Question

(a) Show that for a gas, the mean free path between collisions is related to the mean distance between nearest neighbors r by the approximate relation 1 r(r2/0) where o is the collision cross- section. (b) Given that the molecular radius of a gas molecule such as O2, N2, or CO2 is about 0.15 nm, estimate the value of r and for air at STP (standard temperature and pressure, T = 273 K, p = 1.00 atm = 1.01 X 10ʻN/m²). (c) Why, qualita- tively, is 1 » r, for dilute gases?

Answer 1

The mean free path & is the average distance a particle travels between collisims. if two particles radins & come within disneter (22) of each each 7 other then they &=(2872 Collide the collisim Cross sectional 482 area the mean free path & is related to the cross sostino and the numbel density in f particle SOLCL molecular radius is gine so distance between nearest neighbur=20.0.15 when der=o. zum d KT 2 7d² p -23 +38*10* 273 V2 X3.14 (0.3X1699x101xe -23 930.2 x 100m -18 da 376.74 X 10 4.5x10X0-09 X 16" de 93.02x109 ha d = 93nm fc) The collisime betweess gas atent ac molecolel with the wally container are elastic. the average distance between Collisim is known as mean free path. the valuey free path depends in density of gas Particled, their sice and temperature. dillute gas contains very large number of atems so we use statistical any lysis. hence its is assume for dilute gas that a 70