Chapter 19, Problem 19A.2P

(a)

Interpretation Introduction

Interpretation:

The collision frequency made by ${\text{O}}_{\text{2}}$ molecules at $300K$ and pressure at $\text{100 kPa}$ has to be calculated. Also the number of collisions for each second with single surface atom has to be determined.

Concept Introduction:

The surface which is exposed to gas gets constantly attacked that result in formation of freshly prepared surface very quickly. This quick formation can be estimated using kinetic theory of gas which derive expression for collision flux, $Z_w$ as follows,

    $Z_w\text{ = }\frac{p}{\sqrt{\left(2\pi mkT\right)}}$

Here,    p is Pressure

    m is Molar mass

    k is Boltzmann’s constant

    T is Temperature.

Answer to Problem 19A.2P

The collision frequency is $2.69\times {10}^{23}{\text{ cm}}^{\text{-2}}{s}^{-1}$ and the number of collisions is $2\times {10}^8{\text{ s}}^{\text{-1}}$.

Explanation of Solution

Given:

  $\begin{array}{l}T\text{ = 300K}\\ \text{p = 100 kPa}\end{array}$

First using the collision frequency formula the $Z_w$ value at $100\text{ kPa}$ is calculated as shown below,

  $\begin{array}{c}{Z}_w\text{ = }\frac{p}{\sqrt{\left(2\pi mkT\right)}}\\ =\frac{\text{p/Pa}}{{\left[2\pi \left(32\right)\times \left(1.6605\times {10}^{-27}kg\right)\times \left(1.381\times {10}^{-23}{\text{ JK}}^{\text{-1}}\right)\left(300K\right)\right]}^{\text{1/2}}}\\ =\text{ 2}\text{.69}\times {\text{10}}^{\text{22}}{\text{ m}}^{\text{-2}}{\text{ s}}^{\text{-1}}\times \text{p/Pa}\\ =\text{ 2}\text{.69}\times {\text{10}}^{\text{18}}{\text{ cm}}^{\text{-2}}{\text{ s}}^{\text{-1}}\times \text{p/Pa}\\ {\text{Z}}_{\text{w }}{\text{at 10}}^{\text{2}}\text{ kPa = 2}\text{.69}\times {\text{10}}^{\text{18}}{\text{ cm}}^{\text{-2}}{\text{ s}}^{\text{-1}}\times {10}^2\times {10}^5\\ =\text{ 2}\text{.69}\times {\text{10}}^{\text{23}}{\text{ cm}}^{\text{-2}}{\text{ s}}^{\text{-1}}\end{array}$

The given information says that the distance that nearest neighbor found in $Ti$ is $291\text{ pm}$ hence, the atoms number found in $1{\text{ cm}}^{\text{2}}$ is $\text{1}{\text{.4×10}}^{\text{15}}{\text{ cm}}^{\text{-2}}$.

Therefore, the number of collisions is determined by dividing the collision frequency by $\text{1}{\text{.4×10}}^{\text{15}}{\text{ cm}}^{\text{-2}}$.

  $\begin{array}{c}\text{No}\text{. of collisions = }\frac{\text{2}\text{.69}\times {\text{10}}^{\text{23}}{\text{ cm}}^{\text{-2}}{\text{ s}}^{\text{-1}}}{\left(1.4\times {10}^{15}{\text{ cm}}^{\text{-2}}\right)}\\ =\text{ 2}\times {\text{10}}^{\text{8}}{\text{ s}}^{\text{-1}}\end{array}$

(a)

Interpretation Introduction

Interpretation:

The collision frequency made by ${\text{O}}_{\text{2}}$ molecules at $300K$ and pressure at $\text{1Pa}$ has to be calculated. Also the number of collisions for each second with single surface atom has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 19A.2P

The collision frequency is $2.69\times {10}^{18}{\text{ cm}}^{\text{-2}}{s}^{-1}$ and the number of collisions is $2\times {10}^3{\text{ s}}^{\text{-1}}$.

Explanation of Solution

Given:

  $\begin{array}{l}T\text{ = 300K}\\ \text{p = 1Pa}\end{array}$

First using the collision frequency formula the $Z_w$ value at $1\text{Pa}$ is calculated as shown below,

  $\begin{array}{c}{Z}_w\text{ = }\frac{p}{\sqrt{\left(2\pi mkT\right)}}\\ =\frac{\text{p/Pa}}{{\left[2\pi \left(32\right)\times \left(1.6605\times {10}^{-27}kg\right)\times \left(1.381\times {10}^{-23}{\text{ JK}}^{\text{-1}}\right)\left(300K\right)\right]}^{\text{1/2}}}\\ =\text{ 2}\text{.69}\times {\text{10}}^{\text{22}}{\text{ m}}^{\text{-2}}{\text{ s}}^{\text{-1}}\times \text{p/Pa}\\ =\text{ 2}\text{.69}\times {\text{10}}^{\text{18}}{\text{ cm}}^{\text{-2}}{\text{ s}}^{\text{-1}}\times \text{p/Pa}\\ {\text{Z}}_{\text{w }}\text{at 1 Pa = 2}\text{.69}\times {\text{10}}^{\text{18}}{\text{ cm}}^{\text{-2}}{\text{ s}}^{\text{-1}}\times {10}^2\times 1\\ =\text{ 2}\text{.69}\times {\text{10}}^{\text{18}}{\text{ cm}}^{\text{-2}}{\text{ s}}^{\text{-1}}\end{array}$

The given information says that the distance that nearest neighbor found in $Ti$ is $291\text{ pm}$ hence, the atoms number found in $1{\text{ cm}}^{\text{2}}$ is $\text{1}{\text{.4×10}}^{\text{15}}{\text{ cm}}^{\text{-2}}$.

Therefore, the number of collisions is determined by dividing the collision frequency by $\text{1}{\text{.4×10}}^{\text{15}}{\text{ cm}}^{\text{-2}}$.

  $\begin{array}{c}\text{No}\text{. of collisions = }\frac{\text{2}\text{.69}\times {\text{10}}^{\text{18}}{\text{ cm}}^{\text{-2}}{\text{ s}}^{\text{-1}}}{\left(1.4\times {10}^{15}{\text{ cm}}^{\text{-2}}\right)}\\ =\text{ 2}\times {\text{10}}^{\text{3}}{\text{ s}}^{\text{-1}}\end{array}$