Download 3.2 Ideal gas- Boltzman constant

Topic 3.2 Modelling a gas
The Ideal Gas Law and
Kinetic Theory
The Mole, Avogadro's
Number, and Molecular Mass
Atomic Mass Unit, U
By international agreement, the reference element is chosen to be
the most abundant type of carbon, called carbon-12, and its
atomic mass is defined to be exactly twelve atomic mass units, or
12 u.
Molecular Mass
The molecular mass of a molecule is the sum of the atomic
masses of its atoms.
For instance, hydrogen and oxygen have atomic masses of
1.007 94 u and 15.9994 u, respectively.
The molecular mass of a water molecule (H2O) is:
2(1.007 94 u) + 15.9994 u = 18.0153 u.
Avogadro's Number NA
The number of atoms per mole is known as Avogadro's number
NA, after the Italian scientist Amedeo Avogadro (1776–1856):
An example
• The molar mass of helium is 4.0 g.
Determine the mass of a single atom of
helium in kilograms.
An example
• The molar mass of helium is 4.0 g.
Determine the mass of a single atom of
helium in kilograms.
There are 6.02 × 1023 atoms in 4.0 g of helium.
0.004
mass of atom 
6.02  10 23
= 6.645 × 10–27 kg » 6.6 × 10–27 kg
An example 2
3
The molar mass of uranium is 238 g.
a) Calculate the mass of one atom of uranium.
b) A small rock contains 0.12 g of uranium.
For this rock, calculate the number of:
i
moles of uranium
ii
atoms of uranium.
3 a
There are 6.02 × 1023 atoms in 0.238 kg of uranium.
mass of atom 0.238 23
6.02  10
bi
= 3.95 × 10−25 kg  4.0 × 10−25 kg
number of moles 
number of moles 
mass of uranium
molar mass of uranium
0.12
238
 5.04 × 10–4 » 5.0 × 10–4
ii number of atoms  number of moles × NA
number of atoms  5.04 × 10–4 × 6.02 × 1023  3.06 × 1020 » 3.1 × 1020
Number of Moles, n
The number of moles n contained in any sample is the number
of particles N in the sample divided by the number of particles
per mole NA (Avogadro's number):
The number of moles contained in a sample can also be found from
its mass.
The Ideal Gas Law
An ideal gas is an idealized model for real gases that have
sufficiently low densities.
The Ideal Gas Law
An ideal gas is an idealized model for real gases that have
sufficiently low densities.
The condition of low density means that the molecules of the
gas are so far apart that they do not interact (except during
collisions that are effectively elastic).
The Ideal Gas Law
An ideal gas is an idealized model for real gases that have
sufficiently low densities.
The condition of low density means that the molecules of the
gas are so far apart that they do not interact (except during
collisions that are effectively elastic).
The ideal gas law expresses the relationship between the
absolute pressure (P), the Kelvin temperature (T), the volume
(V), and the number of moles (n) of the gas.
PV  nRT
Where R is the universal gas constant. R = 8.31 J/(mol · K).
Ideal gas assumptions
• The particles of gas (atoms or molecules)
obey Newton’s laws of motion.
You should know
these by now!
Ideal gas assumptions
• The particles in a gas move with a range of
speeds
Ideal gas assumptions
• The volume of the individual gas particles is
very small compared to the volume of the
gas
Ideal gas assumptions
• The collisions between the particles and the
walls of the container and between the
particles themselves are elastic (no kinetic
energy lost)
Ideal gas assumptions
• There are no forces between the particles
(except when colliding). This means that the
particles only have kinetic energy (no
potential)
Do you remember what
internal energy is?
Ideal gas assumptions
• The duration of a collision is small
compared to the time between collisions.
An example
At the top of Mount Everest the
temperature is around 250K, with
atmospheric pressure around 3.3 x 104 Pa
At seas level these values are 300K and 1.
x 105 Pa respectively. If the density of air
sea level is 1.2 kg.m-3, what is the density
the air on Mount Everest?
“Physics”, Patrick Fullick, Heinemann
Example question
At the top of Mount Everest the
temperature is around 250K, with
atmospheric pressure around 3.3 x
104 Pa. At seas level these values are
300K and 1.0 x 105 Pa respectively. If
the density of air at sea level is 1.2
kg.m-3, what is the density of the air
on Mount Everest?
An example
At the top of Mount Everest the temperature is around 250K, with atmospheric pressure
around 3.3 x 104 Pa. At seas level these values are 300K and 1.0 x 105 Pa respectively. If
the density of air at sea level is 1.2 kg.m-3, what is the density of the air on Mount
Everest?
Take 1kg of air at sea level
Volume = mass/density = 1/1.2 = 0.83 m3.
Therefore at sea level
p1 = 1.0 x 105 Pa, V1 = 0.83 m3, T1 = 300K.
An example
At the top of Mount Everest the temperature is around 250K, with atmospheric pressure
around 3.3 x 104 Pa. At seas level these values are 300K and 1.0 x 105 Pa respectively. If
the density of air at sea level is 1.2 kg.m-3, what is the density of the air on Mount
Everest?
Therefore at sea level
p1 = 1.0 x 105 Pa, V1 = 0.83 m3, T1 = 300K.
At the top of Mount Everest
p2 = 3.3 x 104 Pa, V2 = ? m3, T1 = 250K.
An example
At the top of Mount Everest the temperature is around 250K, with atmospheric pressure
around 3.3 x 104 Pa. At seas level these values are 300K and 1.0 x 105 Pa respectively. If
the density of air at sea level is 1.2 kg.m-3, what is the density of the air on Mount
Everest?
Therefore at sea level p1 = 1.0 x 105 Pa, V1 = 0.83 m3, T1 = 300K.
At the top of Mount Everest
p2 = 3.3 x 104 Pa, V2 = ? m3, T1 = 250K.
p1V1/T1 = p2V2/T2
(1.0 x 105 Pa x 0.83 m3)/300K = (3.3 x 104 Pa x V2)/250K
V2 = 2.1 m3,
This is the volume of 1kg of air on Everest
Density = mass/volume = 1/2.1 = 0.48 kg.m-3.
Sample question
• A container of hydrogen of volume 0.1m3
and temperature 25°C contains 3.20 x 1023
molecules. What is the pressure in the
container?
K.A.Tsokos “Physics for the IB Diploma” 5th Edition
Sample question
• A container of hydrogen of volume 0.1m3
and temperature 25°C contains 3.20 x 1023
molecules. What is the pressure in the
container?
# moles = 3.20 x 1023/6.02 x 1023 = 0.53
K.A.Tsokos “Physics for the IB Diploma” 5th Edition
Sample question
• A container of hydrogen of volume 0.1m3
and temperature 25°C contains 3.20 x 1023
molecules. What is the pressure in the
container?
# moles = 3.20 x 1023/6.02 x 1023 = 0.53
P = RnT/V = (8.31 x 0.53 x 298)/0.1 = 1.3 x 104 N.m-2
K.A.Tsokos “Physics for the IB Diploma” 5th Edition
• The kinetic theory now links temperature with the microscopic
energies of the gas molecules
• The equation resembles the kinetic energy formula.
• Adjusting for N molecules gives 3/2 NkBT
• This represents the total internal energy of an ideal gas (only
considering translational motion of molecules of monoatomic gases)
This means; No intermolecular forces between molecules between
collisions i.e. energy is completely kinetic
• Gas consists of large number of identical tiny particles- molecules in
constant random motion
• Molecules undergo perfectly elastic collisions between each other and
with the walls of the container; momentum is reversed during collision
• Duration of collision negligible compared with the time between
collisions
• Each molecule produces a force on the wall of the container
• The forces of individual molecules will average out to produce a
uniform pressure throughout the gas- ignoring the effect of gravity
The Ideal Gas Law
The constant term R/NA is referred to as Boltzmann's
constant, in honor of the Austrian physicist Ludwig
Boltzmann (1844–1906), and is represented by the symbol k:
PV = NkT
Kinetic Theory of Gases
Kinetic Theory
of Gases
The pressure that a gas exerts is
caused by the impact of its molecules
on the walls of the container.
Kinetic Theory
of Gases
The pressure that a gas exerts is
caused by the impact of its molecules
on the walls of the container.
It can be shown that the average translational kinetic
energy of a molecule of an ideal gas is given by,
where k is Boltzmann's constant and T is the Kelvin temperature.
Derivation of,
Consider a gas molecule colliding elastically with the right
wall of the container and rebounding from it.
The force on the molecule is obtained using Newton’s second
law as follows,
F 
P
,
t
The force on one of the molecule,
According to Newton's law of action–reaction, the force on
the wall is equal in magnitude to this value, but oppositely
directed.
The force exerted on the wall by one molecule,
mv 2

L
If N is the total number of molecules, since these particles
move randomly in three dimensions, one-third of them on the
average strike the right wall. Therefore, the total force is:
Vrms = root-mean-square velocity.
Pressure is force per unit area, so the pressure P acting on a wall
of area L2 is
Pressure is force per unit area, so the pressure P acting on a wall
of area L2 is
Since the volume of the box is V = L3, the equation above can
be written as,
PV = NkT
An example
The Speed of Molecules in Air
Air is primarily a mixture of nitrogen N2 (molecular mass = 28.0 u)
and oxygen O2 (molecular mass = 32.0 u). Assume that each behaves
as an ideal gas and determine the rms speed of the nitrogen and
oxygen molecules when the temperature of the air is 293 K.
The Internal Energy of a
Monatomic Ideal Gas