The kinetic theory of matter
The ancients such as the philosophers Democritus and Lucretius held that matter was
composed of minute particles. They also Maintained that these particles were in a state of
continuous random motion within solids, liquids and gases. The theory was therefore called
the kinetic theory of matter, after the Greek work kinema - motion.
Strong evidence for the existence of molecules is provided by the
following observations:
(a) the diffusion of gases and liquids - diffusion in solids has actually
also been observed: a slab of lead was clamped to a slab of gold
for some years, and diffusion of each metal a few millimetres into
the other was demonstrated by chemical analysis.
(b) the mixing of two liquids to give a final volume which is less than
the sum of their original volumes;
(c) dissolving a solid in a liquid.
Figure 1
Various phenomena dealt with elsewhere in detail may be explained in terms of the kinetic
theory:
(a) evaporation - molecules in the surface of a liquid gain sufficient energy to escape from it
(b) saturated vapour pressure - there is a dynamic equilibrium between molecules entering
and leaving a liquid surface in an enclosed space
(c) surface tension - the intermolecular forces at a liquid surface explain capillary rise, liquid
drop shapes, the wetting of surfaces and so on
(d ) latent heat - energy is required to overcome the intermolecular attraction to change the
state of a substance
(e) viscosity - the movement of molecules between adjacent layers of a moving fluid and their
attraction gives the effect of viscous drag
(f) the behaviour of gases, both ideal and real, may be explained in terms of molecules.
Derivation of the kinetic theory formula
Remember that what follows applies to ideal gases only; the assumptions that we make
certainly do not all apply to solids and liquids.
This proof was originally proposed by Maxwell in 1860. He considered a gas to be a
collection of molecules and made the following assumptions about these molecules:
molecules behave as if they were hard, smooth, elastic spheres
molecules are in continuous random motion;
the average kinetic energy of the molecules is proportional to the
absolute temperature of the gas;
the molecules do not exert any appreciable attraction on each other
the volume of the molecules is infinitesimal when compared with
the volume of the gas
the time spent in collisions is small compared with the time between
collisions.
Figure 2
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Consider a volume of gas V enclosed by a cubical box of sides L. Let the box contain N
molecules of gas each of mass m, and let the density of the gas be . Let the velocities of the
molecules be u1, u2, u3 . . . un. (Figure 2)
Consider a molecule moving in the x-direction towards face A
with velocity u1. On collision with face A the molecule will
experience a change of momentum equal to 2mu1. (Figure 3)
It will then travel back across the box, collide with the opposite
face and hit face A again after a time t, where t = 2L/u1.
L
u1
L
The number of impacts per second on face A will therefore be
1/t = u1/2L.
Face A
L
Therefore rate of change of momentum = [mu12]/L = force on
face A due to one molecule.
But the area of face A = L2, so
Figure 3
pressure on face A = [mu12]/L3
But there are N molecules in the box and if they were all travelling along the x-direction then
total pressure on face A = [m/L3](u12 + u22 +...+ uN2)
But on average only one-third of the molecules will be travelling along the x-direction.
Therefore:
pressure = 1/3 [m/L3](u12 + u22 +...+ uN2)
If we rewrite Nc2 = [u12 + u22 + …+ uN2 ] where c is the mean square velocity of the molecules:
pressure = 1/3 [m/L3]Nc2
But L3 is the volume of the gas and therefore:
Pressure (P) = ⅓ [m/V]Nc2 and so PV = ⅓ [mNc2]
and this is the kinetic theory equation.
Now the total mass of the gas M = mN, and since  = M/V we can write
Pressure (P) = ⅓ [c2]
The root mean square velocity or r.m.s. velocity is written as c r.m.s. and is given by the
equation:
r.m.s. velocity = c r.m.s.= √c2 = √[u12 + u22 + …+ uN2 ]/N
We can use this equation to calculate the root mean square velocity of gas molecules at any
given temperature and pressure.
Example
The density of nitrogen at s.t.p. = 1.251 kg m -3. Calculate the r.m.s. velocity of nitrogen molecules.
c2= 3p/ = [3x9.81x13600x0.76]/ 1.251 = 2.432 x 105 Therefore: c r.m.s = 493 ms-1
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The table below gives some further values of the root mean square velocity at s.t.p. for other
gases.
r.m.s. vetocity (ms-1)
18.39 x 102
13.10 x 102
4.61 x 102
3.92 x 102
2.06 X 102
Gas
Hydrogen
Helium
Oxygen
Carbon dioxide
Bromine
Use of the kinetic theory formula to prove the gas laws
(i) Boyle’s law
We have PV =1/3 [mNc2] but c2 is proportional to the absolute temperature T.
For a given mass of gas at constant temperature 1/3 [mNc2] 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 L molecules where L
is the Avogadro constant (6.02x1023) and therefore:
PV =⅓ [mLc2] = RT
Therefore:
1/3 [mc2] = RT/L = kT
and the quantity k (= R/L) is known as Boltzmann’s constant k .
The equation for one molecule may therefore be rewritten as:
⅓ [mc2] = kT
(ii) Charles’s law
The total mass of gas is mN = N, therefore:
PV = ⅓ [Mc2]
But if the temperature of the gas is changed from T1 to T2 with a resulting change in volume
from V1 to V2, the pressure being kept constant:
PV1 = ⅓ [Mc12] and
PV2 = ⅓ [Mc22]
Therefore:
PV1/PV2= [Mc12]/[Mc22] and so
But kinetic energy = ½ mc2 and therefore:
V1/V2= [c12]/[c22]
V1/V2= [kinetic energy1]/[kinetic energy2]
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:
V1/V2= [c12]/[c22] = T1/T2
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So the volume is directly proportional to the absolute temperature, and this is Charles’s law.
the absolute
(iii) Avogadro's law
Maxwell showed that the average kinetic energies of molecules are equal at the same
temperature, that is:
½ [m1c12] = kT = ½ [m2c22] and so [m1c12] = [m2c22]
But P1V1 = 1/3 [m1N1c12] and P2V2 = 1/3 [m2N2c22]
Now if P1 = P2 and V1 = V2, [m1N1c12] = [m2N2c22] and therefore:
N1 = N2
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 ([m1N1c12] + 1/3 [m2N2c22] + …………)
P = 1/3 ([m1N1c12]/V + 1/3 [m2N2c22]/V + …………) = P1 + P2 + …..
where P1, P2 ... 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 = = P1 + P2 + …..
(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 √c2.
Rate of diffusion of gas A/Rate of diffusion of gas = uA/uB = √[cA2/cB2] = √[B/A]
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 ( 238U
and 238U) 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 propagation 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.
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