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