Chapter 1, Problem 1B.7E

(a)

Interpretation Introduction

Interpretation:

Number of collisions per second for argon atom when the pressure is $\text{10}\text{bar}$ has to be calculated using the given data.

Concept introduction:

Mean free path:

The average distance travelled by a molecule between the collisions is called the mean free path.  The mean free path is calculated by the formula,

    $\text{λ=}\frac{\text{kT}}{\sqrt{2}\text{σp}}$

    $\boxed{\begin{array}{l}\text{where,}\\ \text{λ}\text{=}\text{mean}\text{free}\text{path}\\ \text{P}\text{=}\text{Pressure}\\ \text{σ}\text{=}collisioncross-\sec tionofthemolecules\\ \text{T}\text{=}\text{Temperature}\end{array}}$

Collision frequency:

$\begin{array}{c}\text{Collision}\text{frequency,}\text{z}\text{=}\frac{\text{2σp}}{\text{kT}}{\left(\frac{\text{8RT}}{\text{πM}}\right)}^{\text{1/2}}\\ \\ \text{where,}\\ \text{σ}\text{-}\text{collision cross-section of the molecules}\\ \text{p}\text{-}\text{pressure}\\ \text{k-}\text{Boltzmann constant}\text{.}\\ \text{T}\text{-}\text{Temperature}\\ \text{M}\text{-}\text{Molar}\text{mass}\end{array}$

Explanation of Solution

Given data is as follows:

  $\begin{array}{l}\text{σ}\text{=}\text{0}\text{.36}{\text{nm}}^{\text{2}}\\ \\ \text{P}\text{=}10bar\text{ = }1.0\times {10}^6\text{Pa}\\ \\ \text{T}\text{=}\text{298}\text{K}\end{array}$

The expression for the mean free path is,

    $\text{λ=}\frac{\text{kT}}{\sqrt{2}\text{σp}}$

Where,

• $\lambda$ is the mean free path.
• $k$ is the Boltzmann constant.
• $T$ is the temperature.
• $p$ is the pressure.
• $\sigma$ the collision cross-section of the molecules.

The mean free path, $\lambda$ can be calculated as follows:

  $\begin{array}{c}\text{λ=}\frac{\left(\text{1}\text{.38}\text{×}{\text{10}}^{\text{-23}}\text{Pa}{\text{m}}^{\text{3}}{\text{K}}^{\text{-1}}\right)\text{×298K}}{\sqrt{2}\times \text{0}\text{.36}\text{×}{\text{10}}^{\text{-18}}{\text{m}}^{\text{2}}\text{×}1.0\times {10}^6\text{Pa}}\\ \\ \text{=}\underset{\_}{\text{8}\text{.07×}{\text{10}}^{\text{-9}}\text{m}}\end{array}$

Hence, the mean free path of argon molecules is $\underset{\_}{\text{8}\text{.07×}{\text{10}}^{\text{-9}}\text{m}}$.

The number of collisions per second does an argon molecule make at the given pressure can be calculated by the equation given below:

$\begin{array}{c}\text{z}\text{=}\frac{\text{2σp}}{\text{kT}}{\left(\frac{\text{8RT}}{\text{πM}}\right)}^{\text{1/2}}\\ \\ \text{=}\frac{\text{2×0}\text{.36}\text{×}{\text{10}}^{\text{-18}}{\text{m}}^{\text{2}}{\text{×10}}^{6}Pa}{\left(\text{1}\text{.38}\text{×}{\text{10}}^{\text{-23}}\text{Pa}{\text{m}}^{\text{3}}{\text{K}}^{\text{-1}}\right)\text{×298K}}{\left(\frac{\text{8×8}\text{.314}{\text{m}}^{\text{3}}\text{Pa}\text{.}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\text{×298K}}{{\text{π×40×10}}^{\text{-3}}\text{kg}{\text{.mol}}^{\text{-1}}}\right)}^{\text{1/2}}\\ \\ \text{=6}\text{.9}\times {\text{10}}^{10}{\text{s}}^{\text{-1}}\end{array}$

The number of collisions per second does an argon molecule make at the given pressure is $\text{6}\text{.9}\times {\text{10}}^{10}{\text{s}}^{\text{-1}}$.

(b)

Interpretation Introduction

Interpretation:

Number of collisions per second for argon atom when the pressure is $\text{100}\text{kPa}$ has to be calculated using the given data.

Concept introduction:

Mean free path:

The average distance travelled by a molecule between the collisions is called the mean free path.  The mean free path is calculated by the formula,

    $\text{λ=}\frac{\text{kT}}{\sqrt{2}\text{σp}}$

    $\boxed{\begin{array}{l}\text{where,}\\ \text{λ}\text{=}\text{mean}\text{free}\text{path}\\ \text{P}\text{=}\text{Pressure}\\ \text{σ}\text{=}collisioncross-\sec tionofthemolecules\\ \text{T}\text{=}\text{Temperature}\end{array}}$

Collision frequency:

$\begin{array}{c}\text{Collision}\text{frequency,}\text{z}\text{=}\frac{\text{2σp}}{\text{kT}}{\left(\frac{\text{8RT}}{\text{πM}}\right)}^{\text{1/2}}\\ \\ \text{where,}\\ \text{σ}\text{-}\text{collision cross-section of the molecules}\\ \text{p}\text{-}\text{pressure}\\ \text{k-}\text{Boltzmann constant}\text{.}\\ \text{T}\text{-}\text{Temperature}\\ \text{M}\text{-}\text{Molar}\text{mass}\end{array}$

Explanation of Solution

Given data is as follows:

  $\begin{array}{l}\text{σ}\text{=}\text{0}\text{.36}{\text{nm}}^{\text{2}}\\ \\ \text{P}\text{=}\text{100}\text{kPa = }1.0\times {10}^5\text{Pa}\\ \\ \text{T}\text{=}\text{298}\text{K}\end{array}$

The expression for the mean free path is,

    $\text{λ=}\frac{\text{kT}}{\sqrt{2}\text{σp}}$

Where,

• $\lambda$ is the mean free path.
• $k$ is the Boltzmann constant.
• $T$ is the temperature.
• $p$ is the pressure.
• $\sigma$ the collision cross-section of the molecules.

The mean free path, $\lambda$ can be calculated as follows:

  $\begin{array}{c}\text{λ=}\frac{\left(\text{1}\text{.38}\text{×}{\text{10}}^{\text{-23}}\text{Pa}{\text{m}}^{\text{3}}{\text{K}}^{\text{-1}}\right)\text{×298K}}{\sqrt{2}\times \text{0}\text{.36}\text{×}{\text{10}}^{\text{-18}}{\text{m}}^{\text{2}}\text{×}1.0\times {10}^5\text{Pa}}\\ \\ \text{=}\underset{\_}{\text{8}\text{.07×}{\text{10}}^{\text{-8}}\text{m}}\end{array}$

Hence, the mean free path of argon molecules is $\text{8}\text{.07×}{\text{10}}^{\text{-8}}\text{m}$.

The number of collisions per second does an argon molecule make at the given pressure can be calculated by the equation given below:

$\begin{array}{c}\text{z}\text{=}\frac{\text{2σp}}{\text{kT}}{\left(\frac{\text{8RT}}{\text{πM}}\right)}^{\text{1/2}}\\ \\ \text{=}\frac{\text{2×0}\text{.36}\text{×}{\text{10}}^{\text{-18}}{\text{m}}^{\text{2}}{\text{×10}}^{5}Pa}{\left(\text{1}\text{.38}\text{×}{\text{10}}^{\text{-23}}\text{Pa}{\text{m}}^{\text{3}}{\text{K}}^{\text{-1}}\right)\text{×298K}}{\left(\frac{\text{8×8}\text{.314}{\text{m}}^{\text{3}}\text{Pa}\text{.}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\text{×298K}}{{\text{π×40×10}}^{\text{-3}}\text{kg}{\text{.mol}}^{\text{-1}}}\right)}^{\text{1/2}}\\ \\ \text{=7}\text{.1}\times {\text{10}}^9{\text{s}}^{\text{-1}}\end{array}$

The number of collisions per second does an argon molecule make at the given pressure is $\text{7}\text{.1}\times {\text{10}}^9{\text{s}}^{\text{-1}}$.

(c)

Interpretation Introduction

Interpretation:

Number of collisions per second for argon atom when the pressure is $\text{1 Pa}$ has to be calculated using the given data.

Concept introduction:

Mean free path:

The average distance travelled by a molecule between the collisions is called the mean free path.  The mean free path is calculated by the formula,

    $\text{λ=}\frac{\text{kT}}{\sqrt{2}\text{σp}}$

    $\boxed{\begin{array}{l}\text{where,}\\ \text{λ}\text{=}\text{mean}\text{free}\text{path}\\ \text{P}\text{=}\text{Pressure}\\ \text{σ}\text{=}collisioncross-\sec tionofthemolecules\\ \text{T}\text{=}\text{Temperature}\end{array}}$

Collision frequency:

$\begin{array}{c}\text{Collision}\text{frequency,}\text{z}\text{=}\frac{\text{2σp}}{\text{kT}}{\left(\frac{\text{8RT}}{\text{πM}}\right)}^{\text{1/2}}\\ \\ \text{where,}\\ \text{σ}\text{-}\text{collision cross-section of the molecules}\\ \text{p}\text{-}\text{pressure}\\ \text{k-}\text{Boltzmann constant}\text{.}\\ \text{T}\text{-}\text{Temperature}\\ \text{M}\text{-}\text{Molar}\text{mass}\end{array}$

Explanation of Solution

Given data is as follows:

  $\begin{array}{l}\text{σ}\text{=}\text{0}\text{.36}{\text{nm}}^{\text{2}}\\ \\ \text{P}\text{=}\text{1 Pa}\\ \\ \text{T}\text{=}\text{298}\text{K}\end{array}$

The expression for the mean free path is,

    $\text{λ=}\frac{\text{kT}}{\sqrt{2}\text{σp}}$

Where,

• $\lambda$ is the mean free path.
• $k$ is the Boltzmann constant.
• $T$ is the temperature.
• $p$ is the pressure.
• $\sigma$ the collision cross-section of the molecules.

The mean free path, $\lambda$ can be calculated as follows:

  $\begin{array}{c}\text{λ=}\frac{\left(\text{1}\text{.38}\text{×}{\text{10}}^{\text{-23}}\text{Pa}{\text{m}}^{\text{3}}{\text{K}}^{\text{-1}}\right)\text{×298K}}{\sqrt{2}\times \text{0}\text{.36}\text{×}{\text{10}}^{\text{-18}}{\text{m}}^{\text{2}}\text{×}1.0\text{Pa}}\\ \\ \text{=}\underset{\_}{\text{8}\text{.07×}{\text{10}}^{\text{-3}}\text{m}}\end{array}$

Hence, the mean free path of argon molecules is $\text{8}\text{.07×}{\text{10}}^{\text{-3}}\text{m}$.

The number of collisions per second does an argon molecule make at the given pressure can be calculated by the equation given below:

$\begin{array}{c}\text{z}\text{=}\frac{\text{2σp}}{\text{kT}}{\left(\frac{\text{8RT}}{\text{πM}}\right)}^{\text{1/2}}\\ \\ \text{=}\frac{\text{2×0}\text{.36}\text{×}{\text{10}}^{\text{-18}}{\text{m}}^{\text{2}}\text{×1}\text{.0}Pa}{\left(\text{1}\text{.38}\text{×}{\text{10}}^{\text{-23}}\text{Pa}{\text{m}}^{\text{3}}{\text{K}}^{\text{-1}}\right)\text{×298K}}{\left(\frac{\text{8×8}\text{.314}{\text{m}}^{\text{3}}\text{Pa}\text{.}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\text{×298K}}{{\text{π×40×10}}^{\text{-3}}\text{kg}{\text{.mol}}^{\text{-1}}}\right)}^{\text{1/2}}\\ \\ \text{=7}\text{.0}\times {\text{10}}^4{\text{s}}^{\text{-1}}\end{array}$

The number of collisions per second does an argon molecule make at the given pressure is $\text{7}\text{.0}\times {\text{10}}^4{\text{s}}^{\text{-1}}$.