Question

Derive an expression of the mean free path of a gas molecule assuming a hard sphere collision. Calculate the ratio of the mean free path of CO molecules in a vessel at a pressure P1=10-4 Torr at 300 K to that at a pressure P2=10-9 Torr at the same temperature. (CO molecules d =0.73nm)

Answer 1

Mean free path of a gas molecule is the distance covered by a number of gas molecules subjected to internal collision, in a given span of time.

Mathematically, mean free path = D = (distance travelled by molecules in a certain time / number of collisions in that time)

Or, D = v_a * t / (π * d² * v_r * t * [P/kT])....(i) , where,

v_a = average velocity of gas molecules

t = time-span

d = diameter of gas molecules

v_r = relative velocity of gas molecules

P = pressure to which molecules are subjected

k = Boltzmann Constant

T = absolute temperature

Recognizing that v_r = √ (2) * v_a , from (i) , we have,

D = v_a * t / (π * d² * √ 2 * v_a * t * [P/kT])

Or , D = 1 / (√ 2 π d² * [P/kT])....(ii)

From (ii), we can infer that all factors remaining constant, the mean free path is inversely proportional to the pressure.

Thus, re-ordering (ii), we have,

D = kT / (√2 * π * d² * P)....(iii)

Now, since given that temperature is constant at 300 K and d = constant = 0.73 nm, we consider the parameter

[ kT / (√2 * π * d²)] as a constant K (say).

Thus, from (iii),

D = K / P

Or, D1 / D2 = P2 / P1....(iv)

Now given that P1 = 10^-4 Torr and P2 = 10^-9 Torr

From (iv), D1 / D2 = (10^-9 / 10^-4) = 10^-5

Or, (D1 / D2) = 10^-5

Thus, the mean free path travelled by gas molecules in the second vessel (D2) is 10⁵ times more than that in the first vessel (D1).