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JEE Main 201912 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Evening Shift
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main 2019 · 12 January · Evening Shift · Physics · Kinetic Theory of Gases

An ideal gas is enclosed in a cylinder at pressure of 2 atm and...

An ideal gas is enclosed in a cylinder at pressure of 22 atm and temperature, 300K.300K . The mean time between two successive collisions is 6×10−8s.6 \times {10}^{- 8}s . If the pressure is doubled and temperature is increased to 500K,500K, the mean time between two successive collisions will be close to:

Options

  1. A2x10−7s2 x {10}^{- 7}s
  2. B3x10−6s3 x {10}^{- 6}s
  3. C0.5x10−8s0 . 5 x {10}^{- 8}s
  4. D4x10−8s4 x {10}^{- 8}s

Correct answer

D. 4x10−8s4 x {10}^{- 8}s

Step-by-step solution

τ=12nπd2vavgτ = \dfrac{1}{\sqrt{2}n\pi{d}^{2}{v}_{avg}}

Where n=n = number of molecules per unit volume.

τ∝VT⇒τ∝TPτ ∝ \dfrac{V}{\sqrt{T}} ⇒ τ ∝ \dfrac{\sqrt{T}}{P}

τ2τ1=T2T1.P1P2⇒τ2=6×10−8×5000300×24\dfrac{{τ}_{2}}{{τ}_{1}} = \sqrt{\dfrac{{T}_{2}}{{T}_{1}}} . \dfrac{{P}_{1}}{{P}_{2}} ⇒ {τ}_{2} = 6 \times {10}^{- 8} \times \sqrt{\dfrac{5000}{300}} \times \dfrac{2}{4}

≈4×10−8s\approx 4 \times {10}^{- 8}s
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