physical-chemistry ii chapter-1-molecules in motion · pdf filethe motion of all kinds of...

Physical-Chemistry II

Chapter-1-Molecules in motion

جزيئات في حركة

Dr. El Hassane ANOUAR

Chemistry Department, College of Sciences and Humanities, Prince Sattam bin

Abdulaziz University, P.O. Box 83, Al-Kharij 11942, Saudi Arabia.

3/18/2017 College of Science and humanities, Al-kharj 1

Physical Chemistry II

(Chem 3320)

1437/1438 (Semester 2)

Important :

These slides are prepared in reference to chapter 20 in Physical Chemistry, Ninth

Edition, Peter Atkins, and Julio De Paula

Introduction


Types of molecular motion الجزيئيةالحركة) ):

The random motion الحركة

(العشوائية ) of molecules of a perfect

gas (غاز مثالي)

Uniform motion of ions ( حركة لاليونات

in solution in the presence of an ( منتظمة

electric field (مجال الكهربائي).

Molecular mobility (االنتقال الجزيئي) in

liquids (السوائل).

Derivation of expressions (تعابيراو صيغ) that govern the migration (انتقال) of

properties through matter.

Introduction

Objectives:

Example: Diffusion

equation.

The spread of matter and

energy (المادة والطاقة) through

media (اوساط)

We discuss:

The motion of all kinds of particles in all kinds of fluids (الموائع).

Transport properties (diffusion, thermal conduction, electric conduction,

viscosity and effusion) of a substance (مادة), its ability to transfer matter ( نقل

) energy ,(المادة .or some other property from one place to another ,(الطاقة


3/18/2017 College of Science and humanities, Al-kharj

4

Part-1- Molecular motion in gases

( الغازاتالحركة الجزيئية في )

1.1. The kinetic model of gases (النموذج الجزيئي للغازات)

1.1.1. Pressure and speeds (الظغط والسرعات الجزيئية)

𝐩𝐕 =𝟏

𝟑𝐧𝐌𝐜𝟐 where

𝐜 = 𝐯𝟐 𝟏/𝟐

M: The molar mass of the molecules M=mNA

c: The root mean square speed of the molecules

where v speeds, of the

molecules:

To be the equation of state of a perfect gas

𝟏

𝟑𝐧𝐌𝐜𝟐 = 𝐧𝐑𝐓 𝐜 =

𝟑𝐑𝐓

𝐌

𝟏/𝟐

𝐜 ∝ (𝐓)𝟏/𝟐, 𝒊. 𝒆 𝐓 ↗ ⟹ 𝐜 ↗

𝐚𝐭 𝐓 = 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭,

𝐜 ∝ (𝟏

𝑴)𝟏/𝟐, 𝒊. 𝒆 𝐌 ↗ ⟹ 𝐜 ↘

Justification in pdf document


Example 1.1: The root mean square speed of N2 molecules (M= 28.02 g mol−1) at

298 K is found from eqn 1.3 to be

𝑐 =3𝑅𝑇

𝑀

1/2

=3 × 8.3145 𝐽𝐾−1𝑚𝑜𝑙−1 × 298𝐾

28.02 × 10−3 𝐾𝑔 𝑚𝑜𝑙−1

12

= 515 𝑚𝑠−1

1. Molecular motion in gases ( الغازاتالحركة الجزيئية في )



Note

In an actual gas the speeds of individual molecules span a wide range, and the

collisions in the gas continually redistribute the speeds among the molecules.


Exercise 1

Find a relation between < 𝑣2 >1/2 and < 𝑣4 >1/4 for molecules in a gas at a

temperature T.

Data:

𝑓(𝑣) = 4𝜋𝑀

2𝜋𝑅𝑇

3/2

𝑣2𝑒−𝑀2𝑅𝑇𝑣

2

𝑥6𝑒−𝑎𝑥2𝑑𝑥

+∞

0

=15𝜋1/2

16

1

𝑎

7/2

Solution





The fraction of molecules that have speeds in the

range 𝑣 to 𝑣 + 𝑑𝑣 is proportional to the width of the

range, and is written 𝒇(𝒗)𝒅𝒗, where 𝑓(𝑣) is called the

distribution of speeds or Maxwell distribution of

speeds.




Decaying exponential

function

𝒇(𝒗) = 𝟒𝝅𝑴

𝟐𝝅𝑹𝑻

𝟑/𝟐

𝒗𝟐𝒆−𝑴𝟐𝑹𝑻𝒗

𝟐


To calculate the fraction of molecules in a given narrow range of speeds, 𝜟𝒗, we

evaluate 𝑓(𝑣) at the speed of interest, then multiply it by the width of the range of

speeds of interest, that is:

Fraction of molecules = 𝒇(𝒗)𝜟𝒗.

To use the distribution to calculate the fraction in a range

of speeds that is too wide to be treated as infinitesimal,

we evaluate the integral:




𝐅𝐫𝐚𝐜𝐭𝐢𝐨𝐧 𝐢𝐧 𝐭𝐡𝐞 𝐫𝐚𝐧𝐠𝐞 𝐯𝟏 𝐭𝐨 𝐯𝟐 = 𝐟(𝐯)𝐝𝐯𝐯𝟐

𝐯𝟏


Exercise 2

Use the Maxwell distribution of speeds 𝑓(𝑣) to estimate the fraction of CO2

molecules at 300 K that have speeds in the range 200 to 250 m s−1. Use the constancy

assumption of 𝑓(𝑣) in the range 200 to 250 m s−1.

Data: The fraction of molecules calculated using MathCad software is

𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 CO2 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑎𝑛𝑔𝑒 200 to 250 m s

−1

= 𝑓(𝑣)𝑑𝑣𝑣2

𝑣1

= 0.0954

Solution








Example 1.1

What is the mean speed, 𝑐 , of N2 molecules in air at 25°C?

Solution (See chapter 1, pdf format)

Homework 1.1

Evaluate the root mean square speed of the molecules by integration. You will need

the integral

𝑥4𝑒−𝑎𝑥2𝑑𝑥

+∞

0

=3

8

𝜋

𝑎5

1/2





The mean speed, 𝒄 , of the molecules in a gas:

𝐜 =𝟖𝐑𝐓

𝛑𝐌

𝟏/𝟐

The most probable speed, c*, obtained by

differentiating f with respect to 𝑣 and

looking for the value of 𝑣 at which the

derivative is zero (other than at 𝑣 = 0 and

𝑣 = ∞):

𝒄∗ =𝟐𝑹𝑻

𝑴

𝟏/𝟐





The relative mean speed, 𝑐 𝑟𝑒𝑙, the

mean speed with which one molecule

approaches another:

𝒄 𝒓𝒆𝒍 = 𝟐𝟏/𝟐𝒄

Relative mean speed of 2 dissimilar

molecules withs mA and mB: 𝒄 𝒓𝒆𝒍

=𝟖𝒌𝑻

𝝅𝝁

𝟏/𝟐 ; 𝝁 =

𝒎𝑨𝒎𝑩

𝒎𝑨+𝒎𝑩

𝑐 𝑟𝑒𝑙 of molecules in a gas is related to the mean speed:

Molecules are moving in

the same direction =>

𝑐 𝑟𝑒𝑙 = 0

Molecules are

approaching each other

=> 𝑐 𝑟𝑒𝑙 = 2v

Molecules are approaching

from side =>𝑐 𝑟𝑒𝑙 = 21/2𝑣


Exercise 3 (Ex. 20.1b, p. 777, chapter 20 Atkins book)

Determine the ratios of (a) the mean speeds, (b) the mean kinetic energies of He

atoms and Hg atoms at 25°C.

Data: M(He) = 4.003 g/mol; M(Hg) = 200.6 g/mol

Solution

(Ex. 20.1b, p. 777, chapter 20 exercises solutions)








Exercise 3 (Solution)



1.1.2. The collision frequency


The collision frequency, z: the number of collisions made by one molecule

divided by the time interval during which the collisions are counted, when are N

in a volume V, z is

𝒛 = 𝝈𝒄 𝒓𝒆𝒍𝓝

𝓝 = 𝑵/𝑽 𝒩is the number density

σ is the collision cross-section of the molecules σ

=πd2

In terms of the pressure

𝒛 =𝝈𝒄 𝒓𝒆𝒍𝒑

𝒌𝑻

where

𝑎𝑡 𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, T ↗ ⇒ 𝒄 𝒓𝒆𝒍 => z ↗

𝑎𝑡 𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, p ↗ => z ↗





The mean free path, λ, is the average distance a molecule travels between

collisions.

λ = 𝐜 𝐫𝐞𝐥𝚫𝐭 =𝐜 𝐫𝐞𝐥𝐳=𝐤𝐓

𝛔𝐩

𝛌 =


Exercise 4

The best laboratory vacuum pump can generate a vacuum of about 1 nTorr. At 25°C

and assuming that air consists of N2 molecules with a collision diameter of 395 pm,

calculate

(a) Calculate the mean speed of the molecules?

(b) Calculate the mean free path?

(c) Calculate the collision frequency in the gas?

Solution





1.2. Collisions with walls and surfaces


The collision flux, ZW: The number of collisions with the area in a given time

interval divided by the area and the duration of the interval. The collision frequency,

the number of hits per second, is obtained by multiplication of the collision flux by the

area of interest.

𝐙𝐰 =𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐜𝐨𝐥𝐥𝐢𝐬𝐢𝐨𝐧𝐬

𝐀 (∆𝐭)=

𝐩

𝟐𝛑𝐦𝐤𝐓 𝟏/𝟐


Exercise 5

A solid surface with dimensions 3.5 cm × 4.0 cm is exposed to helium gas at 111 Pa

and 1500 K. How many collisions do the He atoms make with this surface in 10 s?

Data:

R = 8.3145 JK-1mol-1; Atomic mass of He = 4 g/mol

Solution




1.3. The rate of effusion


Graham’s law of effusion: The rate of effusion of molecules through a small

hole of area A0 is inversely proportional to the square root of the molar mass.

𝐑𝐚𝐭𝐞 𝐨𝐟 𝐞𝐟𝐟𝐮𝐬𝐢𝐨𝐧 = 𝐙𝐰𝑨𝟎

=𝒑𝑨𝟎

𝟐𝝅𝒎𝒌𝑻 𝟏/𝟐 =𝒑𝑨𝟎𝑵𝑨

𝟐𝝅𝑴𝑹𝑻 𝟏/𝟐




The rate effusion equation is the basis of the Knudsen method for the determination of

the vapour pressures of liquids and solids, particularly of substances with very low

vapour pressures.

Example 1.2 Calculating the vapour pressure from a mass loss

Caesium (m.p. 29°C, b.p. 686°C) was introduced into a container and heated to 500°C. When

a hole of diameter 0.50 mm was opened in the container for 100 s, a mass loss of 385 mg was

measured. Calculate the vapour pressure of liquid Caesium at 500 K.

Solution (see word document, chapter 1, p. 18)

Homework 1.2

How long would it take 1.0 g of Cs atoms to effuse out of the oven under the same conditions


Exercise 6

An effusion cell has a circular hole of diameter 3.00 mm. If the molar mass of the

solid in the cell is 300 g mol−1 and its vapour pressure is 0.224 Pa at 450 K, by how

much will the mass of the solid decrease in a period of 24.00 h?

Solution




Exercise 7

A solid compound of molar mass 200 g mol−1 was introduced into a container and

heated to 300°C. When a hole of diameter 0.50 mm was opened in the container for

500 s, a mass loss of 277 mg was measured. Calculate the vapour pressure of the

compound at 300°C.

Solution




Exercise 8

A container of internal volume 22.0 m3 was punctured, and a hole of radius 0.050

mm was formed. If the nitrogen pressure within the vehicle is initially 122 kPa and its

temperature 293 K, how long will the pressure take to fall to 105 kPa?

Solution




Exercise 8 (solution)




1.4. Transport properties of a perfect gas


Transport properties are commonly expressed in terms of a number of

‘phenomenological’ equations, or equations that are empirical summaries of

experimental observations.

1.4.1 The phenomenological equations

The rate of migration of a property is measured by its flux J.

Matter

Matter flux (as in diffusion)

Energy

Energy flux (as in thermal conduction)

𝐉 𝐦𝐚𝐭𝐭𝐞𝐫 ∝ 𝐝𝓝

𝐝𝐙 (Fick’s first law of diffusion) 𝐉 𝐞𝐧𝐞𝐫𝐠𝐲 ∝

𝐝𝐓

𝐝𝐙





Because matter flows down a

concentration gradient (i.e., from high

concentration to low concentration)

=> J is positive if dN/dz is negative

𝐉(𝐦𝐚𝐭𝐭𝐞𝐫) = −𝐃 𝐝𝓝

𝐝𝐙

Diffusion coefficient

Energy migrates down a temperature gradient,

and the same reasoning leads to

𝐉 𝐞𝐧𝐞𝐫𝐠𝐲 = −𝛋 𝐝𝐓

𝐝𝐙

coefficient of thermal conductivity





Connection between the flux of momentum and the viscosity

The flux of the x-component of momentum is

proportional to dvx/dz

𝐉 𝐱 − 𝐜𝐨𝐦𝐩𝐨𝐧𝐞𝐧𝐭 𝐨𝐟 𝐭𝐡𝐞 𝐦𝐨𝐦𝐞𝐧𝐭𝐮𝐦 = −𝛈 𝐝𝒗

𝒙

𝐝𝐳




1.4.2. The transport parameters

Diffusion coefficient of a perfect gas

𝐃 =𝟏

𝟑𝛌𝐜

Mean speed, 𝒄 =𝟖𝑹𝑻

𝝅𝑴

𝟏/𝟐

Mean free path , 𝝀 =𝑘𝑇

𝜎𝒑

∗ 𝒑 ↗ ⇒ 𝜆 ↘ ⇒ 𝑫 ↘ ∗ 𝝈 ↘ ⇒ 𝜆 ↗ ⇒ 𝑫 ↗

𝒄 ↗ ⇒ 𝑇 ↗ ⇒ 𝑫 ↗

Thermal conductivity, κ of a perfect gas A

having molar concentration [A] 𝛋 =𝟏

𝟑𝛌𝐜 𝑪𝑽,𝒎 𝑨

Molar heat capacity at

constant volume

* λ = f(1/p) => λ = f(1/[A]) => κ independent of p

* Cv, m ↗ ⇒ κ ↗




1.4.2. The transport parameters

The viscosity, η of the gas molecules

𝛈 =𝟏

𝟑𝐌𝛌𝐜 𝑨

Molar concentration of the gas Molar mass of the gas

Mean speed, 𝐜 =𝟖𝐑𝐓

𝛑𝐌

𝟏/𝟐

𝐜 ↗ ⇒ T ↗ ⇒ η ↗ * λ ∝ 1/p and [A] ∝ p

=> η independent of p

Mean free path , 𝝀 =𝑘𝑇

𝜎𝒑


Exercise 9

Calculate the flux of energy arising from a temperature gradient of 3.5 K.m−1 in a sample of

hydrogen in which the mean temperature is 260 K.

Data:

M(H2) = 2.016 g.mol-1; σ = 0.27 nm2; CV,m = 20.510 JK-1mol-1






Exercise 10

Use the experimental value of the thermal conductivity of nitrogen (𝜅 = 0.0465 J K-1m-1 s-1 at

273 k)) to estimate the collision cross-section of N2 atoms at 298 K.

Data:

M(N2) = 28.04 g.mol-1; CV,m = 20.876 JK-1mol-1


Exercise 11

Calculate the viscosity of benzene vapour at (a) 273 K, (b) 298 K, (c) 1000 K. Take σ ≈ 0.88

nm2.

Data:

M(Benzene) = 78.11 g.mol-1




Exercise 12

Calculate the thermal conductivity of nitrogen (CV,m = 20.8 J K−1 mol−1,σ = 0.43 nm2) at

room temperature (20°C).

solution



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