Two gases-argon (atomic radius 0.07 nm, atomic weight 40) and xenon (atomic radius 0.1 nm, atomic weight 140) have the same number density and are at the same temperature. The raito of their respective mean free times is closest to :

Solution

Physics sir · 3 steps

Mean free time is free path divided by speed, so radius and mass both enter, but in different powers.

1. Build the formula. Mean free time is τ = λ / v. The mean free path is λ = 1/(√2 π d² n), and the molecular speed is v ∝ √(T/M). With the same n and the same T, this gives τ ∝ √M/r².

   τ_(Ar)/τ_(Xe) = (r_(Xe)/r_(Ar))² √(M_(Ar)/M_(Xe))

2. Put in numbers. (0.1/0.07)² = 1/0.49 = 2.04 and √(40/140) = √(2/7) = 0.535. Their product is 2.04 × 0.535 ≈ 1.09.
3. So. The ratio works out to about 1.09. That value is not listed, so pick the nearest option. 1.83 is 0.74 away, while 2.3 is 1.21 away and the others are farther still. The answer is (b).

The idea to keep. Mean free time goes as √M/d²: a bigger atom collides more often, and a heavier atom moves more slowly between collisions. Square the size ratio, but take only the root of the mass ratio.

Answer from the NTA final answer key. Solution by PrepWiser.