Student Solutions Manual to accompany Atkins' Physical Chemistry 11th  edition
Student Solutions Manual to accompany Atkins' Physical Chemistry 11th edition
11th Edition
ISBN: 9780198807773
Author: ATKINS
Publisher: Oxford University Press
Question
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Chapter 19, Problem 19A.2P

(a)

Interpretation Introduction

Interpretation:

The collision frequency made by O2 molecules at 300K and pressure at 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, Zw as follows,

    Zw = p(2πmkT)

Here,    p is Pressure

    m is Molar mass

    k is Boltzmann’s constant

    T is Temperature.

(a)

Expert Solution
Check Mark

Answer to Problem 19A.2P

The collision frequency is 2.69×1023 cm-2s1 and the number of collisions is 2×108 s-1.

Explanation of Solution

Given:

  T = 300Kp = 100 kPa

First using the collision frequency formula the Zw value at 100 kPa is calculated as shown below,

  Zw = p(2πmkT)= p/Pa[2π(32)×(1.6605×1027kg)×(1.381×1023 JK-1)(300K)]1/2= 2.69×1022 m-2 s-1×p/Pa= 2.69×1018 cm-2 s-1×p/PaZat 102 kPa = 2.69×1018 cm-2 s-1×102×105= 2.69×1023 cm-2 s-1

The given information says that the distance that nearest neighbor found in Ti is 291 pm hence, the atoms number found in 1 cm2 is 1.4×1015 cm-2.

Therefore, the number of collisions is determined by dividing the collision frequency by 1.4×1015 cm-2.

  No. of collisions = 2.69×1023 cm-2 s-1(1.4×1015 cm-2)= 2×108 s-1

(a)

Interpretation Introduction

Interpretation:

The collision frequency made by O2 molecules at 300K and pressure at 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).

(a)

Expert Solution
Check Mark

Answer to Problem 19A.2P

The collision frequency is 2.69×1018 cm-2s1 and the number of collisions is 2×103 s-1.

Explanation of Solution

Given:

  T = 300Kp = 1Pa

First using the collision frequency formula the Zw value at 1Pa is calculated as shown below,

  Zw = p(2πmkT)= p/Pa[2π(32)×(1.6605×1027kg)×(1.381×1023 JK-1)(300K)]1/2= 2.69×1022 m-2 s-1×p/Pa= 2.69×1018 cm-2 s-1×p/PaZat 1 Pa = 2.69×1018 cm-2 s-1×102×1= 2.69×1018 cm-2 s-1

The given information says that the distance that nearest neighbor found in Ti is 291 pm hence, the atoms number found in 1 cm2 is 1.4×1015 cm-2.

Therefore, the number of collisions is determined by dividing the collision frequency by 1.4×1015 cm-2.

  No. of collisions = 2.69×1018  cm-2 s-1(1.4×1015 cm-2)= 2×103 s-1

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