Question

Calculate the number density (number per unit volume) for (a) molecules of oxygen gas at $\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{°}\mathit{C}$and 1.0 atm pressure and (b) conduction electrons in copper. (c) What is the ratio of the latter to the former? What is the average distance between (d) the oxygen molecules and (e) the conduction electrons, assuming this distance is the edge length of a cube with a volume equal to the available volume per particle (molecule or electron)?

Short answer

a) The number density for molecules of oxygen gas is $2.7\times {10}^{25}\hspace{0.33em}{m}^{-3}$.

b) The number density for conduction electrons in copper is $8.43\times {10}^{28}\hspace{0.33em}{m}^{-3}$.

c) The ratio of the latter to the former is $3.1\times {10}^3$.

d) The average distance between the oxygen molecules is $3.3\hspace{0.33em}\mathrm{nm}$.

e) The average distance between the conduction electrons is 0.23 nm.

Step-by-step solution

The given data

The given temperature and pressure of oxygen gas are $T=0{\hspace{0.33em}}^{0}Cor273K$and localid="1661510570166" $P=1\hspace{0.33em}atm$respectively.

Understanding the concept of electrical properties of metals

Using the ideal gas equation, we can get the number density (or the number of molecules per unit volume) value for the required element oxygen. Now, combining the number of conduction electrons of copper, we can get the number density of copper using the same number density equation. Further using these number densities we can get their ratio and their molecular or electronic average distance in each case.

Formula:

The number density (number of molecules per unit volume) using ideal gas equation,

$n_0=\frac{P}{kt},where,n_0=N/V$ (i)

a) Calculation of the number density for molecules of oxygen gas

Using the given data in equation (i), we can get the number density for molecules of oxygen gas as follows:

$$\begin{gathered} n_0=\frac{\left(1atm\right)\left(1\times {10}^{5}Pa/atm\right)}{\left(1.38\times {10}^{-23}J/k\right)\left(273K\right)} \\ =2.7\times {10}^{25}{\mathrm{m}}^{-3} \end{gathered}$$

Hence, the value of the number density is $2.7\times {10}^{25}{m}^{-3}$.

b) Calculation of the number density for conduction electrons in copper

Using the combination of the equations, 41-2, 41-3 and 41-4, leads to the conduction electrons per cubic meter in copper using equation (i) as follows:

$$\begin{gathered} n_0=\frac{\left(8.96\times {10}^{3}kg/m^2\right)}{\left(63.54\right)\left(1.67\times {10}^{-27}kg\right)} \\ =8.43\times {10}^{28}{\mathrm{m}}^{-3} \end{gathered}$$

Hence, the value of the number density is $8.43\times {10}^{28}{\mathrm{m}}^{-3}$.

c) Calculation of the ratio of the number densities

Using the values of the number densities of copper and oxygen, we can get the ratio value as follows:

$$\begin{gathered} \frac{n}{n_0}=\frac{\left(8.43\times {10}^{28}{m}^{-3}\right)}{\left(2.7\times {10}^{25}{m}^{-3}\right)} \\ =3.1\times {10}^3 \end{gathered}$$

Hence, the value of the ratio is $3.1\times {10}^3$.

d) Calculation of the average distance between the oxygen molecules

It is given that $d_{avg}=n^{-1/3}...............\left(a\right)$

Thus, for the case of oxygen molecules, we can get the average distance between the oxygen molecules as follows:

$$\begin{gathered} d_{avg}={\left(2.7\times {10}^{25}{m}^{-3}\right)}^{-1/3} \\ =3.3nm \end{gathered}$$

Hence, the value of the average distance is 3.3 nm .

e) Calculation of the average distance between the conduction electrons of copper

Thus, for the case of conduction electrons in copper, we can get the average distance between the electrons using equation (a) as follows:

$$\begin{gathered} d_{avg}={\left(8.43\times {10}^{28}{m}^{-3}\right)}^{-1/3} \\ =0.23nm \end{gathered}$$

Hence, the value of average distance is 0.23 nm.