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JEE Main 2019 (10 Apr, Shift 1) · Mean Free Path & Real Gases

Mean Free Path & Real Gases question from JEE Main 2019 (10 Apr, Shift 1)

Q22JEE Main 2019 · 10 APR · SHIFT 1Medium

Mean Free Path & Real Gases

A 25 × 10–3 m3 volume cylinder is filled with 1 mol of O2 gas at room temperature (300K). The molecular diameter of O2, and its root mean square speed, are found to be 0.3 nm, and 200 m/s, respectively. What is the average collision rate (per second) for an O2 molecule?

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Answer: (b)

  1. Number density n=NAV=6.02×102325×10−3=2.4×1025 m−3n = \dfrac{N_A}{V} = \dfrac{6.02\times10^{23}}{25\times10^{-3}} = 2.4\times10^{25}\text{ m}^{-3}
  2. Collision rate f=2 nπd2v=1.414×2.4×1025×3.14×9×10−20×200f = \sqrt{2}\,n\pi d^2 v = 1.414\times2.4\times10^{25}\times3.14\times9\times10^{-20}\times200
  3. f≈1.9×109 s−1f \approx 1.9\times10^{9}\text{ s}^{-1}, of order 10910^{9}–101010^{10} per second.
  4. Nearest option is (d) ∼1010\sim10^{10}; key gives (b) ∼1012\sim10^{12}.
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