Physics 2 week 7 Chapter 3 The Kinetic Theory of … 8...(a macroscopic quantity) depends on the speed of the molecules (a microscopic quantity) b. Translational Kinetic Energy Consider

Chapter 3 The Kinetic Theory of Gases 3.1. Ideal Gases

3.1.1. Experimental Laws and the Equation of State

3.1.2. Molecular Model of an Ideal Gas

3.2. Mean Free Path

3.3. The Boltzmann Distribution Law and

The Distribution of Molecular Speeds

3.4. The Molar Specific Heats of an Ideal Gas

3.5. The Equipartition-of-Energy Theorem

3.6. The Adiabatic Expansion of an Ideal Gas

Physics 2 – week 7

• Number of moles in a sample:

AN

Nn N is the number of molecules

A

samplesample

mN

M

M

Mn

M is the molar mass (the mass of 1 mol)

m is the mass of one molecule

nRTpV p is the absolute pressure (Pa)

T is the temperature (in K)

11Kmol J 13.8R

• Ideal gas law:

NkTpV

1-23

A

K J1038.1N

Rk The Boltzmann constant (k)

Review

or Gas constant (R)

f

i

V

VpdVW

:)(isochoricconstant V 1) If 0W

:(isobaric) constantp 2) If Vp)Vp(VW if

Work done by the gas:

3) If :l)(isotherma constantT

i

f

V

Vln nRTW

Review

13. A sample of an ideal gas is taken through the cyclic process abca shown in the figure below; at point a, T=200 K. (a) How many moles of gas are in the sample? (b) the temperature of the gas at point b and point c, (c) the net energy added to the gas as heat during the cycle?

RT

pVnnRTpV

(a) Applying the equation of state:

At point a, p=2.5 kN/m2 or 2500 N/m2; V=1 m3.

(mol) 5.120031.8

12500

n

(b) 5.12 nRT

Vp

T

VpnRTpV

b

bb

a

aa

(c) Applying the first law of thermodynamics:

WQE

For a closed cycle, E=0: WQ

W: work done by the system.

))((2

1abcb VVppW

(J) 10520.50002

1 3W

(K) 0605.12

32500

nR

VpT cc

c

Point b, p=7.5 kN/m2; V=3 m3

(K) 01805.12

37500

nR

VpT bb

b

Point c, p=2.5 kN/m2; V=3 m3

3.1.2. Molecular Model for an Ideal Gas

In this model:

1. The molecules obey Newton's laws of motion.

2. The molecules move in all direction with equal probability.

3. There is no interactions between molecules (no collisions between molecules).

4. The molecules undergo elastic collisions with the walls.

a. Pressure, Temperature, and RMS Speed

First, we consider a cubical box of edge length L, containing n moles of an ideal gas. A molecule of mass m and velocity v collide with the shaded wall.

Key question: What is the connection between the pressure p exerted by the gas and the speed of the molecules?

Problem: Let n moles of an ideal gas be confined in a cubical box of volume V. The walls of the box are held at temperature T.

For an elastic collision, the

particle’s momentum (=m.v) along the x axis is reserved and change with an amount:

xxxx 2mv)(mv)mv(Δp

The average rate at which momentum is delivered to the shaded wall by this molecule:

L

mv

2L/v

2mv

Δt

Δp 2

x

x

xx

dt

pd

dt

)vd(m

dt

vdmamF

Recall: L

mvF

2

xx,1

The pressure exerted on the wall by this single molecule:

2

x,1

1L

Fp

For N molecules, the total pressure p:

2

2

Nx,

2

x,2

2

x,1

2

x

L

/Lmv.../Lmv/Lmv

L

Fp

Travel time b/w 2 walls with a speed v

Note: Pressure is the force applied perpendicular to the surface of an object

)v...vv(L

mp 2

Nx,

2

x,2

2

x,13

The average value of the square of the x components of all the molecular speeds:

N

v...vv 2

Nx,

2

x,2

2

x,12

xv

2

x3

A vL

nmNp

gas theof massmolar the:mNM ASince

2

xvV

nMp :LV 3

For any molecule: 2

z

2

y

2

x

2 vvvv

As all molecules move in random directions: 22

x v3

1v

2v3V

nMp

The square root of is called the root-mean-square speed: 2v

rms

2 vv

3V

nMvp

2

rms

Combining with the equation of state: nRTpV

M

3RTvrms

This relationship shows us how the pressure of the gas (a macroscopic quantity) depends on the speed of the molecules (a microscopic quantity)

b. Translational Kinetic Energy

Consider a single molecule of an ideal gas moving around in the box (see Section a) .

2rms

22 mv2

1v

2

1

2

1K mmv

M/m

3RT

2

1

M

3RT

2

1K

m

A2N

3RTK

The Boltzmann constant k: AN

Rk

kT2

3K

22z

2y

2x v

3

1vvv kT

2

1vm

2

1vm

2

1vm

2

1 2z

2y

2x

Kdoes not depend on the mass of the molecule

24. At 273 K and 1.0 x 10-2 atm, the density of a gas is 1.24 x 10-5 g/cm3. (a) Find vrms for the gas molecules. (b) Find the molar mass of the gas and (c) identify the gas (hint: see Table 19-1).

)1(3

M

RTvrms

Root-mean-square speed:

)2(n

VM

V

nM

V

M gas

(1) and (2):

p

V

nRTvrms

33

3235 kg/m1024.1g/cm1024.1

Pa1001.1atm100.1 32 p

m/s494rmsv

(a)

)2(n

VM

Equation of state:

)3(nRTpV

p

RT

n

VM

g/mol28kg/mol028.0 M

From Table 19.1, the gas is nitrogen (N2)

(b)

(c)

Homework: 18, 20, 23, 25, 27 (p. 531-532)

Chapter 3 The Kinetic Theory of Gases 3.1. Ideal Gases

3.1.1. Experimental Laws and the Equation of State

3.1.2. Molecular Model of an Ideal Gas

3.2. Mean Free Path

3.3. The Boltzmann Distribution Law and

The Distribution of Molecular Speeds

3.4. The Molar Specific Heats of an Ideal Gas

3.5. The Equipartition-of-Energy Theorem

3.6. The Adiabatic Expansion of an Ideal Gas

Physics 2 – week 8

3.2. Mean Free Path

3.2.1 Concept

•A molecule traveling through a gas changes both speed and direction as it elastically collides with other molecules in its path.

•Between collisions, the molecules moves in a straight line at constant speed.

•The mean free path is the average

distance traversed by a molecule between

collisions.

V

N

1

density

1λ

V

N is the number of molecules per unit volume or the density of molecules

molecules ofnumber theis N

gas theof volume theis V where

To count the number of collisions: We further consider that this single molecule has an equivalent radius of d and all the other molecules are points (see cartoons next slides for an equivalent problem).

Our goal: Estimate of of a single molecule.

Assumptions: + Our molecule is traveling with a constant speed v and all the other molecules are at rest. + All molecules are spheres of diameter d a collision occurs as the centers of 2 molecules come within a distance d.

= 1 collision

d

= 1 collision . d

Equivalent problem

1st collision

2nd collision

3rd collision

Equivalent problem

d

2d

The number of collisions = the number molecules lie in a cylinder of length vt and cross-sectional area d2

If all the molecules are moving:

V

Nd 22

1

Using the equation of state: pV = NkT

pd

kT22

The average time between collisions (the mean free time):

vpd

kT

vt

22

The average time between collisions (the mean free time):

The frequency of collisions:

tf

1

3.3. The Boltzmann Distribution Law and the Distribution of Molecular Speeds

The Boltzmann distribution law: if the energy is associated with some state or condition of a system is then the frequency with which that state or condition occurs, or the probability of its occurrence is proportional to:

kTe /

constantBoltzmann the:k

Many of the most familiar laws of physical chemistry are special cases of the Boltzmann distribution law:

P(v)dv is the fraction of molecules with speeds in the infinitesimal range (v,v+dv).

1)(0

dvvP

The fraction of molecules with speeds from v1 to v2:

2

1)(frac

v

vdvvP

3.3.1. The distribution of molecular speeds (or the Maxwell speed distribution law):

Let M be the molar mass of the gas, v be the molecular speed, and P(v) be the speed distribution function:

)1(2

4)( 2/22

2/3

kTMvevRT

MvP

Average, RMS, and Most Probable Speeds

)2()(0

dvvvPvThe average speed:

from (1) & (2): M

RTv

8

0

22 )( dvvPvv

M

RTv

32

The root-mean-square speed:

M

RTvvrms

32

The most probable speed is the speed at which P(v) is maximum:

0)(

dv

vdP

M

RTvP

2

3.3.2. The barometric distribution law:

This law gives the number density (h), i.e. number of molecules per unit volume, of an ideal gas of uniform temperature T as a function of height h in the field of the Earth’s gravity.

kThhmgehh

/)0(

0 )()(

where h0 is an arbitrary fixed reference height; m is the mass of a molecule.

Abell 1982 nasa.gov

Homework: 28, 32, 33, 40 (Page. 531-532)

Chapter 3 The Kinetic Theory of Gases 3.1. Ideal Gases

3.1.1. Experimental Laws and the Equation of State

3.1.2. Molecular Model of an Ideal Gas

3.2. Mean Free Path

3.3. The Boltzmann Distribution Law and

The Distribution of Molecular Speeds

3.4. The Molar Specific Heats of an Ideal Gas

3.5. The Equipartition-of-Energy Theorem

3.6. The Adiabatic Expansion of an Ideal Gas

Physics 2 – week 9

3.4. The Molar Specific Heats of an Ideal Gas

Let’s consider our ideal gas of n moles that is a monatomic gas, which has individual atoms, e.g. helium, argon, neon. For a single atom, the average translational KE:

kTK2

3

The internal energy Eint of the gas (no rotational KE for monatomic gases):

nRTnkTKEN

2

3N

2

3A

1int

Recall molar specific heat: TCnQ

a. Molar specific heat at constant volume:

• Consider n moles of an ideal gas at state i: p, T, and fixed V state f: p+p, T+T

TnCQ V

CV is a constant and called the molar specific heat at constant volume.

Þ DEint =3

2nRDT

TnRWTnCWQE V 2

3int

11

V K mol J5.122

3C 0 Since RW

Note: For diatomic and polyatomic gases, their CV is greater than that of monatomic gases.

TnCE V int

So, the change in internal energy can be calculated by:

TnRE 2

3int

or

b. Molar specific heat at constant pressure:

TnCQ p

WQE int

Cp is the molar specific heat at constant pressure.

TnRVpW

TnRTnCTnR p 2

3

RRRCp2

5

2

3

RCC Vp

Checkpoint 4 (p. 522): The figure here shows 5 paths traversed by a gas on a p-V diagram. Rank the paths according to the change in internal energy of the gas, greatest first.

TnRE 2

3int

123 TTT

Example: (Problem 8, page 530) Suppose 1.8 mol of an ideal gas is taken from a volume of 3.0 m3 to a volume of 1.5 m3 via an isothermal compression at 300C. (a) How much energy is transferred as heat during the compression, and (b) is the transfer to or from the gas?

(a) We have:

An isothermal process: T=constant

Work done by the gas for isotherm:

(b) Q<0: heat transferred from the gas

3.5. The Equipartition-of-Energy Theorem

Every kind of molecule has a certain number f of degrees of freedom. For each degree of freedom in which a molecule can store energy, the average internal energy is per molecule. kT

2

1

Molecule Example

Degrees of freedom

Translational Rotational Total (f)

Monatomic He

Diatomic O2

Polyatomic CH4

3 0 3

3 2 5

3 3 6

RCC

Rf

C

Vp

V

2

Six degrees of freedom Technical Aspects of robotics

wac.nsw.edu.au

5 degrees of freedom of a diatomic molecule

3.6. The Adiabatic Expansion of an Ideal Gas

What is an adiabatic process?: a process for which Q = 0

constantpV

Vp/CC where

nRTpV

constant1 TV

Proof of the equations above, see p. 526-527 (homework)

Equation of state:

Free expansions:

0 :Recall WQ

fi TTE 0int

ffii VpVp

Homework: 42, 44, 46, 54, 56, 78 (p. 533-535)