# Use of the kinetic theory formula to prove the gas
laws

### (i) Boyle's law

We have PV =1/3 [mNc^{2}] but
c^{2} is proportional to the absolute temperature T.

For a given mass of gas at
constant temperature 1/3 [mNc^{2}] is constant and therefore PV is constant and this is
Boyle's law.

The Ideal Gas equation for n moles of gas is PV = nRT and so for one
mole of gas we have PV = RT, where R is the gas constant. But in one mole of gas there are
N_{A} molecules where N_{A} is the Avogadro constant (6.02x10^{23}) and therefore:

PV = (1/3)[mN_{A}c^{2}] = RT
Therefore: 1/3
[mc

^{2}] = RT/N

_{A} = kT

and the quantity k (= R/N

_{A}) is known as

**Boltzmann's constant (k)** .

The equation for one molecule may therefore be rewritten as:

(1/3)[mc^{2}] = kT
However the mean kinetic energy of a molecule in the gas is ˝mc

^{2} and so:

Mean kinetic energy of a molecule in an ideal gas = (1/2)mc^{2} = (3/2)kT = (3/2)RT/N_{A}
You should be able to see from this formula that the mean kinetic energy of a molecule in an ideal gas is proportional to the absolute temperature (T) of the gas.

### (ii) Charles's
law

The total mass of gas is mN = M, therefore:

PV = 1/3
[Mc

^{2}]

But if the temperature of the gas is changed from T

_{1} to
T

_{2} with a resulting change in volume from V

_{1} to V

_{2}, the pressure
being kept constant:

PV

_{1} =1/3 [Mc

_{1}^{2}] and
PV

_{2} =1/3 [Mc

_{2}^{2}]

Therefore:

PV

_{1}/PV

_{2}= [Mc

_{1}^{2}]/[Mc

_{2}^{2}] and so
V

_{1}/V

_{2}= [c

_{1}^{2}]/[c

_{2}^{2}]

But kinetic
energy = ˝ mc

^{2} and therefore: V

_{1}/V

_{2}= [kinetic
energy

_{1}]/[kinetic energy

_{2}]

Now one of the assumptions of the
kinetic theory is that the kinetic energy of the molecules is directly proportional to the
absolute temperature of the gas (T) and therefore:

V_{1}/V_{2}= [c_{1}^{2}]/[c_{2}^{2}] = T_{1}/T_{2}
So the volume is directly proportional to the
absolute temperature, and this is Charles's law.

### (iii)
Avogadro's law

Maxwell showed that the average kinetic energies of molecules are
equal at the same temperature, that is:

˝ [m

_{1}c

_{1}^{2}] = kT = ˝
[m

_{2}c

_{2}^{2}] and so [m

_{1}c

_{1}2] = [m

_{2}c

_{2}^{2}]

But
P

_{1}V

_{1} = 1/3 [m

_{1}N

_{1}c

_{1}^{2}] and P

_{2}V

_{2} = 1/3
[m

_{2}N

_{2}c

_{2}^{2}]

Now if P

_{1} = P

_{2} and V

_{1} = V

_{2},
[m

_{1}N

_{1}c

_{1}^{2}] = [m

_{2}N

_{2}c

_{2}^{2}] and therefore:

N_{1} = N_{2}
and this is
Avogadro's law (equal volumes of all gases at the same temperature and pressure contain
equal numbers of molecules).

### (iv) Dalton's law of partial pressures

For a
mixture of gases:

PV = 1/3 ([m

_{1}N

_{1}c

_{1}^{2}] + 1/3
[m

_{2}N

_{2}c

_{2}^{2}] + …………)

P = 1/3 ([m

_{1}N

_{1}c

_{1}^{2}]/V
+ 1/3 [m

_{2}N

_{2}c

_{2}^{2}]/V + …………) = P

_{1} + P

_{2} + …..

where
P

_{1}, P

_{2} ... are the partial pressures of the gases, and this is Dalton's law
(the sum of the partial pressures of all the gases occupying a given volume is equal to the
total pressure).

P = P_{1} + P_{2} + ....
### (v) Graham's law of diffusion

The rate
of diffusion of a gas is directly proportional to the mean velocity of the gas molecules, and
this is also proportional to [c

^{2}]

^{1/2}.

Rate of diffusion of gas
A/Rate of diffusion of gas = u

_{A}/u

_{B} = [c

_{A}^{2}/c

_{B}^{2}]

^{1/2} = [

r_{B}/[

r_{A}]

^{1/2} This is Graham's law (the
rate of diffusion of a gas varies inversely as the square root of the density of the
gas).

The difference in mass and hence in the diffusion rates of the two isotopes of
uranium (

^{235}U and

^{238}U) is used in the nuclear industry to separate them, although
the differences are small, and so a series of diffusion processes have to be used to obtain
sufficient purity.

**Student investigation**

The following investigation is designed to study the absorption of sound in different gases.

Set up a horizontal tube 5 cm in diameter and 3 m long (a piece of water pipe is suitable) with a loudspeaker directed into one end.

(a) Using a signal generator, investigate the propa¬gation of sound of frequency 8000 Hz in the tube when it is filled first with air and then with carbon dioxide.

(b) Reduce the frequency to 2000 Hz and repeat the experiment.

Suggest possible reasons for what you observe.