Deriving the Kinetic Theory Equation

Thermal Physics Lesson 6

Learning Objectives Derive the kinetic theory equation.

Here we go…

We consider a molecule of mass m in a box of dimensions lx, ly and lz with a velocity components u1, v1 and w1 in the x, y and z directions respectively.

      

 

A derivation of two halves…

The strategy is to derive the pressure for one molecule on one face (1st half) and then sum all the pressures from all the molecules (2nd half).

The first part involves Newton’s second law which states that the force on a body is equal to the rate of change of momentum.

Also note… The speed c of the molecule is given by:-

This comes from doing Pythagoras’ theorem in 3 dimensions.

We will use this later in the 2nd half

21

21

21

21 wvuc

Find the force on one molecule…

When the molecule impacts with Face A the x-component of its momentum is changed from mu1 to -mu1.

But we want the rate of change of momentum, so need to divide by the time taken between collisions with Face A.

111 mu 2 )(-mu– mu momentumin Change

Time, t, between collisions…

1

2 velocity theofcomponent x

back and face opposite todistance totalult x

xx lmu

ulmu 2

1

1

1

)/2(2

takentimemomentum of changemoleculeon force

So Newton’s 2nd Law gives us:-

And Newton’s 3rd Law gives us:-

xlmu 2

1moleculeon force on wall force

Pressure Recall that to work out the pressure:-

So the pressure p1 on face A:-

So that’s the first bit done! – the pressure from one molecule in one direction.

areaforce pressure

Vlll)llA( face of areaforce

21

zyx

21

zy1

mumup

Summing the pressures

Np...pppp 321

The total pressure on face A can be calculated by summing the pressures of all of the molecules:-

Where p2, p3… refer to the pressures of all the other molecules up to N molecules.

Vmu...

Vmu

Vmu

Vmup N

223

22

21

Summing the pressures The last line can be rewritten as:-

Where the mean square x-component velocity is given by:-

And similar equations can be derived for the y and z components

VuNmp

2

Nuuuuu N

223

22

212 ...

Nvvvvv N

223

22

212 ...

Almost there… But we want our equation to include the

root mean square speed in all directions. Using Pythagoras’s theorem, the speed for one molecule is given by:-

You can show that the root mean square speed for all the molecules is:-

21

21

21

21 wvuc

2222 wvucrms

Almost there… Because the motions are random we can

write:-

Otherwise there would be a drift of particles in one direction. So using the above two equations:-

So now we can write:-

222 wvu

22

31

rmscu

VNmc

VuNmp rms

22

31

Final Rearrangement

2

31

rmsNmcpV

VNm

The Kinetic Theory Equation:-

Can also be re-written as:-2

31

rmscp

Because:-

Recap Use Newton’s Second Law Use Newton’s Third Law Calculate pressure due to one

molecule (force/area) Sum the pressures of all the

molecules. Rewrite the speed in terms of root

mean square speed.

Deriving Ideal Gas Equation

From Boyle’s Law:

From Pressure Law:

From Avogadro’s Law:

Combining these three:

Rewriting using the gas constant R:

pV 1

pnTV

nV

TV

nRTpV

pnTRV

Therefore:-

Mean kinetic energy

Nccccm N )...( 22

32

22

121

Nccccc N

rms

223

22

212 ...

Nmcmcmcmc N

22

1232

1222

1212

1 ...

22

1rmsmc

The mean kinetic energy of a molecule is the total kinetic energy of all the molecules/total number of molecules.

The root mean square speed is defined as:

…and so the mean kinetic energy of one molecule is gvien by…

Mean Kinetic EnergynRTpV 2

31

rmsNmcpV

2

31

rmsNmcnRT

2

31

rmsAmcnNnRT

2

31

rmsA

mcTNR

2

31

rmsmckT

Note we have two equations with pV on the left hand side.

(N=nNA)

The Boltzmann constant, k, is defined as R/NA.

Mean Kinetic Energy

kTmcrms 23

21 2

2

31

23

23

rmsmckT

nRTmcrms 23

21 2

Multiply both sides by 3/2.

Mean kinetic energy of a molecule of an ideal gas.

Kinetic energy of one mole of gas(because R=NAk)

Kinetic energy of n moles of gas.

RTmcrms 23

21 2