Chapter 13, Problem 13B.4P

Interpretation Introduction

Interpretation:

The temperature at which the value of $q$ is $10$ according to the integral approximation has to be estimated.  The exact partition function at that temperature has to be calculated.

Concept introduction:

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

  $q={\displaystyle \sum _i{g}_i{e}^{-{\epsilon }_i/kT}}$

Where,

• $g_i$ represents the degeneracy.
• ${\epsilon }_i$ represents the energy of $i^{th}$ microstate.
• $k$ represents the Boltzmann constant with value $1.38\times {10}^{-23}\text{J}/\text{K}$.
• $T$ represents the temperature $\left(\text{K}\right)$.

It is also called as canonical ensemble partition function.

Answer to Problem 13B.4P

The temperature at which the value of $q$ is $10$ according to the integral approximation is $\underset{\_}{1.356\times 10^{-3}K}$.

The exact partition function at the calculated temperature is $\underset{\_}{3.0655}$.

Explanation of Solution

The equation for a translational partition function for one-dimensional box is shown below.

    $q^{\text{T}}=\frac{V{\left(2\pi mkT\right)}^{3/2}}{h^3}$        (1)

Where,

• $q^{\text{T}}$ is the translational partition function.
• $V$ is the volume.
• $m$ is the mass.
• $k$ is the Boltzmann constant.
• $T$ is the temperature.
• $h$ is the Planck’s constant.

In order to estimate temperature, equation (1) is modified as shown below.

    $\begin{array}{c}{q}^{\text{T}}=\frac{V{\left(2\pi mkT\right)}^{3/2}}{h^3}\\ q^{\text{T}}{h}^3=V{\left(2\pi mkT\right)}^{3/2}\\ {\left(q^{\text{T}}{h}^3\right)}^{2/3}={\left(V{\left(2\pi mkT\right)}^{3/2}\right)}^{2/3}\\ {\left(q^{\text{T}}{h}^3\right)}^{2/3}=V^{2/3}\times 2\pi mkT\end{array}$

Simplify the above equation.

    $\begin{array}{c}2\pi mkT=\frac{{\left(q^{\text{T}}{h}^3\right)}^{2/3}}{V^{2/3}}\\ T={\left(\frac{q^{\text{T}}}{V}\right)}^{2/3}\frac{h^2}{2\pi mk}\end{array}$

Therefore, the temperature is calculated by using the formula shown below.

    $T=\frac{h^2}{2\pi mk}{\left(\frac{q^{\text{T}}}{V}\right)}^{2/3}$        (2)

The value of Planck’s constant is $6.626\times {10}^{-34}\text{J}\text{s}$.

The value of Boltzmann constant is $1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}$.

The value of translational partition function is $10$.

The mass of hydrogen atom is $1.6735\times {10}^{-27}\text{kg}$.

The side of a box is $100\text{nm}$.

The conversion of nm into m is done as shown below.

    $1\text{nm}={10}^{-9}\text{m}$

Therefore, the conversion of $100\text{nm}$ into m is done as shown below.

    $\begin{array}{c}100\text{nm}=100\times {10}^{-9}\text{m}\\ ={10}^{-7}\text{m}\end{array}$

The volume of a one-dimensional box is calculated by the formula shown below.

    $\text{Volume}\text{of}\text{a}\text{box}={\left(\text{Side}\right)}^3$        (3)

Substitute the value of side in equation (3) to calculate the volume of a box.

    $\begin{array}{c}\text{Volume}\text{of}\text{a}\text{box}={\left({10}^{-7}\text{m}\right)}^3\\ =1\times {10}^{-21}{\text{m}}^3\end{array}$

The volume of a box is $1\times {10}^{-21}{\text{m}}^3$.

Also,

  $\begin{array}{r}1\text{J}=1\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-2}\\ 1\text{kg}=1\text{J}{\text{m}}^{-2}{\text{s}}^2\end{array}$

Substitute the values of $q^{\text{T}}$, $V$, $m$, $k$ and $h$ in equation (2) to calculate the temperature.

    $\begin{array}{c}T=\frac{{\left(6.626\times {10}^{-34}\text{J}\text{s}\right)}^2}{2\times 3.14\times 1.6735\times {10}^{-27}\text{kg}\times 1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}}{\left(\frac{10}{1\times {10}^{-21}{\text{m}}^3}\right)}^{2/3}\\ =\frac{4.390\times {10}^{-67}{\text{J}}^{\text{2}}{\text{s}}^{\text{2}}}{2\times 3.14\times 1.6735\times {10}^{-27}\text{J}{\text{m}}^{-2}{\text{s}}^2\times 1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}}\times \frac{4.634}{1.032\times {10}^{-14}{\text{m}}^{\text{2}}}\\ =\frac{2.034\times {10}^{-66}}{1.50\times {10}^{-63}}\text{K}\\ =\underset{\_}{1.356\times 10^{-3}K}\end{array}$

Therefore, the temperature at which the value of $q$ is $10$ according to the integral approximation is $\underset{\_}{1.356\times 10^{-3}K}$.

The one-dimensional partition is expressed by the formula shown below.

    $q^T={\displaystyle \sum _{n=1}^{\infty }{e}^{-\frac{\left(n^2-1\right)\beta h^2}{8m{X}^2}}}$        (4)

The equation (4) is modified as shown below.

    $q^T={\displaystyle \sum _{n=1}^{\infty }{e}^{-\frac{\left(n^2-1\right)h^2}{8mkT{X}^2}}}$        (5)

Now, the value of $\frac{h^2}{8mkT{X}^2}$ is calculated as shown below.

Substitute the values of $X=100\text{nm}={10}^{-7}\text{m}$, $T$, $m$, $k$ and $h$ in the above formula.

    $\begin{array}{c}\frac{h^2}{8mkT{X}^2}=\frac{{\left(6.626\times {10}^{-34}\text{J}\text{s}\right)}^2}{\begin{array}{l}8\times 1.6735\times {10}^{-27}\text{J}{\text{m}}^{-2}{\text{s}}^2\times 1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\\ \times \text{1}{\text{.356×10}}^{-3}\text{K}\times {\left({10}^{-7}\text{m}\right)}^2\end{array}}\\ =\frac{4.390\times {10}^{-67}}{2.505\times {10}^{-66}}\\ =0.175\end{array}$

Substitute the value of $\frac{h^2}{8mkT{X}^2}$ in equation (5).

    $q^T={\displaystyle \sum _{n=1}^{\infty }{e}^{-0.175\left(n^2-1\right)}}$        (6)

Substitute the value of $n=0$ in equation (6) to calculate the value of partition function.

    $\begin{array}{c}{q}^T=e^{-0.175\left({\left(0\right)}^2-1\right)}\\ q^T=e^{-0.175\times -1}\\ =e^{0.175}\\ =1.191\end{array}$

Similarly, substitute the value of $n=1$ in equation (6) to calculate the value of partition function.

    $\begin{array}{c}{q}^T=e^{-0.175\left({\left(1\right)}^2-1\right)}\\ q^T=e^{-0.175\times 0}\\ =e^{-0}\\ =1\end{array}$

Substitute the value of $n=2$ in equation (6) to calculate the value of partition function.

    $\begin{array}{c}{q}^T=e^{-0.175\left({\left(2\right)}^2-1\right)}\\ q^T=e^{-0.175\times 3}\\ =e^{-0.525}\\ =0.60\end{array}$

Substitute the value of $n=3$ in equation (6) to calculate the value of partition function.

    $\begin{array}{c}{q}^T=e^{-0.175\left({\left(3\right)}^2-1\right)}\\ q^T=e^{-0.175\times 8}\\ =e^{-1.4}\\ =0.25\end{array}$

Substitute the value of $n=4$ in equation (6) to calculate the value of partition function.

    $\begin{array}{c}{q}^T=e^{-0.175\left({\left(4\right)}^2-1\right)}\\ q^T=e^{-0.175\times 15}\\ =e^{-2.625}\\ =0.0072\end{array}$

Substitute the value of $n=5$ in equation (6) to calculate the value of partition function.

    $\begin{array}{c}{q}^T=e^{-0.175\left({\left(5\right)}^2-1\right)}\\ q^T=e^{-0.175\times 24}\\ =e^{-4.2}\\ =0.0149\end{array}$

Substitute the value of $n=6$ in equation (6) to calculate the value of partition function.

    $\begin{array}{c}{q}^T=e^{-0.175\left({\left(6\right)}^2-1\right)}\\ q^T=e^{-0.175\times 35}\\ =e^{-6.125}\\ =2.187\times {10}^{-3}\end{array}$

Substitute the value of $n=7$ in equation (6) to calculate the value of partition function.

    $\begin{array}{c}{q}^T=e^{-0.175\left({\left(7\right)}^2-1\right)}\\ q^T=e^{-0.175\times 48}\\ =e^{-8.4}\\ =2.248\times {10}^{-4}\end{array}$

Substitute the value of $n=8$ in equation (6) to calculate the value of partition function.

    $\begin{array}{c}{q}^T=e^{-0.175\left({\left(8\right)}^2-1\right)}\\ q^T=e^{-0.175\times 63}\\ =e^{-11.025}\\ =1.628\times {10}^{-5}\end{array}$

Substitute the value of $n=9$ in equation (6) to calculate the value of partition function.

    $\begin{array}{c}{q}^T=e^{-0.175\left({\left(9\right)}^2-1\right)}\\ q^T=e^{-0.175\times 80}\\ =e^{-14}\\ =8.315\times {10}^{-7}\end{array}$

Substitute the value of $n=10$ in equation (6) to calculate the value of partition function.

    $\begin{array}{c}{q}^T=e^{-0.175\left({\left(10\right)}^2-1\right)}\\ q^T=e^{-0.175\times 99}\\ =e^{-17.325}\\ =2.99\times {10}^{-8}\end{array}$

Therefore, the values of partition function at different values of $n$ are shown below.

$n$	$q^T$
$0$	$1.191$
$1$	$1$
$2$	$0.60$
$3$	$0.25$
$4$	$0.0072$
$5$	$0.0149$
$6$	$2.187\times {10}^{-3}$
$7$	$2.248\times {10}^{-4}$
$8$	$1.628\times {10}^{-5}$
$9$	$8.315\times {10}^{-7}$
$10$	$2.99\times {10}^{-8}$

The value of partition function is calculated as shown below.

    $\begin{array}{c}{q}^T=\left(\begin{array}{l}1.191+1+0.60+0.25+0.0072+0.0149+2.187\times {10}^{-3}\\ +2.248\times {10}^{-4}+1.628\times {10}^{-5}+8.315\times {10}^{-7}+2.99\times {10}^{-8}\end{array}\right)\\ =\underset{\_}{3.0655}\end{array}$

Therefore, the exact partition function at the calculated temperature is $\underset{\_}{3.0655}$.