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

Answer:

The answer is below

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, $n_a=avogadro\ constant=6.02*10^{23}\ mol^{-1}$

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

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

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