Chapter 2, Problem 2.6P

To determine

(a)

The density of FCC iron.

Answer to Problem 2.6P

The density of FCC iron is $8.027\text{g}/{\text{cm}}^{\text{3}}$.

Explanation of Solution

Given:

The density of BCC iron at room temperature is listed as $7.87\text{g}/{\text{cm}}^{\text{3}}$.

At temperature above $912°\text{C}$ iron has the FCC $\left(\gamma \right)$ structure.

The lattice parameter is $0.3589\text{nm}$.

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 the $\text{Number}\text{of}\text{atoms}$ is the numbers of atoms in a unit cell.

Write the formula to find the density of silver.

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

Here, ${\rho }_{\gamma \text{Fe}}$ is thedensity of FCC iron, $M_{\text{Fe}}$ is the molecular weight of iron, and $N_{\text{A}}$ is the Avogadro’s number.

Calculation:

Gama phase iron has theFace Centered Cubic (FCC) structure with $4$ number of atoms.

Substitute $4\text{atoms}$ for $\text{Number}\text{of}\text{atoms}$ and $0.3589\text{nm}$ for $a$ in the equation (I) to find $n_{\text{a}}$.

$\begin{array}{c}{n}_{\text{a}}=\frac{4\text{atoms}}{{\left(0.3589\text{nm}\right)}^3}\\ =\frac{4\text{atoms}}{{\left(0.3589\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}^3}\\ =\frac{4\text{atoms}}{0.04623\times {10}^{-27}{\text{m}}^3}\\ =86.52\times {10}^{27}\text{atoms}/{\text{m}}^{\text{3}}\end{array}$

The molecular weight of silver is $55.85\times {10}^{-3}\text{kg}/\text{mole}$.

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

Substitute $86.52\times {10}^{27}\text{atoms}/{\text{m}}^{\text{3}}$ for $n_{\text{a}}$, $55.85\times {10}^{-3}\text{kg}/\text{mole}$ for $M_{Fe}$ and $6.02\times {10}^{23}\text{atoms}/\text{mole}$ for $N_{\text{A}}$ in equation (II) to find ${\rho }_{\gamma \text{Fe}}$.

$\begin{array}{c}{\rho }_{\gamma \text{Fe}}=\frac{86.52\times {10}^{27}\text{atoms}/{\text{m}}^{\text{3}}\times 55.85\times {10}^{-3}\text{kg}/\text{mole}}{6.02\times {10}^{23}\text{atoms}/\text{mole}}\\ =8.027\times {10}^3\frac{\text{kg}}{{\text{m}}^{\text{3}}}\times \left(\frac{\text{10}\text{g}}{\text{kg}}\right)\times \left(\frac{{10}^6{\text{cm}}^3}{{\text{m}}^3}\right)\\ =8.027\text{g}/{\text{cm}}^{\text{3}}\end{array}$

Conclusion:

Therefore, the density of FCC iron is $8.027\text{g}/{\text{cm}}^{\text{3}}$.

To determine

(b)

The difference in the density of the FCC phase relative to the BCC phase of iron.

Answer to Problem 2.6P

The density of the obtained FCC phase is higher than the BCC phase of iron.

Explanation of Solution

Calculation:

The obtained density of FCC iron is $8.027\text{g}/{\text{cm}}^{\text{3}}$, and the density of BCC iron at room temperature is listed as $7.87\text{g}/{\text{cm}}^{\text{3}}$.

Compare the obtained density of FCC iron at $912°\text{C}$ temperature with the density of iron at room temperature.The obtained density of Face Centered Cubic (FCC) iron is greater than the density of Body Centered Cubic (BCC) iron at room temperature. This implies that theBCCiron structure is loosely packed when compared with FCC gamma phase iron.

Conclusion:

Therefore, the density of the obtainedFCC phase is higherthan the BCC phaseof iron.