**Effect of pressure and temperature on the value of mean free path**

**(a) Effect of pressure:** For is given quantity of gas n, i.e., the number of molecules per unit volume, the mean free path decreases with increase of volume (i.e. decrease of pressure) so that increases with decrease of pressure.

**(b) Effect of temperature:** Again for a given quantity of gas when the temperature is increased at constant pressure the volume Increases Hence the n, the number cat molecules in unit volume, decreases. From the expression above with decrease of n, the value of ** l** increases.

**Example:** Calculate the mean free path of nitrogen molecules at 1.0 atm pressure and 25°C. Collision diameter for nitrogen is 3.50 x 10^{-8} cm.

**Solution:** It is necessary first to calculate the number of molecules per cubic centimeter.

Volume of 1 mole of gas at 25^{0}C = (22,414/273.16) x 298.16 ml = 24474 ml.

n = [6.02×10^{23}/24474] = 2.46×10^{19}

We know

** l** =

**[(1/√2) x 1/(πσ**

^{2}n)]**= **[(1/√2) x [1/(3.142).( 3.50 x 10^{-8}).( 2.46×10^{19})]

= 7.5×10^{-6}

At 1 atm pressure and 25^{0 }C the value of ** l** is of the order of 10

^{-5}cm. At a pressure of 10

^{-5}atm, therefore, the mean free path becomes about 1 cm.