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CAMBRIDGE A LEVEL
PHYSICS
IDEAL GASES
L E A R N I N G O U T C O M E S
NUMBER LEARNING OUTCOME
i L e a r n a n e w S I b a s e q u a n t i t y , t h e a m o u n t o f
s u b s t a n c e .
ii Understand the difference between macroscopic and microscopic
properties of a substance. Know what is the meaning of state variables.
iii L e a r n t h e i d e a l g a s e q u a t i o n a n d u s e i t . U n d e r s t a n d
w h a t i d e a l g a s e s a r e .
iv Understand the importance of Brownian motion to the kinetic model.
v W h a t i s i n t e r n a l e n e r g y ?
vi L e a r n t h e a s s u m p t i o n s o f t h e k i n e t i c t h e o r y f o r
i d e a l g a s e s a n d l e a r n h o w g a s e s e x e r t p r e s s u r e .
vii D e r i v e e q u a t i o n s t h a t r e l a t e a m a c r o s c o p i c p r o p e r t y ,
t e m p e r a t u r e a n d a m i c r o s c o p i c o n e , a v e r a g e s p e e d .
THE MOLETHE MOLE
The amount of substance is one of
the 6 S.I. base quantities.
Units of amount of substance is the
mole (symbol = mol).
THE MOLETHE MOLE
1 mol of a substance is defined as
the amount of that substance that
has a equal number of particles to
the number of atoms in 0.012 kg of
carbon -12.
THE MOLETHE MOLE Avogadros constant is equal to
Avogadros constant is equal tothe number of atoms in 0.012 kgof carbon 12.
0.012 kg of carbon 12 has 6.022 1023 atoms.
Hence, Avogadros constant, is equal to .
THE MOLETHE MOLE Hence, 1 mole of a substance is the
Hence, 1 mole of a substance is theamount of substance that has anumber of particles equal to .
The number of moles, , can becalculated from the mass of substance,m by using the equation
, where
M = molar mass of the substance.
THE MOLETHE MOLE
The molar mass, M of a
substance is the mass (in g) of 1
mol of a substance.
THE MOLETHE MOLE We can use Avogadros constant We can use Avogadros constant
and the number of moles to findthe number of particles presentin a sample of a substance.
How? Use ,wherethe number of elementary
particles.
M I C R O S C O P I C v s .
M A C R O S C O P I C
M I C R O S C O P I C v s .
M A C R O S C O P I C
Substances (solid, liquids or gases) are
is made up of
Substances (solid, liquids or gases) are made up of the elementary units of the substance.
For example, gaseous is made up of
molecules.
Microscopic properties are properties ofthe elementary particles that make upthe substance.
M I C R O S C O P I C v s .
M A C R O S C O P I C
M I C R O S C O P I C v s .
M A C R O S C O P I C
For example, a sample of gaseous For example, a sample of gaseous would have molecules, and each of itsmolecules would have momentum, velocity(or speed), mass and kinetic energy.
It is difficult to measure the microscopicproperties of all the elementary particles ina substance due to the large number ofparticles that make up the substance.
M I C R O S C O P I C v s .
M A C R O S C O P I C
M I C R O S C O P I C v s .
M A C R O S C O P I C
Macroscopic properties are properties Macroscopic properties are properties of the substance of the whole.
For example, a sample of gaseous
would have a temperature, pressure, volume, mass, density and number of moles.
Macroscopic properties define the state the of the substance.
STATE VARIABLESSTATE VARIABLES The state variables are the variables The state variables are the variables
that define the state of a matter.
The state variables we will encounter arepressure, temperature, volume, densityand amount of substance.
State variables are related to oneanother via an equation of state.
IDEAL GASESIDEAL GASES
Ideal gases are gases that precisely
obey the ideal gas equation at all
temperatures, volumes and
pressures.
Real gases obey this law only at low pressures
and high temperatures, when they are furthest
apart and moving the fastest. However, we can
use this equation for approximate calculations.
T H E I D E A L G A S E Q UAT I O N
If this hypothetical experiment was
carried out using a gas, what would
be observed are:
I. The volume of the gas, V, is
directly proportional to the
amount (number of moles) of
the gas, n (if temperature, T and
pressure, p are kept constant).
II. The pressure of the gas, p is
inversely proportional to the
volume, V of the gas (for
constant T and n).
Figure 18.1, page 591: Chapter 18: THERMAL PROPERTIES OF MATTER; SEARS AND ZEMANSKYS
UNIVERSITY PHYSICS WITH MODERN PHYSICS; Young, Hugh D. and Freedman, Roger A., Addison Wesley,
San Francisco, 2012.
T H E I D E A L G A S E Q UAT I O N
III. The pressure, p of the gas is
directly proportional to the
thermodynamic temperature
of the gas, T, for constant V
and n.
IV. The thermodynamic
temperature of a gas is the
temperature of the gas
expressed in Kelvin.
For conversion, use
.
Figure 18.1, page 591: Chapter 18: THERMAL PROPERTIES OF MATTER; SEARS AND ZEMANSKYS
UNIVERSITY PHYSICS WITH MODERN PHYSICS; Young, Hugh D. and Freedman, Roger A., Addison Wesley,
San Francisco, 2012.
T H E I D E A L G A S E Q UAT I O N
.
All of those previous relationships can be
summed up into one equation
This equation is known as the ideal gas
equation
, the proportionality constant, is the
universal gas constant
.
What are the units of p, V, n and T?
T H E I D E A L G A S E Q UAT I O N
The ideal gas equation can be
The ideal gas equation can bemanipulated to obtain otherforms of it.
I. pV #$
, or
II. &'
(), where * density, or,
III.
where the subscripts 1 and
2 represent different states.
T H E I D E A L G A S E Q UAT I O N
Recall that
.Hence,
o +, #$
, or
o -,
o where -
. . //
k is known as the Boltzmann constant.
E X A M P L E S
Example 18.1, page 593: Chapter 18: THERMAL PROPERTIES OF MATTER; SEARS AND
ZEMANSKYS UNIVERSITY PHYSICS WITH MODERN PHYSICS; Young, Hugh D. and
Freedman, Roger A., Addison Wesley, San Francisco, 2012.
E X A M P L E S
Example; Section 6.3 The Gas Laws, Page 145, Chapter 6: Thermal Physics ,
International A/AS Level Physics, by Mee, Crundle, Arnold and Brown, Hodder
Education, United Kingdom, 2008.
E X A M P L E S
Exercises; Section 6.3 The Gas Laws, Page 146, Chapter 6: Thermal Physics ,
International A/AS Level Physics, by Mee, Crundle, Arnold and Brown, Hodder
Education, United Kingdom, 2008.
E X A M P L E S
Exercises; Section 6.3 The Gas Laws, Page 146, Chapter 6: Thermal Physics ,
International A/AS Level Physics, by Mee, Crundle, Arnold and Brown, Hodder
Education, United Kingdom, 2008.
INTERNAL ENERGY
Each of the gas molecule has a certain
amount of energy.
This energy is the kinetic energy,
associated with its movement, and the
potential energy due to the forces that
exist between the gas molecules.
INTERNAL ENERGY
In addition all the molecules will have
different kinetic energies as some are
moving faster and some slower and also
different amount of potential energies
as this energy is dependent on the
position of the molecule in a given
container.
INTERNAL ENERGY
We can now say that the kinetic energies
of each of the molecules are randomly
distributed, and the potential energy of
each of the molecule also follows a
random distribution.
INTERNAL ENERGY
When we add the kinetic energies and
potential energies of all the gas
molecules, we remove the random
nature of the energies.
What we get is known as the internal
energy of the gas.
INTERNAL ENERGY
Definition: The internal energy of a
substance is the sum of the random
distribution of kinetic and potential
energies of all the molecules associated
with the system.
BROWNIAN MOTION
Robert Brown, an English botanist, put forward his Robert Brown, an English botanist, put forward hisobservation of tiny pollen grains floating on waterundergoing a constant , random , haphazard motion,even though the water appeared still when viewed underthe microscope.
This movement is now known as Brownian motion.
This motion is only possible if the water molecules are in astate of rapid and random motion. These watermolecules randomly collide with the pollen grains fromall directions causing the pollen grains to experience thisBrownian motion.
BROWNIAN MOTION BROWNIAN MOTION
Figure 12a, Chapter 17 : Atoms, Molecules and Atomic Processes, page 6; PHYSICS
2000 ; E.R. HUGGINS; Moose Mountain Digital Press, New Hampshire 2000.
Diagram shows
how a simple set
up can be used to
show Brownian
motion.Laser beam can be
replaced by another
coherent source of
light.
BROWNIAN MOTION
Figure 6.14: Observing Brownian Motion, Page 148, Chapter 6: Thermal Physics ,
International A/AS Level Physics, by Mee, Crundle, Arnold and Brown, Hodder
Education, United Kingdom, 2008.
K I N E T I C T H E O R Y O F I D E A L
G A S E S
K I N E T I C T H E O R Y O F I D E A L
G A S E S
We also know that samples of gases have
macroscopic and microscopic properties.
We can relate the macroscopic and
microscopic properties of gases.
However, we must make some
assumptions about the atoms/
molecules of the gases first.
K I N E T I C T H E O R Y O F I D E A L
G A S E S
K I N E T I C T H E O R Y O F I D E A L
G A S E S
These assumptions are known as the kinetic
theory of ideal gases.
We will look at these assumptions first, and
then derive a very important relationship
between a microscopic property, 1 2 3
(average kinetic energy of one molecule)
with a macroscopic (state) property , T
(temperature).
K I N E T I C T H E O R Y O F I D E A L
G A S E S
K I N E T I C T H E O R Y O F I D E A L
G A S E S
The assumptions are:
Section 6.4 A microscopic model of a gas, Page 149, Chapter 6: Thermal Physics ,
International A/AS Level Physics, by Mee, Crundle, Arnold and Brown, Hodder
Education, United Kingdom, 2008.
K I N E T I C T H EO RY O F
I D EA L G A S E S
K I N E T I C T H EO RY O F
I D EA L G A S E S
Ideal gas particles / molecules also exert
pressure on the inner walls of the
container.
Question: How and why?
We can use the kinetic theory of ideal
gases to explain this.
K I N E T I C T H EO RY O F
I D EA L G A S E S
K I N E T I C T H EO RY O F
I D EA L G A S E S USING SIMPLE KINETIC MODEL TO EXPLAIN PRESSURE EXERTED
BY GASES
Gas particles / molecules are in a state of continuous, random motion.
As a result, the gas particles / molecules are constantly involved in
collisions with the inner walls of the container
The collisions are assumed to be elastic.
The momentum of the colliding particles / molecules change and the
particles / molecules have lower momentum after collision.
This change of momentum over a short time produces a force acting on
the particular area of the surface of the container.
This force acting per unit of area is the pressure that is exerted by the gas
particles on the inner surface of the container.
K I N E T I C T H EO RY O F
I D EA L G A S E S
K I N E T I C T H EO RY O F
I D EA L G A S E S We will try to derive an equation for We will try to derive an equation forthe pressure exerted by gas moleculeson the walls of a cubic container oflength = L.
We assume that all molecules have thesame speed in the x - direction, bothbefore and after collision.
The mass of each gas molecule = m
K I N E T I C T H EO RY O F
I D EA L G A S E S
K I N E T I C T H EO RY O F
I D EA L G A S E S
Figure 19.14, page 614: Chapter 19: THE IDEAL GAS; Physics for Scientists and Engineers ,
Volume 1; 3rd edition Ohanian, Hans C. and Markert, John T., W.W Norton and Company Inc,
New York, 2007.
45
K I N E T I C T H EO RY O F
I D EA L G A S E S
K I N E T I C T H EO RY O F
I D EA L G A S E S
The change in momentum of the one of
the gas molecules = 2745 The time taken for one gas molecule to
move from one wall to the other and
back , t 9
:;
Hence, the force exerted by one gas
molecule on the wall, < =:;
>
9
K I N E T I C T H EO RY O F
I D EA L G A S E S
K I N E T I C T H EO RY O F
I D EA L G A S E S
We assume there are N gas molecules,
hence ?@@A BC
D
The pressure exerted by the gas
molecules on one face of the wall,
?@@A
D
BC
D
BC
K I N E T I C T H EO RY O F
I D EA L G A S E S
K I N E T I C T H EO RY O F
I D EA L G A S E S
We assumed all molecules have the
same speed, but some molecules move
slower, others faster. Hence use the
average square of the speeds, 45
Our equation now becomes + E= :;
>
F
K I N E T I C T H EO RY O F
I D EA L G A S E S
K I N E T I C T H EO RY O F
I D EA L G A S E S
The molecules are also equally likely to move
in the x, y and z directions.
We obtain B
We can rearrange the above equation to yield
B
K I N E T I C T H EO RY O F
I D EA L G A S E S
K I N E T I C T H EO RY O F
I D EA L G A S E S
The equation above is one equation that
relates the macroscopic properties of +, ,
with microscopic properties of 7,1 4 3.
We will now move on to relate the
macroscopic quantity of with 2.
This will be done by using the equations
- and B
.
K I N E T I C T H EO RY O F
I D EA L G A S E S
K I N E T I C T H EO RY O F
I D EA L G A S E S
Recall 2-
B where 2-
the average kinetic energy of a gas
molecule.
Hence, +,
HI
J
7 4
HI KL .
K I N E T I C T H EO RY O F
I D EA L G A S E S
K I N E T I C T H EO RY O F
I D EA L G A S E S
We already know that +, IM$
Hence,
2- -, or 2-
-
We now have an equation to help us
calculate the average kinetic energy
of a gas molecule by just knowing its
temperature.
K I N E T I C T H EO RY O F
I D EA L G A S E S
K I N E T I C T H EO RY O F
I D EA L G A S E S
+,
Some analysis of the equation(s):
+, actually gives us the work done on or by
the gas. This work done will change the
kinetic energy of an ideal gas, but not the
potential energy of an ideal gas.
In other words, when we change the internal
energy of an ideal gas, we are changing only
the average kinetic energy of each molecule
K I N E T I C T H EO RY O F
I D EA L G A S E S
K I N E T I C T H EO RY O F
I D EA L G A S E S
p, I K
Some analysis of the equation(s) (contd):
We can use p,
HI KL to calculate the
change in the average kinetic energy of an
ideal gas molecule.
K I N E T I C T H EO RY O F
I D EA L G A S E S
K I N E T I C T H EO RY O F
I D EA L G A S E S
K J7 4
HM$
Some analysis of the equation(s) (contd):
From KL J
7 4
H
M$, we can get
4 HLO
=, and 4
HLO
=, where
B root mean squared (r.m.s.) of
the speed.
E X A M P L E S
Example; Section 6.4 A Microscopic Model of a Gas, Page 152, Chapter 6: Thermal
Physics , International A/AS Level Physics, by Mee, Crundle, Arnold and Brown,
Hodder Education, United Kingdom, 2008.
E X A M P L E S
Exercises; Section 6.4 A Microscopic Model of a Gas, Page 152, Chapter 6: Thermal
Physics , International A/AS Level Physics, by Mee, Crundle, Arnold and Brown,
Hodder Education, United Kingdom, 2008.
E X A M P L E S
May/June 2008 Paper 4, Question 2.
E X A M P L E S
May/June 2008 Paper 4, Question 2 (contd).
E X A M P L E S
May/June 2008 Paper 4, Question 2 (contd).
E X A M P L E S
May/June 2008 Paper 4, Question 2 (contd).
E X A M P L E S
May/June 2009 Paper 4, Question 2.
E X A M P L E S
May/June 2009 Paper 4, Question 2 (contd).
H O M E W O R K
1. Winter 2008, Paper 4, question 5.1. Winter 2008, Paper 4, question 5.
2. Winter 2009, Paper 41, question 2.
3. Summer 2010, Paper 41, question 2.
4. Summer 2011, Paper 41, question 2.
5. Summer 2011, Paper 42, question 2.
6. Winter 2011, Paper 41, question 2.
7. Winter 2011, Paper 43, question 1.
8. Summer 2012, Paper 41, question 2.
H O M E W O R K
9. Winter 2012, Paper 41, question 2.
10.Winter 2012, Paper 43, question 1.