Download Kinetic Theory of Gases – A2 level notes – LOJ

Kinetic Theory of Gases – A2 level notes – LOJ - 2010
Data Sheet Extract
The theory for ideal gases makes the following assumptions:
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The gas consists of very small atoms or molecules (spoken of as particles), each of which has
an identical mass and are perfectly spherical in shape, and elastic in nature.
The number of molecules in the gas is so large that the mathematics of statistics can be
applied to it.
The molecules are in constant, random motion. The rapidly moving particles constantly
collide with each other and with the walls of the container with no preferred direction.
This means that we have to ignore the force due to gravity and any electrostatic attraction
between them so that the interactions between molecules are negligible. They exert no
forces on one another except during collisions.
The collisions of gas particles with the walls of the container holding them are perfectly
elastic (kinetic energy is conserved).
The total volume of the individual gas molecules added up is negligible compared to the
volume of the container. Another way of saying this is that the average distance separating
the gas particles is relatively large compared to their size.
The average kinetic energy of the gas particles depends only on the temperature of the
system.
Relativistic effects are negligible (relativity comes into play when speed approach the speed
of light - ignore this). In other words Newtonian mechanics apply!
The time during collision of molecule with the container's wall is negligible as comparable
to the time between successive collisions.
An ideal gas does not exist - the nearest we have to it is helium remote from its liquefying point.
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Kinetic Theory of Gases – A2 level notes – LOJ - 2010
The kinetic theory of gases (derivation of the equation relating pressure to
mean square speed and density)
Let’s try to explain experimentally some observed properties of gases by considering the motion of
the particles (molecules or atoms) which they are made up of.
To do that we need to make a number of assumptions:
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The molecules of a particular gas are identical.
Collisions between the molecules and with the container are (perfectly) elastic. That means
there is no loss of kinetic energy on collision.
The molecules only have kinetic energy (no potential energy) so they exert no forces on
each other except during impacts (which are assumed to have negligible duration anyway)
and the effect of gravity is ignored so that:
o between collisions the molecules move in straight lines at constant speed, and
o the motion is random.
There are a sufficiently large number of molecules for the mathematics of statistics to be
meaningfully applied.
The size of the molecules is negligible compared to their separation. They are point particles
rather than real atoms and molecules
The laws of Newtonian mechanics apply.
Some of these assumptions run through the entire analysis; others are used more specifically.
We need to think about the
velocity of the particles. To
make the mathematical
treatment easier to
understand we need to break
all of those random velocities
into components that are
mutually at right angles to
each other. That will make it
easier later on to add them all
up or average them!
Consider a fixed mass of gas enclosed in a cubical container
of side ‘L’.
Let each particle of the gas have mass ‘m’ and let there be
‘N’ particles
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Kinetic Theory of Gases – A2 level notes – LOJ - 2010
Let’s start with just looking at one of the millions of particles in the gas.
Look at the diagram. The particle it represents is one of millions, but we will forget those for the
moment!
The diagram shows the u component of a single particle moving towards the wall shaded in grey. We
will call this particle 1
The x-component of velocity is u1 towards the wall
The x-component of momentum is mu1 towards the wall.
When the particle hits the wall it will bounce back with the same speed but in the opposite direction
(because kinetic energy is conserved – the collision being elastic!). The x-component of momentum
will then be -mu1
The change in momentum -
p - will therefore be mu1 - (-mu1 ) = 2mu1
The particle has to travel a distance 2L (from the grey wall to the opposite face and back again)
before it next collides with the grey wall. Now, velocity = distance/time so
so the rate of change of momentum due to a collision with a wall will be:
By Newton's second law, the rate of change of momentum is equal to force, so mu12/L is the force
exerted on the particle by the wall during the collision. By Newton's third law, the particle exerts an
equal but oppositely directed force on the wall, and so we can say that:
The force on the wall during the collision = mu12/L
Now it isn’t only one particle that hits the wall – there are N particles so their x-components of
velocity are going to be u1, u2,…. uN
The total force, F, will therefore be = mu12/L + mu22/L + mu32/L….. + muN2/L
F = m/ L ( u12+ u22+ u32... + uN2)
Now, pressure on that wall will be force/contact area = F/A, but A = L 2
Therefore pressure on the wall – the pressure of the gas p = m/ L 3 ( u12+ u22+ u32... + uN2)
We can work out the mean square velocity is the x direction from this:
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Kinetic Theory of Gases – A2 level notes – LOJ - 2010
So now we know that the total pressure due to the u-components of the velocities of the particles
will be:
So far we have only looked at one of the component velocities. It is now time to look at the other
two and to add them into the equation.
By Pythagoras
c2=u2+v2+w2
So, it follows that this is true for the mean square velocities also.
Since there are large numbers of molecules and they are moving randomly the mean square
components are equal to each other (random movement means there is no preferred direction). We
can therefore say that
and
We can now replace our mean square velocity component with one related to the mean square
velocity itself.
and as L3 = V we get
Now, density is mass/volume, Nm is the total mass of gas and L3 is the volume; we can therefore say
that:
To investigate the relationship between molecular kinetic energy and temperature
From the kinetic theory of gases proof above we have:
Let us multiply both sides by V to get an expression for any volume V of gas:
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Kinetic Theory of Gases – A2 level notes – LOJ - 2010
Now
So,
Where M is the total mass of the gas of volume V.
Now, M = Nm ( where m is the mass of one particle in the gas and N is the number of particles in the
gas)
So
A bit of mathematical magic can make that last part of the equation represent kinetic energy. To do
that we need a ½ factor before the m… to make it fair we have to put a factor of 2 in front of the N:
Now the Equation of State or Ideal Gas Equation (which you should know by heart!) is
pV = nRT
where
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R = the universal molar gas constant,
T = the temperature in kelvin, and
n = number of moles of gas
We can therefore put:
Rearranging the equation we get that:
Now, as N/n is the number of molecules per mole, i.e. NA, the Avogadro constant, so we can get rid
of the factors that require us to know the details of the gas and replace them with a constant.
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Kinetic Theory of Gases – A2 level notes – LOJ - 2010
Both R and NA are universal constants, and therefore so also is R/NA; it is called Boltzmann's
constant, ‘k’ and it is the gas constant per particle. The left-hand term is the average translational
kinetic energy of a single molecule, and therefore
Notes
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‘T’ must be in kelvin – remember to convert!
The Ek is in joules and is for one average particle – the question often asks you for the energy
of a number of moles of gas.
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