question that the energy from vibration is given by $\frac{\text{Δε}}{\text{1}\text{-}{\text{e}}^{\frac{\text{-}\text{Δε}}{\text{kT}}}}$.

Now the exponential form given is,

  ${\text{e}}^{\text{x}}\text{=}\text{1}\text{+}\text{x}\text{+}\frac{{\text{x}}^{\text{2}}}{\text{2!}}\text{+}\frac{{\text{x}}^{\text{3}}}{\text{3!}}\text{+}\mathrm{....}$

Hence expanding ${\text{e}}^{\frac{\text{-}\text{Δε}}{\text{kT}}}$ like the given exponential form,

  ${\text{e}}^{\frac{\text{-}\text{Δε}}{\text{kT}}}=1-\frac{\text{Δε}}{\text{kT}}+{\left(\frac{\text{Δε}}{\text{kT}}\right)}^2-{\left(\frac{\text{Δε}}{\text{kT}}\right)}^3+\mathrm{....}$

Now using this expansion

  $\begin{array}{l}{\text{E}}_{\text{v}}\text{=}\frac{\text{Δε}}{\text{1}\text{-}{\text{e}}^{\frac{\text{-}\text{Δε}}{\text{kT}}}}\\ \\ {\text{E}}_{\text{v}}\text{=}\frac{\text{Δε}}{\text{1}\text{-}\left(1-\frac{\text{Δε}}{\text{kT}}+{\left(\frac{\text{Δε}}{\text{kT}}\right)}^2-{\left(\frac{\text{Δε}}{\text{kT}}\right)}^3+\mathrm{....}\right)}\end{array}$

Now only at higher temperature vibrational degrees of freedom gives full contribution to the total energy.

Now at higher temperature all the higher terms of the expansion will be very much lesser than one and become negligible

Hence the equation becomes,

  $\begin{array}{l}{\text{E}}_{\text{v}}\text{=}\frac{\text{Δε}}{\text{1}\text{-}\left(1-\frac{\text{Δε}}{\text{kT}}\right)}\\ \\ {\text{E}}_{\text{v}}=\frac{\text{Δε}}{\text{1}\text{-}1+\frac{\text{Δε}}{\text{kT}}}\\ \\ {\text{E}}_{\text{v}}=\frac{\text{Δε}}{\frac{\text{Δε}}{\text{kT}}}\\ \\ {\text{E}}_{\text{v}}=\text{kT}\end{array}$

Thus at higher temperature the exact expression for vibrational degrees of freedom reduces to the result obtained from equipartition theorem.