Mean Free Path - UNSOLVED PRACTICE SET

Q7. Define mean free path. Write its expression and explain each symbol.

Q8. Explain why the mean free path increases with decreasing pressure at constant temperature.

Q9. Show that the mean free path can also be written as λ = kT/(√2 πd²P).

Q10. Why is the factor √2 present in the denominator of the mean free path formula? Explain the physical reasoning.

Q11. The mean free path of a gas is 10⁻⁷ m at STP. What will be the mean free path if the pressure is reduced to 0.1 atm at the same temperature?

Q12. Explain why the concept of mean free path is important in understanding transport phenomena like viscosity, thermal conductivity, and diffusion.

Q13. Derive the expression for the mean free path of gas molecules: λ = 1/(√2 πd²n). Start by considering:

A molecule of diameter d moving through a gas of number density n

The collision cylinder swept out by the moving molecule

The effective collision cross-section

The average number of collisions per unit distance

The final expression for mean free path

Also derive the alternative form λ = kT/(√2 πd²P).

Q14. Discuss the factors affecting the mean free path of gas molecules. Explain how λ depends on:

(i) Number density (n)

(ii) Temperature (T)

(iii) Pressure (P)

(iv) Molecular diameter (d)

(v) Molecular mass (indirectly)

Draw graphs showing λ vs P (at constant T) and λ vs T (at constant P). Explain the physical significance of each graph.

Q15. The mean free path of nitrogen molecules at STP is approximately 0.1 μm.

(i) Calculate the molecular diameter of nitrogen.

(ii) Calculate the mean free path at 0.01 atm and 300 K.

(iii) Calculate the collision frequency at STP if the rms speed is 515 m/s.

(iv) At what pressure would the mean free path become comparable to the size of a typical container (say, 0.1 m)? What does this imply?

Q16. For oxygen gas at STP:

Number density n = 2.7 × 10²⁵ m⁻³

Molecular diameter d = 3.6 × 10⁻¹⁰ m

rms speed v_rms = 460 m/s

Calculate:

(i) The mean free path of oxygen molecules

(ii) The collision frequency

(iii) The average time between collisions

(iv) The total distance traveled by a molecule in 1 second

(v) The number of collisions made by a molecule in 1 second

Q17. The mean free path of hydrogen molecules at 1 atm and 300 K is 1.5 × 10⁻⁷ m. The molecular diameter of H₂ is 2.9 × 10⁻¹⁰ m.

(i) Calculate the number density of hydrogen molecules at these conditions.

(ii) Calculate the mean free path at 0.5 atm and the same temperature.

(iii) Calculate the mean free path at 1 atm and 600 K.

(iv) Calculate the collision frequency at the original conditions if v_rms = 1920 m/s.

(v) In a vacuum chamber where the pressure is 10⁻⁶ atm, calculate the mean free path. Compare this to the size of the chamber (0.5 m) and comment on the significance.

Q18. In a school science project, a student wants to demonstrate the effect of mean free path on gas behaviour. She has a container of volume 1 L containing air at 1 atm and 300 K.

(i) Calculate the number density of air molecules.

(ii) Calculate the mean free path assuming the molecular diameter of air molecules is 3.7 × 10⁻¹⁰ m.

(iii) She now pumps out air until the pressure is 10⁻³ atm. Calculate the new mean free path.

(iv) At what pressure would the mean free path equal the size of the container (0.1 m)?

(v) Explain why this pressure is called the "molecular flow regime" and why normal gas laws don't apply there.

(Given: k = 1.38 × 10⁻²³ J/K)