Chapter 27, Problem 27.24P

(a)

To determine

The amount of iron in $1.00\text{mole}$ of iron.

Answer to Problem 27.24P

$1.00\text{mole}$ of iron consists $0.05585\text{kg}$ of iron.

Explanation of Solution

Given Info: The cross-sectional area of iron wire is $5.00\times {10}^{-6}{\text{m}}^2$ and the electric current is $30.0\text{A}$.

Explanation:

Amount of iron in one mole is given as,

$M\text{=55}\text{.85}\text{g}/\text{mol}$

It is the standard value for iron wire.

Convert the amount of iron into $\text{kg}/\text{mol}$.

$\begin{array}{c}M\text{=55}\text{.85}\text{g}/\text{mol}\times \left(\frac{{10}^{-3}\text{kg}/\text{mol}}{1\text{g}/\text{mol}}\right)\\ =0.05585\end{array}$

Thus, $1.00\text{mole}$ of iron consists $0.05585\text{kg}$ of iron.

Conclusion:

Therefore, $1.00\text{mole}$ of iron consists $0.05585\text{kg}$ of iron.

(b)

To determine

The molar density of iron.

Answer to Problem 27.24P

The molar density of iron is $1.41\times {10}^5\text{mol}/{\text{m}}^3$.

Explanation of Solution

Given Info: The cross-sectional area of iron wire is $5.00\times {10}^{-6}{\text{m}}^2$ and the electric current is $30.0\text{A}$.

Explanation:

The formula for the molar density is,

$M_{\text{D}}=\frac{{\rho }_{\text{i}}}{M}$

Here,

${\rho }_{\text{i}}$ is the density of iron.

$M$ is the moles.

Substitute $7.86\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{i}}$ and $0.05585\text{kg}/\text{mol}$ for $M$ in above equation to find the $M_{\text{D}}$.

$\begin{array}{c}{M}_{\text{D}}=\frac{7.86\text{kg}/{\text{m}}^3}{0.05585\text{kg}/\text{mol}}\\ =1.41\times {10}^5\text{mol}/{\text{m}}^3\end{array}$

Thus, the molar density of iron is $1.41\times {10}^5\text{mol}/{\text{m}}^3$.

Conclusion:

Therefore, the molar density of iron is $1.41\times {10}^5\text{mol}/{\text{m}}^3$.

(c)

To determine

The number density of iron atoms.

Answer to Problem 27.24P

The number density of iron atoms is $8.49\times {10}^{28}\text{atoms}/{\text{m}}^{\text{3}}$

Explanation of Solution

Given Info: The cross-sectional area of iron wire is $5.00\times {10}^{-6}{\text{m}}^2$ and the electric current is $30.0\text{A}$.

Explanation:

The formula for the number density is,

$n=A{M}_{\text{D}}$

Here,

$A$ is the Avogadro number.

$M_{\text{D}}$ is the molar density.

Substitute $1.41\times {10}^5\text{mol}/{\text{m}}^3$ for $M_{\text{D}}$ and $6.02\times {10}^{23}\text{atoms}/\text{mol}$ for $A$ in above equation to find $n$.

$\begin{array}{c}n=\left(6.02\times {10}^{23}\text{atoms}/\text{mol}\right)\left(1.41\times {10}^5\text{mol}/{\text{m}}^3\right)\\ =8.49\times {10}^{28}\text{atoms}/{\text{m}}^{\text{3}}\end{array}$

Thus, the number density of iron atoms is $8.49\times {10}^{28}\text{atoms}/{\text{m}}^{\text{3}}$.

Conclusion:

Therefore, the number density of iron atoms is $8.49\times {10}^{28}\text{atoms}/{\text{m}}^{\text{3}}$.

(d)

To determine

The number density of two conduction iron atoms.

Answer to Problem 27.24P

The number density of two conduction iron atoms is $1.7\times {10}^{29}\text{atoms}/{\text{m}}^{\text{3}}$

Explanation of Solution

Given Info: The cross-sectional area of iron wire is $5.00\times {10}^{-6}{\text{m}}^2$ and the electric current is $30.0\text{A}$.

Explanation:

The formula for the number density of two conduction atoms is,

$n=2A{M}_{\text{D}}$

Here,

$A$ is the Avogadro number.

$M_{\text{D}}$ is the molar density.

Substitute $1.41\times {10}^5\text{mol}/{\text{m}}^3$ for $M_{\text{D}}$ and $6.02\times {10}^{23}\text{atoms}/\text{mol}$ for $A$ in above equation to find $n$.

$\begin{array}{c}n=2\left(6.02\times {10}^{23}\text{atoms}/\text{mol}\right)\cdot \left(1.41\times {10}^5\text{mol}/{\text{m}}^3\right)\\ =1.7\times {10}^{29}\text{atoms}/{\text{m}}^{\text{3}}\end{array}$

Thus, the number density of two conduction iron atoms is $1.7\times {10}^{29}\text{atoms}/{\text{m}}^{\text{3}}$

Conclusion:

Therefore, the number density of two conduction iron atoms is $1.7\times {10}^{29}\text{atoms}/{\text{m}}^{\text{3}}$

(e)

To determine

The drift speed of the conduction electrons.

Answer to Problem 27.24P

The drift speed of the conduction electrons is. $2.2\times {10}^{-4}\text{m}/\text{s}$.

Explanation of Solution

Given Info: The cross-sectional area of iron wire is $5.00\times {10}^{-6}{\text{m}}^2$ and the electric current is $30.0\text{A}$.

Explanation:

Formula to calculate the drift speed is,

$V_{\text{d}}=\frac{I}{nqA}$

Substitute $30.0\text{A}$ for $I$, $1.6\times {10}^{-19}\text{C}$ for $q$, $1.7\times {10}^{29}\text{atoms}/{\text{m}}^{\text{3}}$ for $n$ and $5.00\times {10}^{-6}{\text{m}}^2$ for $A$ in above equation to find the $V_{\text{d}}$

$\begin{array}{c}{V}_{\text{d}}=\frac{30.0\text{A}}{\left(1.7\times {10}^{29}\text{atoms}/{\text{m}}^{\text{3}}\right)\times \left(1.6\times {10}^{-19}C\right)\times \left(5.00\times {10}^{-6}{\text{m}}^2\right)}\\ =2.2\times {10}^{-4}\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the total drift speed of the conduction electrons is. $2.2\times {10}^{-4}\text{m}/\text{s}$.