Chapter 13, Problem 38P

(a)

To determine

The number density of the gas.

Answer to Problem 38P

The number density of the gas is $2.69\times {10}^{25}{\text{m}}^{\text{-3}}$.

Explanation of Solution

Write the expression to find the number density of the gas.

${\rho }_N=\frac{N}{V}$

Here, $N$ is the number of molecules, $V$ is the volume, ${\rho }_N$ is the number density.

Substitute $n{N}_A$ for $N$ in the above expression.

${\rho }_N=\frac{n{N}_A}{V}$

Here, $N_A$ is Avogadro number, $n$ is the number of moles.

Conclusion:

Substitute $1.00\text{mol}$ for $n$, $0.0224{\text{m}}^{\text{3}}$ for $V$ and $6.023\times {10}^{26}{\left(\text{kmol}\right)}^{-1}$ for $N_A$ to find the number density.

$\begin{array}{c}{\rho }_N=\frac{\left(1.00\text{mol}\right)6.023\times {10}^{23}{\left(\text{mol}\right)}^{-1}}{0.0224{\text{m}}^{\text{3}}}\\ =2.69\times {10}^{25}{\text{m}}^{\text{-3}}\end{array}$

(b)

To determine

The average distance between the molecules.

Answer to Problem 38P

The average distance between the molecules is $4\times {10}^{-9}\text{m}$.

Explanation of Solution

Consider every molecule situates at the center of sphere.

Write the expression to find the number radius of the sphere.

$\begin{array}{c}\frac{V}{N}=\frac{1}{\frac{N}{V}}\\ =\frac{1}{{\rho }_N}\end{array}$

Substitute $2.69\times {10}^{25}{\text{m}}^{\text{-3}}$ for ${\rho }_N$ to find the radius of the sphere.

$\begin{array}{c}\frac{V}{N}=\frac{1}{2.69\times {10}^{25}{\text{m}}^{\text{-3}}}\\ =3.72\times {10}^{-26}{\text{m}}^{\text{3}}\end{array}$

Equate the expression with a volume of a sphere to find the radius of the sphere.

$\begin{array}{c}\frac{V}{N}=\frac{4}{3}\pi r^3\\ \approx 4r^3\end{array}$

Re-arrange the above expression to find the radius of the sphere.

$r={\left(\frac{V}{4N}\right)}^{\frac{1}{3}}$ (I)

The distance between the molecules can be defined as the diameter of the sphere.

Write the expression to find the average distance between the molecules using equation (I).

$\begin{array}{c}d=2r\\ =2\left({\left(\frac{V}{4N}\right)}^{\frac{1}{3}}\right)\end{array}$

Conclusion:

Substitute $3.72\times {10}^{-26}{\text{m}}^{\text{3}}$ for $\frac{V}{N}$ to find the average distance between the molecules.

$\begin{array}{c}d=2{\left(\frac{3.72\times {10}^{-26}{\text{m}}^{\text{3}}}{4}\right)}^{\frac{1}{3}}\\ \approx 4\times {10}^{-9}\text{m}\end{array}$

(c)

To determine

The total mass and mass density of the gas if the gas is ${\text{N}}_2$.

Answer to Problem 38P

The total mass and mass density of the gas if the gas is ${\text{N}}_2$ are $28.0\text{g}$ and $1.25{\text{kg/m}}^{\text{3}}$.

Explanation of Solution

Write the expression to find the mass of ${\text{N}}_2$.

$M=nm$

Here, $m$ is the molar mass of ${\text{N}}_2$.

Write the expression to find the mass density of ${\text{N}}_2$.

$\rho =\frac{M}{V}$

Conclusion:

Substitute $1.00\text{mol}$ for $n$ and $2\left(14.00674\text{g/mol}\right)$ for $m$ to find $M$.

$\begin{array}{c}M=\left(1.00\text{mol}\right)\left(2\left(14.00674\text{g/mol}\right)\right)\\ =28.0\text{g}\end{array}$

Substitute $28.0\text{g}$ for $M$ and $0.0224{\text{m}}^{\text{3}}$ for $V$ to find the mass density of ${\text{N}}_2$.

$\begin{array}{c}\rho =\frac{28.0\text{g}\left(\frac{1.0\text{kg}}{{10}^3\text{g}}\right)}{0.0224{\text{m}}^{\text{3}}}\\ =1.25{\text{kg/m}}^{\text{3}}\end{array}$