The molecules of a gas move with high speeds at a given temperature but even then a molecule of the gas takes a very long time to go from one point to another point in the container of the gas. This is due to the fact that a gas molecule suffers a number of collisions with other gas molecules surrounding it. As a result of these collisions, the path followed by a gas molecule in the container of the gas is zig – zag as shown in the figure. During two successive collisions, a molecule of a gas moves in a straight line with constant velocity and the distance travelled by a gas molecule between two successive collisions is known as free path.The distance travelled by a gas molecule between two successive collisions is not constant and hence the average distance travelled by a molecule during all collisions is to be calculated. This average distance travelled by a gas molecule is known as mean free path.

Let λ₁, λ₂, λ₃, …λ_{n} be the distance travelled by a gas molecule during n collisions respectively, then the mean free path of a gas molecule is given by λ = (λ₁ + λ₂ + λ₃ + … + λ_{n})/ n

(1) λ= 1/ (√2πnd)²; where d = Diameter of the molecule, n = Number of molecules per unit volume

(2) As PV = m RT = m NkT Þ N/ V = P/ kT = n = Number of molecule per unit volume S λ = (1/√2) (kT/ πd²P)

(3) From λ= 1/ (√2πnd)² = m/ √2π (mn) d² = m/ √2πd²ρ [As mn = Mass per unit volume = Density =r].

(4) If average speed of molecule is v then λ = v x t/ N = v x T [As N = Number of collision in time t, T = time interval between two collisions].

(5) Every gas consists of extremely small particles known as molecules. The molecules of a given gas are all identical but are different than those of another gas.

(6) The molecules of a gas are identical, spherical, rigid and perfectly elastic point masses.

(7) Their size is negligible in comparison to intermolecular distance (10^{–9} m).

(8) The gas molecules keep on colliding among themselves as well as with the walls of containing vessel. These collisions are perfectly elastic. (i.e. the total energy before collision = total energy after the collision).

(9) Molecules move in a straight line with constant speeds during successive collisions.

(10) The distance covered by the molecules between two successive collisions is known as free path and mean of all free paths is known as mean free path.

(11) No attractive or repulsive force acts between gas molecules.

(12) Gravitational attraction among the molecules is ineffective due to extremely small masses and very high speed of molecules.

(13) Molecules constantly collide with the walls of container due to which their momentum changes. The change in momentum is transferred to the walls of the container. Consequently pressure is exerted by gas molecules on the walls of container.

(1) **Root mean square speed: **It is defined as the square root of mean of squares of the speed of different molecules i.e. v_{rms} = √(v²₁ + v²₂ + v²₃ + v²₄ + …)/N

(i) From the expression for pressure of ideal gas P = 1/3 mN/V v²_{rms}

v_{rms} = √ [3PV/ mN] = √ [3PV/ Mass of gas] = √ [3P/ ρ] [As ρ = Mass of gas/ V]

(ii) v_{rms} = √ [3PV/ Mass of gas] = √ [3μRT/ μM] = √ [3RT/ M]

[As if M is the molecular weight of gas PV = μRT and Mass of gas = μM]

(iii) v_{rms} = √ [3RT/ M] = √ [3 N_{A}kT/ N_{A}M] = √ [3kT/ m] [As M = N_{A}m and R = N_{A}k]

∴ Root mean square velocity v_{rms} = √ [3P/ ρ] = √ [3 RT/ M] = √ [3kT/ m]

(i) With rise in temperature rms speed of gas molecules increases as v_{rms} α √T.

(ii) With increase in molecular weight rms speed of gas molecule decreases as v_{rms} α 1/ √M.

e.g., rms speed of hydrogen molecules is four times that of oxygen molecules at the same temperature.

(iii) rms speed of gas molecules is of the order of km/s

e.g., At NTP for hydrogen gas (v_{rms}) = √ [3 RT/ M] = √ [3 x 8.31 x 273/ 2 x 10^{3}] = 1840 m/s.

(iv) rms speed of gas molecules is √3/γ times that of speed of sound in gas

As v_{rms} = √ [3 RT/ M] and v_{s} = √ [γRT/ M] ∴ As v_{rms} = √3/γ (v_{s})

(v) rms speed of gas molecules does not depends on the pressure of gas (if temperature remains constant) because P µ r (Boyle’s law) if pressure is increased n times then density will also increases by n times but v_{rms} remains constant.

(vi) Moon has no atmosphere because v_{rms} of gas molecules is more than escape velocity (v_{e}).

A planet or satellite will have atmosphere only and only if v_{rms} < v_{e}

(vii) At T = 0; v_{rms} = 0 i.e. the rms speed of molecules of a gas is zero at 0 K. This temperature is called absolute zero.

(2) **Most probable speed: **The particles of a gas have a range of speeds. This is defined as the speed which is possessed by maximum fraction of total number of molecules of the gas. e.g., if speeds of 10 molecules of a gas are 1, 2, 2, 3, 3, 3, 4, 5, 6, 6 km/s, then the most probable speed is 3 km/s, as maximum fraction of total molecules possess this speed.

Most probable speed v_{mp} = √2P/ρ = √2RT/M = √2kT/m

(3) **Average speed: **It is the arithmetic mean of the speeds of molecules in a gas at given temperature.

v_{av} = (v₁ + v₂ + v₃ + v₄ + …)/ N

And according to kinetic theory of gases

Average speed v_{av} = √ (8/π) (P/ρ) = √ (8/π) (RT/M) = √ (8/π) (kT/m)

v_{rms} > v_{av} > v_{mp} (order remembering trick) (RAM)

v_{rms }: v_{av} : v_{mp} = √3 : √8/π :√2 = √3 : √2.5 : √2

(i) P = 1/3 mN/V (v²_{rms}) Or P α (mN) T/ V [Asv²_{rms} α T]

(a) If volume and temperature of a gas are constant P µ mN i.e. Pressure µ (Mass of gas).

i.e. if mass of gas is increased, number of molecules and hence number of collision per second increases i.e. pressure will increase.

(b) If mass and temperature of a gas are constant. P µ (1/V), i.e., if volume decreases, number of collisions per second will increase due to lesser effective distance between the walls resulting in greater pressure.

(c) If mass and volume of gas are constant, P α (v_{rms})² α T

i.e., if temperature increases, the mean square speed of gas molecules will increase and as gas molecules are moving faster, they will collide with the walls more often with greater momentum resulting in greater pressure.

(ii) P = 1/3 mN/V (v²rms) = 1/3 M/V (v²_{rms}) [As M = mN = Total mass of the gas]

∴ P = 1/3 ρv^{2}_{rms} [As ρ = M/V]

(iii) Relation between pressure and kinetic energy

Kinetic energy = ½ Mv²_{rms}

∴ Kinetic energy per unit volume (E) = ½ (M/V) v^{2}_{rms} = ½ ρv^{2}_{rms} … (i)

And we know P = 1/3 ρv^{2}_{rms} … (ii)

From (i) and (ii), we get P = 2/3 E

i.e. the pressure exerted by an ideal gas is numerically equal to the two third of the mean kinetic energy of translation per unit volume of the gas.