Chapter 2, Problem 2.5P

To determine

The comparison of theoretical density of silver at $960°\text{C}$ to that at room temperature.

Answer to Problem 2.5P

The obtained density of silver at $960°\text{C}$ temperature $9.842\text{g}/{\text{cm}}^{\text{3}}$ is less than the density of silver at room temperature.

Explanation of Solution

Given:

The density of silver at room temperature is $10.49\text{g}/{\text{cm}}^{\text{3}}$.

The lattice parameter for is $0.4176\text{nm}$.

Formula used:

Write the expression to find the number of atoms per unit volume.

$n_{\text{a}}=\frac{\text{Number}\text{of}\text{atoms}}{a^3}$ …… (I)

Here, $n_{\text{a}}$ is the number of atoms per unit volume and $\text{Number}\text{of}\text{atoms}$ is the numbers of atoms in a unit cell.

Write the formula to find the density of silver.

${\rho }_{\text{Ag}}=\frac{n_{\text{a}}\times M_{\text{Ag}}}{N_{\text{A}}}$ …… (II)

Here, ${\rho }_{\text{Ag}}$ is the density of silver, $M_{\text{Ag}}$ is the molecular weight of silver, and $N_{\text{A}}$ is the Avogadro’s number.

Calculation:

Silver has the Face Centered Cubic (FCC) structure with $4$ number of atoms.

Substitute $4\text{atoms}$ for $\text{Number}\text{of}\text{atoms}$ and $0.4176\text{nm}$ for $a$ in equation (I) to find $n_{\text{a}}$.

$\begin{array}{c}{n}_{\text{a}}=\frac{4\text{atoms}}{{\left(0.4176\text{nm}\right)}^3}\\ =\frac{4\text{atoms}}{{\left(0.4176\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}^3}\\ =\frac{4\text{atoms}}{0.0728\times {10}^{-27}{\text{m}}^3}\\ =54.94\times {10}^{27}\text{atoms}/{\text{m}}^{\text{3}}\end{array}$

The molecular weight of silver is $107.87\times {10}^{-3}\text{kg}/\text{mole}$.

Avogadro’s number is $6.02\times {10}^{23}\text{atoms}/\text{mole}$.

Substitute $54.94\times {10}^{27}\text{atoms}/{\text{m}}^{\text{3}}$ for $n_{\text{a}}$, $107.87\times {10}^{-3}\text{kg}/\text{mole}$ for $M_{\text{Ag}}$ and $6.02\times {10}^{23}\text{atoms}/\text{mole}$ for $N_{\text{A}}$ in equation (II) to find ${\rho }_{\text{Ag}}$ at temperature $960°\text{C}$.

$\begin{array}{c}{\rho }_{\text{Ag}}=\frac{54.94\times {10}^{27}\text{atoms}/{\text{m}}^{\text{3}}\times 107.87\times {10}^{-3}\text{kg}/\text{mole}}{6.02\times {10}^{23}\text{atoms}/\text{mole}}\\ =9.842\times {10}^3\frac{\text{kg}}{{\text{m}}^{\text{3}}}\times \left(\frac{{\text{10}}^3\text{g}}{\text{kg}}\right)\times \left(\frac{{10}^{-6}{\text{cm}}^3}{{\text{m}}^3}\right)\\ =9.842\text{g}/{\text{cm}}^{\text{3}}\end{array}$

Compare the obtained density of silver at $960°\text{C}$ temperature with the density of silver at room temperature is $10.49\text{g}/{\text{cm}}^{\text{3}}$.

The obtained density of silver is less than the density of silver at room temperature.

Conclusion:

Therefore, the obtained density of silver at $960°\text{C}$ temperature $9.842\text{g}/{\text{cm}}^{\text{3}}$ is less than the density of silver at room temperature.