An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature 300 K. The mean time between two successive collisions is 6 × 10–8 s. If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions will be close to (a) 0.5 × 10 - 8 s (b) 4 × 10 - 8 s (c) 3 × 10 - 6 s (d) 2 × 10 - 7 s

Answer: (b). τ = (λ)/( v) with λ ∝ T/P and v ∝ √(T), so τ ∝ √(T)/P. (τ₂)/(τ₁) = √((500)/(300)) × 2/4 = 1.29 × 0.5 = 0.645 τ₂ = 0.645 × 6×10⁻⁸ ≈ 3.9×10⁻⁸ s So the answer is (b) 4×10⁻⁸ s.

An ideal gas is enclosed in a cylinder at pressure of 2 atm and… | JEE Main 2019 (12 Jan, Shift 2)

Q23JEE Main 2019 · 12 JAN · SHIFT 2Medium

Mean Free Path & Real Gases

An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature 300 K. The mean time between two successive collisions is 6 $\times$ 10–8 s. If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions will be close to

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Answer: (b)

$\tau = \dfrac{\lambda}{\bar v}$ with $\lambda \propto \dfrac{T}{P}$ and $\bar v \propto \sqrt{T}$ , so $\tau \propto \dfrac{\sqrt{T}}{P}$ .
$\dfrac{\tau_2}{\tau_1} = \sqrt{\dfrac{500}{300}} \times \dfrac{2}{4} = 1.29 \times 0.5 = 0.645$
$\tau_2 = 0.645 \times 6\times10^{-8} \approx 3.9\times10^{-8}$ s
So the answer is (b) $4\times10^{-8}$ s.

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Mean Free Path & Real Gases question from JEE Main 2019 (12 Jan, Shift 2)

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