# Mean Free Path

##### Mean Free Path (l)

The average distance covered by a molecule between two successive collisions is called mean free path and is denoted by l .

$\large l = \frac{1}{\sqrt{2} \pi \sigma^2 n}$

where n = no. of molecules/cc.

Again, if P & T denote the pressure and temperature of the gas, from kinetic theory.

$\large P \times 10^{-3} = \frac{n R T}{N_o}$

$\large n = \frac{N_o \times 10^{-3}}{R} \times \frac{P}{T}$

$\large n = K’ \frac{P}{T}$

$\large l = \frac{T}{\sqrt{2}\pi \sigma^2 K’ P}$

Thus mean free path is directly proportional to temperature and inversely to pressure

##### Collision Frequency (z)

It is the number of collisions taking place per second

z = π σ2 N uav

where σ = collision diameter. It is the distance between the centres of two molecule without collision.

N = number of molecules per unit volume,

And uav = average velocity

#### Relative Humidity (RH)

At a given temperature it is given by equation

$\large R_H = \frac{Partial \; pressure \; of \; water \; in\; air}{Vapour \; pressure \; of \; water }$

##### Boyle’s Temperature (Tb)

Temperature at which real gas obeys the gas laws over a wide range of pressure is called Boyle’s Temperature. Gases which are easily liquefied have a high Boyle’s temperature [Tb(O2)] = 46 K] whereas the gases which are difficult to liquefy have a low Boyle’s temperature [Tb(He) = 26K]

$\large T_b = \frac{a}{R b} = \frac{1}{2}T_i$

where Ti is called Inversion Temperature and a, b are called van der Waals constant.

#### Critical Constants

(i) Critical Temperature (Tc): It (Tc) is the maximum temperature at which a gas can be liquefied i.e. the temperature above which a gas can’t exist as liquid.

$\large T_b = \frac{8 a}{27 R b}$

(ii) Critical Pressure (Pc): It is the minimum pressure required to cause liquefaction at Tc

$\large P_c = \frac{a}{27 b^2}$

(iii) Critical Volume: It is the volume occupied by one mol of a gas at Tc and Pc

Vc = 3b

Illustration :   Calculate Vander Waals constants for ethylene

TC = 282.8 K; PC = 50 atm

Solution : $\large b = \frac{1}{8}\frac{R T_c}{P_c}$

$\large b = \frac{1}{8} \times \frac{0.082 \times 282.8}{50}$

= 0.057 litres/mole

$\large a = \frac{27}{64} R^2 \times \frac{T_c^2}{P_c}$

$\large a = \frac{27}{64} (0.082)^2 \times \frac{(282.8)^2}{50}$

= 4.47 lit2 atm mole–2