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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.

Sagot :

ajeigbeibraheem

Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density [tex]\rho_{avg}[/tex]; we have:

[tex]\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }[/tex]

The average atomic weight is:

[tex]A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }[/tex]

So; in terms of vanadium, the Concentration of iron is:

[tex]C_{Fe} = 100 - C_v[/tex]

From a unit cell volume [tex]V_c[/tex]

[tex]V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}[/tex]

where;

[tex]N_A[/tex] = number of Avogadro constant.

SO; replacing [tex]V_c[/tex] with [tex]a^3[/tex] ; [tex]\rho_{avg}[/tex] with [tex]\dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }[/tex] ; [tex]A_{avg}[/tex] with [tex]\dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }[/tex] and

[tex]C_{Fe}[/tex] with [tex]100-C_v[/tex]

Then:

[tex]a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }[/tex]

[tex]a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }[/tex]

[tex]a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }[/tex]

Replacing the values; we have:

[tex](0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol) } }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3) } }[/tex]

[tex]2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} } \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }[/tex]

[tex]2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})[/tex]

[tex]\mathbf{C_v = 9.1 \ wt\%}[/tex]

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