JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Heat & Thermodynamics

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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 1 0^{- 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) $2 \times 1 0^{- 7} s$
(B) $4 \times 1 0^{- 8} s$ ✓ Correct
(C) $0.5 \times 1 0^{- 8} s$
(D) $3 \times 1 0^{- 6} s$

Answer: (B)

Solution

The mean time between collisions follows

τ \propto \frac{T}{P}

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raising the pressure crowds the molecules and shortens it, while heating speeds them up and lengthens it a little through the mean free path. Taking the ratio,

τ_{2} = τ_{1} \frac{P _{1}}{P _{2}} \frac{T _{2}}{T _{1}} = 6 \times 1 0^{- 8} \times \frac{1}{2} \times \frac{500}{300} \approx 3.9 \times 1 0^{- 8}

s, which is close to

4 \times 1 0^{- 8}

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Heat & Thermodynamics

Solution

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