Thermal Physics Tutorial 3 - Molecular Kinetic Theory

Molecular Kinetic Theory Model

The model, which we will study in this tutorial, tells us how gases behave.  The topic fits quite well with mechanics because you need to understand about collisions to make sense of the Kinetic Theory model.  It may be worth revising these ideas in Units 2 and 4.

In mechanics we have been looked at big things, like tennis balls, cars, people, to name but a few.  This can be referred to as the macroscopic level.  In this piece of work we will go down to the molecular level, where things are very small.  We call this the microscopic level.

 

A Scottish botanist, Robert Brown, first suggested that gases were in a state of random motion. He observed that pollen particles seem to jiggle about for no apparent reason. We can see the same by observing smoke particles under a microscope.  The random jerky movement of the particles is called Brownian motion and is due to the continuous and random bombardment by air molecules. The work of Einstein, Maxwell, and Boltzmann gave us the kinetic theory, as we know it today.

 
Pressure Exerted by an Ideal Gas

Pressure is exerted by the impact of molecules on the sides of a container. We can use the idea to derive an expression for gas pressure. We must make the following assumptions:

The gas consists of a large number of identical molecules, such that the statistical treatments are meaningful.

Health Warning:  The derivation of the kinetic theory equation is rather tedious but it's on the syllabus. 

Here we go:

Stage 1 Momentum Change

Consider a large box, side l, full of molecules:  

Now consider one molecule of mass m moving towards face 1 and the x- component of its velocity is cx.

            Ž its momentum in the x direction is mcx

At face 1 there is a perfectly elastic collision so that the momentum is reversed to -mcx.

This results in a momentum change of:

mcx - (-mcx) = 2mcx

After colliding with face 1, the molecule travels a distance 2l before it collides again with face 1 again.  This takes a time t, which is given by:

t = 2l/cx

 

Stage 2 Impulse

So that we can get the pressure, we need to know the force.  This can be worked out by Newton II, considering the impulse, which is the change of momentum.

impulse = FDt

 

Rearranging gives us:

F = change in momentum

Dt

 Time interval is the time taken for the molecule to move up the box to the far end, bounce off and come back again. Time  = distance / speed = 2l / cx

Ž F = 2mcx  = m(cx)2

       2l/cx         l

 

Stage 3 Working out the pressure

Since pressure = force/area, we can work out the pressure on face 1.

The area of face 1 = l2.

p = F = mcx2/l = mcx2

     A       l2          l3

Suppose we have N molecules of gas. The pressure will be:

p = m ( cx12 + cx22 + cx32....+ cxN2) = m(Ncx2)

l3                                                     l3

 

Now l3 is the volume of the gas, so we can rewrite the equation:

Stage 4 The Statistics

The term <cx 2> ("cx-squared bar") is called the mean square speed of the molecules in the x direction.  The bar is written over the c2, as in the yellow box.  (Notice the different way of writing it in the text.  You may see this in some books.  It is also almost impossible to write c-bar in this web-page editor!)  However the molecules are moving randomly in the container and very few would be moving exactly parallel to the x-axis.  However we can consider each molecule’s velocity to be the resultant of three components, cx, cy, and cz.

As in two dimensions, the three components can be combined by Pythagoras to give the resultant velocity.

Similarly we can combine the mean square velocities:

Since there are a large number of molecules we can assume that there are equal numbers moving in each of the co-ordinate directions:

So we can write:

So our final equation becomes:

Stage 5 Density

However we can go further.  Since Nm is the total mass of the gas and V is the volume, we can express the pressure in terms of the density.

       

density =   mass      Nm

               volume       V

This gives us the equation:

Be careful of the difference between the square of the mean (<c>)2 ["c-bar squared"] and the mean square <c2> ["c-squared bar"].  Consider 1 + 2 + 3 = 6.  The average is 2 and the square of the mean is 4.  (12 + 22 + 32 ) ÷ 3 = (1 + 4 + 9) ÷ 3 = 14 ÷ 3 = 4.67, so there is quite a difference.

The most probable speed is that at which the greatest number of molecules are moving. The mean speed is the average value of all the speeds. The root mean square speed is the square root of the mean square speed of the molecules. They are slightly but significantly different.

Question 1

The observed speed of ten particles at a particular instant are shown in the table:

Number of Particles

1

2

4

1

1

1

Speed (m/s)

5.0

7.0

9.0

12.0

14.0

15.0

            (a)  What is the most probable speed?                                                

            (b) What is the mean speed? 

            (c) What is the rms speed of the molecules?

Answer
Question 2

Nitrogen gas is kept in a closed container at a temperature of 27 oC and a pressure of 1.0 × 105 Pa.  The density of nitrogen is 1.25 kg m-3.  Calculate:

(a) the rms speed of the molecules;                                                 

(b) the temperature at which the molecules travel twice as fast.   

Answer
 
 
Internal Energy and Kinetic Energy of Molecules and Temperature

The internal energy of a material is the sum of its potential and kinetic energies.  In an ideal gas the molecules are so far apart that intermolecular forces can be ignored.  Therefore there is no potential energy.  So all the energy is kinetic.  The total kinetic energy is shared randomly throughout the molecules in the gas.  Therefore there is a random distribution of the speeds of the molecules, which we can show on a graph.

If no energy is being transferred as heat between an object and its surroundings, we say that it is in thermal equilibrium.  If a gas is in thermal equilibrium and it is not being compressed or expanded, the average kinetic energy will remain constant, so will the temperature.

We have seen that:

 pV = nRT

  and:

    pV = 1/3 Nmc2

It does not take a genius to see that:

      nRT = 1/3 Nmc2

We also know that kinetic energy,

Ek = 1/2 mv2 (or 1/2 mc2).

We can now rewrite our expression as (since 2/3 ×1/2 = 1/3):

nRT = 2/3 N[1/2mc2]

We can rearrange this to write:

    1/2mc2 = 3/2nRT = 3/2RT

                    N         N/n

Since N is the total number of molecules and n is the number of moles, N/n will always be NA, which is Avogadro's number, 6.02 ×1023.

 Ž 1/2mc2 = 3/2RT

                      NA

Now R/NA is called Boltzmann's constant and is given the code k.  We can easily work out the value for k.

                   k =     8.31    = 1.38 × 10-23 J K-1

6.02 × 1023

So the equation now becomes:

     

The translational kinetic energy of an ideal gas molecule is NOT dependent on what the gas is.  It is only dependent on the temperature.  So the kinetic energy of helium atoms is the same as the kinetic energy of xenon atoms at a given temperature.

Remember the bear trap. Temperature must be in Kelvin.

 

Question 3

A cylinder of volume 0.25 m3 contains nitrogen gas at a temperature of 17 oC and a pressure of 1.0 × 105 Pa.  Molar mass of nitrogen = 0.028 kg mol-1.  Calculate:

a)      The number of moles of gas in the cylinder                                                               

b)      The rms speed of the gas molecules at 17 oC                                                              

c)      The average translational KE of a nitrogen molecule;                                        

d)   The total kinetic energy of the gas in the  cylinder.

Answer