Transcript CHAPTER 15 : KINETIC THEORY OF GASSES

CHAPTER 15 : KINETIC
THEORY OF GASSES
15.1 : Ideal Gas Equation
i.
Pressure (P) inversely proportional with
Volume (V) at constant Temperature
: Boyle’s law.
ii. Volume (V) directly proportional with
Temperature (T) at constant Pressure
: Charle’s law.
iii. Pressure (p) directly proportional with
Temperature (T) at constant Volume
: Gay-Lussac’s law.
Iv. Relationships between three equations
can be combine into single equation:PV = nRT (Ideal gas equation)
Unit R, for 1kg gas is Jkg-1K-1
Definition
Definition of ideal gas
• Obey the gas equation in any
pressure and temperature
condition.
• Practically, ideal gas do not exist
• Real gas at low pressure and high
temperature nearly an ideal gas.
Examples
1. An ideal gas is held in a
container at constant volume.
Initially, its temperature is 10.0
oC and its pressure is 2.50 atm.
What is it s pressure when its
temperature is 80.0 oC?
15.2 : Kinetic Theory Of
Gases
Introduction :
A gas contains large molecules that
moves in random directions :• velocity
• momentum
• continuous collisions that exerts
forces per unit area is called
pressure.
The assumptions of the kinetic of
an ideal gas for molecular model
are:
• A container with volume V contains a very
large number N of identical molecules,
each of mass m.
• The size of each molecule is small
compared with the average distance
between them and the dimensions of the
container.
• The molecules are in constant motion.
The molecules undergo perfectly elastic
collisions with each other and also with
the wall of the container.
• The container walls are perfectly rigid and
do not move as a result of the collisions.
Newton’s laws of
motion
The molecules are in
constant Newton’s laws of
motion
• Elastic collisions.
• The container walls are
perfectly :=} rigid and infinitely massive
and do not move.
15.3 : Gas Pressure
Derivation of ideal gas
equation
pV= N/3 Nm<v2>
and
p = 1/3<v2>
Examples
In a period of 1.00 s, 5.00 x
1023 nitrogen molecules strike
a wall with an area of 8.00 cm2.
If the molecules move with the
speed of 300 m/s and strike the
wall head-on in a perfectly
elastic collisions, what is the
pressure exerted on the wall? (
The mass of one N2 is 4.68 x
10-26 kg.)
15.4 : Root Mean Square
(Rms) Speed Of Gas
Molecule
As known pV = 2/3 N [ ½ m<v2>]...............(1)
Derivation of vrms;
By multiplying (4) with M= NAm, we obtain
the average translational kinetic energy
per mole,
NA1/2 m<v2> = ½ M <v2> = 3/2 RT.............(5)
From equation (5), we can obtain the root
mean square speed (vrms)
Root mean square speed
(vrms)
Vrms =
3kT
3RT
 v 

m
M
2
15.5 : Molecular Kinetic
Energy (Average Translational
Kinetic Energy)
• Each molecules of gas has a true speed
of its own due to moving randomly.
• We use V rms to determine kinetic energies
of all those molecules, which is a kind of
an average value
• Hence <K> = ½mo<c2>
where mo = mass of molecule
<c2> = square of the V rms
Relationship Between Thermodynamic
Heat and Kinetic Energy of Molecule
From the ideal gas equation,
pV = nRT
with substitution of n = N/ NA
the ideal gas equation will be
pV = (N/NA) RT ..............................(2)
we know that R/ NA = k
thus, pV = NkT .............................................(3)
As known pV = 2/3 N [ ½ m<v2>]...............(1)
(1)= (3)
2/3 N [ ½ m<v2>] = NkT
we obtain
[ ½ m<v2>]= 3/2 kT...................(4)
where Ktr = ½ m<v2>
therefore Ktr = 3/2 kT
Examples
• What is the average
translational kinetic energy of a
molecule of an ideal gas at a
temperature of 20 oC.
15.6 : Principle Of
Equipartition Of Energy
Definition of degrees of
freedom
The number of velocity components
needed to describe the motion of a
molecule completely.
Law of equipartition of
energy
•
•
Average energy that relate with every rotational
and translational degree of freedom will have the
same average value that is 1/2kT and this energy
depends only to the absolute temperature.
Therefore, K = ( f T+ f R) ½kT
Where as f = degree of freedom
T = absolute temperature
k = Boltzmann constant
Rotational kinetic energy per mole (at absolute
temperature)
E = f x 1/2RT
whereas R = gas constant = kNA
Translational degree of
freedom(Monatomic)
In ideal gas molecule motion,
every molecule contribute
translational kinetic energy that
is Ktr=3/2kT
•Monatomic gas such as
Helium, Neon and Argon.
•It has kinetic energy at 3
degree of freedom in the
direction of x, y and z axes.
Rotational degree of
freedom(Diatomic)
− Involving diatomic gas such as H2, O2, and
Cl2.
− Undergoes additional rotational motion at
the x and y axes.
− It has 5 total degrees of freedom with 3
translational kinetic energy of degree of
freedom and 2 rotational kinetic energy of
degree of freedom.
Vibrational degree of
freedom
− Involving the diatomic and
polyatomic which contribute the
additional 2 degree of freedom.
− The vibrational factor is very
small and can be
neglected/ignored.
Polyatomic
−
−
−
Polyatomic gas such as H2O, CO2, NH3
and N2O4.
Undergoes rotational motion at x, y and z
axes.
It has 6 total degrees of freedom with 3
translational kinetic energy of degree of
freedom and 3 rotational kinetic energy
of degree of freedom.
Examples
15.7 : Internal Energy
Principle of equipartition of energy
Any amount of energy absorb
by the molecule of gas will be
distributed equally between
every translational and
rotational degree of freedom
of the molecule
Internal energy of ideal
gas
Definition :
•
Internal energy of a gas U is
equal to the total amount of
average kinetic energy and
potential energy which
contains in any gasses.
Potential energy
For the ideal gas:
• Potential energy can be
neglected because
A) the force of the exertion
between the molecules do not
exist so
B) U = Total of average kinetic
energy of the molecule = K.
Examples
A tank of volume 0.300 m3,
contains 1.50 mol of Neon gas
at 220C. Assuming the Neon
behaves like an ideal gas, find
the total internal energy of the
gas.
The end