Answer to Question #175035 in Mechanical Engineering for Varun

Question

Metallic iron changes from BCC to FCC form at 9100C. At this temperature,

the atomic radii of the iron atom in the two structures are 0.1258 nm and

0.1292 nm respectively. Calculate the volume change in percentage during

this structural change. Also, calculate the percentage change in density.

Answer 1

$$\rho=\dfrac{M}{V}, \rho=\dfrac{nA}{V_cN_{avg}}$$

At the transformation temperature the mass of the material is not changing,therefore

$$M=\rho_{BCC}V_{BCC}-\rho_{FCC}V_{FCC}$$

 

$$V_{FCC}=\dfrac{\rho_{BCC}}{\rho_{FCC}}V_{BCC}$$

$$\dfrac{\rho_{BCC}}{\rho_{FCC}}=\dfrac{\dfrac{n_{BCC}A}{V_{BCC}N_{avg}}}{\dfrac{n_{FCC}A}{V_{FCC}N_{avg}}}=\dfrac{n_{BCC}V_{FCC}}{n_{FCC}V_{BCC}}$$

Then

$$\dfrac{V_{FCC}-V_{BCC}}{V_{BCC}}\times100\%=$$

$$=\dfrac{\dfrac{\rho_{BCC}}{\rho_{FCC}}V_{BCC}-V_{BCC}}{V_{BCC}}\times100\%=$$

$$=(\dfrac{n_{BCC}V_{FCC}}{n_{FCC}V_{BCC}}-1)\times 100\%$$

The relation between radius(r) and lattice constant (a) is given as,

For BCC structure: $a_{BCC}=\dfrac{4r}{\sqrt{3}}=0.2905 \ nm$

For FCC structure: $a_{FCC}=\dfrac{4r}{\sqrt{2}}=0.3654 \ nm$

The volume change in percentage during this structural change is

$$\dfrac{V_{FCC}-V_{BCC}}{V_{BCC}}\times100\%=$$

$$=(\dfrac{2(0.3654\ nm)^3}{4(0.2905\ nm)^3}-1)\times 100\%\approx-0.49676\%$$

Percentage change in density is given by,

$$\dfrac{\rho_{FCC}-\rho_{BCC}}{\rho_{BCC}}\times100\%=$$

$$=(\dfrac{n_{FCC}V_{BCC}}{n_{BCC}V_{FCC}}-1)\times10\%$$

$$=(\dfrac{4(0.2905\ nm)^3}{2(0.3654\ nm)^3}-1)\times 100\%\approx0.49924\%$$