Chapter 5, Problem 5.133AP

(a)

Interpretation Introduction

Interpretation: The unfilled rows are to be completed by multiplying each molar mass and specific heat pair. The units of the resulting values in column $4$ are to be stated.

The average of the values of column $4$ is to be calculated.

The missing atomic masses and their link with corresponding elements given in the table with the use of mean value from part (b) are to be stated.

The missing specific heat values with the use of average value from part (b) are to be stated.

Concept introduction: As per Dulong and Petit, the product of atomic masses $\left(\mu \right)$ and specific heat $\left(c_s\right)$ of metals are nearly constant.

$\mu \times c_s=\text{constant}$

The average of the given values is calculated by the formula,

$\text{Average}=\frac{{\displaystyle \sum \text{Observed}\text{values}\left(\mu \times {\text{c}}_s\right)}}{\text{Number}\text{of}\text{observation}\left(\mu \times {\text{c}}_s\right)}$

The missing atomic mass from mean of $\mu \times {\text{c}}_s$ is calculated by the formula,

$\text{Missing}\text{atomic}\text{mass}=\frac{\text{Mean}\text{of}\mu \times {\text{c}}_s}{\text{Specific}\text{heat}}$

The missing specific heat values with the use of average value is calculated by formula,

$\text{Missing}\text{Specific}\text{heat}\text{value}=\frac{\text{Mean}\text{of}\mu \times {\text{c}}_s}{\text{Atomic}\text{mass}}$

To determine: The unfilled rows by multiplying each molar mass and specific heat pair.

Answer to Problem 5.133AP

Solution

The solution is stated below.

Explanation of Solution

Explanation

Given

The atomic mass $\left({\mu }_{\text{Pb}}\right)$ of element Lead is $207.2\text{g}/\text{mol}$.

The atomic mass $\left({\mu }_{\text{Au}}\right)$ of element Gold is $197.0\text{g}/\text{mol}$.

The atomic mass $\left({\mu }_{\text{Sn}}\right)$ of element Tin is $118.7\text{g}/\text{mol}$.

The atomic mass $\left({\mu }_{\text{Zn}}\right)$ of element Zinc is $65.4\text{g}/\text{mol}$.

The atomic mass $\left({\mu }_{\text{Cu}}\right)$ of element Copper is $63.5\text{g}/\text{mol}$.

The atomic mass $\left({\mu }_{\text{Fe}}\right)$ of element Iron is $55.8\text{g}/\text{mol}$.

The specific heat $\left(c_{s,\text{Pb}}\right)$ of element Lead is $0.123\text{J}/\left(\text{g}\cdot °\text{C}\right)$.

The specific heat $\left(c_{s\text{Au}}\right)$ of element Gold is $0.125\text{J}/\left(\text{g}\cdot °\text{C}\right)$.

The specific heat $\left(c_{s,\text{Sn}}\right)$ of element Tin is $0.215\text{J}/\left(\text{g}\cdot °\text{C}\right)$.

The specific heat $\left(c_{s,\text{Zn}}\right)$ of element Zinc is $0.388\text{J}/\left(\text{g}\cdot °\text{C}\right)$.

The specific heat $\left(c_{s,\text{Cu}}\right)$ of element Copper is $0.397\text{J}/\left(\text{g}\cdot °\text{C}\right)$.

The specific heat $\left(c_{s,\text{Fe}}\right)$ of element Iron is $0.460\text{J}/\left(\text{g}\cdot °\text{C}\right)$.

As per Dulong and Petit, the product of atomic masses $\left(\mu \right)$ and specific heat $\left(c_s\right)$ of metals are nearly constant.

$\mu \times c_s=\text{constant}$

The product of specific heat and atomic mass of metal Lead is,

$\begin{array}{c}=207.2\text{g}/\text{mol}\times 0.123\text{J}/\left(\text{g}\cdot °\text{C}\right)\\ =25.48\text{J}/\text{mol}°\text{C}\end{array}$

The product of specific heat and atomic mass of metal Gold is,

$\begin{array}{c}=197.0\text{g}/\text{mol}\times 0.125\text{J}/\left(\text{g}\cdot °\text{C}\right)\\ =24.62\text{J}/\text{mol}°\text{C}\end{array}$

The product of specific heat and atomic mass of metal Tin is,

$\begin{array}{c}=118.7\text{g}/\text{mol}\times 0.215\text{J}/\left(\text{g}\cdot °\text{C}\right)\\ =25.52\text{J}/\text{mol}°\text{C}\end{array}$

The product of specific heat and atomic mass of metal Zinc is,

$\begin{array}{c}=65.4\text{g}/\text{mol}\times 0.388\text{J}/\left(\text{g}\cdot °\text{C}\right)\\ =25.37\text{J}/\text{mol}°\text{C}\end{array}$

The product of specific heat and atomic mass of metal Copper is,

$\begin{array}{c}=63.5\text{g}/\text{mol}\times 0.397\text{J}/\left(\text{g}\cdot °\text{C}\right)\\ =25.20\text{J}/\text{mol}°\text{C}\end{array}$

The product of specific heat and atomic mass of metal Iron is,

$\begin{array}{c}=55.8\text{g}/\text{mol}\times 0.460\text{J}/\left(\text{g}\cdot °\text{C}\right)\\ =25.66\text{J}/\text{mol}°\text{C}\end{array}$

The product of specific heat and atomic mass of every metal has common unit $\text{J}/\text{mol}°\text{C}$.

All of the values of atomic mass, specific heat, and their product are summarized in Table $1$.

Element	Atomic mass $\left(\mu \right)$	Specific heat $\left(c_s\right)$	$\left(\mu \times c_s\right)$
Lead	$207.2\text{g}/\text{mol}$	$0.123\text{J}/\left(\text{g}\cdot °\text{C}\right)$	$25.48\text{J}/\text{mol}°\text{C}$
Gold	$197.0\text{g}/\text{mol}$	$0.125\text{J}/\left(\text{g}\cdot °\text{C}\right)$	$24.62\text{J}/\text{mol}°\text{C}$
Tin	$118.7\text{g}/\text{mol}$	$0.215\text{J}/\left(\text{g}\cdot °\text{C}\right)$	$25.52\text{J}/\text{mol}°\text{C}$
Zinc	$65.4\text{g}/\text{mol}$	$0.388\text{J}/\left(\text{g}\cdot °\text{C}\right)$	$25.37\text{J}/\text{mol}°\text{C}$
Copper	$63.5\text{g}/\text{mol}$	$0.397\text{J}/\left(\text{g}\cdot °\text{C}\right)$	$25.20\text{J}/\text{mol}°\text{C}$
Iron	$55.8\text{g}/\text{mol}$	$0.460\text{J}/\left(\text{g}\cdot °\text{C}\right)$	$25.66\text{J}/\text{mol}°\text{C}$

Table $1$.

(b)

Interpretation Introduction

To determine: The average of the values of column $4$ is to be calculated.

Answer to Problem 5.133AP

Solution

The average of the values of column $4$ is $\underset{\_}{25.30J/mol°C}$.

Explanation of Solution

Explanation

The average of the given values is calculated by the formula,

$\text{Average}=\frac{{\displaystyle \sum \text{Observed}\text{values}\left(\mu \times {\text{c}}_s\right)}}{\text{Number}\text{of}\text{observation}\left(\mu \times {\text{c}}_s\right)}$

The number of observation of $\left(\mu \times {\text{c}}_s\right)$ are $6$.

Substitute the values of $\left(\mu \times {\text{c}}_s\right)$ present in column $4$ in above expression.

$\begin{array}{c}\text{Average}=\frac{\left(25.48+24.62+25.52+25.37+25.20+25.66\right)\text{J}/\text{mol}°\text{C}}{6}\\ =\frac{151.858}{6}\\ =25.30\text{J}/\text{mol}°\text{C}\end{array}$

Hence, the average of the values of column $4$ is $\underset{\_}{25.30J/mol°C}$.

(c)

Interpretation Introduction

To determine: The missing atomic masses and their link with corresponding elements given in the table with the use of mean value from part (b).

Answer to Problem 5.133AP

Solution

The solution is stated in explanation.

Explanation of Solution

Explanation

The missing atomic mass from mean of $\mu \times {\text{c}}_s$ is calculated by the formula,

$\text{Missing}\text{atomic}\text{mass}=\frac{\text{Mean}\text{of}\mu \times {\text{c}}_s}{\text{Specific}\text{heat}}$ (1)

The mean of $\mu \times {\text{c}}_s$ is $25.30\text{J}/\text{mol}°\text{C}$. The missing atomic masses of elements are calculated below.

Bismuth element with specific heat of $0.120\text{J}/\left(\text{g}\cdot °\text{C}\right)$.

Substitute the values of specific heat and the mean of $\mu \times {\text{c}}_s$ in equation (1).

$\begin{array}{c}\text{Missing}\text{atomic}\text{mass}=\frac{25.30\text{J}/\text{mol}°\text{C}}{0.120\text{J}/\left(\text{g}\cdot °\text{C}\right)}\\ =210.83\text{g}/\text{mol}\end{array}$

The atomic mass of Bismuth in modern periodic table is $208.98\text{g}/\text{mol}$. Thus, it indicates that calculated value is close to actual value but not $100\%$ accurate.

An unknown element with specific heat of $0.233\text{J}/\left(\text{g}\cdot °\text{C}\right)$.

Substitute the values of specific heat and the mean of $\mu \times {\text{c}}_s$ in equation (1).

$\begin{array}{c}\text{Missing}\text{atomic}\text{mass}=\frac{25.30\text{J}/\text{mol}°\text{C}}{0.233\text{J}/\left(\text{g}\cdot °\text{C}\right)}\\ =108.6\text{g}/\text{mol}\end{array}$

The calculated atomic mass of unknown element is close to Silver $\left(\text{Ag}\right)$ which has atomic mass of $107.86\text{g}/\text{mol}$ in modern periodic table. Thus, it indicates that calculated value is close to actual value, but not $100\%$ accurate.

An unknown element with specific heat of $0.433\text{J}/\left(\text{g}\cdot °\text{C}\right)$.

Substitute the values of specific heat and the mean of $\mu \times {\text{c}}_s$ in equation (1).

$\begin{array}{c}\text{Missing}\text{atomic}\text{mass}=\frac{25.30\text{J}/\text{mol}°\text{C}}{0.433\text{J}/\left(\text{g}\cdot °\text{C}\right)}\\ =58.4\text{g}/\text{mol}\end{array}$

The calculated atomic mass of unknown element is close to Silver $\left(\text{Ag}\right)$ which has atomic mass of $58.69\text{g}/\text{mol}$ in modern periodic table. Thus, it indicates that calculated value is close to actual value, but not $100\%$ accurate.

Hence, the Dulong and Pettit assumptions about atomic masses are not $100\%$ accurate.

(d)

Interpretation Introduction

To determine: The missing specific heat values with the use of average value from part (b).

Answer to Problem 5.133AP

Solution

The missing specific heat values of Platinum and Sulfur are $\underset{\_}{0.129J/\left(g\times °C\right)}$ and $\underset{\_}{0.788J/\left(g\times °C\right)}$.

Explanation of Solution

Explanation

The missing specific heat values with the use of average value is calculated by formula,

$\text{Missing}\text{Specific}\text{heat}\text{value}=\frac{\text{Mean}\text{of}\mu \times {\text{c}}_s}{\text{Atomic}\text{mass}}$ (2)

The mean of $\mu \times {\text{c}}_s$ is $25.30\text{J}/\text{mol}°\text{C}$. The specific heat values are calculated below.

Platinum with atomic mass of $195.1\text{g}/\text{mol}$.

Substitute the values of atomic mass and the mean of $\mu \times {\text{c}}_s$ in equation (2).

$\begin{array}{c}\text{Missing}\text{Specific}\text{heat}\text{value}=\frac{25.30\text{J}/\text{mol}°\text{C}}{195.1\text{g}/\text{mol}}\\ =0.129\text{J}/\left(\text{g}\cdot °\text{C}\right)\end{array}$

Sulfur with atomic mass of $32.1\text{g}/\text{mol}$.

Substitute the values of atomic mass and the mean of $\mu \times {\text{c}}_s$ in equation (2).

$\begin{array}{c}\text{Missing}\text{Specific}\text{heat}\text{value}=\frac{25.30\text{J}/\text{mol}°\text{C}}{32.1\text{g}/\text{mol}}\\ =0.788\text{J}/\left(\text{g}\cdot °\text{C}\right)\end{array}$

Hence, the missing specific heat values of Platinum and Sulfur are $\underset{\_}{0.129J/\left(g\times °C\right)}$ and $\underset{\_}{0.788J/\left(g\times °C\right)}$.

Conclusion

1. a. The product of specific heat and atomic mass of every metal has common unit $\text{J}/\text{mol}°\text{C}$.
2. b. The average of the values of column $4$ is $\underset{\_}{25.30J/mol°C}$.
3. c. The Dulong and Pettit assumptions about atomic masses are not $100\%$ accurate.
4. d. The missing specific heat values of Platinum and Sulfur are $\underset{\_}{0.129J/\left(g\times °C\right)}$ and $\underset{\_}{0.788J/\left(g\times °C\right)}$