An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature 300K. The mean time between two successive collisions is 6 × 10⁻¹³ s. If the pressure is doubled and temperature is increased to 500K, the mean time between two successive collisions will be close to:

The mean free time (τ) between collisions is given by:
τ = 1 / (√2πd²nv)
where:
d = diameter of the molecule
n = number density of molecules
v = average speed of molecules
The average speed is proportional to √(T/M), where T is temperature and M is molar mass. The number density n is proportional to P/T (from the ideal gas law, PV = nRT). Therefore:
τ ∝ T/(P√T) = √T/P
Let τ₁ be the initial mean time and τ₂ be the final mean time. Then:
τ₂/τ₁ = √(T₂/T₁) * (P₁/P₂)
Given:
τ₁ = 6 × 10⁻¹³ s
T₁ = 300 K
P₁ = 2 atm
T₂ = 500 K
P₂ = 4 atm
Therefore:
τ₂/τ₁ = √(500/300) * (2/4) = √(5/3) * (1/2) ≈ 0.645
τ₂ = τ₁ * 0.645 ≈ 6 × 10⁻¹³ s * 0.645 ≈ 3.87 × 10⁻¹³ s
This is closest to 4 × 10⁻¹³ s.