Chapter 4, Problem 4.14P

Interpretation Introduction

Interpretation:

The density and the packing factor of the given alloy is to be determined.

Concept introduction:

The relationship between the theoretical density $\left(\rho \right)$ of an alloy, number of atoms of all the constituent species present in the alloy $\left(n_i\right)$ present in a unit cell and volume of a unit cell $\left(V\right)$ is:

  $\rho =\frac{{\displaystyle \sum n_i\times {\left(\text{A}\text{.M}\right)}_i}}{V\times N_{\text{A}}}$ ....... (1)

Here, ${\left(\text{A}\text{.M}\right)}_i$ is the atomic mass of the species $i$ in the alloy and $N_{\text{A}}$ is the Avogadro's number.

Volume of a unit cell $\left(V\right)$ is calculated using the lattice parameter $\left(a_0\right)$ as:

  $V={\left(a_0\right)}^3$ ....... (2)

The formula to calculate the packing factor of an alloy is:

  $\text{Packing factor}=\frac{{\displaystyle \sum n_i\times V_i}}{V}$ ....... (3)

Here, $V_i$ is the volume of each atom of the species $i$ in the alloy.

The formula to be used to calculate the volume of a single atom is:

  $V_i=\frac{4}{3}\pi r_i^3$ ....... (4)

Here, $r_i$ is the atomic radius of each atom of the species $i$ in the alloy.

Answer to Problem 4.14P

The density of the given alloy is $7.89{\text{ g/cm}}^3$.

The packing factor of the alloy is $0.681$.

Explanation of Solution

Given information:

  $1$ carbon atom for every $100$ iron atoms at the interstitial positions in BCC iron. The lattice parameter for it is $0.2867\text{ nm}$.

Use equation (2) to calculate the volume of a unit cell from its lattice parameter as:

  $\begin{array}{c}V={\left(a_0\right)}^3\\ ={\left(0.2867\times {10}^{-7}\text{ cm}\right)}^3\\ =2.357\times {10}^{-23}{\text{ cm}}^3\end{array}$

The number of iron atoms present in a unit cell having BCC structure is $2$. It is given that there is $1$ carbon atom for every $100$ iron atoms, so for every cell there is $1/50$ carbon atoms in one cell.

Atomic weights of iron and carbon are $55.847\text{ g/mol}$ and $12\text{ g/mol}$ respectively.

Use equation (1) to calculate the density of the alloy as:

  $\begin{array}{c}\rho =\frac{\left(n_{\text{Fe}}\times {\left(\text{A}\text{.M}\right)}_{\text{Fe}}\right)+\left(n_{\text{C}}\times {\left(\text{A}\text{.M}\right)}_{\text{C}}\right)}{V\times N_{\text{A}}}\\ =\frac{\left(2\times 55.847\right)+\left(\left(1/50\right)\times 12\right)}{\left(2.357\times {10}^{-23}\right)\times \left(6.022\times {10}^{23}\right)}\\ =7.89{\text{ g/cm}}^3\end{array}$

Atomic radius of iron and carbon atoms are $1.241\times {10}^{-7}\text{ cm}$ and $0.77\times {10}^{-7}\text{ cm}$ respectively.

Use equation (4) to calculate the volume of a single atom as:

  $\begin{array}{c}{V}_{\text{Fe}}=\frac{4}{3}\pi r_{\text{Fe}}^3\\ =\frac{4}{3}\left(3.14\right){\left(1.241\times {10}^{-8}\text{ cm}\right)}^3\\ =8.002\times {10}^{-24}{\text{ cm}}^3\\ V_{\text{C}}=\frac{4}{3}\pi r_{\text{C}}^3\\ =\frac{4}{3}\left(3.14\right){\left(0.77\times {10}^{-7}\text{ cm}\right)}^3\\ =1.911\times {10}^{-24}{\text{ cm}}^3\end{array}$

Now, use equation (3) to calculate its packing factor as:

  $\begin{array}{c}\text{Packing factor}=\frac{\left(n_{\text{Fe}}\times V_{\text{Fe}}\right)+\left(n_{\text{C}}\times V_{\text{C}}\right)}{V}\\ =\frac{\left(2\times 8.002\times {10}^{-24}\right)+\left(\left(1/50\right)\times 1.911\times {10}^{-24}\right)}{2.357\times {10}^{-23}}\\ =0.681\end{array}$

Conclusion

The density of the given alloy is calculated as $7.89{\text{ g/cm}}^3$ and the packing factor of the alloy is calculated as $0.681$.