Question

[10 pts]Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to approximately 20 wt.% V at room temperature. Determine the concentration in weight percent of V that must be added to iron to yield a unit cell edge length of 0.289 nm.

Answer 1

The answer is below

Step-by-step explanation:

The unit cell edge length (a) = 0.298 nm = 0.289 * 10⁻⁷ cm

The unit volume (V) = a³ = (0.289 * 10⁻⁷ cm)³ = 24 * 10⁻²⁴ cm³

There are 2 atoms per cell, hence N = 2. Also,
`n_a=avogadro\ constant=6.02*10^(23)\ mol^(-1)`

`atomic\ weight\ of\ iron(A_f)=55.85,atomic\ weight\ of\ vanadium(A_V)=50.94,`

`density\ of\ iron(\rho_f)=7.87,density\ of\ vanadium(\rho_v)=6.1.\\\\V=(nA_(av))/(n_a\rho_(av)) \\\\A_(av)=(100)/(C_v/A_v+C_f/A_f) \\\\\rho_(av)=(100)/(C_v/\rho_v+C_f/\rho_f) \\\\C_v=concentration\ of\ vanadium,C_f=concentration\ of\ iron.\\\\V=(nA_(av))/(n_a\rho_(av))=(n*(100)/(C_v/A_v+C_f/A_f))/(n_a*(100)/(C_v/\rho_v+C_f/\rho_f))\\\\`

`24*10^(-24)=(2*(100)/(C_v/50.94+C_f/55.85))/(6.02*10^(23)*(100)/(C_v/6.1+C_f/7.87))\\\\7.2=(C_v/6.1+C_f/7.87)/(C_v/50.94+C_f/55.85)\ \ \ (1) \\\\Also:\\\\C_v+C_f=100\%\ \ \ (2)\\\\\\solving\ equation\ 1\ and\ 2\ simultaneously\ gives:\\\\C_v=10\% \ and\ C_f=90\%\\\\`