Question

Above 882oC, zirconium has a BCC crystal structure with a = 0.332 nm. Below this temperature, zirconium has an HCP structure with a = 0.2978 nm and c = 0.4735 nm. Determine the percent volume change when BCC zirconium transforms to HCP zirconium. Is that contraction or expansion?

Answer 1

`\mathbf{\Delta V = -0.63 \%}`

Contraction

Step-by-step explanation:

From the given information;

Above 88° C

zirconium has a BCC crystal structure with a = 0.332 nm

Below this temperature

zirconium has an HCP structure with a = 0.2978 nm and c = 0.4735 nm

the volume of BCC can now be:

`V_(BCC) = (a)^3`

`V_(BCC) = (0.332 \ nm)^3`

`V_(BCC) =0.03660 \ nm^3`

the volume of HCP can now be:

`V_(HCP) = (a)^2 (c) cos \ 30^0`

`V_(HCP) = (0.2978)^2 (0.4735) \ cos 30`

`V_(HCP) =0.03637 \ nm^3`

Ths; the volume percent change when BCC zirconium transforms to HCP zirconium can be calculated as:

`\Delta V = (V_(HCP)-V_(BCC))/(V_(BCC)) * 100 \%`

`\Delta V = (0.03637 \ nm^3-0.03660\ nm^3)/(0.03660\ nm^3)} * 100 \%`

`\mathbf{\Delta V = -0.63 \%}`

Hence; it is contraction due to what the negative sign portray, The negative sign signifies that there is contraction during cooling