Transcript Kinetic Theory of an Ideal Gas

Unit 4: Module 3 – Kinetic Theory
The kinetic model of a gas
There are two ways of describing a gas: macroscopic and
microscopic.
Macroscopic:
Considers a gas on a large scale. It has mass and volume. A
change in state/phase occurs once a liquid has reached its
boiling point.
Microscopic:
Considers a large number of moving molecules that collide
with one another and the walls of the container holding
them.
Unit 4: Module 3 – Kinetic Theory
Kinetic Theory of an Ideal Gas
What is it?
It explains the movement of gas molecules by
relating macroscopic and microscopic
descriptions.
Real gases do NOT meet the
conditions assumed for an ideal gas.
An ideal gas has internal energy
ONLY in the form of random kinetic
energy
Unit 4: Module 3 – Kinetic Theory
Assumptions of an Ideal Gas
•All gas molecules are perfectly elastic. They bounce
off each other and the walls of the container without
losing any KE. Molecules never come to a stop or
settle at the bottom of a container.
• All gases consist of identical molecules in
continuous random motion (Brownian Motion).
Unit 4: Module 3 – Kinetic Theory
Assumptions of an Ideal Gas
(Continued)
•Molecules exert forces on each other only when they
collide. Therefore they must be a relatively long way
apart.
•The gravitational force on the molecules is negligible
• Molecules are so tiny that they take up no space at all.
Volume of the molecule is negligible compared to the
volume of the container.
Unit 4: Module 3 – Kinetic Theory
Kinetic Theory
A gas exerts pressure on the walls of a container
due to its collisions.
Consider ONE gas molecule of mass, m travelling at a
velocity, v towards the wall of a container. As it
collides with the container wall it elastically collides.
d
This means its change in momentum is 2mv.
When the molecule next collides with the wall it would have travelled a
distance of 2d. As speed = distance/time.
Time between collisions = distance / speed
Using the above info: Time between collisions = 2d / v
Unit 4: Module 3 – Kinetic Theory
Kinetic Theory Cont.
As we know Force is proportional to the rate of change of momentum
Therefore F ∝ change in momentum/ time
F ∝ 2mv/(2d/v)
F ∝ mv2/d
If the mass of the molecule decreases then the force exerted on the
walls of the container decrease.
This is hard to consider for just ONE
molecule but remember a real gas would
have millions of molecules each exerting
a force - PRESSURE
Unit 4: Module 3 – Kinetic Theory
Kinetic Theory Cont.(again)
If the force due to one molecule is: F ∝ mv2/d
Consider n molecules: F ∝ nmv2/d
The pressure on a face of the cube container (of area d2) is
2
or
3
d
Pressure = nmv /d
(as pressure = force/area)
Pressure = nmv2/V
(where V is volume)
As these particles can move in 3 dimensions, the total number of
molecules in the box is 3n = N
P = Nmv2/3V = Mv2/3V =
1/3 ρv2
Where M = total mass of gas and ρ its density
Unit 4: Module 3 – Kinetic Theory
Molecular Speed
The molecules in a gas do not all travel at the same
speed. The average value of molecular speed, c is not
the mean but a mean square speed. We write this as
2
c
Therefore we write: P = 1/3 ρv2
as
2
P = 1/3 ρc
The square root of the average (mean) of the squares
of the speeds is known as the r.m.s. speed.
Unit 4: Module 3 – Kinetic Theory
Worked Example
Example Qu:
Four molecules have speeds of 300, 400, 500 and 600ms-1.
Calculate a) the mean (average) speed and b) the mean square speed, c2
c) the r.m.s. speed.
a)
Mean speed = 300+400+500+600
= 450ms-1.
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b) Mean square speed = 3002+ 4002+ 5002+ 6002 = 215000 m2s-2
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c)
r.m.s. speed (square root of mean speed) =  (215000) = 460 ms-1
Unit 4: Module 3 – Kinetic Theory
An Ideal Gas – The Theoretical Equation
The total mass of gas in the box = N x m
as density = mass/ volume
density = N x m
V
If we replace
ρ
2
in P = 1/3 ρv with Nm/V
We can rewrite the equation as:
pV = 1 Nm <c2>
3
Unit 4: Module 3 – Kinetic Theory
An Ideal Gas – The Theoretical Equation
Using Kinetic Theory it is possible to derive the
equations that describe and explain things like
pressure and temperature in terms of the movement
of individual molecules.
2
Where :
pV = 1 Nm <c >
3
p =
V =
N =
m =
<c2>=
pressure
volume
No of atoms/molecules
mass of one atom/molecule
mean square speed of the atoms
Unit 4: Module 3 – Kinetic Theory
Kinetic Energy
We now know: pV = 1 Nm <c2> and pV = nRT
3
If we equate both (for 1 mole of gas):
1 NAm <c2> = RT
(rearrange
3
We get: 1 m<c2> = 3RT
2
2NA
and multiply both sides by ½)
Where K =R
NA
K = Boltzmann constant (JK-1)
As 1 m<c2> = Mean Kinetic Energy of a gas molecule
2
Ek = 3 KT
2
Where Ek = kinetic energy
K = Boltzmann Constant
T = Kelvin temperature