Chapter 13, Problem 76P

(a)

To determine

The mean free path of nitrogen molecule at sea level.

Answer to Problem 76P

The mean free path of nitrogen molecule at sea level is $100\text{nm}$.

Explanation of Solution

The diameter of nitrogen molecule is $0.3\text{nm}$, the temperature is $290\text{K}$ and the pressure is $100\text{ kPa}$.

Write the expression for rms speed of atoms and molecules

$\lambda =\frac{1}{\sqrt{2}\pi d^2\left(N/V\right)}$                                                           (I)

Here, $\lambda$ is the mean free path, $d$ is the diameter of the molecule, $N$ is the number of molecules and $V$ is the volume.

Write the ideal gas equation

$PV=NkT$                                                                        (II)

Here, $P$ is the pressure, $k$ is the Boltzmann’s constant and $T$ is the temperature.

Rearrange

$\frac{N}{V}=\frac{P}{kT}$                                                                            (III)

Substitute (III) in (I)

$\lambda =\frac{kT}{\sqrt{2}\pi d^{2}P}$                                                           (IV)

Substitute $0.3\text{nm}$ for $d$, $100\text{ kPa}$ for $P$, $1.38\times {10}^{-23}\text{}{\text{JK}}^{-1}$ for $k$ and $290\text{}\text{K}$ for $T$ in (IV) to find $\lambda$ of nitrogen molecule.

$\begin{array}{c}\lambda =\frac{1.38\times {10}^{-23}\text{}{\text{JK}}^{-1}\times 290\text{}\text{K}}{\sqrt{2}\times \pi {\left(0.3\text{nm}\right)}^2\left(100\text{ kPa}\right)}\\ =\frac{1.38\times {10}^{-23}\text{}\text{J}\cdot \frac{{\text{kgm}}^2{\text{s}}^{-2}}{\text{J}}\times 290}{\sqrt{2}\times \pi {\left(0.3\times {10}^{-9}\text{m}\right)}^2\left(100\times {10}^3\text{ Pa}\cdot \frac{{\text{kgm}}^{-1}{\text{s}}^{-2}}{\text{Pa}}\right)}\\ =\frac{1.38\times {10}^{-23}\text{}{\text{kgm}}^2{\text{s}}^{-2}\times 290}{\sqrt{2}\times \pi {\left(0.3\times {10}^{-9}\text{m}\right)}^2\left(100\times {10}^3{\text{ kgm}}^{-1}{\text{s}}^{-2}\right)}\\ =100\times {10}^{-9}\text{m}\end{array}$

$\lambda =100\text{nm}$

Thus, the mean free path of nitrogen molecule is $100\text{nm}$.

(b)

To determine

The mean free path of nitrogen molecule at Mount Everest.

Answer to Problem 76P

The mean free path of nitrogen molecule at Mount Everest is $159\text{nm}$.

Explanation of Solution

The diameter of nitrogen molecule is $0.3\text{nm}$, the temperature is $230\text{K}$ and the pressure is $50\text{ kPa}$.

Substitute $0.3\text{nm}$ for $d$, $\text{50 kPa}$ for $P$, $1.38\times {10}^{-23}\text{}{\text{JK}}^{-1}$ for $k$ and $230\text{}\text{K}$ for $T$ in (IV) to find $\lambda$ of nitrogen molecule.

$\begin{array}{c}\lambda =\frac{1.38\times {10}^{-23}\text{}{\text{JK}}^{-1}\times 230\text{}\text{K}}{\sqrt{2}\times \pi {\left(0.3\text{nm}\right)}^2\left(50\text{ kPa}\right)}\\ =\frac{1.38\times {10}^{-23}\text{}\text{J}\cdot \frac{{\text{kgm}}^2{\text{s}}^{-2}}{\text{J}}\times 230}{\sqrt{2}\times \pi {\left(0.3\times {10}^{-9}\text{m}\right)}^2\left(50\times {10}^3\text{ Pa}\cdot \frac{{\text{kgm}}^{-1}{\text{s}}^{-2}}{\text{Pa}}\right)}\\ =\frac{1.38\times {10}^{-23}\text{}{\text{kgm}}^2{\text{s}}^{-2}\times 230}{\sqrt{2}\times \pi {\left(0.3\times {10}^{-9}\text{m}\right)}^2\left(50\times {10}^3{\text{ kgm}}^{-1}{\text{s}}^{-2}\right)}\\ =159\times {10}^{-9}\text{m}\end{array}$

$\lambda =159\text{nm}$

Thus, the mean free path of nitrogen molecule is $159\text{nm}$.

(c)

To determine

The mean free path of nitrogen molecule at an altitude of $30\text{km}$.

Answer to Problem 76P

The mean free path of nitrogen molecule at an altitude of $30\text{km}$ is $8\text{μm}$.

Explanation of Solution

The diameter of nitrogen molecule is $0.3\text{nm}$, the temperature is $230\text{K}$ and the pressure is $1\text{ kPa}$.

Substitute $0.3\text{nm}$ for $d$, $100\text{ kPa}$ for $P$, $1.38\times {10}^{-23}\text{}{\text{JK}}^{-1}$ for $k$ and $290\text{}\text{K}$ for $T$ in (IV) to find $\lambda$ of nitrogen molecule.

$\begin{array}{c}\lambda =\frac{1.38\times {10}^{-23}\text{}{\text{JK}}^{-1}\times 230\text{}\text{K}}{\sqrt{2}\times \pi {\left(0.3\text{nm}\right)}^2\left(1\text{ kPa}\right)}\\ =\frac{1.38\times {10}^{-23}\text{}\text{J}\cdot \frac{{\text{kgm}}^2{\text{s}}^{-2}}{\text{J}}\times 230}{\sqrt{2}\times \pi {\left(0.3\times {10}^{-9}\text{m}\right)}^2\left(1\times {10}^3\text{ Pa}\cdot \frac{{\text{kgm}}^{-1}{\text{s}}^{-2}}{\text{Pa}}\right)}\\ =\frac{1.38\times {10}^{-23}\text{}{\text{kgm}}^2{\text{s}}^{-2}\times 230}{\sqrt{2}\times \pi {\left(0.3\times {10}^{-9}\text{m}\right)}^2\left(1\times {10}^3{\text{ kgm}}^{-1}{\text{s}}^{-2}\right)}\\ =8\times {10}^{-6}\text{m}\end{array}$

$\lambda =8\text{μm}$

Thus, the mean free path of nitrogen molecule is $8\text{μm}$.