Chapter 3, Problem 3C.9P

(a)

Interpretation Introduction

Interpretation: The reason of the plot of $\frac{C_{p,\text{m}}\left(T\right)}{T}$ against $T^2$ is expected to be a straight line with a slope $a$ and intercept $b$ has to be stated.

Concept introduction: $C_{p,\text{m}}$ represents the molar heat capacity.  At low temperatures, the value of $C_{p,\text{m}}$ for solids can be calculated by the Debye’s law.  At low temperatures, the value of $C_{p,\text{m}}$ for metals is calculated by the Debye contribution as well as the electronic contributions.

Answer to Problem 3C.9P

The plot between $\frac{C_{p,\text{m}}\left(T\right)}{T}$ against $T^2$ follows the straight line equation with a slope $a$ and intercept with $y$ axis as $b$.

Explanation of Solution

The heat capacity at low temperature is calculated by the following formula.

    $C_{p,\text{m}}\left(T\right)=a{T}^3+bT$

Divide the above equation by $T$.

    $\frac{C_{p,\text{m}}\left(T\right)}{T}=a{T}^2+b$                                                                                        (1)

Where,

• $\frac{C_{p,\text{m}}\left(T\right)}{T}$ is the molar heat capacity with respect to the temperature $T$.
• $a$ and $b$ are the Debye’s constant.

Compare the above equation with the straight line equation, $y=mx+c$.

The vertical axis, $y$ represents $\frac{C_{p,\text{m}}\left(T\right)}{T}$

The horizontal axis, $x$ represents $T^2$.

The slope of the plot, $m$ represents the Debye’s conatant $a$.

The intercept with $y$ axis $c$, represents the Debye’s constant $b$.

The plot between $\frac{C_{p,\text{m}}\left(T\right)}{T}$ against $T^2$ follows the straight line equation with a slope $a$ and intercept with $y$ axis as $b$.  Therefore, the plot of $\frac{C_{p,\text{m}}\left(T\right)}{T}$ against $T^2$ is expected to be a straight line with a slope $a$ and intercept $b$.

(b)

Interpretation Introduction

Interpretation: The values of the constants $a$ and $b$ has to be calculated.

Concept introduction: $C_{p,\text{m}}$ represents the molar heat capacity.  At low temperatures, the value of $C_{p,\text{m}}$ for solids can be calculated by the Debye’s law.  At low temperatures, the value of $C_{p,\text{m}}$ for metals is calculated by the Debye contribution as well as the electronic contributions.

Answer to Problem 3C.9P

The value of $a$ and $b$ are $\underset{\_}{2.44Jmo{l}^{-1}{K}^{-4}}$ and $\underset{\_}{2.087Jmo{l}^{-1}{K}^{-2}}$ respectively.

Explanation of Solution

The given data for the variation of molar heat capacity at very low temperature is shown below.

$T/\text{K}$	$0.20$	$0.25$	$0.30$	$0.35$	$0.40$	$0.45$	$0.50$	$0.55$
$C_{p,\text{m}}/\left(\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)$	$0.437$	$0.560$	$0.693$	$0.838$	$0.996$	$1.170$	$1.361$	$1.572$

The table for the value of $\frac{C_{p,\text{m}}\left(T\right)}{T}$ against $T^2$ is shown below.

$T^2$	$0.04$	$0.0625$	$0.09$	$0.1225$	$0.16$	$0.2025$	$0.25$	$0.3025$
$\frac{C_{p,\text{m}}\left(T\right)}{T}$	$2.185$	$2.24$	$2.31$	$2.394$	$2.49$	$2.6$	$2.722$	$2.85$

The plot of $\frac{C_{p,\text{m}}\left(T\right)}{T}$ against $T^2$ is shown below.

From the above table, values of $T^2$ are $0.04$ and $0.0625$ and their corresponding values of $\frac{C_{p,\text{m}}\left(T\right)}{T}$ are $2.185$ and $2.24$ respectively.

Substitute the values of corresponding values of $\frac{C_{p,\text{m}}\left(T\right)}{T}$ and $T^2$ in equation (1).

    $2.185=0.04a+b$                                                                                          (2)

    $2.24=0.0625a+b$                                                                                        (3)

Solve the linear equation (2) and (3).

    $a=\underset{\_}{2.44Jmo{l}^{-1}{K}^{-4}}$ and $b=\underset{\_}{2.087Jmo{l}^{-1}{K}^{-2}}$

Hence, the value of $a$ and $b$ are $\underset{\_}{2.44Jmo{l}^{-1}{K}^{-4}}$ and $\underset{\_}{2.087Jmo{l}^{-1}{K}^{-2}}$ respectively.

(c)

Interpretation Introduction

Interpretation: The expression for the molar entropy has to be predicted.

Concept introduction: $C_{p,\text{m}}$ represents the molar heat capacity.  At low temperatures, the value of $C_{p,\text{m}}$ for solids can be calculated by the Debye’s law.  At low temperatures, the value of $C_{p,\text{m}}$ for metals is calculated by the Debye contribution as well as the electronic contributions.

Answer to Problem 3C.9P

The expression for the molar entropy is $S_{\text{m}}=a\left[\frac{T_2^3}{3}-\frac{T_1{}^3}{3}\right]+b\left[T_2-T_1\right]$.

Explanation of Solution

The molar entropy is calculated by the following formula.

    $S_{\text{m}}=C_{p,\text{m}}\left(T\right)\ln \left(\frac{T_2}{T_1}\right)$                                                                                    (4)

The value of $C_{p,\text{m}}\ln \left(\frac{T_2}{T_1}\right)$ is calculated by integrating the equation (1) in between the limits $T_2$ and $T_1$ with respect to the temperature.

Integrate the equation (1).

    $\begin{array}{c}{\displaystyle {\int }_{T_1}^{T_2}\frac{C_{p,\text{m}}\left(T\right)}{T}}\text{d}T={\displaystyle {\int }_{T_1}^{T_2}a{T}^2\text{d}T}+{\displaystyle {\int }_{T_1}^{T_2}b}\text{d}T\\ {\left[C_{p,\text{m}}\left(T\right)\ln \left(T\right)\right]}_{T_1}^{T_2}=a{\left[\frac{T^3}{3}\right]}_{T_1}^{T_2}+b{\left[T\right]}_{T_1}^{T_2}\\ C_{p,\text{m}}\left(T\right)\ln \left(\frac{T_2}{T_1}\right)=a\left[\frac{T_2^3}{3}-\frac{T_1{}^3}{3}\right]+b\left[T_2-T_1\right]\end{array}$

Substitute the value of $C_{p,\text{m}}\left(T\right)\ln \left(\frac{T_2}{T_1}\right)$ in equation (4).

    $S_{\text{m}}=a\left[\frac{T_2^3}{3}-\frac{T_1{}^3}{3}\right]+b\left[T_2-T_1\right]$

Hence the expression for molar entropy is $S_{\text{m}}=a\left[\frac{T_2^3}{3}-\frac{T_1{}^3}{3}\right]+b\left[T_2-T_1\right]$.

(d)

Interpretation Introduction

Interpretation: The molar entropy of potassium at $2.0\text{K}$ has to be calculated.

Concept introduction: Standard entropy of reaction predicts the feasibility of the reaction.  For a reaction to be feasible, the change in the standard entropy of the reaction must be positive. Standard entropy of the reaction increases as the numbers of gaseous constituents in the reaction increases.

Answer to Problem 3C.9P

The molar entropy of potassium at $2.0\text{K}$ is $\underset{\_}{21.8106J{K}^{-1}mo{l}^{-1}}$.

Explanation of Solution

The molar entropy at $2.0\text{K}$ is calculated by the following formula.

    $S_T=\frac{C_{p,\text{m}}\Delta T}{T}$                                                                                                 (5)

Where,

• $S_T$ is the entropy at $T\text{K}$.
• $C_{p,\text{m}}$ is the molar heat capacity at constant pressure.
• $\Delta T$ is the change in temperature.

The change in temperature is calculated by the following formula.

    $\Delta T=T_2-T_1$                                                                                                   (6)

It is given that the temperature $T_1$ is $0.20\text{K}$ and temperature $T_2$ is $2.0\text{K}$.

Substitute the values of $T_1$ and $T_2$ in equation (6).

    $\begin{array}{c}\Delta T=2.0\text{K}-0.20\text{K}\\ \text{=1}\text{.8}\text{K}\end{array}$

Therefore, the change in temperature is $1.8\text{K}$.

The molar heat capacity is calculated by the given expression.

    $C_{p,\text{m}}\left(T\right)=a{T}^3+bT$                                                                                        (7)

It is given that the temperature $T$ is $2.0\text{K}$.

The value of $a$ is $2.44\text{}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-4}}$.

The value of $b$ is $2.087\text{}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-2}}$.

$\begin{array}{c}{C}_{p,\text{m}}\left(2.0\text{K}\right)=2.44\text{}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-4}}{\left(2\text{K}\right)}^3+2.087\text{}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-2}}\left(2\text{K}\right)\\ =19.52\text{}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}+4.174\text{}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\\ =24.234\text{}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\end{array}$

Therefore the value of $C_{p,\text{m}}\left(2.0\text{K}\right)$ is $24.234\text{}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}$.

Substitute the value of $T,\Delta T$ and $C_{p,\text{m}}\left(2.0\text{K}\right)$ in equation (1).

    $\begin{array}{c}{S}_{\text{m},2.0\text{K}}=\frac{24.234\text{}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\times 1.8\text{K}}{2.0\text{}\text{K}}\\ =\underset{\_}{21.8106J{K}^{-1}mo{l}^{-1}}\end{array}$

Hence, the molar entropy at $2.0\text{K}$ is $\underset{\_}{21.8106J{K}^{-1}mo{l}^{-1}}$.